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583 | A | Asphalting Roads | PROGRAMMING | 1,000 | [
"implementation"
] | null | null | City X consists of *n* vertical and *n* horizontal infinite roads, forming *n*<=×<=*n* intersections. Roads (both vertical and horizontal) are numbered from 1 to *n*, and the intersections are indicated by the numbers of the roads that form them.
Sand roads have long been recognized out of date, so the decision was made to asphalt them. To do this, a team of workers was hired and a schedule of work was made, according to which the intersections should be asphalted.
Road repairs are planned for *n*2 days. On the *i*-th day of the team arrives at the *i*-th intersection in the list and if none of the two roads that form the intersection were already asphalted they asphalt both roads. Otherwise, the team leaves the intersection, without doing anything with the roads.
According to the schedule of road works tell in which days at least one road will be asphalted. | The first line contains integer *n* (1<=≤<=*n*<=≤<=50) — the number of vertical and horizontal roads in the city.
Next *n*2 lines contain the order of intersections in the schedule. The *i*-th of them contains two numbers *h**i*,<=*v**i* (1<=≤<=*h**i*,<=*v**i*<=≤<=*n*), separated by a space, and meaning that the intersection that goes *i*-th in the timetable is at the intersection of the *h**i*-th horizontal and *v**i*-th vertical roads. It is guaranteed that all the intersections in the timetable are distinct. | In the single line print the numbers of the days when road works will be in progress in ascending order. The days are numbered starting from 1. | [
"2\n1 1\n1 2\n2 1\n2 2\n",
"1\n1 1\n"
] | [
"1 4 \n",
"1 \n"
] | In the sample the brigade acts like that:
1. On the first day the brigade comes to the intersection of the 1-st horizontal and the 1-st vertical road. As none of them has been asphalted, the workers asphalt the 1-st vertical and the 1-st horizontal road; 1. On the second day the brigade of the workers comes to the intersection of the 1-st horizontal and the 2-nd vertical road. The 2-nd vertical road hasn't been asphalted, but as the 1-st horizontal road has been asphalted on the first day, the workers leave and do not asphalt anything; 1. On the third day the brigade of the workers come to the intersection of the 2-nd horizontal and the 1-st vertical road. The 2-nd horizontal road hasn't been asphalted but as the 1-st vertical road has been asphalted on the first day, the workers leave and do not asphalt anything; 1. On the fourth day the brigade come to the intersection formed by the intersection of the 2-nd horizontal and 2-nd vertical road. As none of them has been asphalted, the workers asphalt the 2-nd vertical and the 2-nd horizontal road. | 500 | [
{
"input": "2\n1 1\n1 2\n2 1\n2 2",
"output": "1 4 "
},
{
"input": "1\n1 1",
"output": "1 "
},
{
"input": "2\n1 1\n2 2\n1 2\n2 1",
"output": "1 2 "
},
{
"input": "2\n1 2\n2 2\n2 1\n1 1",
"output": "1 3 "
},
{
"input": "3\n2 2\n1 2\n3 2\n3 3\n1 1\n2 3\n1 3\n3 1\n2 1",
"output": "1 4 5 "
},
{
"input": "3\n1 3\n3 1\n2 1\n1 1\n1 2\n2 2\n3 2\n3 3\n2 3",
"output": "1 2 6 "
},
{
"input": "4\n1 3\n2 3\n2 4\n4 4\n3 1\n1 1\n3 4\n2 1\n1 4\n4 3\n4 1\n3 2\n1 2\n4 2\n2 2\n3 3",
"output": "1 3 5 14 "
},
{
"input": "4\n3 3\n4 2\n2 3\n3 4\n4 4\n1 2\n3 2\n2 2\n1 4\n3 1\n4 1\n2 1\n1 3\n1 1\n4 3\n2 4",
"output": "1 2 9 12 "
},
{
"input": "9\n4 5\n2 3\n8 3\n5 6\n9 3\n4 4\n5 4\n4 7\n1 7\n8 4\n1 4\n1 5\n5 7\n7 8\n7 1\n9 9\n8 7\n7 5\n3 7\n6 6\n7 3\n5 2\n3 6\n7 4\n9 6\n5 8\n9 7\n6 3\n7 9\n1 2\n1 1\n6 2\n5 3\n7 2\n1 6\n4 1\n6 1\n8 9\n2 2\n3 9\n2 9\n7 7\n2 8\n9 4\n2 5\n8 6\n3 4\n2 1\n2 7\n6 5\n9 1\n3 3\n3 8\n5 5\n4 3\n3 1\n1 9\n6 4\n3 2\n6 8\n2 6\n5 9\n8 5\n8 8\n9 5\n6 9\n9 2\n3 5\n4 9\n4 8\n2 4\n5 1\n4 6\n7 6\n9 8\n1 3\n4 2\n8 1\n8 2\n6 7\n1 8",
"output": "1 2 4 9 10 14 16 32 56 "
},
{
"input": "8\n1 1\n1 2\n1 3\n1 4\n1 5\n8 6\n1 7\n1 8\n2 1\n8 5\n2 3\n2 4\n2 5\n2 6\n4 3\n2 2\n3 1\n3 2\n3 3\n3 4\n3 5\n3 6\n5 6\n3 8\n4 1\n4 2\n2 7\n4 4\n8 8\n4 6\n4 7\n4 8\n5 1\n5 2\n5 3\n6 5\n5 5\n3 7\n5 7\n5 8\n6 1\n6 2\n6 3\n6 4\n5 4\n6 6\n6 7\n6 8\n7 1\n7 2\n7 3\n7 4\n7 5\n7 6\n7 7\n7 8\n8 1\n8 2\n8 3\n8 4\n2 8\n1 6\n8 7\n4 5",
"output": "1 6 11 18 28 36 39 56 "
},
{
"input": "9\n9 9\n5 5\n8 8\n3 3\n2 2\n6 6\n4 4\n1 1\n7 7\n8 4\n1 4\n1 5\n5 7\n7 8\n7 1\n1 7\n8 7\n7 5\n3 7\n5 6\n7 3\n5 2\n3 6\n7 4\n9 6\n5 8\n9 7\n6 3\n7 9\n1 2\n4 5\n6 2\n5 3\n7 2\n1 6\n4 1\n6 1\n8 9\n2 3\n3 9\n2 9\n5 4\n2 8\n9 4\n2 5\n8 6\n3 4\n2 1\n2 7\n6 5\n9 1\n8 3\n3 8\n9 3\n4 3\n3 1\n1 9\n6 4\n3 2\n6 8\n2 6\n5 9\n8 5\n4 7\n9 5\n6 9\n9 2\n3 5\n4 9\n4 8\n2 4\n5 1\n4 6\n7 6\n9 8\n1 3\n4 2\n8 1\n8 2\n6 7\n1 8",
"output": "1 2 3 4 5 6 7 8 9 "
}
] | 1,523,093,671 | 2,147,483,647 | PyPy 3 | OK | TESTS | 39 | 202 | 23,142,400 | n = int(input())
r = [False]*n
l = [False]*n
c = ''
for i in range(n*n):
k, m= [int(i) for i in input().split()]
if not r[k - 1] and not l[m - 1]:
r[k - 1] = True
l[m - 1] = True
c += str(i + 1) + ' '
print(c) | Title: Asphalting Roads
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
City X consists of *n* vertical and *n* horizontal infinite roads, forming *n*<=×<=*n* intersections. Roads (both vertical and horizontal) are numbered from 1 to *n*, and the intersections are indicated by the numbers of the roads that form them.
Sand roads have long been recognized out of date, so the decision was made to asphalt them. To do this, a team of workers was hired and a schedule of work was made, according to which the intersections should be asphalted.
Road repairs are planned for *n*2 days. On the *i*-th day of the team arrives at the *i*-th intersection in the list and if none of the two roads that form the intersection were already asphalted they asphalt both roads. Otherwise, the team leaves the intersection, without doing anything with the roads.
According to the schedule of road works tell in which days at least one road will be asphalted.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=50) — the number of vertical and horizontal roads in the city.
Next *n*2 lines contain the order of intersections in the schedule. The *i*-th of them contains two numbers *h**i*,<=*v**i* (1<=≤<=*h**i*,<=*v**i*<=≤<=*n*), separated by a space, and meaning that the intersection that goes *i*-th in the timetable is at the intersection of the *h**i*-th horizontal and *v**i*-th vertical roads. It is guaranteed that all the intersections in the timetable are distinct.
Output Specification:
In the single line print the numbers of the days when road works will be in progress in ascending order. The days are numbered starting from 1.
Demo Input:
['2\n1 1\n1 2\n2 1\n2 2\n', '1\n1 1\n']
Demo Output:
['1 4 \n', '1 \n']
Note:
In the sample the brigade acts like that:
1. On the first day the brigade comes to the intersection of the 1-st horizontal and the 1-st vertical road. As none of them has been asphalted, the workers asphalt the 1-st vertical and the 1-st horizontal road; 1. On the second day the brigade of the workers comes to the intersection of the 1-st horizontal and the 2-nd vertical road. The 2-nd vertical road hasn't been asphalted, but as the 1-st horizontal road has been asphalted on the first day, the workers leave and do not asphalt anything; 1. On the third day the brigade of the workers come to the intersection of the 2-nd horizontal and the 1-st vertical road. The 2-nd horizontal road hasn't been asphalted but as the 1-st vertical road has been asphalted on the first day, the workers leave and do not asphalt anything; 1. On the fourth day the brigade come to the intersection formed by the intersection of the 2-nd horizontal and 2-nd vertical road. As none of them has been asphalted, the workers asphalt the 2-nd vertical and the 2-nd horizontal road. | ```python
n = int(input())
r = [False]*n
l = [False]*n
c = ''
for i in range(n*n):
k, m= [int(i) for i in input().split()]
if not r[k - 1] and not l[m - 1]:
r[k - 1] = True
l[m - 1] = True
c += str(i + 1) + ' '
print(c)
``` | 3 |
|
958 | E1 | Guard Duty (easy) | PROGRAMMING | 1,600 | [
"brute force",
"geometry",
"greedy",
"math"
] | null | null | The Rebel fleet is afraid that the Empire might want to strike back again. Princess Heidi needs to know if it is possible to assign *R* Rebel spaceships to guard *B* bases so that every base has exactly one guardian and each spaceship has exactly one assigned base (in other words, the assignment is a perfect matching). Since she knows how reckless her pilots are, she wants to be sure that any two (straight) paths – from a base to its assigned spaceship – do not intersect in the galaxy plane (that is, in 2D), and so there is no risk of collision. | The first line contains two space-separated integers *R*,<=*B*(1<=≤<=*R*,<=*B*<=≤<=10). For 1<=≤<=*i*<=≤<=*R*, the *i*<=+<=1-th line contains two space-separated integers *x**i* and *y**i* (|*x**i*|,<=|*y**i*|<=≤<=10000) denoting the coordinates of the *i*-th Rebel spaceship. The following *B* lines have the same format, denoting the position of bases. It is guaranteed that no two points coincide and that no three points are on the same line. | If it is possible to connect Rebel spaceships and bases so as satisfy the constraint, output Yes, otherwise output No (without quote). | [
"3 3\n0 0\n2 0\n3 1\n-2 1\n0 3\n2 2\n",
"2 1\n1 0\n2 2\n3 1\n"
] | [
"Yes\n",
"No\n"
] | For the first example, one possible way is to connect the Rebels and bases in order.
For the second example, there is no perfect matching between Rebels and bases. | 0 | [
{
"input": "3 3\n0 0\n2 0\n3 1\n-2 1\n0 3\n2 2",
"output": "Yes"
},
{
"input": "2 1\n1 0\n2 2\n3 1",
"output": "No"
},
{
"input": "1 1\n3686 4362\n-7485 5112",
"output": "Yes"
},
{
"input": "1 2\n1152 -7324\n-5137 -35\n-6045 -5271",
"output": "No"
},
{
"input": "1 3\n-8824 -9306\n-5646 -9767\n8123 9355\n-6203 -1643",
"output": "No"
},
{
"input": "1 5\n-8775 6730\n-3806 -6906\n-9256 -8240\n-1320 6849\n8155 746\n8284 -7317",
"output": "No"
},
{
"input": "1 8\n8741 3638\n381 -9191\n7534 8792\n-8848 -414\n2926 -7444\n9475 559\n6938 2359\n2570 4721\n3329 -9365",
"output": "No"
},
{
"input": "1 9\n6207 1655\n2728 8520\n9142 3418\n-1258 -8727\n5370 3161\n-5167 -7691\n517 2242\n3702 -9904\n-6862 -2997\n2524 -5492",
"output": "No"
},
{
"input": "1 10\n9424 3979\n-8582 9252\n-2432 -3287\n-4247 1932\n-9491 5544\n-635 5689\n8260 -6790\n8841 3067\n-5624 -1990\n1569 1045\n-8844 -8462",
"output": "No"
},
{
"input": "2 1\n2893 -5488\n-5087 -5042\n-8928 -9453",
"output": "No"
},
{
"input": "2 2\n359 -29\n6964 -7332\n2384 -4529\n4434 2253",
"output": "Yes"
},
{
"input": "2 3\n-9617 845\n4195 -2181\n-6305 -9903\n-535 -6060\n9417 -8419",
"output": "No"
},
{
"input": "2 5\n-9568 -3121\n-1408 2942\n-827 -7497\n4348 2432\n-7958 231\n6440 1896\n2647 -1305",
"output": "No"
},
{
"input": "2 8\n7948 3490\n2779 3512\n3403 -3024\n-3180 -4831\n6815 4601\n7631 9772\n-6320 -1060\n5592 362\n-785 4040\n8030 3272",
"output": "No"
},
{
"input": "2 9\n5414 -8195\n-5171 -1634\n5012 4161\n-5888 -585\n9258 2646\n5548 1523\n7259 -8619\n9580 5738\n-8715 706\n-2232 -3280\n1866 1775",
"output": "No"
},
{
"input": "2 10\n-5118 -3971\n-1169 -9140\n-7807 -3139\n9702 -5328\n8555 3460\n-1442 -733\n701 -2802\n-5784 2578\n8186 -4810\n-2722 -1013\n-9437 4021\n-5403 -1331",
"output": "No"
},
{
"input": "3 1\n-8199 -7896\n7015 -4898\n-499 -8710\n9953 -6411",
"output": "No"
},
{
"input": "3 2\n9268 -9879\n4245 2515\n-9188 -3786\n-2458 -2165\n3420 463",
"output": "No"
},
{
"input": "3 3\n-8149 697\n6593 7667\n2123 -9160\n-5165 9523\n747 -8933\n-1536 -2691",
"output": "Yes"
},
{
"input": "3 5\n-658 7030\n990 3086\n-4958 -6755\n7159 -1986\n5634 -7726\n1740 -1450\n1947 7835\n-2755 -4709",
"output": "No"
},
{
"input": "3 8\n-3143 -6360\n-5121 -6641\n-727 -9723\n-369 454\n-9298 4086\n5787 -1016\n2683 -9660\n-1089 1121\n-4898 7743\n418 5485\n7425 -6644",
"output": "No"
},
{
"input": "3 9\n6882 -8342\n4669 -8932\n882 4904\n-220 4700\n587 -5311\n3704 -1823\n6559 -6921\n-7399 6497\n-5387 -5890\n-9844 -1067\n5367 -7237\n-8314 -939",
"output": "No"
},
{
"input": "3 10\n-7100 -1623\n-3459 2172\n9676 1595\n-6053 4558\n-842 8819\n-9691 3144\n3440 -9112\n7034 4946\n4851 -2513\n430 4372\n-7175 -3497\n5719 381\n-8859 -1347",
"output": "No"
},
{
"input": "5 1\n9621 -154\n6694 -2348\n944 -7225\n-1568 -5543\n-3805 -872\n1204 -2651",
"output": "No"
},
{
"input": "5 2\n-355 -9579\n-1256 -4638\n-4890 7402\n-1420 -1297\n-1362 2290\n-879 9101\n9514 -6689",
"output": "No"
},
{
"input": "5 3\n9670 8440\n1091 -9784\n6422 4884\n3314 -9610\n8523 -7107\n-2963 8293\n3092 -3950\n-4093 -6502",
"output": "No"
},
{
"input": "5 5\n-2840 4475\n2931 -6923\n-659 -8125\n8197 -1118\n851 -5899\n313 6679\n-9751 6115\n-6415 4250\n-227 -9732\n-6282 5041",
"output": "Yes"
},
{
"input": "5 8\n-5325 1383\n-5441 3351\n-3870 1465\n669 -8381\n-4377 5913\n4360 -329\n8725 8620\n7810 -2479\n4019 4850\n8052 9911\n4130 -4668\n3744 2537\n-7171 -3933",
"output": "No"
},
{
"input": "5 9\n-2742 -600\n6609 8502\n-5118 6389\n-4300 5568\n-1934 -3484\n9719 -1137\n2303 -8641\n1500 2897\n-6172 -8783\n-2210 -6939\n9514 -5262\n-3773 -4081\n1983 -4032\n4503 -3496",
"output": "No"
},
{
"input": "5 10\n1493 7658\n-598 7650\n9226 -964\n2439 -3114\n366 2391\n-1008 -2258\n6063 8568\n7488 6824\n-4674 9523\n9590 9960\n-8361 -8234\n520 -1312\n-3878 -1142\n-8261 -239\n-2346 -2362",
"output": "No"
},
{
"input": "8 1\n-796 -1\n3591 -2510\n-6330 4706\n-7422 -9093\n7860 -7002\n5375 -5310\n3538 3108\n-9851 -9798\n-8884 -170",
"output": "No"
},
{
"input": "8 2\n-3330 -1983\n-6621 -4800\n-4721 9630\n9871 -4847\n-2256 -8957\n3292 -6118\n4558 -6712\n-5863 5282\n-9373 3938\n-5179 -8073",
"output": "No"
},
{
"input": "8 3\n6695 8593\n-7129 352\n6590 -5447\n-2540 -3457\n7630 1647\n8651 5634\n-1864 -6829\n7828 -1901\n-7005 -9695\n4561 -4921\n-4782 -6478",
"output": "No"
},
{
"input": "8 5\n6744 2367\n-5290 -7085\n-491 6662\n2343 -2407\n-43 2855\n-8075 6875\n-7265 -4206\n-4197 8851\n7433 780\n4038 -8321\n-1455 -7665\n3139 -1225\n9884 -167",
"output": "No"
},
{
"input": "8 8\n4260 1536\n-8545 6045\n-3702 3693\n-5185 -2228\n-5271 -5335\n-4027 4453\n-8790 8598\n7172 -5320\n-880 -4638\n-1630 -3452\n2076 8296\n-9116 -5599\n2461 9832\n4268 5116\n-7582 -805\n3548 3776",
"output": "Yes"
},
{
"input": "8 9\n-5716 6995\n1245 3754\n7610 8617\n-451 -5424\n-2828 5270\n-6111 6502\n-2653 1039\n3718 7498\n-8810 -7973\n667 -300\n-2838 -2001\n3367 5523\n-8386 -2827\n6929 -6260\n3247 1167\n1873 6265\n4376 -8781",
"output": "No"
},
{
"input": "8 10\n5844 -8156\n9676 -8121\n-6302 -1050\n-4823 -8343\n4736 -3859\n9129 5920\n-3990 2792\n3615 -8930\n-7831 -8703\n-5542 931\n7599 -7930\n8705 -8735\n-6438 1724\n-7568 -8351\n5893 2316\n2574 -9723\n2416 3827\n856 -4877",
"output": "No"
},
{
"input": "9 1\n8114 -9851\n872 -9807\n9541 5449\n7948 -3808\n8892 -7517\n-6767 3903\n-18 -311\n-3973 5845\n-3295 3533\n-4790 -4426",
"output": "No"
},
{
"input": "9 2\n5580 8167\n-7078 -4655\n3707 -9628\n2980 438\n1632 -9472\n-8850 -4346\n-6440 2428\n-2841 923\n6515 -2658\n-2492 -8716\n8219 5104",
"output": "No"
},
{
"input": "9 3\n8163 6185\n-4731 2757\n-4982 -4704\n3128 4684\n-8483 1132\n6807 2288\n4878 2311\n-6295 6299\n8882 -5992\n-195 4733\n6162 4510\n-7264 -1020",
"output": "No"
},
{
"input": "9 5\n-4347 -5222\n-2891 5618\n-4621 7404\n-4548 -6825\n3846 2340\n2640 3530\n-7965 4934\n-8617 -2950\n-9240 4483\n-718 6451\n-8251 -6379\n558 3484\n9861 -6432\n483 -7331",
"output": "No"
},
{
"input": "9 8\n-6832 -872\n1295 -4109\n-7832 -8123\n-2373 -6646\n-1383 -5849\n3832 -6334\n-7229 -2263\n-6951 -9678\n4709 1326\n-6386 -1239\n2721 -8159\n-4255 -890\n9880 3567\n3349 5921\n2487 -828\n-783 2422\n-5497 -8399",
"output": "No"
},
{
"input": "9 9\n3193 -2855\n787 -6399\n3479 9360\n5217 -9842\n1061 4755\n1748 -7142\n-6209 -2380\n6740 -4302\n-5482 5433\n3353 -5529\n664 1546\n8228 -9769\n-8409 -1650\n893 9365\n-9542 8585\n7245 -9972\n-475 -6359\n-3775 2139",
"output": "Yes"
},
{
"input": "9 10\n-3581 3894\n7385 3191\n-8820 6540\n-577 -5900\n2781 -5943\n8322 -7944\n-1251 -5779\n-3567 3140\n8835 -6406\n-2390 -1126\n7006 4553\n-174 -7023\n-6538 1530\n3318 2477\n7864 -9657\n-2379 -6961\n4456 9852\n1462 -5871\n-9931 6466",
"output": "No"
},
{
"input": "10 1\n3362 3137\n-6006 -2168\n-9207 8006\n-6284 -114\n4617 -4997\n-4360 3540\n-6423 2328\n-8768 8468\n2899 1032\n-7561 -3623\n6979 653",
"output": "No"
},
{
"input": "10 2\n5945 8596\n-3658 -4459\n-7598 -7071\n3567 4132\n7060 -1835\n-6443 -4709\n4895 2211\n-4780 3546\n5266 7400\n2178 -472\n4922 -9643\n4163 6030",
"output": "No"
},
{
"input": "10 3\n3411 6614\n8392 693\n-8846 7555\n-1402 -4181\n-3055 -3789\n4033 -5516\n-1527 4950\n-792 8922\n-4925 4065\n4475 5536\n-9695 9764\n6943 -2849\n7022 1986",
"output": "No"
},
{
"input": "10 5\n3460 5504\n529 -6744\n4075 9961\n-3961 4311\n-7871 9977\n7308 -4275\n-6928 7573\n-3114 -327\n-3046 -5461\n3953 4398\n-4106 -3981\n-8092 -8048\n7590 9228\n9433 -4\n-8808 -6742",
"output": "No"
},
{
"input": "10 8\n8417 -444\n-5582 6386\n863 6992\n-4047 6751\n-5658 1788\n-1204 5862\n-6192 -2480\n813 -7056\n-9098 -1176\n-1715 -3292\n6866 -2905\n-7788 137\n7609 -774\n-7702 -6753\n-6622 -3090\n3089 -7006\n-9374 1882\n-481 -5698",
"output": "No"
},
{
"input": "10 9\n-9001 -9868\n4207 1240\n-7826 1618\n-6755 3555\n-3214 -167\n4155 -4648\n-2316 259\n4801 -1679\n-6730 8048\n-4535 -9843\n4809 -5759\n4695 -8742\n9321 -5991\n2401 4133\n6468 6324\n1414 -9103\n-6613 3922\n5544 -5092\n-6777 -788",
"output": "No"
},
{
"input": "10 10\n8530 -3814\n-9330 -6035\n3951 -217\n-9276 8291\n636 -3118\n5024 -2403\n4601 7977\n-3620 -1428\n4954 -9632\n-9852 6553\n-3457 5430\n-8866 -7343\n1020 -5748\n5043 -3820\n-2832 1528\n-5058 -825\n2406 -3530\n9152 -7463\n-8547 7108\n2492 8953",
"output": "Yes"
}
] | 1,523,693,135 | 3,635 | PyPy 3 | OK | TESTS | 51 | 233 | 26,009,600 | def ccw(A, B, C):
return (C[1] - A[1]) * (B[0] - A[0]) > (B[1] - A[1]) * (C[0] - A[0])
def intersect(A, B, C, D):
return ccw(A, C, D) != ccw(B, C, D) and ccw(A, B, C) != ccw(A, B, D)
R, B = map(int, input().split())
rs = []
bs = []
for r in range(R):
rs.append(list(map(int, input().split())))
for r in range(B):
bs.append(list(map(int, input().split())))
if R != B:
print('No')
else:
def rec(at, done, remain):
if at >= B:
return True
for b in remain:
for r, d in zip(rs, done):
if intersect(r, bs[d], rs[at], bs[b]):
break
else:
ok = rec(at + 1, done + [b], remain - {b})
if ok:
return True
return False
print(['NO', 'YES'][rec(0, [], set(range(B)))])
| Title: Guard Duty (easy)
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Rebel fleet is afraid that the Empire might want to strike back again. Princess Heidi needs to know if it is possible to assign *R* Rebel spaceships to guard *B* bases so that every base has exactly one guardian and each spaceship has exactly one assigned base (in other words, the assignment is a perfect matching). Since she knows how reckless her pilots are, she wants to be sure that any two (straight) paths – from a base to its assigned spaceship – do not intersect in the galaxy plane (that is, in 2D), and so there is no risk of collision.
Input Specification:
The first line contains two space-separated integers *R*,<=*B*(1<=≤<=*R*,<=*B*<=≤<=10). For 1<=≤<=*i*<=≤<=*R*, the *i*<=+<=1-th line contains two space-separated integers *x**i* and *y**i* (|*x**i*|,<=|*y**i*|<=≤<=10000) denoting the coordinates of the *i*-th Rebel spaceship. The following *B* lines have the same format, denoting the position of bases. It is guaranteed that no two points coincide and that no three points are on the same line.
Output Specification:
If it is possible to connect Rebel spaceships and bases so as satisfy the constraint, output Yes, otherwise output No (without quote).
Demo Input:
['3 3\n0 0\n2 0\n3 1\n-2 1\n0 3\n2 2\n', '2 1\n1 0\n2 2\n3 1\n']
Demo Output:
['Yes\n', 'No\n']
Note:
For the first example, one possible way is to connect the Rebels and bases in order.
For the second example, there is no perfect matching between Rebels and bases. | ```python
def ccw(A, B, C):
return (C[1] - A[1]) * (B[0] - A[0]) > (B[1] - A[1]) * (C[0] - A[0])
def intersect(A, B, C, D):
return ccw(A, C, D) != ccw(B, C, D) and ccw(A, B, C) != ccw(A, B, D)
R, B = map(int, input().split())
rs = []
bs = []
for r in range(R):
rs.append(list(map(int, input().split())))
for r in range(B):
bs.append(list(map(int, input().split())))
if R != B:
print('No')
else:
def rec(at, done, remain):
if at >= B:
return True
for b in remain:
for r, d in zip(rs, done):
if intersect(r, bs[d], rs[at], bs[b]):
break
else:
ok = rec(at + 1, done + [b], remain - {b})
if ok:
return True
return False
print(['NO', 'YES'][rec(0, [], set(range(B)))])
``` | 3 |
|
965 | A | Paper Airplanes | PROGRAMMING | 800 | [
"math"
] | null | null | To make a paper airplane, one has to use a rectangular piece of paper. From a sheet of standard size you can make $s$ airplanes.
A group of $k$ people decided to make $n$ airplanes each. They are going to buy several packs of paper, each of them containing $p$ sheets, and then distribute the sheets between the people. Each person should have enough sheets to make $n$ airplanes. How many packs should they buy? | The only line contains four integers $k$, $n$, $s$, $p$ ($1 \le k, n, s, p \le 10^4$) — the number of people, the number of airplanes each should make, the number of airplanes that can be made using one sheet and the number of sheets in one pack, respectively. | Print a single integer — the minimum number of packs they should buy. | [
"5 3 2 3\n",
"5 3 100 1\n"
] | [
"4\n",
"5\n"
] | In the first sample they have to buy $4$ packs of paper: there will be $12$ sheets in total, and giving $2$ sheets to each person is enough to suit everyone's needs.
In the second sample they have to buy a pack for each person as they can't share sheets. | 500 | [
{
"input": "5 3 2 3",
"output": "4"
},
{
"input": "5 3 100 1",
"output": "5"
},
{
"input": "10000 10000 1 1",
"output": "100000000"
},
{
"input": "1 1 10000 10000",
"output": "1"
},
{
"input": "300 300 21 23",
"output": "196"
},
{
"input": "300 2 37 51",
"output": "6"
},
{
"input": "2 400 23 57",
"output": "1"
},
{
"input": "1 1 1 1",
"output": "1"
},
{
"input": "10000 10000 3 2",
"output": "16670000"
},
{
"input": "1 2 1 2",
"output": "1"
},
{
"input": "1 1 2 2",
"output": "1"
},
{
"input": "1 1 10 10",
"output": "1"
},
{
"input": "5324 5439 32 13",
"output": "69622"
},
{
"input": "9000 1 2432 1",
"output": "9000"
},
{
"input": "230 1234 9124 23",
"output": "10"
},
{
"input": "11 1 1 1",
"output": "11"
},
{
"input": "6246 8489 1227 9",
"output": "4858"
},
{
"input": "9 20 5 7",
"output": "6"
}
] | 1,548,683,957 | 2,147,483,647 | Python 3 | OK | TESTS | 18 | 108 | 0 | from math import ceil
X = list(map(int, input().split()))
Math = ceil(X[1] / X[2])
Math = ceil(Math * X[0])
print(ceil(Math / X[3]))
| Title: Paper Airplanes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
To make a paper airplane, one has to use a rectangular piece of paper. From a sheet of standard size you can make $s$ airplanes.
A group of $k$ people decided to make $n$ airplanes each. They are going to buy several packs of paper, each of them containing $p$ sheets, and then distribute the sheets between the people. Each person should have enough sheets to make $n$ airplanes. How many packs should they buy?
Input Specification:
The only line contains four integers $k$, $n$, $s$, $p$ ($1 \le k, n, s, p \le 10^4$) — the number of people, the number of airplanes each should make, the number of airplanes that can be made using one sheet and the number of sheets in one pack, respectively.
Output Specification:
Print a single integer — the minimum number of packs they should buy.
Demo Input:
['5 3 2 3\n', '5 3 100 1\n']
Demo Output:
['4\n', '5\n']
Note:
In the first sample they have to buy $4$ packs of paper: there will be $12$ sheets in total, and giving $2$ sheets to each person is enough to suit everyone's needs.
In the second sample they have to buy a pack for each person as they can't share sheets. | ```python
from math import ceil
X = list(map(int, input().split()))
Math = ceil(X[1] / X[2])
Math = ceil(Math * X[0])
print(ceil(Math / X[3]))
``` | 3 |
|
1 | A | Theatre Square | PROGRAMMING | 1,000 | [
"math"
] | A. Theatre Square | 1 | 256 | Theatre Square in the capital city of Berland has a rectangular shape with the size *n*<=×<=*m* meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size *a*<=×<=*a*.
What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square. | The input contains three positive integer numbers in the first line: *n*,<=<=*m* and *a* (1<=≤<=<=*n*,<=*m*,<=*a*<=≤<=109). | Write the needed number of flagstones. | [
"6 6 4\n"
] | [
"4\n"
] | none | 0 | [
{
"input": "6 6 4",
"output": "4"
},
{
"input": "1 1 1",
"output": "1"
},
{
"input": "2 1 1",
"output": "2"
},
{
"input": "1 2 1",
"output": "2"
},
{
"input": "2 2 1",
"output": "4"
},
{
"input": "2 1 2",
"output": "1"
},
{
"input": "1 1 3",
"output": "1"
},
{
"input": "2 3 4",
"output": "1"
},
{
"input": "1000000000 1000000000 1",
"output": "1000000000000000000"
},
{
"input": "12 13 4",
"output": "12"
},
{
"input": "222 332 5",
"output": "3015"
},
{
"input": "1000 1000 10",
"output": "10000"
},
{
"input": "1001 1000 10",
"output": "10100"
},
{
"input": "100 10001 1000000000",
"output": "1"
},
{
"input": "1000000000 1000000000 1000000000",
"output": "1"
},
{
"input": "1000000000 1000000000 999999999",
"output": "4"
},
{
"input": "1000000000 1000000000 192",
"output": "27126743055556"
},
{
"input": "1000000000 987654321 1",
"output": "987654321000000000"
},
{
"input": "456784567 1000000000 51",
"output": "175618850864484"
},
{
"input": "39916800 134217728 40320",
"output": "3295710"
}
] | 1,690,919,297 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 1 | 31 | 0 | n,m,a=map(int, input().split())
if n/a > int(n/a) : x=int(n/a)+1
if m/a > int (m/a) : y=int (n/a)+1
print (x+y)
| Title: Theatre Square
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Theatre Square in the capital city of Berland has a rectangular shape with the size *n*<=×<=*m* meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size *a*<=×<=*a*.
What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square.
Input Specification:
The input contains three positive integer numbers in the first line: *n*,<=<=*m* and *a* (1<=≤<=<=*n*,<=*m*,<=*a*<=≤<=109).
Output Specification:
Write the needed number of flagstones.
Demo Input:
['6 6 4\n']
Demo Output:
['4\n']
Note:
none | ```python
n,m,a=map(int, input().split())
if n/a > int(n/a) : x=int(n/a)+1
if m/a > int (m/a) : y=int (n/a)+1
print (x+y)
``` | -1 |
886 | A | ACM ICPC | PROGRAMMING | 1,000 | [
"brute force"
] | null | null | In a small but very proud high school it was decided to win ACM ICPC. This goal requires to compose as many teams of three as possible, but since there were only 6 students who wished to participate, the decision was to build exactly two teams.
After practice competition, participant number *i* got a score of *a**i*. Team score is defined as sum of scores of its participants. High school management is interested if it's possible to build two teams with equal scores. Your task is to answer that question. | The single line contains six integers *a*1,<=...,<=*a*6 (0<=≤<=*a**i*<=≤<=1000) — scores of the participants | Print "YES" (quotes for clarity), if it is possible to build teams with equal score, and "NO" otherwise.
You can print each character either upper- or lowercase ("YeS" and "yes" are valid when the answer is "YES"). | [
"1 3 2 1 2 1\n",
"1 1 1 1 1 99\n"
] | [
"YES\n",
"NO\n"
] | In the first sample, first team can be composed of 1st, 2nd and 6th participant, second — of 3rd, 4th and 5th: team scores are 1 + 3 + 1 = 2 + 1 + 2 = 5.
In the second sample, score of participant number 6 is too high: his team score will be definitely greater. | 500 | [
{
"input": "1 3 2 1 2 1",
"output": "YES"
},
{
"input": "1 1 1 1 1 99",
"output": "NO"
},
{
"input": "1000 1000 1000 1000 1000 1000",
"output": "YES"
},
{
"input": "0 0 0 0 0 0",
"output": "YES"
},
{
"input": "633 609 369 704 573 416",
"output": "NO"
},
{
"input": "353 313 327 470 597 31",
"output": "NO"
},
{
"input": "835 638 673 624 232 266",
"output": "NO"
},
{
"input": "936 342 19 398 247 874",
"output": "NO"
},
{
"input": "417 666 978 553 271 488",
"output": "NO"
},
{
"input": "71 66 124 199 67 147",
"output": "YES"
},
{
"input": "54 26 0 171 239 12",
"output": "YES"
},
{
"input": "72 8 186 92 267 69",
"output": "YES"
},
{
"input": "180 179 188 50 75 214",
"output": "YES"
},
{
"input": "16 169 110 136 404 277",
"output": "YES"
},
{
"input": "101 400 9 200 300 10",
"output": "YES"
},
{
"input": "101 400 200 9 300 10",
"output": "YES"
},
{
"input": "101 200 400 9 300 10",
"output": "YES"
},
{
"input": "101 400 200 300 9 10",
"output": "YES"
},
{
"input": "101 200 400 300 9 10",
"output": "YES"
},
{
"input": "4 4 4 4 5 4",
"output": "NO"
},
{
"input": "2 2 2 2 2 1",
"output": "NO"
},
{
"input": "1000 1000 999 1000 1000 1000",
"output": "NO"
},
{
"input": "129 1 10 29 8 111",
"output": "NO"
},
{
"input": "1000 1000 1000 999 999 1000",
"output": "YES"
},
{
"input": "101 200 300 400 9 10",
"output": "YES"
},
{
"input": "101 400 200 300 10 9",
"output": "YES"
},
{
"input": "101 200 400 300 10 9",
"output": "YES"
},
{
"input": "101 200 300 400 10 9",
"output": "YES"
},
{
"input": "101 200 300 10 400 9",
"output": "YES"
},
{
"input": "1 1 1 1 1 5",
"output": "NO"
},
{
"input": "8 1 1 3 3 0",
"output": "NO"
},
{
"input": "1 1 2 2 3 3",
"output": "YES"
},
{
"input": "1 2 2 5 2 5",
"output": "NO"
},
{
"input": "1 2 3 6 6 6",
"output": "NO"
},
{
"input": "36 91 7 86 51 89",
"output": "NO"
},
{
"input": "10 1 1 1 23 24",
"output": "NO"
},
{
"input": "1 1 1 10 23 24",
"output": "NO"
},
{
"input": "20 10 1 2 3 44",
"output": "NO"
},
{
"input": "7 0 14 11 8 6",
"output": "NO"
},
{
"input": "100 496 1 1 1 1",
"output": "NO"
},
{
"input": "5 4 2 5 11 3",
"output": "NO"
},
{
"input": "1 3 7 8 8 9",
"output": "YES"
},
{
"input": "1 3 4 5 18 19",
"output": "YES"
},
{
"input": "5 5 1 2 2 15",
"output": "NO"
},
{
"input": "2 1 0 0 0 5",
"output": "NO"
},
{
"input": "1 6 6 1 20 2",
"output": "NO"
},
{
"input": "2 10 0 0 0 0",
"output": "NO"
},
{
"input": "1 1 3 1 1 11",
"output": "NO"
},
{
"input": "10 10 1 1 1 37",
"output": "NO"
},
{
"input": "1 1 0 0 0 4",
"output": "NO"
},
{
"input": "1 1 10 1 1 28",
"output": "NO"
},
{
"input": "1 5 5 5 6 8",
"output": "YES"
},
{
"input": "0 2 3 4 4 5",
"output": "YES"
}
] | 1,510,511,049 | 8,349 | Python 3 | RUNTIME_ERROR | PRETESTS | 0 | 30 | 0 | a = input().split()
c = 0
for i in range(len(a)):
a[i] = int(a[i])
c += a[i]
c = c/2
t = 0
for i in range(6):
if a[i] == c:
t = 1
for i in range(6):
for j in range(6):
if a[i]+a[i1] == c:
t = 1
for i in range(6):
for j in range(6):
for z in range(6):
if a[i]+a[i1]+a[2] == c:
t = 1
if t == 1:
print("Yes")
if t == 0:
print("No")
| Title: ACM ICPC
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In a small but very proud high school it was decided to win ACM ICPC. This goal requires to compose as many teams of three as possible, but since there were only 6 students who wished to participate, the decision was to build exactly two teams.
After practice competition, participant number *i* got a score of *a**i*. Team score is defined as sum of scores of its participants. High school management is interested if it's possible to build two teams with equal scores. Your task is to answer that question.
Input Specification:
The single line contains six integers *a*1,<=...,<=*a*6 (0<=≤<=*a**i*<=≤<=1000) — scores of the participants
Output Specification:
Print "YES" (quotes for clarity), if it is possible to build teams with equal score, and "NO" otherwise.
You can print each character either upper- or lowercase ("YeS" and "yes" are valid when the answer is "YES").
Demo Input:
['1 3 2 1 2 1\n', '1 1 1 1 1 99\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first sample, first team can be composed of 1st, 2nd and 6th participant, second — of 3rd, 4th and 5th: team scores are 1 + 3 + 1 = 2 + 1 + 2 = 5.
In the second sample, score of participant number 6 is too high: his team score will be definitely greater. | ```python
a = input().split()
c = 0
for i in range(len(a)):
a[i] = int(a[i])
c += a[i]
c = c/2
t = 0
for i in range(6):
if a[i] == c:
t = 1
for i in range(6):
for j in range(6):
if a[i]+a[i1] == c:
t = 1
for i in range(6):
for j in range(6):
for z in range(6):
if a[i]+a[i1]+a[2] == c:
t = 1
if t == 1:
print("Yes")
if t == 0:
print("No")
``` | -1 |
|
767 | A | Snacktower | PROGRAMMING | 1,100 | [
"data structures",
"implementation"
] | null | null | According to an old legeng, a long time ago Ankh-Morpork residents did something wrong to miss Fortune, and she cursed them. She said that at some time *n* snacks of distinct sizes will fall on the city, and the residents should build a Snacktower of them by placing snacks one on another. Of course, big snacks should be at the bottom of the tower, while small snacks should be at the top.
Years passed, and once different snacks started to fall onto the city, and the residents began to build the Snacktower.
However, they faced some troubles. Each day exactly one snack fell onto the city, but their order was strange. So, at some days the residents weren't able to put the new stack on the top of the Snacktower: they had to wait until all the bigger snacks fell. Of course, in order to not to anger miss Fortune again, the residents placed each snack on the top of the tower immediately as they could do it.
Write a program that models the behavior of Ankh-Morpork residents. | The first line contains single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the total number of snacks.
The second line contains *n* integers, the *i*-th of them equals the size of the snack which fell on the *i*-th day. Sizes are distinct integers from 1 to *n*. | Print *n* lines. On the *i*-th of them print the sizes of the snacks which the residents placed on the top of the Snacktower on the *i*-th day in the order they will do that. If no snack is placed on some day, leave the corresponding line empty. | [
"3\n3 1 2\n",
"5\n4 5 1 2 3\n"
] | [
"3\n \n2 1",
"5 4\n \n \n3 2 1\n"
] | In the example a snack of size 3 fell on the first day, and the residents immediately placed it. On the second day a snack of size 1 fell, and the residents weren't able to place it because they were missing the snack of size 2. On the third day a snack of size 2 fell, and the residents immediately placed it. Right after that they placed the snack of size 1 which had fallen before. | 500 | [
{
"input": "3\n3 1 2",
"output": "3 \n\n2 1 "
},
{
"input": "5\n4 5 1 2 3",
"output": "5 4 \n\n\n3 2 1 "
},
{
"input": "1\n1",
"output": "1 "
},
{
"input": "2\n1 2",
"output": "2 1 "
},
{
"input": "10\n5 1 6 2 8 3 4 10 9 7",
"output": "10 \n9 8 \n7 6 5 4 3 2 1 "
},
{
"input": "30\n16 10 4 29 5 28 12 21 11 30 18 6 14 3 17 22 20 15 9 1 27 19 24 26 13 25 2 23 8 7",
"output": "30 29 28 \n\n\n\n\n\n\n\n\n\n\n27 \n\n\n26 \n\n25 24 \n\n23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 \n8 \n7 6 5 4 3 2 1 "
},
{
"input": "100\n98 52 63 2 18 96 31 58 84 40 41 45 66 100 46 71 26 48 81 20 73 91 68 76 13 93 17 29 64 95 79 21 55 75 19 85 54 51 89 78 15 87 43 59 36 1 90 35 65 56 62 28 86 5 82 49 3 99 33 9 92 32 74 69 27 22 77 16 44 94 34 6 57 70 23 12 61 25 8 11 67 47 83 88 10 14 30 7 97 60 42 37 24 38 53 50 4 80 72 39",
"output": "100 \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n99 98 \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n97 96 95 94 93 92 91 90 89 88 87 86 85 84 83 82 81 \n\n\n\n\n\n\n\n\n80 79 78 77 76 75 74 73 \n72 71 70 69 68 67 66 65 64 63 62 61 60 59 58 57 56 55 54 53 52 51 50 49 48 47 46 45 44 43 42 41 40 \n39 38 37 36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 "
},
{
"input": "2\n2 1",
"output": "2 \n1 "
}
] | 1,595,654,506 | 2,147,483,647 | Python 3 | OK | TESTS | 25 | 389 | 17,817,600 | import sys
import heapq
def main(arr):
visited = [False] * len(arr)
last = len(visited) - 1
for n in arr:
visited[n - 1] = True
doing = []
while last >= 0 and visited[last]:
doing.append(last + 1)
last -= 1
print(" ".join(str(d) for d in doing))
if __name__ == "__main__":
for e, line in enumerate(sys.stdin.readlines()):
if e == 0:
continue
else:
arr = list(map(int, line.strip().split()))
# Proof by induction that sorting is that
# same as solving this problem.
main(arr)
| Title: Snacktower
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
According to an old legeng, a long time ago Ankh-Morpork residents did something wrong to miss Fortune, and she cursed them. She said that at some time *n* snacks of distinct sizes will fall on the city, and the residents should build a Snacktower of them by placing snacks one on another. Of course, big snacks should be at the bottom of the tower, while small snacks should be at the top.
Years passed, and once different snacks started to fall onto the city, and the residents began to build the Snacktower.
However, they faced some troubles. Each day exactly one snack fell onto the city, but their order was strange. So, at some days the residents weren't able to put the new stack on the top of the Snacktower: they had to wait until all the bigger snacks fell. Of course, in order to not to anger miss Fortune again, the residents placed each snack on the top of the tower immediately as they could do it.
Write a program that models the behavior of Ankh-Morpork residents.
Input Specification:
The first line contains single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the total number of snacks.
The second line contains *n* integers, the *i*-th of them equals the size of the snack which fell on the *i*-th day. Sizes are distinct integers from 1 to *n*.
Output Specification:
Print *n* lines. On the *i*-th of them print the sizes of the snacks which the residents placed on the top of the Snacktower on the *i*-th day in the order they will do that. If no snack is placed on some day, leave the corresponding line empty.
Demo Input:
['3\n3 1 2\n', '5\n4 5 1 2 3\n']
Demo Output:
['3\n\xa0\n2 1', '5 4\n\xa0\n\xa0\n3 2 1\n']
Note:
In the example a snack of size 3 fell on the first day, and the residents immediately placed it. On the second day a snack of size 1 fell, and the residents weren't able to place it because they were missing the snack of size 2. On the third day a snack of size 2 fell, and the residents immediately placed it. Right after that they placed the snack of size 1 which had fallen before. | ```python
import sys
import heapq
def main(arr):
visited = [False] * len(arr)
last = len(visited) - 1
for n in arr:
visited[n - 1] = True
doing = []
while last >= 0 and visited[last]:
doing.append(last + 1)
last -= 1
print(" ".join(str(d) for d in doing))
if __name__ == "__main__":
for e, line in enumerate(sys.stdin.readlines()):
if e == 0:
continue
else:
arr = list(map(int, line.strip().split()))
# Proof by induction that sorting is that
# same as solving this problem.
main(arr)
``` | 3 |
|
466 | A | Cheap Travel | PROGRAMMING | 1,200 | [
"implementation"
] | null | null | Ann has recently started commuting by subway. We know that a one ride subway ticket costs *a* rubles. Besides, Ann found out that she can buy a special ticket for *m* rides (she can buy it several times). It costs *b* rubles. Ann did the math; she will need to use subway *n* times. Help Ann, tell her what is the minimum sum of money she will have to spend to make *n* rides? | The single line contains four space-separated integers *n*, *m*, *a*, *b* (1<=≤<=*n*,<=*m*,<=*a*,<=*b*<=≤<=1000) — the number of rides Ann has planned, the number of rides covered by the *m* ride ticket, the price of a one ride ticket and the price of an *m* ride ticket. | Print a single integer — the minimum sum in rubles that Ann will need to spend. | [
"6 2 1 2\n",
"5 2 2 3\n"
] | [
"6\n",
"8\n"
] | In the first sample one of the optimal solutions is: each time buy a one ride ticket. There are other optimal solutions. For example, buy three *m* ride tickets. | 500 | [
{
"input": "6 2 1 2",
"output": "6"
},
{
"input": "5 2 2 3",
"output": "8"
},
{
"input": "10 3 5 1",
"output": "4"
},
{
"input": "1000 1 1000 1000",
"output": "1000000"
},
{
"input": "1000 3 1000 1000",
"output": "334000"
},
{
"input": "1 1 1 1",
"output": "1"
},
{
"input": "10 2 1 1",
"output": "5"
},
{
"input": "1 1000 1 2",
"output": "1"
},
{
"input": "1 1000 3 2",
"output": "2"
},
{
"input": "10 3 1 2",
"output": "7"
},
{
"input": "995 1 2 1",
"output": "995"
},
{
"input": "556 2 16 15",
"output": "4170"
},
{
"input": "477 2 16 14",
"output": "3346"
},
{
"input": "101 110 1 100",
"output": "100"
},
{
"input": "9 3 3 10",
"output": "27"
},
{
"input": "100 8 10 1",
"output": "13"
},
{
"input": "6 4 1 3",
"output": "5"
},
{
"input": "8 5 2 8",
"output": "14"
},
{
"input": "1000 2 1 1000",
"output": "1000"
}
] | 1,693,628,928 | 2,147,483,647 | Python 3 | OK | TESTS | 19 | 46 | 0 | import math
numbers = input('')
numbers = numbers.split(' ')
numbers = [int(i) for i in numbers]
n = numbers[0]
m = numbers[1]
a = numbers[2]
b = numbers[3]
without_special = n * a
with_special = ((math.ceil(n / m) * m) / m) * b
hybrid = (math.floor(n / m) * b) + ((n % m) * a)
print(int(min(without_special, with_special, hybrid))) | Title: Cheap Travel
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Ann has recently started commuting by subway. We know that a one ride subway ticket costs *a* rubles. Besides, Ann found out that she can buy a special ticket for *m* rides (she can buy it several times). It costs *b* rubles. Ann did the math; she will need to use subway *n* times. Help Ann, tell her what is the minimum sum of money she will have to spend to make *n* rides?
Input Specification:
The single line contains four space-separated integers *n*, *m*, *a*, *b* (1<=≤<=*n*,<=*m*,<=*a*,<=*b*<=≤<=1000) — the number of rides Ann has planned, the number of rides covered by the *m* ride ticket, the price of a one ride ticket and the price of an *m* ride ticket.
Output Specification:
Print a single integer — the minimum sum in rubles that Ann will need to spend.
Demo Input:
['6 2 1 2\n', '5 2 2 3\n']
Demo Output:
['6\n', '8\n']
Note:
In the first sample one of the optimal solutions is: each time buy a one ride ticket. There are other optimal solutions. For example, buy three *m* ride tickets. | ```python
import math
numbers = input('')
numbers = numbers.split(' ')
numbers = [int(i) for i in numbers]
n = numbers[0]
m = numbers[1]
a = numbers[2]
b = numbers[3]
without_special = n * a
with_special = ((math.ceil(n / m) * m) / m) * b
hybrid = (math.floor(n / m) * b) + ((n % m) * a)
print(int(min(without_special, with_special, hybrid)))
``` | 3 |
|
985 | B | Switches and Lamps | PROGRAMMING | 1,200 | [
"implementation"
] | null | null | You are given *n* switches and *m* lamps. The *i*-th switch turns on some subset of the lamps. This information is given as the matrix *a* consisting of *n* rows and *m* columns where *a**i*,<=*j*<==<=1 if the *i*-th switch turns on the *j*-th lamp and *a**i*,<=*j*<==<=0 if the *i*-th switch is not connected to the *j*-th lamp.
Initially all *m* lamps are turned off.
Switches change state only from "off" to "on". It means that if you press two or more switches connected to the same lamp then the lamp will be turned on after any of this switches is pressed and will remain its state even if any switch connected to this lamp is pressed afterwards.
It is guaranteed that if you push all *n* switches then all *m* lamps will be turned on.
Your think that you have too many switches and you would like to ignore one of them.
Your task is to say if there exists such a switch that if you will ignore (not use) it but press all the other *n*<=-<=1 switches then all the *m* lamps will be turned on. | The first line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=2000) — the number of the switches and the number of the lamps.
The following *n* lines contain *m* characters each. The character *a**i*,<=*j* is equal to '1' if the *i*-th switch turns on the *j*-th lamp and '0' otherwise.
It is guaranteed that if you press all *n* switches all *m* lamps will be turned on. | Print "YES" if there is a switch that if you will ignore it and press all the other *n*<=-<=1 switches then all *m* lamps will be turned on. Print "NO" if there is no such switch. | [
"4 5\n10101\n01000\n00111\n10000\n",
"4 5\n10100\n01000\n00110\n00101\n"
] | [
"YES\n",
"NO\n"
] | none | 0 | [
{
"input": "4 5\n10101\n01000\n00111\n10000",
"output": "YES"
},
{
"input": "4 5\n10100\n01000\n00110\n00101",
"output": "NO"
},
{
"input": "1 5\n11111",
"output": "NO"
},
{
"input": "10 1\n1\n0\n0\n0\n0\n0\n0\n0\n0\n1",
"output": "YES"
},
{
"input": "1 1\n1",
"output": "NO"
},
{
"input": "3 4\n1010\n0100\n1101",
"output": "YES"
},
{
"input": "2 5\n10101\n11111",
"output": "YES"
},
{
"input": "5 5\n10000\n11000\n11100\n11110\n11111",
"output": "YES"
},
{
"input": "2 5\n10000\n11111",
"output": "YES"
},
{
"input": "4 5\n01000\n10100\n00010\n10101",
"output": "YES"
},
{
"input": "2 2\n10\n11",
"output": "YES"
},
{
"input": "2 5\n00100\n11111",
"output": "YES"
},
{
"input": "4 5\n00000\n11000\n00110\n00011",
"output": "YES"
},
{
"input": "4 3\n000\n010\n001\n100",
"output": "YES"
},
{
"input": "4 5\n10000\n10101\n01000\n00111",
"output": "YES"
},
{
"input": "4 5\n10000\n01000\n10101\n00111",
"output": "YES"
},
{
"input": "2 2\n01\n11",
"output": "YES"
},
{
"input": "3 3\n010\n101\n000",
"output": "YES"
},
{
"input": "2 2\n11\n00",
"output": "YES"
},
{
"input": "3 5\n10110\n11000\n00111",
"output": "YES"
},
{
"input": "3 8\n00111111\n01011100\n11000000",
"output": "YES"
},
{
"input": "4 6\n100000\n110000\n001100\n000011",
"output": "YES"
},
{
"input": "2 5\n11111\n00000",
"output": "YES"
},
{
"input": "2 3\n101\n111",
"output": "YES"
},
{
"input": "2 5\n01000\n11111",
"output": "YES"
},
{
"input": "2 2\n00\n11",
"output": "YES"
},
{
"input": "4 15\n111110100011010\n111111011010110\n101000001011001\n100110000111011",
"output": "YES"
},
{
"input": "2 3\n010\n111",
"output": "YES"
},
{
"input": "4 5\n10100\n11000\n00110\n00101",
"output": "YES"
},
{
"input": "4 4\n1111\n0000\n0000\n0000",
"output": "YES"
},
{
"input": "3 5\n11100\n00110\n00011",
"output": "YES"
},
{
"input": "2 1\n0\n1",
"output": "YES"
},
{
"input": "4 4\n1000\n1001\n0010\n0100",
"output": "YES"
},
{
"input": "3 5\n00110\n10011\n01100",
"output": "YES"
},
{
"input": "3 5\n10101\n00111\n01000",
"output": "NO"
},
{
"input": "4 5\n00101\n00011\n01000\n10010",
"output": "YES"
},
{
"input": "3 3\n100\n110\n111",
"output": "YES"
},
{
"input": "2 2\n11\n01",
"output": "YES"
},
{
"input": "3 3\n100\n100\n111",
"output": "YES"
},
{
"input": "4 2\n10\n01\n10\n01",
"output": "YES"
},
{
"input": "3 3\n111\n000\n000",
"output": "YES"
},
{
"input": "3 3\n010\n100\n011",
"output": "YES"
},
{
"input": "2 3\n111\n000",
"output": "YES"
},
{
"input": "3 4\n0001\n1101\n1010",
"output": "YES"
},
{
"input": "3 4\n1010\n0101\n1000",
"output": "YES"
},
{
"input": "3 4\n0001\n1101\n0110",
"output": "YES"
},
{
"input": "3 3\n111\n101\n001",
"output": "YES"
},
{
"input": "4 5\n10001\n10010\n01010\n00101",
"output": "YES"
},
{
"input": "3 3\n000\n000\n111",
"output": "YES"
},
{
"input": "2 3\n100\n111",
"output": "YES"
},
{
"input": "3 10\n1111011100\n0001100011\n1111010101",
"output": "YES"
},
{
"input": "3 4\n0110\n1010\n0101",
"output": "YES"
},
{
"input": "3 3\n100\n001\n011",
"output": "YES"
},
{
"input": "3 3\n100\n010\n001",
"output": "NO"
},
{
"input": "3 3\n010\n100\n001",
"output": "NO"
},
{
"input": "3 5\n10101\n01010\n01010",
"output": "YES"
},
{
"input": "2 6\n111111\n000000",
"output": "YES"
},
{
"input": "3 5\n00000\n10101\n01010",
"output": "YES"
},
{
"input": "4 6\n111000\n100100\n010010\n001001",
"output": "YES"
}
] | 1,526,985,800 | 2,147,483,647 | Python 3 | OK | TESTS | 67 | 2,698 | 18,636,800 | #!/usr/bin/env python3
# -*- coding: utf-8 -*-
"""
Created on Mon May 21 22:31:07 2018
@author: thomas
"""
integers=input()
[n,m]=[int(x) for x in integers.split()]
a=[]
for i in range(n):
row_i=input()
a_i=[]
for j in range(m):
a_i.append(int(row_i[j]))
a.append(a_i)
indicator=False
#all_sum=[0]*m
#for i in range(n):
# for j in range(m):
# all_sum[j]+=a[i][j]
all_sum=[sum(x) for x in zip(*a)]
for i in range(n):
consider=[x-y for x, y in zip(all_sum, a[i])]
indicator=(sum([x>0 for x in consider])==m)
if (indicator==True): break
if (indicator==True): print("YES")
else: print("NO") | Title: Switches and Lamps
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given *n* switches and *m* lamps. The *i*-th switch turns on some subset of the lamps. This information is given as the matrix *a* consisting of *n* rows and *m* columns where *a**i*,<=*j*<==<=1 if the *i*-th switch turns on the *j*-th lamp and *a**i*,<=*j*<==<=0 if the *i*-th switch is not connected to the *j*-th lamp.
Initially all *m* lamps are turned off.
Switches change state only from "off" to "on". It means that if you press two or more switches connected to the same lamp then the lamp will be turned on after any of this switches is pressed and will remain its state even if any switch connected to this lamp is pressed afterwards.
It is guaranteed that if you push all *n* switches then all *m* lamps will be turned on.
Your think that you have too many switches and you would like to ignore one of them.
Your task is to say if there exists such a switch that if you will ignore (not use) it but press all the other *n*<=-<=1 switches then all the *m* lamps will be turned on.
Input Specification:
The first line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=2000) — the number of the switches and the number of the lamps.
The following *n* lines contain *m* characters each. The character *a**i*,<=*j* is equal to '1' if the *i*-th switch turns on the *j*-th lamp and '0' otherwise.
It is guaranteed that if you press all *n* switches all *m* lamps will be turned on.
Output Specification:
Print "YES" if there is a switch that if you will ignore it and press all the other *n*<=-<=1 switches then all *m* lamps will be turned on. Print "NO" if there is no such switch.
Demo Input:
['4 5\n10101\n01000\n00111\n10000\n', '4 5\n10100\n01000\n00110\n00101\n']
Demo Output:
['YES\n', 'NO\n']
Note:
none | ```python
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
"""
Created on Mon May 21 22:31:07 2018
@author: thomas
"""
integers=input()
[n,m]=[int(x) for x in integers.split()]
a=[]
for i in range(n):
row_i=input()
a_i=[]
for j in range(m):
a_i.append(int(row_i[j]))
a.append(a_i)
indicator=False
#all_sum=[0]*m
#for i in range(n):
# for j in range(m):
# all_sum[j]+=a[i][j]
all_sum=[sum(x) for x in zip(*a)]
for i in range(n):
consider=[x-y for x, y in zip(all_sum, a[i])]
indicator=(sum([x>0 for x in consider])==m)
if (indicator==True): break
if (indicator==True): print("YES")
else: print("NO")
``` | 3 |
|
432 | A | Choosing Teams | PROGRAMMING | 800 | [
"greedy",
"implementation",
"sortings"
] | null | null | The Saratov State University Olympiad Programmers Training Center (SSU OPTC) has *n* students. For each student you know the number of times he/she has participated in the ACM ICPC world programming championship. According to the ACM ICPC rules, each person can participate in the world championship at most 5 times.
The head of the SSU OPTC is recently gathering teams to participate in the world championship. Each team must consist of exactly three people, at that, any person cannot be a member of two or more teams. What maximum number of teams can the head make if he wants each team to participate in the world championship with the same members at least *k* times? | The first line contains two integers, *n* and *k* (1<=≤<=*n*<=≤<=2000; 1<=≤<=*k*<=≤<=5). The next line contains *n* integers: *y*1,<=*y*2,<=...,<=*y**n* (0<=≤<=*y**i*<=≤<=5), where *y**i* shows the number of times the *i*-th person participated in the ACM ICPC world championship. | Print a single number — the answer to the problem. | [
"5 2\n0 4 5 1 0\n",
"6 4\n0 1 2 3 4 5\n",
"6 5\n0 0 0 0 0 0\n"
] | [
"1\n",
"0\n",
"2\n"
] | In the first sample only one team could be made: the first, the fourth and the fifth participants.
In the second sample no teams could be created.
In the third sample two teams could be created. Any partition into two teams fits. | 500 | [
{
"input": "5 2\n0 4 5 1 0",
"output": "1"
},
{
"input": "6 4\n0 1 2 3 4 5",
"output": "0"
},
{
"input": "6 5\n0 0 0 0 0 0",
"output": "2"
},
{
"input": "3 4\n0 1 0",
"output": "1"
},
{
"input": "3 4\n0 2 0",
"output": "0"
},
{
"input": "6 5\n0 0 0 0 0 0",
"output": "2"
},
{
"input": "12 2\n0 1 2 3 4 0 1 2 3 4 0 1",
"output": "3"
},
{
"input": "15 2\n0 1 2 3 4 0 1 2 3 4 0 1 2 3 4",
"output": "4"
},
{
"input": "13 1\n5 0 5 0 1 5 0 4 1 1 1 4 1",
"output": "3"
},
{
"input": "20 1\n5 0 4 2 2 3 2 1 2 4 1 3 5 5 5 4 4 1 3 0",
"output": "5"
},
{
"input": "20 2\n1 4 5 1 3 4 0 5 0 3 3 0 4 0 0 1 4 2 0 4",
"output": "4"
},
{
"input": "20 2\n1 4 5 1 3 4 0 5 0 3 3 0 4 0 0 1 4 2 0 4",
"output": "4"
},
{
"input": "20 2\n1 4 5 1 3 4 0 5 0 3 3 0 4 0 0 1 4 2 0 4",
"output": "4"
},
{
"input": "20 2\n1 4 5 1 3 4 0 5 0 3 3 0 4 0 0 1 4 2 0 4",
"output": "4"
},
{
"input": "6 3\n4 4 4 4 4 4",
"output": "0"
},
{
"input": "1 4\n1",
"output": "0"
},
{
"input": "1 1\n0",
"output": "0"
},
{
"input": "2 3\n0 0",
"output": "0"
}
] | 1,691,347,758 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 15 | 0 | n, k = map(int, input().split())
participation = list(map(int, input().split()))
max_teams = 0
for i in range(n):
available_students = min(5, k - participation[i])
max_teams += available_students // 3
print(max_teams)
| Title: Choosing Teams
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Saratov State University Olympiad Programmers Training Center (SSU OPTC) has *n* students. For each student you know the number of times he/she has participated in the ACM ICPC world programming championship. According to the ACM ICPC rules, each person can participate in the world championship at most 5 times.
The head of the SSU OPTC is recently gathering teams to participate in the world championship. Each team must consist of exactly three people, at that, any person cannot be a member of two or more teams. What maximum number of teams can the head make if he wants each team to participate in the world championship with the same members at least *k* times?
Input Specification:
The first line contains two integers, *n* and *k* (1<=≤<=*n*<=≤<=2000; 1<=≤<=*k*<=≤<=5). The next line contains *n* integers: *y*1,<=*y*2,<=...,<=*y**n* (0<=≤<=*y**i*<=≤<=5), where *y**i* shows the number of times the *i*-th person participated in the ACM ICPC world championship.
Output Specification:
Print a single number — the answer to the problem.
Demo Input:
['5 2\n0 4 5 1 0\n', '6 4\n0 1 2 3 4 5\n', '6 5\n0 0 0 0 0 0\n']
Demo Output:
['1\n', '0\n', '2\n']
Note:
In the first sample only one team could be made: the first, the fourth and the fifth participants.
In the second sample no teams could be created.
In the third sample two teams could be created. Any partition into two teams fits. | ```python
n, k = map(int, input().split())
participation = list(map(int, input().split()))
max_teams = 0
for i in range(n):
available_students = min(5, k - participation[i])
max_teams += available_students // 3
print(max_teams)
``` | 0 |
|
940 | B | Our Tanya is Crying Out Loud | PROGRAMMING | 1,400 | [
"dp",
"greedy"
] | null | null | Right now she actually isn't. But she will be, if you don't solve this problem.
You are given integers *n*, *k*, *A* and *B*. There is a number *x*, which is initially equal to *n*. You are allowed to perform two types of operations:
1. Subtract 1 from *x*. This operation costs you *A* coins. 1. Divide *x* by *k*. Can be performed only if *x* is divisible by *k*. This operation costs you *B* coins. | The first line contains a single integer *n* (1<=≤<=*n*<=≤<=2·109).
The second line contains a single integer *k* (1<=≤<=*k*<=≤<=2·109).
The third line contains a single integer *A* (1<=≤<=*A*<=≤<=2·109).
The fourth line contains a single integer *B* (1<=≤<=*B*<=≤<=2·109). | Output a single integer — the minimum amount of coins you have to pay to make *x* equal to 1. | [
"9\n2\n3\n1\n",
"5\n5\n2\n20\n",
"19\n3\n4\n2\n"
] | [
"6\n",
"8\n",
"12\n"
] | In the first testcase, the optimal strategy is as follows:
- Subtract 1 from *x* (9 → 8) paying 3 coins. - Divide *x* by 2 (8 → 4) paying 1 coin. - Divide *x* by 2 (4 → 2) paying 1 coin. - Divide *x* by 2 (2 → 1) paying 1 coin.
The total cost is 6 coins.
In the second test case the optimal strategy is to subtract 1 from *x* 4 times paying 8 coins in total. | 1,250 | [
{
"input": "9\n2\n3\n1",
"output": "6"
},
{
"input": "5\n5\n2\n20",
"output": "8"
},
{
"input": "19\n3\n4\n2",
"output": "12"
},
{
"input": "1845999546\n999435865\n1234234\n2323423",
"output": "1044857680578777"
},
{
"input": "1604353664\n1604353665\n9993432\n1",
"output": "16032999235141416"
},
{
"input": "777888456\n1\n98\n43",
"output": "76233068590"
},
{
"input": "1162261467\n3\n1\n2000000000",
"output": "1162261466"
},
{
"input": "1000000000\n1999999999\n789987\n184569875",
"output": "789986999210013"
},
{
"input": "2000000000\n2\n1\n2000000000",
"output": "1999999999"
},
{
"input": "1999888325\n3\n2\n2000000000",
"output": "3333258884"
},
{
"input": "1897546487\n687\n89798979\n879876541",
"output": "110398404423"
},
{
"input": "20\n1\n20\n1",
"output": "380"
},
{
"input": "16\n5\n17\n3",
"output": "54"
},
{
"input": "19\n19\n19\n1",
"output": "1"
},
{
"input": "18\n2\n3\n16",
"output": "40"
},
{
"input": "1\n11\n8\n9",
"output": "0"
},
{
"input": "9\n10\n1\n20",
"output": "8"
},
{
"input": "19\n10\n19\n2",
"output": "173"
},
{
"input": "16\n9\n14\n2",
"output": "100"
},
{
"input": "15\n2\n5\n2",
"output": "21"
},
{
"input": "14\n7\n13\n1",
"output": "14"
},
{
"input": "43\n3\n45\n3",
"output": "189"
},
{
"input": "99\n1\n98\n1",
"output": "9604"
},
{
"input": "77\n93\n100\n77",
"output": "7600"
},
{
"input": "81\n3\n91\n95",
"output": "380"
},
{
"input": "78\n53\n87\n34",
"output": "2209"
},
{
"input": "80\n3\n15\n1",
"output": "108"
},
{
"input": "97\n24\n4\n24",
"output": "40"
},
{
"input": "100\n100\n1\n100",
"output": "99"
},
{
"input": "87\n4\n17\n7",
"output": "106"
},
{
"input": "65\n2\n3\n6",
"output": "36"
},
{
"input": "1000000\n1435\n3\n999999",
"output": "1005804"
},
{
"input": "783464\n483464\n2\n966928",
"output": "1566926"
},
{
"input": "248035\n11\n3\n20",
"output": "202"
},
{
"input": "524287\n2\n945658\n999756",
"output": "34963354"
},
{
"input": "947352\n78946\n85\n789654",
"output": "790589"
},
{
"input": "1000000\n1\n999899\n60",
"output": "999898000101"
},
{
"input": "753687\n977456\n6547\n456",
"output": "4934382242"
},
{
"input": "1000000\n500000\n1\n999997",
"output": "999998"
},
{
"input": "997458\n843596\n1\n843596",
"output": "997457"
},
{
"input": "821109\n92\n6547\n98787",
"output": "394566"
},
{
"input": "1073741823\n2\n9543\n8923453",
"output": "188412866"
},
{
"input": "1000999777\n1934999345\n2356346\n34534565",
"output": "2358701818178496"
},
{
"input": "2000000000\n1\n2000000000\n98",
"output": "3999999998000000000"
},
{
"input": "1999324353\n978435356\n1\n978435356",
"output": "1020888998"
},
{
"input": "2000000000\n2\n2000000000\n2000000000",
"output": "84000000000"
},
{
"input": "241375690\n17\n2\n1998789654",
"output": "482751378"
},
{
"input": "171507000\n350\n789\n6548687",
"output": "14216965"
},
{
"input": "1100220011\n10001\n2\n1999778654",
"output": "1999998674"
},
{
"input": "1867622656\n43216\n789644\n12315468",
"output": "24630936"
},
{
"input": "1867622656\n43216\n1\n1879865413",
"output": "1867622655"
},
{
"input": "1999999999\n1000000000\n789987\n184569875",
"output": "789987183779888"
},
{
"input": "1987987897\n103546\n7\n98754563",
"output": "98946650"
},
{
"input": "10\n2\n2\n5",
"output": "13"
},
{
"input": "7\n2\n1\n100000",
"output": "6"
},
{
"input": "7\n2\n3\n1",
"output": "8"
},
{
"input": "2000000000\n666666667\n1\n1",
"output": "666666668"
},
{
"input": "1999999997\n666666666\n2\n2",
"output": "1333333334"
}
] | 1,618,682,481 | 2,147,483,647 | PyPy 3 | OK | TESTS | 58 | 93 | 102,400 | n=int(input())
k=int(input())
a=int(input())
b=int(input())
if(k==1):
print ((n-1)*a)
else:
t=(b*k)/(a*(k-1))
ans=n
c=0
while(ans>1):
if(ans>=t and ans%k==0):
c+=b
ans=ans//k
else:
if(ans>=t and ans>k):
h=ans%k
ans-=h
c+=a*h
else:
c+=a*(ans-1)
ans=1
print(c)
| Title: Our Tanya is Crying Out Loud
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Right now she actually isn't. But she will be, if you don't solve this problem.
You are given integers *n*, *k*, *A* and *B*. There is a number *x*, which is initially equal to *n*. You are allowed to perform two types of operations:
1. Subtract 1 from *x*. This operation costs you *A* coins. 1. Divide *x* by *k*. Can be performed only if *x* is divisible by *k*. This operation costs you *B* coins.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=2·109).
The second line contains a single integer *k* (1<=≤<=*k*<=≤<=2·109).
The third line contains a single integer *A* (1<=≤<=*A*<=≤<=2·109).
The fourth line contains a single integer *B* (1<=≤<=*B*<=≤<=2·109).
Output Specification:
Output a single integer — the minimum amount of coins you have to pay to make *x* equal to 1.
Demo Input:
['9\n2\n3\n1\n', '5\n5\n2\n20\n', '19\n3\n4\n2\n']
Demo Output:
['6\n', '8\n', '12\n']
Note:
In the first testcase, the optimal strategy is as follows:
- Subtract 1 from *x* (9 → 8) paying 3 coins. - Divide *x* by 2 (8 → 4) paying 1 coin. - Divide *x* by 2 (4 → 2) paying 1 coin. - Divide *x* by 2 (2 → 1) paying 1 coin.
The total cost is 6 coins.
In the second test case the optimal strategy is to subtract 1 from *x* 4 times paying 8 coins in total. | ```python
n=int(input())
k=int(input())
a=int(input())
b=int(input())
if(k==1):
print ((n-1)*a)
else:
t=(b*k)/(a*(k-1))
ans=n
c=0
while(ans>1):
if(ans>=t and ans%k==0):
c+=b
ans=ans//k
else:
if(ans>=t and ans>k):
h=ans%k
ans-=h
c+=a*h
else:
c+=a*(ans-1)
ans=1
print(c)
``` | 3 |
|
26 | A | Almost Prime | PROGRAMMING | 900 | [
"number theory"
] | A. Almost Prime | 2 | 256 | A number is called almost prime if it has exactly two distinct prime divisors. For example, numbers 6, 18, 24 are almost prime, while 4, 8, 9, 42 are not. Find the amount of almost prime numbers which are between 1 and *n*, inclusive. | Input contains one integer number *n* (1<=≤<=*n*<=≤<=3000). | Output the amount of almost prime numbers between 1 and *n*, inclusive. | [
"10\n",
"21\n"
] | [
"2\n",
"8\n"
] | none | 500 | [
{
"input": "10",
"output": "2"
},
{
"input": "21",
"output": "8"
},
{
"input": "1",
"output": "0"
},
{
"input": "2",
"output": "0"
},
{
"input": "4",
"output": "0"
},
{
"input": "3",
"output": "0"
},
{
"input": "8",
"output": "1"
},
{
"input": "19",
"output": "6"
},
{
"input": "40",
"output": "19"
},
{
"input": "77",
"output": "41"
},
{
"input": "222",
"output": "125"
},
{
"input": "987",
"output": "501"
},
{
"input": "1000",
"output": "508"
},
{
"input": "2000",
"output": "958"
},
{
"input": "3000",
"output": "1375"
},
{
"input": "2999",
"output": "1375"
},
{
"input": "2998",
"output": "1375"
},
{
"input": "2997",
"output": "1374"
},
{
"input": "1429",
"output": "706"
},
{
"input": "1673",
"output": "808"
},
{
"input": "1500",
"output": "732"
},
{
"input": "500",
"output": "266"
},
{
"input": "856",
"output": "439"
}
] | 1,521,676,691 | 2,147,483,647 | PyPy 3 | OK | TESTS | 23 | 278 | 23,142,400 | a,p = [],[2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101,103,107,109,113,127,131,137,139,149,151,157,163,167,173,179,181,191,193,197,199,211,223,227,229,233,239,241,251,257,263,269,271,277,281,283,293,307,311,313,317,331,337,347,349,353,359,367,373,379,383,389,397,401,409,419,421,431,433,439,443,449,457,461,463,467,479,487,491,499,503,509,521,523,541,547,557,563,569,571,577,587,593,599,601,607,613,617,619,631,641,643,647,653,659,661,673,677,683,691,701,709,719,727,733,739,743,751,757,761,769,773,787,797,809,811,821,823,827,829,839,853,857,859,863,877,881,883,887,907,911,919,929,937,941,947,953,967,971,977,983,991,997,1009,1013,1019,1021,1031,1033,1039,1049,1051,1061,1063,1069,1087,1091,1093,1097,1103,1109,1117,1123,1129,1151,1153,1163,1171,1181,1187,1193,1201,1213,1217,1223,1229,1231,1237,1249,1259,1277,1279,1283,1289,1291,1297,1301,1303,1307,1319,1321,1327,1361,1367,1373,1381,1399,1409,1423,1427,1429,1433,1439,1447,1451,1453,1459,1471,1481,1483,1487,1489,1493,1499]
for i in range(0, len(p)):
for j in range(i+1, len(p)):
k = 1
while p[j]*(p[i]**k) < 3000:
l = 1
while (p[i]**k)*(p[j]**l) < 3000:
a.append((p[i]**k)*(p[j]**l))
l+=1
k+=1
a, i, x= sorted(set(a)), 0, int(input())
while i < len(a) and x >= a[i]: i+=1
print(i) | Title: Almost Prime
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
A number is called almost prime if it has exactly two distinct prime divisors. For example, numbers 6, 18, 24 are almost prime, while 4, 8, 9, 42 are not. Find the amount of almost prime numbers which are between 1 and *n*, inclusive.
Input Specification:
Input contains one integer number *n* (1<=≤<=*n*<=≤<=3000).
Output Specification:
Output the amount of almost prime numbers between 1 and *n*, inclusive.
Demo Input:
['10\n', '21\n']
Demo Output:
['2\n', '8\n']
Note:
none | ```python
a,p = [],[2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101,103,107,109,113,127,131,137,139,149,151,157,163,167,173,179,181,191,193,197,199,211,223,227,229,233,239,241,251,257,263,269,271,277,281,283,293,307,311,313,317,331,337,347,349,353,359,367,373,379,383,389,397,401,409,419,421,431,433,439,443,449,457,461,463,467,479,487,491,499,503,509,521,523,541,547,557,563,569,571,577,587,593,599,601,607,613,617,619,631,641,643,647,653,659,661,673,677,683,691,701,709,719,727,733,739,743,751,757,761,769,773,787,797,809,811,821,823,827,829,839,853,857,859,863,877,881,883,887,907,911,919,929,937,941,947,953,967,971,977,983,991,997,1009,1013,1019,1021,1031,1033,1039,1049,1051,1061,1063,1069,1087,1091,1093,1097,1103,1109,1117,1123,1129,1151,1153,1163,1171,1181,1187,1193,1201,1213,1217,1223,1229,1231,1237,1249,1259,1277,1279,1283,1289,1291,1297,1301,1303,1307,1319,1321,1327,1361,1367,1373,1381,1399,1409,1423,1427,1429,1433,1439,1447,1451,1453,1459,1471,1481,1483,1487,1489,1493,1499]
for i in range(0, len(p)):
for j in range(i+1, len(p)):
k = 1
while p[j]*(p[i]**k) < 3000:
l = 1
while (p[i]**k)*(p[j]**l) < 3000:
a.append((p[i]**k)*(p[j]**l))
l+=1
k+=1
a, i, x= sorted(set(a)), 0, int(input())
while i < len(a) and x >= a[i]: i+=1
print(i)
``` | 3.887394 |
614 | A | Link/Cut Tree | PROGRAMMING | 1,500 | [
"brute force",
"implementation"
] | null | null | Programmer Rostislav got seriously interested in the Link/Cut Tree data structure, which is based on Splay trees. Specifically, he is now studying the *expose* procedure.
Unfortunately, Rostislav is unable to understand the definition of this procedure, so he decided to ask programmer Serezha to help him. Serezha agreed to help if Rostislav solves a simple task (and if he doesn't, then why would he need Splay trees anyway?)
Given integers *l*, *r* and *k*, you need to print all powers of number *k* within range from *l* to *r* inclusive. However, Rostislav doesn't want to spent time doing this, as he got interested in playing a network game called Agar with Gleb. Help him! | The first line of the input contains three space-separated integers *l*, *r* and *k* (1<=≤<=*l*<=≤<=*r*<=≤<=1018, 2<=≤<=*k*<=≤<=109). | Print all powers of number *k*, that lie within range from *l* to *r* in the increasing order. If there are no such numbers, print "-1" (without the quotes). | [
"1 10 2\n",
"2 4 5\n"
] | [
"1 2 4 8 ",
"-1"
] | Note to the first sample: numbers 2<sup class="upper-index">0</sup> = 1, 2<sup class="upper-index">1</sup> = 2, 2<sup class="upper-index">2</sup> = 4, 2<sup class="upper-index">3</sup> = 8 lie within the specified range. The number 2<sup class="upper-index">4</sup> = 16 is greater then 10, thus it shouldn't be printed. | 500 | [
{
"input": "1 10 2",
"output": "1 2 4 8 "
},
{
"input": "2 4 5",
"output": "-1"
},
{
"input": "18102 43332383920 28554",
"output": "28554 815330916 "
},
{
"input": "19562 31702689720 17701",
"output": "313325401 "
},
{
"input": "11729 55221128400 313",
"output": "97969 30664297 9597924961 "
},
{
"input": "5482 100347128000 342",
"output": "116964 40001688 13680577296 "
},
{
"input": "3680 37745933600 10",
"output": "10000 100000 1000000 10000000 100000000 1000000000 10000000000 "
},
{
"input": "17098 191120104800 43",
"output": "79507 3418801 147008443 6321363049 "
},
{
"input": "10462 418807699200 2",
"output": "16384 32768 65536 131072 262144 524288 1048576 2097152 4194304 8388608 16777216 33554432 67108864 134217728 268435456 536870912 1073741824 2147483648 4294967296 8589934592 17179869184 34359738368 68719476736 137438953472 274877906944 "
},
{
"input": "30061 641846400000 3",
"output": "59049 177147 531441 1594323 4782969 14348907 43046721 129140163 387420489 1162261467 3486784401 10460353203 31381059609 94143178827 282429536481 "
},
{
"input": "1 1000000000000000000 2",
"output": "1 2 4 8 16 32 64 128 256 512 1024 2048 4096 8192 16384 32768 65536 131072 262144 524288 1048576 2097152 4194304 8388608 16777216 33554432 67108864 134217728 268435456 536870912 1073741824 2147483648 4294967296 8589934592 17179869184 34359738368 68719476736 137438953472 274877906944 549755813888 1099511627776 2199023255552 4398046511104 8796093022208 17592186044416 35184372088832 70368744177664 140737488355328 281474976710656 562949953421312 1125899906842624 2251799813685248 4503599627370496 900719925474099..."
},
{
"input": "32 2498039712000 4",
"output": "64 256 1024 4096 16384 65536 262144 1048576 4194304 16777216 67108864 268435456 1073741824 4294967296 17179869184 68719476736 274877906944 1099511627776 "
},
{
"input": "1 2576683920000 2",
"output": "1 2 4 8 16 32 64 128 256 512 1024 2048 4096 8192 16384 32768 65536 131072 262144 524288 1048576 2097152 4194304 8388608 16777216 33554432 67108864 134217728 268435456 536870912 1073741824 2147483648 4294967296 8589934592 17179869184 34359738368 68719476736 137438953472 274877906944 549755813888 1099511627776 2199023255552 "
},
{
"input": "5 25 5",
"output": "5 25 "
},
{
"input": "1 90 90",
"output": "1 90 "
},
{
"input": "95 2200128528000 68",
"output": "4624 314432 21381376 1453933568 98867482624 "
},
{
"input": "64 426314644000 53",
"output": "2809 148877 7890481 418195493 22164361129 "
},
{
"input": "198765 198765 198765",
"output": "198765 "
},
{
"input": "42 2845016496000 12",
"output": "144 1728 20736 248832 2985984 35831808 429981696 5159780352 61917364224 743008370688 "
},
{
"input": "6 6 3",
"output": "-1"
},
{
"input": "1 10 11",
"output": "1 "
},
{
"input": "2 10 11",
"output": "-1"
},
{
"input": "87 160 41",
"output": "-1"
},
{
"input": "237171123124584251 923523399718980912 7150",
"output": "-1"
},
{
"input": "101021572000739548 453766043506276015 8898",
"output": "-1"
},
{
"input": "366070689449360724 928290634811046396 8230",
"output": "-1"
},
{
"input": "438133886369772308 942612870269666780 7193",
"output": "-1"
},
{
"input": "10 10 10",
"output": "10 "
},
{
"input": "16 16 256",
"output": "-1"
},
{
"input": "1 1000000000000000000 1000000000",
"output": "1 1000000000 1000000000000000000 "
},
{
"input": "1000000000000000000 1000000000000000000 1000000000",
"output": "1000000000000000000 "
},
{
"input": "1000000000 1000000000000000000 1000000000",
"output": "1000000000 1000000000000000000 "
},
{
"input": "1 1 4",
"output": "1 "
},
{
"input": "1 999999999999999999 1000000000",
"output": "1 1000000000 "
},
{
"input": "1 1000000000000000000 999999990",
"output": "1 999999990 999999980000000100 "
},
{
"input": "1 1000000000000000000 999999984",
"output": "1 999999984 999999968000000256 "
},
{
"input": "1 1000000000000000000 324325",
"output": "1 324325 105186705625 34114678301828125 "
},
{
"input": "1 1000000000000000000 999999523",
"output": "1 999999523 999999046000227529 "
},
{
"input": "1 243 3",
"output": "1 3 9 27 81 243 "
},
{
"input": "62769392426654367 567152589733560993 688813",
"output": "326816522793383797 "
},
{
"input": "1 1000000000000000000 690852001",
"output": "1 690852001 477276487285704001 "
},
{
"input": "1 1000000000000000000 918745157",
"output": "1 918745157 844092663510954649 "
},
{
"input": "1 1000000000000000000 131299843",
"output": "1 131299843 17239648771824649 "
},
{
"input": "2861381721051425 2861381721051425 1234",
"output": "-1"
}
] | 1,453,543,279 | 2,147,483,647 | PyPy 3 | OK | TESTS | 44 | 140 | 0 | def x():
l, r, k = [int(s) for s in input().split()]
res = k
if l == 1:
yield 1
while k<=r:
if l <= k <= r:
yield k
k*=res
if k>r:
break
d = [i for i in x()]
if len(d):
print(*d)
else:
print(-1)
| Title: Link/Cut Tree
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Programmer Rostislav got seriously interested in the Link/Cut Tree data structure, which is based on Splay trees. Specifically, he is now studying the *expose* procedure.
Unfortunately, Rostislav is unable to understand the definition of this procedure, so he decided to ask programmer Serezha to help him. Serezha agreed to help if Rostislav solves a simple task (and if he doesn't, then why would he need Splay trees anyway?)
Given integers *l*, *r* and *k*, you need to print all powers of number *k* within range from *l* to *r* inclusive. However, Rostislav doesn't want to spent time doing this, as he got interested in playing a network game called Agar with Gleb. Help him!
Input Specification:
The first line of the input contains three space-separated integers *l*, *r* and *k* (1<=≤<=*l*<=≤<=*r*<=≤<=1018, 2<=≤<=*k*<=≤<=109).
Output Specification:
Print all powers of number *k*, that lie within range from *l* to *r* in the increasing order. If there are no such numbers, print "-1" (without the quotes).
Demo Input:
['1 10 2\n', '2 4 5\n']
Demo Output:
['1 2 4 8 ', '-1']
Note:
Note to the first sample: numbers 2<sup class="upper-index">0</sup> = 1, 2<sup class="upper-index">1</sup> = 2, 2<sup class="upper-index">2</sup> = 4, 2<sup class="upper-index">3</sup> = 8 lie within the specified range. The number 2<sup class="upper-index">4</sup> = 16 is greater then 10, thus it shouldn't be printed. | ```python
def x():
l, r, k = [int(s) for s in input().split()]
res = k
if l == 1:
yield 1
while k<=r:
if l <= k <= r:
yield k
k*=res
if k>r:
break
d = [i for i in x()]
if len(d):
print(*d)
else:
print(-1)
``` | 3 |
|
914 | B | Conan and Agasa play a Card Game | PROGRAMMING | 1,200 | [
"games",
"greedy",
"implementation"
] | null | null | Edogawa Conan got tired of solving cases, and invited his friend, Professor Agasa, over. They decided to play a game of cards. Conan has *n* cards, and the *i*-th card has a number *a**i* written on it.
They take turns playing, starting with Conan. In each turn, the player chooses a card and removes it. Also, he removes all cards having a number strictly lesser than the number on the chosen card. Formally, if the player chooses the *i*-th card, he removes that card and removes the *j*-th card for all *j* such that *a**j*<=<<=*a**i*.
A player loses if he cannot make a move on his turn, that is, he loses if there are no cards left. Predict the outcome of the game, assuming both players play optimally. | The first line contains an integer *n* (1<=≤<=*n*<=≤<=105) — the number of cards Conan has.
The next line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=105), where *a**i* is the number on the *i*-th card. | If Conan wins, print "Conan" (without quotes), otherwise print "Agasa" (without quotes). | [
"3\n4 5 7\n",
"2\n1 1\n"
] | [
"Conan\n",
"Agasa\n"
] | In the first example, Conan can just choose the card having number 7 on it and hence remove all the cards. After that, there are no cards left on Agasa's turn.
In the second example, no matter which card Conan chooses, there will be one one card left, which Agasa can choose. After that, there are no cards left when it becomes Conan's turn again. | 1,000 | [
{
"input": "3\n4 5 7",
"output": "Conan"
},
{
"input": "2\n1 1",
"output": "Agasa"
},
{
"input": "10\n38282 53699 38282 38282 38282 38282 38282 38282 38282 38282",
"output": "Conan"
},
{
"input": "10\n50165 50165 50165 50165 50165 50165 50165 50165 50165 50165",
"output": "Agasa"
},
{
"input": "10\n83176 83176 83176 23495 83176 8196 83176 23495 83176 83176",
"output": "Conan"
},
{
"input": "10\n32093 36846 32093 32093 36846 36846 36846 36846 36846 36846",
"output": "Conan"
},
{
"input": "3\n1 2 3",
"output": "Conan"
},
{
"input": "4\n2 3 4 5",
"output": "Conan"
},
{
"input": "10\n30757 30757 33046 41744 39918 39914 41744 39914 33046 33046",
"output": "Conan"
},
{
"input": "10\n50096 50096 50096 50096 50096 50096 28505 50096 50096 50096",
"output": "Conan"
},
{
"input": "10\n54842 54842 54842 54842 57983 54842 54842 57983 57983 54842",
"output": "Conan"
},
{
"input": "10\n87900 87900 5761 87900 87900 87900 5761 87900 87900 87900",
"output": "Agasa"
},
{
"input": "10\n53335 35239 26741 35239 35239 26741 35239 35239 53335 35239",
"output": "Agasa"
},
{
"input": "10\n75994 64716 75994 64716 75994 75994 56304 64716 56304 64716",
"output": "Agasa"
},
{
"input": "1\n1",
"output": "Conan"
},
{
"input": "5\n2 2 1 1 1",
"output": "Conan"
},
{
"input": "5\n1 4 4 5 5",
"output": "Conan"
},
{
"input": "3\n1 3 3",
"output": "Conan"
},
{
"input": "3\n2 2 2",
"output": "Conan"
},
{
"input": "5\n1 1 1 2 2",
"output": "Conan"
},
{
"input": "4\n1 2 1 2",
"output": "Agasa"
},
{
"input": "7\n7 7 7 7 6 6 6",
"output": "Conan"
},
{
"input": "3\n2 3 3",
"output": "Conan"
},
{
"input": "3\n1 1 100000",
"output": "Conan"
},
{
"input": "1\n100000",
"output": "Conan"
},
{
"input": "5\n3 3 3 4 4",
"output": "Conan"
},
{
"input": "3\n1 2 2",
"output": "Conan"
},
{
"input": "3\n4 4 5",
"output": "Conan"
},
{
"input": "1\n2",
"output": "Conan"
},
{
"input": "3\n97 97 100",
"output": "Conan"
},
{
"input": "5\n100000 100000 100000 1 1",
"output": "Conan"
},
{
"input": "7\n7 7 6 6 5 5 4",
"output": "Conan"
},
{
"input": "5\n100000 100000 100000 2 2",
"output": "Conan"
},
{
"input": "4\n3 3 2 1",
"output": "Conan"
},
{
"input": "1\n485",
"output": "Conan"
},
{
"input": "3\n4 4 100000",
"output": "Conan"
},
{
"input": "3\n1 1 2",
"output": "Conan"
},
{
"input": "3\n1 1 1",
"output": "Conan"
},
{
"input": "5\n1 1 2 2 2",
"output": "Conan"
}
] | 1,619,258,765 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 93 | 7,168,000 | n = int(input())
a = [int(i) for i in input().split()]
max = a[0]
for i in a:
if max < i:
max = i
count = 0
for i in a:
if max == i:
count +=1
if count == 1:
print("Conan")
elif (n - count)%2 == 0 and n%2 !=0:
print("Conan")
elif (n - count)%2 !=0 and n%2 == 0:
print("Agasa")
else:
print("Agasa")
| Title: Conan and Agasa play a Card Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Edogawa Conan got tired of solving cases, and invited his friend, Professor Agasa, over. They decided to play a game of cards. Conan has *n* cards, and the *i*-th card has a number *a**i* written on it.
They take turns playing, starting with Conan. In each turn, the player chooses a card and removes it. Also, he removes all cards having a number strictly lesser than the number on the chosen card. Formally, if the player chooses the *i*-th card, he removes that card and removes the *j*-th card for all *j* such that *a**j*<=<<=*a**i*.
A player loses if he cannot make a move on his turn, that is, he loses if there are no cards left. Predict the outcome of the game, assuming both players play optimally.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=105) — the number of cards Conan has.
The next line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=105), where *a**i* is the number on the *i*-th card.
Output Specification:
If Conan wins, print "Conan" (without quotes), otherwise print "Agasa" (without quotes).
Demo Input:
['3\n4 5 7\n', '2\n1 1\n']
Demo Output:
['Conan\n', 'Agasa\n']
Note:
In the first example, Conan can just choose the card having number 7 on it and hence remove all the cards. After that, there are no cards left on Agasa's turn.
In the second example, no matter which card Conan chooses, there will be one one card left, which Agasa can choose. After that, there are no cards left when it becomes Conan's turn again. | ```python
n = int(input())
a = [int(i) for i in input().split()]
max = a[0]
for i in a:
if max < i:
max = i
count = 0
for i in a:
if max == i:
count +=1
if count == 1:
print("Conan")
elif (n - count)%2 == 0 and n%2 !=0:
print("Conan")
elif (n - count)%2 !=0 and n%2 == 0:
print("Agasa")
else:
print("Agasa")
``` | 0 |
|
50 | A | Domino piling | PROGRAMMING | 800 | [
"greedy",
"math"
] | A. Domino piling | 2 | 256 | You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions. | In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16). | Output one number — the maximal number of dominoes, which can be placed. | [
"2 4\n",
"3 3\n"
] | [
"4\n",
"4\n"
] | none | 500 | [
{
"input": "2 4",
"output": "4"
},
{
"input": "3 3",
"output": "4"
},
{
"input": "1 5",
"output": "2"
},
{
"input": "1 6",
"output": "3"
},
{
"input": "1 15",
"output": "7"
},
{
"input": "1 16",
"output": "8"
},
{
"input": "2 5",
"output": "5"
},
{
"input": "2 6",
"output": "6"
},
{
"input": "2 7",
"output": "7"
},
{
"input": "2 14",
"output": "14"
},
{
"input": "2 15",
"output": "15"
},
{
"input": "1 4",
"output": "2"
},
{
"input": "2 16",
"output": "16"
},
{
"input": "3 5",
"output": "7"
},
{
"input": "3 6",
"output": "9"
},
{
"input": "3 10",
"output": "15"
},
{
"input": "3 14",
"output": "21"
},
{
"input": "3 15",
"output": "22"
},
{
"input": "3 16",
"output": "24"
},
{
"input": "5 7",
"output": "17"
},
{
"input": "16 16",
"output": "128"
},
{
"input": "15 16",
"output": "120"
},
{
"input": "2 3",
"output": "3"
},
{
"input": "15 15",
"output": "112"
},
{
"input": "14 16",
"output": "112"
},
{
"input": "11 13",
"output": "71"
},
{
"input": "5 16",
"output": "40"
},
{
"input": "8 15",
"output": "60"
},
{
"input": "2 2",
"output": "2"
},
{
"input": "3 4",
"output": "6"
},
{
"input": "4 4",
"output": "8"
},
{
"input": "1 1",
"output": "0"
},
{
"input": "1 2",
"output": "1"
},
{
"input": "1 3",
"output": "1"
},
{
"input": "14 15",
"output": "105"
}
] | 1,626,617,083 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 92 | 6,758,400 | integers = (input().split(" "))
print(int(integers[0] * integers[1]) // 2)
| Title: Domino piling
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions.
Input Specification:
In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16).
Output Specification:
Output one number — the maximal number of dominoes, which can be placed.
Demo Input:
['2 4\n', '3 3\n']
Demo Output:
['4\n', '4\n']
Note:
none | ```python
integers = (input().split(" "))
print(int(integers[0] * integers[1]) // 2)
``` | -1 |
22 | A | Second Order Statistics | PROGRAMMING | 800 | [
"brute force"
] | A. Second Order Statistics | 2 | 256 | Once Bob needed to find the second order statistics of a sequence of integer numbers. Lets choose each number from the sequence exactly once and sort them. The value on the second position is the second order statistics of the given sequence. In other words it is the smallest element strictly greater than the minimum. Help Bob solve this problem. | The first input line contains integer *n* (1<=≤<=*n*<=≤<=100) — amount of numbers in the sequence. The second line contains *n* space-separated integer numbers — elements of the sequence. These numbers don't exceed 100 in absolute value. | If the given sequence has the second order statistics, output this order statistics, otherwise output NO. | [
"4\n1 2 2 -4\n",
"5\n1 2 3 1 1\n"
] | [
"1\n",
"2\n"
] | none | 0 | [
{
"input": "4\n1 2 2 -4",
"output": "1"
},
{
"input": "5\n1 2 3 1 1",
"output": "2"
},
{
"input": "1\n28",
"output": "NO"
},
{
"input": "2\n-28 12",
"output": "12"
},
{
"input": "3\n-83 40 -80",
"output": "-80"
},
{
"input": "8\n93 77 -92 26 21 -48 53 91",
"output": "-48"
},
{
"input": "20\n-72 -9 -86 80 7 -10 40 -27 -94 92 96 56 28 -19 79 36 -3 -73 -63 -49",
"output": "-86"
},
{
"input": "49\n-74 -100 -80 23 -8 -83 -41 -20 48 17 46 -73 -55 67 85 4 40 -60 -69 -75 56 -74 -42 93 74 -95 64 -46 97 -47 55 0 -78 -34 -31 40 -63 -49 -76 48 21 -1 -49 -29 -98 -11 76 26 94",
"output": "-98"
},
{
"input": "88\n63 48 1 -53 -89 -49 64 -70 -49 71 -17 -16 76 81 -26 -50 67 -59 -56 97 2 100 14 18 -91 -80 42 92 -25 -88 59 8 -56 38 48 -71 -78 24 -14 48 -1 69 73 -76 54 16 -92 44 47 33 -34 -17 -81 21 -59 -61 53 26 10 -76 67 35 -29 70 65 -13 -29 81 80 32 74 -6 34 46 57 1 -45 -55 69 79 -58 11 -2 22 -18 -16 -89 -46",
"output": "-91"
},
{
"input": "100\n34 32 88 20 76 53 -71 -39 -98 -10 57 37 63 -3 -54 -64 -78 -82 73 20 -30 -4 22 75 51 -64 -91 29 -52 -48 83 19 18 -47 46 57 -44 95 89 89 -30 84 -83 67 58 -99 -90 -53 92 -60 -5 -56 -61 27 68 -48 52 -95 64 -48 -30 -67 66 89 14 -33 -31 -91 39 7 -94 -54 92 -96 -99 -83 -16 91 -28 -66 81 44 14 -85 -21 18 40 16 -13 -82 -33 47 -10 -40 -19 10 25 60 -34 -89",
"output": "-98"
},
{
"input": "2\n-1 -1",
"output": "NO"
},
{
"input": "3\n-2 -2 -2",
"output": "NO"
},
{
"input": "100\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0",
"output": "NO"
},
{
"input": "100\n100 100 100 100 100 100 100 100 100 100 100 100 -100 100 100 100 100 100 100 100 100 100 100 100 -100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 -100 100 100 100 100 100 100 100 100 100 100 -100 100 100 100 100 -100 100 100 100 100 100 100 100 100 100 100 100",
"output": "100"
},
{
"input": "10\n40 71 -85 -85 40 -85 -85 64 -85 47",
"output": "40"
},
{
"input": "23\n-90 -90 -41 -64 -64 -90 -15 10 -43 -90 -64 -64 89 -64 36 47 38 -90 -64 -90 -90 68 -90",
"output": "-64"
},
{
"input": "39\n-97 -93 -42 -93 -97 -93 56 -97 -97 -97 76 -33 -60 91 7 82 17 47 -97 -97 -93 73 -97 12 -97 -97 -97 -97 56 -92 -83 -93 -93 49 -93 -97 -97 -17 -93",
"output": "-93"
},
{
"input": "51\n-21 6 -35 -98 -86 -98 -86 -43 -65 32 -98 -40 96 -98 -98 -98 -98 -86 -86 -98 56 -86 -98 -98 -30 -98 -86 -31 -98 -86 -86 -86 -86 -30 96 -86 -86 -86 -60 25 88 -86 -86 58 31 -47 57 -86 37 44 -83",
"output": "-86"
},
{
"input": "66\n-14 -95 65 -95 -95 -97 -90 -71 -97 -97 70 -95 -95 -97 -95 -27 35 -87 -95 -5 -97 -97 87 34 -49 -95 -97 -95 -97 -95 -30 -95 -97 47 -95 -17 -97 -95 -97 -69 51 -97 -97 -95 -75 87 59 21 63 56 76 -91 98 -97 6 -97 -95 -95 -97 -73 11 -97 -35 -95 -95 -43",
"output": "-95"
},
{
"input": "77\n-67 -93 -93 -92 97 29 93 -93 -93 -5 -93 -7 60 -92 -93 44 -84 68 -92 -93 69 -92 -37 56 43 -93 35 -92 -93 19 -79 18 -92 -93 -93 -37 -93 -47 -93 -92 -92 74 67 19 40 -92 -92 -92 -92 -93 -93 -41 -93 -92 -93 -93 -92 -93 51 -80 6 -42 -92 -92 -66 -12 -92 -92 -3 93 -92 -49 -93 40 62 -92 -92",
"output": "-92"
},
{
"input": "89\n-98 40 16 -87 -98 63 -100 55 -96 -98 -21 -100 -93 26 -98 -98 -100 -89 -98 -5 -65 -28 -100 -6 -66 67 -100 -98 -98 10 -98 -98 -70 7 -98 2 -100 -100 -98 25 -100 -100 -98 23 -68 -100 -98 3 98 -100 -98 -98 -98 -98 -24 -100 -100 -9 -98 35 -100 99 -5 -98 -100 -100 37 -100 -84 57 -98 40 -47 -100 -1 -92 -76 -98 -98 -100 -100 -100 -63 30 21 -100 -100 -100 -12",
"output": "-98"
},
{
"input": "99\n10 -84 -100 -100 73 -64 -100 -94 33 -100 -100 -100 -100 71 64 24 7 -100 -32 -100 -100 77 -100 62 -12 55 45 -100 -100 -80 -100 -100 -100 -100 -100 -100 -100 -100 -100 -39 -48 -100 -34 47 -100 -100 -100 -100 -100 -77 -100 -100 -100 -100 -100 -100 -52 40 -55 -100 -44 -100 72 33 70 -100 -100 -78 -100 -3 100 -77 22 -100 95 -30 -100 10 -69 -100 -100 -100 -100 52 -39 -100 -100 -100 7 -100 -98 -66 95 -17 -100 52 -100 68 -100",
"output": "-98"
},
{
"input": "100\n-99 -98 -64 89 53 57 -99 29 -78 18 -3 -54 76 -98 -99 -98 37 -98 19 -47 89 73 -98 -91 -99 -99 -98 -48 -99 22 -99 -98 -99 -99 -98 -60 84 67 -99 -98 20 -98 88 -98 46 -98 -99 -98 -99 -71 -99 -98 -98 -39 83 95 -98 63 -98 -99 32 -98 -99 -64 57 -30 -53 -83 -4 -99 58 20 -98 -10 -99 -44 -99 -99 -99 -99 75 34 -98 -52 -98 -30 -98 -99 -98 -98 51 -99 -99 -99 -98 -99 -99 -82 -90 92",
"output": "-98"
},
{
"input": "3\n1 2 3",
"output": "2"
},
{
"input": "3\n1 3 2",
"output": "2"
},
{
"input": "3\n2 1 3",
"output": "2"
},
{
"input": "3\n2 3 1",
"output": "2"
},
{
"input": "3\n3 1 2",
"output": "2"
},
{
"input": "3\n3 2 1",
"output": "2"
},
{
"input": "9\n99 99 99 99 100 100 100 100 100",
"output": "100"
},
{
"input": "5\n-100 -100 -100 -100 -100",
"output": "NO"
}
] | 1,609,265,914 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 0 | 248 | 0 | n=int(input())
a=list(map(int,input().split()))
l=list(set(a))
if len(l) >=2:
print(l[1])
else:
print("NO") | Title: Second Order Statistics
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Once Bob needed to find the second order statistics of a sequence of integer numbers. Lets choose each number from the sequence exactly once and sort them. The value on the second position is the second order statistics of the given sequence. In other words it is the smallest element strictly greater than the minimum. Help Bob solve this problem.
Input Specification:
The first input line contains integer *n* (1<=≤<=*n*<=≤<=100) — amount of numbers in the sequence. The second line contains *n* space-separated integer numbers — elements of the sequence. These numbers don't exceed 100 in absolute value.
Output Specification:
If the given sequence has the second order statistics, output this order statistics, otherwise output NO.
Demo Input:
['4\n1 2 2 -4\n', '5\n1 2 3 1 1\n']
Demo Output:
['1\n', '2\n']
Note:
none | ```python
n=int(input())
a=list(map(int,input().split()))
l=list(set(a))
if len(l) >=2:
print(l[1])
else:
print("NO")
``` | 0 |
443 | A | Anton and Letters | PROGRAMMING | 800 | [
"constructive algorithms",
"implementation"
] | null | null | Recently, Anton has found a set. The set consists of small English letters. Anton carefully wrote out all the letters from the set in one line, separated by a comma. He also added an opening curved bracket at the beginning of the line and a closing curved bracket at the end of the line.
Unfortunately, from time to time Anton would forget writing some letter and write it again. He asks you to count the total number of distinct letters in his set. | The first and the single line contains the set of letters. The length of the line doesn't exceed 1000. It is guaranteed that the line starts from an opening curved bracket and ends with a closing curved bracket. Between them, small English letters are listed, separated by a comma. Each comma is followed by a space. | Print a single number — the number of distinct letters in Anton's set. | [
"{a, b, c}\n",
"{b, a, b, a}\n",
"{}\n"
] | [
"3\n",
"2\n",
"0\n"
] | none | 500 | [
{
"input": "{a, b, c}",
"output": "3"
},
{
"input": "{b, a, b, a}",
"output": "2"
},
{
"input": "{}",
"output": "0"
},
{
"input": "{a, a, c, b, b, b, c, c, c, c}",
"output": "3"
},
{
"input": "{a, c, b, b}",
"output": "3"
},
{
"input": "{a, b}",
"output": "2"
},
{
"input": "{a}",
"output": "1"
},
{
"input": "{b, a, b, a, b, c, c, b, c, b}",
"output": "3"
},
{
"input": "{e, g, c, e}",
"output": "3"
},
{
"input": "{a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a}",
"output": "1"
},
{
"input": "{a, a, b}",
"output": "2"
},
{
"input": "{a, b, b, b, a, b, a, a, a, a, a, a, b, a, b, a, a, a, a, a, b, a, b, a}",
"output": "2"
},
{
"input": "{j, u, a, c, f, w, e, w, x, t, h, p, v, n, i, l, x, n, i, b, u, c, a, a}",
"output": "16"
},
{
"input": "{x, i, w, c, p, e, h, z, k, i}",
"output": "9"
},
{
"input": "{t, k, o, x, r, d, q, j, k, e, z, w, y, r, z, s, s, e, s, b, k, i}",
"output": "15"
},
{
"input": "{y}",
"output": "1"
},
{
"input": "{x}",
"output": "1"
},
{
"input": "{b, z, a, z}",
"output": "3"
},
{
"input": "{z}",
"output": "1"
},
{
"input": "{a, z}",
"output": "2"
},
{
"input": "{a, b, z}",
"output": "3"
},
{
"input": "{s, q, z, r, t, a, b, h, j, i, o, z, r, q}",
"output": "11"
}
] | 1,693,333,952 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 31 | 0 | n = set((str(input())[1:-1]).split(', '))
print(len(n)) | Title: Anton and Letters
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Recently, Anton has found a set. The set consists of small English letters. Anton carefully wrote out all the letters from the set in one line, separated by a comma. He also added an opening curved bracket at the beginning of the line and a closing curved bracket at the end of the line.
Unfortunately, from time to time Anton would forget writing some letter and write it again. He asks you to count the total number of distinct letters in his set.
Input Specification:
The first and the single line contains the set of letters. The length of the line doesn't exceed 1000. It is guaranteed that the line starts from an opening curved bracket and ends with a closing curved bracket. Between them, small English letters are listed, separated by a comma. Each comma is followed by a space.
Output Specification:
Print a single number — the number of distinct letters in Anton's set.
Demo Input:
['{a, b, c}\n', '{b, a, b, a}\n', '{}\n']
Demo Output:
['3\n', '2\n', '0\n']
Note:
none | ```python
n = set((str(input())[1:-1]).split(', '))
print(len(n))
``` | 0 |
|
747 | A | Display Size | PROGRAMMING | 800 | [
"brute force",
"math"
] | null | null | A big company decided to launch a new series of rectangular displays, and decided that the display must have exactly *n* pixels.
Your task is to determine the size of the rectangular display — the number of lines (rows) of pixels *a* and the number of columns of pixels *b*, so that:
- there are exactly *n* pixels on the display; - the number of rows does not exceed the number of columns, it means *a*<=≤<=*b*; - the difference *b*<=-<=*a* is as small as possible. | The first line contains the positive integer *n* (1<=≤<=*n*<=≤<=106) — the number of pixels display should have. | Print two integers — the number of rows and columns on the display. | [
"8\n",
"64\n",
"5\n",
"999999\n"
] | [
"2 4\n",
"8 8\n",
"1 5\n",
"999 1001\n"
] | In the first example the minimum possible difference equals 2, so on the display should be 2 rows of 4 pixels.
In the second example the minimum possible difference equals 0, so on the display should be 8 rows of 8 pixels.
In the third example the minimum possible difference equals 4, so on the display should be 1 row of 5 pixels. | 500 | [
{
"input": "8",
"output": "2 4"
},
{
"input": "64",
"output": "8 8"
},
{
"input": "5",
"output": "1 5"
},
{
"input": "999999",
"output": "999 1001"
},
{
"input": "716539",
"output": "97 7387"
},
{
"input": "1",
"output": "1 1"
},
{
"input": "2",
"output": "1 2"
},
{
"input": "3",
"output": "1 3"
},
{
"input": "4",
"output": "2 2"
},
{
"input": "6",
"output": "2 3"
},
{
"input": "7",
"output": "1 7"
},
{
"input": "9",
"output": "3 3"
},
{
"input": "10",
"output": "2 5"
},
{
"input": "11",
"output": "1 11"
},
{
"input": "12",
"output": "3 4"
},
{
"input": "15",
"output": "3 5"
},
{
"input": "100",
"output": "10 10"
},
{
"input": "101",
"output": "1 101"
},
{
"input": "169",
"output": "13 13"
},
{
"input": "179",
"output": "1 179"
},
{
"input": "190",
"output": "10 19"
},
{
"input": "1000",
"output": "25 40"
},
{
"input": "10000",
"output": "100 100"
},
{
"input": "10001",
"output": "73 137"
},
{
"input": "100000",
"output": "250 400"
},
{
"input": "100001",
"output": "11 9091"
},
{
"input": "1000000",
"output": "1000 1000"
},
{
"input": "999983",
"output": "1 999983"
},
{
"input": "524288",
"output": "512 1024"
},
{
"input": "954493",
"output": "971 983"
},
{
"input": "966289",
"output": "983 983"
},
{
"input": "944663",
"output": "961 983"
},
{
"input": "912673",
"output": "97 9409"
},
{
"input": "732641",
"output": "679 1079"
},
{
"input": "232897",
"output": "343 679"
},
{
"input": "16807",
"output": "49 343"
},
{
"input": "999958",
"output": "2 499979"
},
{
"input": "990151",
"output": "1 990151"
},
{
"input": "997002",
"output": "998 999"
},
{
"input": "20",
"output": "4 5"
},
{
"input": "20261",
"output": "1 20261"
},
{
"input": "999123",
"output": "3 333041"
},
{
"input": "901841",
"output": "1 901841"
}
] | 1,482,300,745 | 2,147,483,647 | Python 3 | OK | TESTS | 43 | 62 | 4,608,000 | n = int(input())
i = 1
ans = 1
while i*i<=n:
if n%i==0 : ans = i
i+=1
print (ans, n//ans)
| Title: Display Size
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A big company decided to launch a new series of rectangular displays, and decided that the display must have exactly *n* pixels.
Your task is to determine the size of the rectangular display — the number of lines (rows) of pixels *a* and the number of columns of pixels *b*, so that:
- there are exactly *n* pixels on the display; - the number of rows does not exceed the number of columns, it means *a*<=≤<=*b*; - the difference *b*<=-<=*a* is as small as possible.
Input Specification:
The first line contains the positive integer *n* (1<=≤<=*n*<=≤<=106) — the number of pixels display should have.
Output Specification:
Print two integers — the number of rows and columns on the display.
Demo Input:
['8\n', '64\n', '5\n', '999999\n']
Demo Output:
['2 4\n', '8 8\n', '1 5\n', '999 1001\n']
Note:
In the first example the minimum possible difference equals 2, so on the display should be 2 rows of 4 pixels.
In the second example the minimum possible difference equals 0, so on the display should be 8 rows of 8 pixels.
In the third example the minimum possible difference equals 4, so on the display should be 1 row of 5 pixels. | ```python
n = int(input())
i = 1
ans = 1
while i*i<=n:
if n%i==0 : ans = i
i+=1
print (ans, n//ans)
``` | 3 |
|
429 | A | Xor-tree | PROGRAMMING | 1,300 | [
"dfs and similar",
"trees"
] | null | null | Iahub is very proud of his recent discovery, propagating trees. Right now, he invented a new tree, called xor-tree. After this new revolutionary discovery, he invented a game for kids which uses xor-trees.
The game is played on a tree having *n* nodes, numbered from 1 to *n*. Each node *i* has an initial value *init**i*, which is either 0 or 1. The root of the tree is node 1.
One can perform several (possibly, zero) operations on the tree during the game. The only available type of operation is to pick a node *x*. Right after someone has picked node *x*, the value of node *x* flips, the values of sons of *x* remain the same, the values of sons of sons of *x* flips, the values of sons of sons of sons of *x* remain the same and so on.
The goal of the game is to get each node *i* to have value *goal**i*, which can also be only 0 or 1. You need to reach the goal of the game by using minimum number of operations. | The first line contains an integer *n* (1<=≤<=*n*<=≤<=105). Each of the next *n*<=-<=1 lines contains two integers *u**i* and *v**i* (1<=≤<=*u**i*,<=*v**i*<=≤<=*n*; *u**i*<=≠<=*v**i*) meaning there is an edge between nodes *u**i* and *v**i*.
The next line contains *n* integer numbers, the *i*-th of them corresponds to *init**i* (*init**i* is either 0 or 1). The following line also contains *n* integer numbers, the *i*-th number corresponds to *goal**i* (*goal**i* is either 0 or 1). | In the first line output an integer number *cnt*, representing the minimal number of operations you perform. Each of the next *cnt* lines should contain an integer *x**i*, representing that you pick a node *x**i*. | [
"10\n2 1\n3 1\n4 2\n5 1\n6 2\n7 5\n8 6\n9 8\n10 5\n1 0 1 1 0 1 0 1 0 1\n1 0 1 0 0 1 1 1 0 1\n"
] | [
"2\n4\n7\n"
] | none | 500 | [
{
"input": "10\n2 1\n3 1\n4 2\n5 1\n6 2\n7 5\n8 6\n9 8\n10 5\n1 0 1 1 0 1 0 1 0 1\n1 0 1 0 0 1 1 1 0 1",
"output": "2\n4\n7"
},
{
"input": "15\n2 1\n3 2\n4 3\n5 4\n6 5\n7 6\n8 7\n9 8\n10 9\n11 10\n12 11\n13 12\n14 13\n15 14\n0 1 0 0 1 1 1 1 1 1 0 0 0 1 1\n1 1 1 1 0 0 1 1 0 1 0 0 1 1 0",
"output": "7\n1\n4\n7\n8\n9\n11\n13"
},
{
"input": "20\n2 1\n3 2\n4 3\n5 4\n6 4\n7 1\n8 2\n9 4\n10 2\n11 6\n12 9\n13 2\n14 12\n15 14\n16 8\n17 9\n18 13\n19 2\n20 17\n1 0 0 1 0 0 0 1 0 1 0 0 0 1 1 0 1 0 1 0\n1 0 0 1 0 0 1 1 0 0 1 0 0 1 0 0 0 1 0 1",
"output": "8\n11\n15\n17\n20\n10\n18\n19\n7"
},
{
"input": "30\n2 1\n3 2\n4 3\n5 3\n6 5\n7 3\n8 3\n9 2\n10 3\n11 2\n12 11\n13 6\n14 4\n15 5\n16 11\n17 9\n18 14\n19 6\n20 2\n21 19\n22 9\n23 19\n24 20\n25 14\n26 22\n27 1\n28 6\n29 13\n30 27\n1 0 1 1 1 1 0 1 0 0 1 0 0 0 1 0 0 1 0 1 0 0 1 0 0 1 1 1 1 0\n0 1 0 1 1 1 0 0 1 1 0 1 1 1 0 1 1 1 1 0 1 0 0 1 0 1 1 0 0 0",
"output": "15\n1\n2\n4\n5\n6\n13\n29\n19\n21\n23\n28\n7\n22\n26\n30"
},
{
"input": "15\n2 1\n3 1\n4 1\n5 1\n6 3\n7 1\n8 1\n9 1\n10 5\n11 9\n12 3\n13 5\n14 5\n15 4\n1 1 0 0 0 0 1 1 1 0 1 1 1 0 0\n1 0 1 1 0 1 1 1 1 1 1 1 1 1 0",
"output": "6\n2\n3\n6\n4\n10\n14"
},
{
"input": "20\n2 1\n3 1\n4 2\n5 2\n6 3\n7 1\n8 6\n9 2\n10 3\n11 6\n12 2\n13 3\n14 2\n15 1\n16 8\n17 15\n18 2\n19 14\n20 14\n0 0 0 0 1 1 1 1 1 1 0 1 1 1 0 0 0 1 1 1\n0 1 0 1 1 1 0 0 1 0 1 1 0 1 0 1 0 0 1 0",
"output": "10\n2\n4\n19\n18\n8\n16\n11\n10\n13\n7"
},
{
"input": "23\n2 1\n3 2\n4 1\n5 1\n6 5\n7 3\n8 2\n9 8\n10 5\n11 6\n12 9\n13 3\n14 11\n15 5\n16 2\n17 3\n18 10\n19 16\n20 14\n21 19\n22 17\n23 7\n0 1 0 1 1 1 0 1 1 0 0 0 1 0 0 1 1 0 1 0 0 0 0\n0 0 0 0 1 1 1 1 1 0 1 0 1 0 0 1 1 0 0 0 0 0 1",
"output": "8\n2\n23\n13\n17\n9\n4\n11\n20"
},
{
"input": "1\n0\n0",
"output": "0"
},
{
"input": "10\n1 10\n1 9\n10 2\n10 3\n3 7\n3 8\n9 4\n9 5\n5 6\n1 0 1 1 0 1 0 1 0 1\n0 0 0 0 0 0 0 0 0 0",
"output": "6\n1\n10\n2\n7\n5\n6"
}
] | 1,465,745,856 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 12 | 716 | 39,321,600 | from collections import defaultdict, deque, Counter, OrderedDict
from bisect import insort, bisect_right, bisect_left
import threading, sys
def main():
n = int(input())
adj = [[] for i in range(n + 1)]
for i in range(n - 1):
a, b = map(int, input().split())
a, b = a - 1, b - 1
adj[a].append(b)
adj[b].append(a)
init = [int(i) for i in input().split()]
goal = [int(i) for i in input().split()]
visited = [0] * n
par = [[] for i in range(n)]
def dfs(s, p):
if visited[s]: return
visited[s] = 1
par[p].append(s)
for i in adj[s]:
dfs(i, s)
dfs(0, 0)
par[0] = par[0][1:]
ans = []
def dfs2(s, l, fo, fe):
if l % 2 == 0:
if fe % 2 == 1:
init[s] = 1 - init[s]
else:
if fo % 2 == 1:
init[s] = 1 - init[s]
if init[s] != goal[s]:
ans.append(s + 1)
if l % 2:
fo += 1
else:
fe += 1
for j in par[s]:
dfs2(j, l + 1, fo, fe)
dfs2(0, 0, 0, 0)
print(len(ans))
print("\n".join(map(str, ans)))
if __name__ == "__main__":
sys.setrecursionlimit(200000)
threading.stack_size(10240000)
thread = threading.Thread(target=main)
thread.start()
| Title: Xor-tree
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Iahub is very proud of his recent discovery, propagating trees. Right now, he invented a new tree, called xor-tree. After this new revolutionary discovery, he invented a game for kids which uses xor-trees.
The game is played on a tree having *n* nodes, numbered from 1 to *n*. Each node *i* has an initial value *init**i*, which is either 0 or 1. The root of the tree is node 1.
One can perform several (possibly, zero) operations on the tree during the game. The only available type of operation is to pick a node *x*. Right after someone has picked node *x*, the value of node *x* flips, the values of sons of *x* remain the same, the values of sons of sons of *x* flips, the values of sons of sons of sons of *x* remain the same and so on.
The goal of the game is to get each node *i* to have value *goal**i*, which can also be only 0 or 1. You need to reach the goal of the game by using minimum number of operations.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=105). Each of the next *n*<=-<=1 lines contains two integers *u**i* and *v**i* (1<=≤<=*u**i*,<=*v**i*<=≤<=*n*; *u**i*<=≠<=*v**i*) meaning there is an edge between nodes *u**i* and *v**i*.
The next line contains *n* integer numbers, the *i*-th of them corresponds to *init**i* (*init**i* is either 0 or 1). The following line also contains *n* integer numbers, the *i*-th number corresponds to *goal**i* (*goal**i* is either 0 or 1).
Output Specification:
In the first line output an integer number *cnt*, representing the minimal number of operations you perform. Each of the next *cnt* lines should contain an integer *x**i*, representing that you pick a node *x**i*.
Demo Input:
['10\n2 1\n3 1\n4 2\n5 1\n6 2\n7 5\n8 6\n9 8\n10 5\n1 0 1 1 0 1 0 1 0 1\n1 0 1 0 0 1 1 1 0 1\n']
Demo Output:
['2\n4\n7\n']
Note:
none | ```python
from collections import defaultdict, deque, Counter, OrderedDict
from bisect import insort, bisect_right, bisect_left
import threading, sys
def main():
n = int(input())
adj = [[] for i in range(n + 1)]
for i in range(n - 1):
a, b = map(int, input().split())
a, b = a - 1, b - 1
adj[a].append(b)
adj[b].append(a)
init = [int(i) for i in input().split()]
goal = [int(i) for i in input().split()]
visited = [0] * n
par = [[] for i in range(n)]
def dfs(s, p):
if visited[s]: return
visited[s] = 1
par[p].append(s)
for i in adj[s]:
dfs(i, s)
dfs(0, 0)
par[0] = par[0][1:]
ans = []
def dfs2(s, l, fo, fe):
if l % 2 == 0:
if fe % 2 == 1:
init[s] = 1 - init[s]
else:
if fo % 2 == 1:
init[s] = 1 - init[s]
if init[s] != goal[s]:
ans.append(s + 1)
if l % 2:
fo += 1
else:
fe += 1
for j in par[s]:
dfs2(j, l + 1, fo, fe)
dfs2(0, 0, 0, 0)
print(len(ans))
print("\n".join(map(str, ans)))
if __name__ == "__main__":
sys.setrecursionlimit(200000)
threading.stack_size(10240000)
thread = threading.Thread(target=main)
thread.start()
``` | -1 |
|
1 | A | Theatre Square | PROGRAMMING | 1,000 | [
"math"
] | A. Theatre Square | 1 | 256 | Theatre Square in the capital city of Berland has a rectangular shape with the size *n*<=×<=*m* meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size *a*<=×<=*a*.
What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square. | The input contains three positive integer numbers in the first line: *n*,<=<=*m* and *a* (1<=≤<=<=*n*,<=*m*,<=*a*<=≤<=109). | Write the needed number of flagstones. | [
"6 6 4\n"
] | [
"4\n"
] | none | 0 | [
{
"input": "6 6 4",
"output": "4"
},
{
"input": "1 1 1",
"output": "1"
},
{
"input": "2 1 1",
"output": "2"
},
{
"input": "1 2 1",
"output": "2"
},
{
"input": "2 2 1",
"output": "4"
},
{
"input": "2 1 2",
"output": "1"
},
{
"input": "1 1 3",
"output": "1"
},
{
"input": "2 3 4",
"output": "1"
},
{
"input": "1000000000 1000000000 1",
"output": "1000000000000000000"
},
{
"input": "12 13 4",
"output": "12"
},
{
"input": "222 332 5",
"output": "3015"
},
{
"input": "1000 1000 10",
"output": "10000"
},
{
"input": "1001 1000 10",
"output": "10100"
},
{
"input": "100 10001 1000000000",
"output": "1"
},
{
"input": "1000000000 1000000000 1000000000",
"output": "1"
},
{
"input": "1000000000 1000000000 999999999",
"output": "4"
},
{
"input": "1000000000 1000000000 192",
"output": "27126743055556"
},
{
"input": "1000000000 987654321 1",
"output": "987654321000000000"
},
{
"input": "456784567 1000000000 51",
"output": "175618850864484"
},
{
"input": "39916800 134217728 40320",
"output": "3295710"
}
] | 1,673,424,413 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 20 | 62 | 0 | n, m, a = map(int, input().split())
row, col = (n//a) + 1 if n % a else (n//a), (m//a) + 1 if m % a else (m//a)
print(row*col)
| Title: Theatre Square
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Theatre Square in the capital city of Berland has a rectangular shape with the size *n*<=×<=*m* meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size *a*<=×<=*a*.
What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square.
Input Specification:
The input contains three positive integer numbers in the first line: *n*,<=<=*m* and *a* (1<=≤<=<=*n*,<=*m*,<=*a*<=≤<=109).
Output Specification:
Write the needed number of flagstones.
Demo Input:
['6 6 4\n']
Demo Output:
['4\n']
Note:
none | ```python
n, m, a = map(int, input().split())
row, col = (n//a) + 1 if n % a else (n//a), (m//a) + 1 if m % a else (m//a)
print(row*col)
``` | 3.969 |
779 | C | Dishonest Sellers | PROGRAMMING | 1,200 | [
"constructive algorithms",
"greedy",
"sortings"
] | null | null | Igor found out discounts in a shop and decided to buy *n* items. Discounts at the store will last for a week and Igor knows about each item that its price now is *a**i*, and after a week of discounts its price will be *b**i*.
Not all of sellers are honest, so now some products could be more expensive than after a week of discounts.
Igor decided that buy at least *k* of items now, but wait with the rest of the week in order to save money as much as possible. Your task is to determine the minimum money that Igor can spend to buy all *n* items. | In the first line there are two positive integer numbers *n* and *k* (1<=≤<=*n*<=≤<=2·105, 0<=≤<=*k*<=≤<=*n*) — total number of items to buy and minimal number of items Igor wants to by right now.
The second line contains sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=104) — prices of items during discounts (i.e. right now).
The third line contains sequence of integers *b*1,<=*b*2,<=...,<=*b**n* (1<=≤<=*b**i*<=≤<=104) — prices of items after discounts (i.e. after a week). | Print the minimal amount of money Igor will spend to buy all *n* items. Remember, he should buy at least *k* items right now. | [
"3 1\n5 4 6\n3 1 5\n",
"5 3\n3 4 7 10 3\n4 5 5 12 5\n"
] | [
"10\n",
"25\n"
] | In the first example Igor should buy item 3 paying 6. But items 1 and 2 he should buy after a week. He will pay 3 and 1 for them. So in total he will pay 6 + 3 + 1 = 10.
In the second example Igor should buy right now items 1, 2, 4 and 5, paying for them 3, 4, 10 and 3, respectively. Item 3 he should buy after a week of discounts, he will pay 5 for it. In total he will spend 3 + 4 + 10 + 3 + 5 = 25. | 1,000 | [
{
"input": "3 1\n5 4 6\n3 1 5",
"output": "10"
},
{
"input": "5 3\n3 4 7 10 3\n4 5 5 12 5",
"output": "25"
},
{
"input": "1 0\n9\n8",
"output": "8"
},
{
"input": "2 0\n4 10\n1 2",
"output": "3"
},
{
"input": "4 2\n19 5 17 13\n3 18 8 10",
"output": "29"
},
{
"input": "5 3\n28 17 20 45 45\n39 12 41 27 9",
"output": "101"
},
{
"input": "10 5\n87 96 19 81 10 88 7 49 36 21\n11 75 28 28 74 17 64 19 81 31",
"output": "243"
},
{
"input": "50 45\n80 125 152 122 85 62 123 195 157 73 8 127 47 193 70 132 131 38 98 97 185 98 68 166 114 53 144 106 141 106 191 123 44 199 35 18 84 169 46 52 129 102 155 130 63 169 116 145 154 41\n155 187 79 106 136 173 17 146 9 47 173 21 37 183 110 74 185 14 82 156 147 25 162 195 26 25 22 80 171 180 7 92 53 18 100 17 159 184 37 127 116 192 163 49 94 95 118 102 8 30",
"output": "4680"
},
{
"input": "5 2\n1 1 1 1 1\n2 2 2 2 2",
"output": "5"
},
{
"input": "5 2\n2 4 4 4 4\n3 1 1 1 1",
"output": "9"
},
{
"input": "6 3\n9 9 9 2 2 2\n6 6 6 1 1 1",
"output": "24"
},
{
"input": "5 3\n10 17 23 26 35\n7 18 18 29 32",
"output": "103"
},
{
"input": "2 1\n10 12\n5 6",
"output": "16"
}
] | 1,488,258,463 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 9 | 405 | 21,401,600 | n, k = map(int, input().strip().split(" "))
a = list(map(int, input().strip().split(" ")))
b = list(map(int, input().strip().split(" ")))
lista = []
for i in range(n):
aux = float(a[i] / b[i])
tupla = (aux, a[i], b[i])
lista.append(tupla)
lista.sort(key = lambda tup: tup[0])
suma = 0
for tup in lista:
if(k > 0):
suma += tup[1]
k -= 1
else:
suma += min([tup[1], tup[2]])
print(suma) | Title: Dishonest Sellers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Igor found out discounts in a shop and decided to buy *n* items. Discounts at the store will last for a week and Igor knows about each item that its price now is *a**i*, and after a week of discounts its price will be *b**i*.
Not all of sellers are honest, so now some products could be more expensive than after a week of discounts.
Igor decided that buy at least *k* of items now, but wait with the rest of the week in order to save money as much as possible. Your task is to determine the minimum money that Igor can spend to buy all *n* items.
Input Specification:
In the first line there are two positive integer numbers *n* and *k* (1<=≤<=*n*<=≤<=2·105, 0<=≤<=*k*<=≤<=*n*) — total number of items to buy and minimal number of items Igor wants to by right now.
The second line contains sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=104) — prices of items during discounts (i.e. right now).
The third line contains sequence of integers *b*1,<=*b*2,<=...,<=*b**n* (1<=≤<=*b**i*<=≤<=104) — prices of items after discounts (i.e. after a week).
Output Specification:
Print the minimal amount of money Igor will spend to buy all *n* items. Remember, he should buy at least *k* items right now.
Demo Input:
['3 1\n5 4 6\n3 1 5\n', '5 3\n3 4 7 10 3\n4 5 5 12 5\n']
Demo Output:
['10\n', '25\n']
Note:
In the first example Igor should buy item 3 paying 6. But items 1 and 2 he should buy after a week. He will pay 3 and 1 for them. So in total he will pay 6 + 3 + 1 = 10.
In the second example Igor should buy right now items 1, 2, 4 and 5, paying for them 3, 4, 10 and 3, respectively. Item 3 he should buy after a week of discounts, he will pay 5 for it. In total he will spend 3 + 4 + 10 + 3 + 5 = 25. | ```python
n, k = map(int, input().strip().split(" "))
a = list(map(int, input().strip().split(" ")))
b = list(map(int, input().strip().split(" ")))
lista = []
for i in range(n):
aux = float(a[i] / b[i])
tupla = (aux, a[i], b[i])
lista.append(tupla)
lista.sort(key = lambda tup: tup[0])
suma = 0
for tup in lista:
if(k > 0):
suma += tup[1]
k -= 1
else:
suma += min([tup[1], tup[2]])
print(suma)
``` | 0 |
|
887 | B | Cubes for Masha | PROGRAMMING | 1,300 | [
"brute force",
"implementation"
] | null | null | Absent-minded Masha got set of *n* cubes for her birthday.
At each of 6 faces of each cube, there is exactly one digit from 0 to 9. Masha became interested what is the largest natural *x* such she can make using her new cubes all integers from 1 to *x*.
To make a number Masha can rotate her cubes and put them in a row. After that, she looks at upper faces of cubes from left to right and reads the number.
The number can't contain leading zeros. It's not required to use all cubes to build a number.
Pay attention: Masha can't make digit 6 from digit 9 and vice-versa using cube rotations. | In first line integer *n* is given (1<=≤<=*n*<=≤<=3) — the number of cubes, Masha got for her birthday.
Each of next *n* lines contains 6 integers *a**i**j* (0<=≤<=*a**i**j*<=≤<=9) — number on *j*-th face of *i*-th cube. | Print single integer — maximum number *x* such Masha can make any integers from 1 to *x* using her cubes or 0 if Masha can't make even 1. | [
"3\n0 1 2 3 4 5\n6 7 8 9 0 1\n2 3 4 5 6 7\n",
"3\n0 1 3 5 6 8\n1 2 4 5 7 8\n2 3 4 6 7 9\n"
] | [
"87",
"98"
] | In the first test case, Masha can build all numbers from 1 to 87, but she can't make 88 because there are no two cubes with digit 8. | 1,000 | [
{
"input": "3\n0 1 2 3 4 5\n6 7 8 9 0 1\n2 3 4 5 6 7",
"output": "87"
},
{
"input": "3\n0 1 3 5 6 8\n1 2 4 5 7 8\n2 3 4 6 7 9",
"output": "98"
},
{
"input": "3\n0 1 2 3 4 5\n0 1 2 3 4 5\n0 1 2 3 4 5",
"output": "5"
},
{
"input": "3\n1 2 3 7 8 9\n9 8 7 1 2 3\n7 9 2 3 1 8",
"output": "3"
},
{
"input": "1\n5 2 2 5 6 7",
"output": "0"
},
{
"input": "1\n7 6 5 8 9 0",
"output": "0"
},
{
"input": "1\n2 5 9 6 7 9",
"output": "0"
},
{
"input": "1\n6 3 1 9 4 9",
"output": "1"
},
{
"input": "1\n1 9 8 3 7 8",
"output": "1"
},
{
"input": "2\n1 7 2 0 4 3\n5 2 3 6 1 0",
"output": "7"
},
{
"input": "2\n6 0 1 7 2 9\n1 3 4 6 7 0",
"output": "4"
},
{
"input": "2\n8 6 4 1 2 0\n7 8 5 3 2 1",
"output": "8"
},
{
"input": "2\n0 8 6 2 1 3\n5 2 7 1 0 9",
"output": "3"
},
{
"input": "2\n0 9 5 7 6 2\n8 6 2 7 1 4",
"output": "2"
},
{
"input": "3\n5 0 7 6 2 1\n2 7 4 6 1 9\n0 2 6 1 7 5",
"output": "2"
},
{
"input": "3\n0 6 2 9 5 4\n3 8 0 1 6 9\n6 9 0 1 5 2",
"output": "6"
},
{
"input": "3\n5 6 2 9 3 5\n5 4 1 5 9 8\n4 4 2 0 3 5",
"output": "6"
},
{
"input": "3\n0 1 9 1 0 8\n9 9 3 5 6 2\n9 3 9 9 7 3",
"output": "3"
},
{
"input": "3\n2 5 7 4 2 7\n1 5 5 9 0 3\n8 2 0 1 5 1",
"output": "5"
},
{
"input": "1\n4 6 9 8 2 7",
"output": "0"
},
{
"input": "1\n5 3 8 0 2 6",
"output": "0"
},
{
"input": "1\n7 9 5 0 4 6",
"output": "0"
},
{
"input": "1\n4 0 9 6 3 1",
"output": "1"
},
{
"input": "1\n7 9 2 5 0 4",
"output": "0"
},
{
"input": "1\n0 7 6 3 2 4",
"output": "0"
},
{
"input": "1\n9 8 1 6 5 7",
"output": "1"
},
{
"input": "1\n7 3 6 9 8 1",
"output": "1"
},
{
"input": "1\n3 9 1 7 4 5",
"output": "1"
},
{
"input": "1\n8 6 0 9 4 2",
"output": "0"
},
{
"input": "1\n8 2 7 4 1 0",
"output": "2"
},
{
"input": "1\n8 3 5 4 2 9",
"output": "0"
},
{
"input": "1\n0 8 7 1 3 2",
"output": "3"
},
{
"input": "1\n6 2 8 5 1 3",
"output": "3"
},
{
"input": "1\n6 0 7 5 4 8",
"output": "0"
},
{
"input": "1\n6 2 8 4 5 1",
"output": "2"
},
{
"input": "1\n4 3 8 9 2 3",
"output": "0"
},
{
"input": "1\n8 1 9 2 9 7",
"output": "2"
},
{
"input": "1\n3 7 7 6 4 2",
"output": "0"
},
{
"input": "1\n1 4 5 7 0 5",
"output": "1"
},
{
"input": "2\n6 6 4 7 9 0\n2 1 2 8 6 4",
"output": "2"
},
{
"input": "2\n5 3 2 9 8 2\n0 7 4 8 1 8",
"output": "5"
},
{
"input": "2\n5 7 4 2 1 9\n2 2 7 1 1 8",
"output": "2"
},
{
"input": "2\n9 3 3 6 7 2\n6 2 9 1 5 9",
"output": "3"
},
{
"input": "2\n2 0 5 7 0 8\n4 5 1 5 4 9",
"output": "2"
},
{
"input": "2\n2 6 8 1 3 1\n2 1 3 8 6 7",
"output": "3"
},
{
"input": "2\n4 3 8 6 0 1\n4 7 1 8 9 0",
"output": "1"
},
{
"input": "2\n0 2 9 1 8 5\n0 7 4 3 2 5",
"output": "5"
},
{
"input": "2\n1 7 6 9 2 5\n1 6 7 0 9 2",
"output": "2"
},
{
"input": "2\n0 2 9 8 1 7\n6 7 4 3 2 5",
"output": "9"
},
{
"input": "2\n3 6 8 9 5 0\n6 7 0 8 2 3",
"output": "0"
},
{
"input": "2\n5 1 2 3 0 8\n3 6 7 4 9 2",
"output": "9"
},
{
"input": "2\n7 8 6 1 4 5\n8 6 4 3 2 5",
"output": "8"
},
{
"input": "2\n2 3 5 1 9 6\n1 6 8 7 3 9",
"output": "3"
},
{
"input": "2\n1 7 8 6 0 9\n3 2 1 7 4 9",
"output": "4"
},
{
"input": "2\n2 4 0 3 7 6\n3 2 8 7 1 5",
"output": "8"
},
{
"input": "2\n6 5 2 7 1 3\n3 7 8 1 0 9",
"output": "3"
},
{
"input": "2\n5 8 4 7 1 2\n0 8 6 2 4 9",
"output": "2"
},
{
"input": "2\n8 0 6 5 1 4\n7 1 0 8 3 4",
"output": "1"
},
{
"input": "2\n2 3 9 1 6 7\n2 5 4 3 0 6",
"output": "7"
},
{
"input": "3\n9 4 3 0 2 6\n7 0 5 3 3 9\n1 0 7 4 6 7",
"output": "7"
},
{
"input": "3\n3 8 5 1 5 5\n1 5 7 2 6 9\n4 3 4 8 8 9",
"output": "9"
},
{
"input": "3\n7 7 2 5 3 2\n3 0 0 6 4 4\n1 2 1 1 9 1",
"output": "7"
},
{
"input": "3\n8 1 6 8 6 8\n7 0 2 5 8 4\n5 2 0 3 1 9",
"output": "32"
},
{
"input": "3\n2 7 4 0 7 1\n5 5 4 9 1 4\n2 1 7 5 1 7",
"output": "2"
},
{
"input": "3\n4 4 5 0 6 6\n7 1 6 9 5 4\n5 0 4 0 3 9",
"output": "1"
},
{
"input": "3\n9 4 3 3 9 3\n1 0 3 4 5 3\n2 9 6 2 4 1",
"output": "6"
},
{
"input": "3\n3 8 3 5 5 5\n3 0 1 6 6 3\n0 4 3 7 2 4",
"output": "8"
},
{
"input": "3\n4 1 0 8 0 2\n1 5 3 5 0 7\n7 7 2 7 2 2",
"output": "5"
},
{
"input": "3\n8 1 8 2 7 1\n9 1 9 9 4 7\n0 0 9 0 4 0",
"output": "2"
},
{
"input": "3\n4 6 0 3 9 2\n8 6 9 0 7 2\n6 9 3 2 5 7",
"output": "0"
},
{
"input": "3\n5 1 2 9 6 4\n9 0 6 4 2 8\n4 6 2 8 3 7",
"output": "10"
},
{
"input": "3\n9 3 1 8 4 6\n6 9 1 2 0 7\n8 9 1 5 0 3",
"output": "21"
},
{
"input": "3\n7 1 3 0 2 4\n2 4 3 0 9 5\n1 9 8 0 6 5",
"output": "65"
},
{
"input": "3\n9 4 6 2 7 0\n3 7 1 9 6 4\n6 1 0 8 7 2",
"output": "4"
},
{
"input": "3\n2 7 3 6 4 5\n0 2 1 9 4 8\n8 6 9 5 4 0",
"output": "10"
},
{
"input": "3\n2 6 3 7 1 0\n9 1 2 4 7 6\n1 4 8 7 6 2",
"output": "4"
},
{
"input": "3\n5 4 8 1 6 7\n0 9 3 5 8 6\n2 4 7 8 1 3",
"output": "21"
},
{
"input": "3\n7 2 1 3 6 9\n0 3 8 4 7 6\n1 4 5 8 7 0",
"output": "21"
},
{
"input": "3\n8 6 0 5 4 9\n1 8 5 3 9 7\n7 4 5 1 6 8",
"output": "1"
},
{
"input": "1\n0 1 2 3 4 5",
"output": "5"
},
{
"input": "3\n0 1 1 2 2 3\n4 5 6 7 8 9\n3 4 5 6 7 8",
"output": "9"
},
{
"input": "2\n0 1 2 3 4 5\n6 7 8 9 1 2",
"output": "29"
},
{
"input": "3\n0 1 2 3 4 5\n6 7 8 9 1 2\n3 4 5 6 7 8",
"output": "98"
},
{
"input": "3\n0 1 1 2 2 3\n4 5 6 7 8 9\n3 4 5 6 7 1",
"output": "19"
},
{
"input": "2\n0 1 2 3 4 5\n6 7 8 9 6 6",
"output": "9"
},
{
"input": "2\n0 1 2 3 4 5\n4 5 6 7 8 9",
"output": "9"
},
{
"input": "2\n1 8 9 1 1 0\n2 3 4 5 6 7",
"output": "9"
},
{
"input": "2\n0 1 2 3 4 5\n9 8 7 6 5 4",
"output": "9"
},
{
"input": "3\n2 3 4 5 6 7\n3 4 5 6 7 8\n9 1 2 3 4 5",
"output": "9"
},
{
"input": "3\n1 1 2 3 4 5\n6 7 8 9 0 2\n3 4 5 6 7 8",
"output": "10"
},
{
"input": "3\n1 1 1 0 2 3\n4 5 6 7 8 9\n0 0 0 0 0 0",
"output": "10"
},
{
"input": "3\n0 1 2 2 4 5\n6 7 8 9 0 1\n3 3 4 5 6 7",
"output": "21"
},
{
"input": "3\n2 0 1 3 4 5\n6 7 8 9 1 1\n3 4 5 6 6 7",
"output": "19"
},
{
"input": "3\n1 1 1 1 1 1\n0 2 3 4 5 6\n7 8 9 2 3 4",
"output": "10"
},
{
"input": "2\n0 1 2 3 4 5\n6 6 6 7 8 9",
"output": "9"
},
{
"input": "3\n3 4 5 6 8 9\n1 1 1 1 1 1\n1 2 4 5 7 0",
"output": "19"
}
] | 1,559,871,883 | 2,147,483,647 | Python 3 | OK | TESTS | 96 | 124 | 307,200 | ##a = list(map(int, input().split()))
##print(' '.join(map(str, res)))
def list_input():
return list(map(int, input().split()))
n = int(input())
c = list()
for i in range(n):
c.append(list_input())
s = set()
for i in range(n):
for x in c[i]:
s.add(x)
if n >= 2:
for i in range(n):
for j in range(n):
if i == j:
continue
for x in c[i]:
for y in c[j]:
if x != 0:
s.add(x*10+y)
if n >= 3:
for i in range(n):
for j in range(n):
if i == j:
continue
for k in range(n):
if i == k or j == k:
continue
for x in c[i]:
for y in c[j]:
for z in c[k]:
if x != 0:
s.add(x*100+y*10+z)
if 0 in s:
s.remove(0)
res = 0
for x in s:
if x == res+1:
res += 1
else:
print(res)
exit(0)
print(res) | Title: Cubes for Masha
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Absent-minded Masha got set of *n* cubes for her birthday.
At each of 6 faces of each cube, there is exactly one digit from 0 to 9. Masha became interested what is the largest natural *x* such she can make using her new cubes all integers from 1 to *x*.
To make a number Masha can rotate her cubes and put them in a row. After that, she looks at upper faces of cubes from left to right and reads the number.
The number can't contain leading zeros. It's not required to use all cubes to build a number.
Pay attention: Masha can't make digit 6 from digit 9 and vice-versa using cube rotations.
Input Specification:
In first line integer *n* is given (1<=≤<=*n*<=≤<=3) — the number of cubes, Masha got for her birthday.
Each of next *n* lines contains 6 integers *a**i**j* (0<=≤<=*a**i**j*<=≤<=9) — number on *j*-th face of *i*-th cube.
Output Specification:
Print single integer — maximum number *x* such Masha can make any integers from 1 to *x* using her cubes or 0 if Masha can't make even 1.
Demo Input:
['3\n0 1 2 3 4 5\n6 7 8 9 0 1\n2 3 4 5 6 7\n', '3\n0 1 3 5 6 8\n1 2 4 5 7 8\n2 3 4 6 7 9\n']
Demo Output:
['87', '98']
Note:
In the first test case, Masha can build all numbers from 1 to 87, but she can't make 88 because there are no two cubes with digit 8. | ```python
##a = list(map(int, input().split()))
##print(' '.join(map(str, res)))
def list_input():
return list(map(int, input().split()))
n = int(input())
c = list()
for i in range(n):
c.append(list_input())
s = set()
for i in range(n):
for x in c[i]:
s.add(x)
if n >= 2:
for i in range(n):
for j in range(n):
if i == j:
continue
for x in c[i]:
for y in c[j]:
if x != 0:
s.add(x*10+y)
if n >= 3:
for i in range(n):
for j in range(n):
if i == j:
continue
for k in range(n):
if i == k or j == k:
continue
for x in c[i]:
for y in c[j]:
for z in c[k]:
if x != 0:
s.add(x*100+y*10+z)
if 0 in s:
s.remove(0)
res = 0
for x in s:
if x == res+1:
res += 1
else:
print(res)
exit(0)
print(res)
``` | 3 |
|
0 | none | none | none | 0 | [
"none"
] | null | null | A new innovative ticketing systems for public transport is introduced in Bytesburg. Now there is a single travel card for all transport. To make a trip a passenger scan his card and then he is charged according to the fare.
The fare is constructed in the following manner. There are three types of tickets:
1. a ticket for one trip costs 20 byteland rubles, 1. a ticket for 90 minutes costs 50 byteland rubles, 1. a ticket for one day (1440 minutes) costs 120 byteland rubles.
Note that a ticket for *x* minutes activated at time *t* can be used for trips started in time range from *t* to *t*<=+<=*x*<=-<=1, inclusive. Assume that all trips take exactly one minute.
To simplify the choice for the passenger, the system automatically chooses the optimal tickets. After each trip starts, the system analyses all the previous trips and the current trip and chooses a set of tickets for these trips with a minimum total cost. Let the minimum total cost of tickets to cover all trips from the first to the current is *a*, and the total sum charged before is *b*. Then the system charges the passenger the sum *a*<=-<=*b*.
You have to write a program that, for given trips made by a passenger, calculates the sum the passenger is charged after each trip. | The first line of input contains integer number *n* (1<=≤<=*n*<=≤<=105) — the number of trips made by passenger.
Each of the following *n* lines contains the time of trip *t**i* (0<=≤<=*t**i*<=≤<=109), measured in minutes from the time of starting the system. All *t**i* are different, given in ascending order, i. e. *t**i*<=+<=1<=><=*t**i* holds for all 1<=≤<=*i*<=<<=*n*. | Output *n* integers. For each trip, print the sum the passenger is charged after it. | [
"3\n10\n20\n30\n",
"10\n13\n45\n46\n60\n103\n115\n126\n150\n256\n516\n"
] | [
"20\n20\n10\n",
"20\n20\n10\n0\n20\n0\n0\n20\n20\n10\n"
] | In the first example, the system works as follows: for the first and second trips it is cheaper to pay for two one-trip tickets, so each time 20 rubles is charged, after the third trip the system understands that it would be cheaper to buy a ticket for 90 minutes. This ticket costs 50 rubles, and the passenger had already paid 40 rubles, so it is necessary to charge 10 rubles only. | 0 | [
{
"input": "3\n10\n20\n30",
"output": "20\n20\n10"
},
{
"input": "10\n13\n45\n46\n60\n103\n115\n126\n150\n256\n516",
"output": "20\n20\n10\n0\n20\n0\n0\n20\n20\n10"
},
{
"input": "7\n100\n138\n279\n308\n396\n412\n821",
"output": "20\n20\n20\n20\n20\n20\n0"
},
{
"input": "8\n0\n2\n51\n63\n69\n75\n80\n90",
"output": "20\n20\n10\n0\n0\n0\n0\n20"
},
{
"input": "8\n0\n3\n5\n6\n7\n8\n9\n10",
"output": "20\n20\n10\n0\n0\n0\n0\n0"
},
{
"input": "1\n0",
"output": "20"
}
] | 1,608,823,111 | 2,147,483,647 | PyPy 3 | OK | TESTS | 31 | 358 | 14,950,400 | from sys import stdin
from bisect import *
rint = lambda: int(stdin.readline())
rint_2d = lambda n: [rint() for _ in range(n)]
n = int(input())
a, mem, out = rint_2d(n), [0] * (n + 1), []
for i in range(n):
vals = [mem[i] + 20]
ix1 = max(bisect_left(a, a[i] - 90 + 1), 0)
vals.append(mem[ix1] + 50)
ix2 = max(bisect_left(a, a[i] - 1440 + 1), 0)
vals.append(mem[ix2] + 120)
mem[i + 1] = min(vals)
out.append(mem[i + 1] - mem[i])
print('\n'.join(map(str, out)))
| Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A new innovative ticketing systems for public transport is introduced in Bytesburg. Now there is a single travel card for all transport. To make a trip a passenger scan his card and then he is charged according to the fare.
The fare is constructed in the following manner. There are three types of tickets:
1. a ticket for one trip costs 20 byteland rubles, 1. a ticket for 90 minutes costs 50 byteland rubles, 1. a ticket for one day (1440 minutes) costs 120 byteland rubles.
Note that a ticket for *x* minutes activated at time *t* can be used for trips started in time range from *t* to *t*<=+<=*x*<=-<=1, inclusive. Assume that all trips take exactly one minute.
To simplify the choice for the passenger, the system automatically chooses the optimal tickets. After each trip starts, the system analyses all the previous trips and the current trip and chooses a set of tickets for these trips with a minimum total cost. Let the minimum total cost of tickets to cover all trips from the first to the current is *a*, and the total sum charged before is *b*. Then the system charges the passenger the sum *a*<=-<=*b*.
You have to write a program that, for given trips made by a passenger, calculates the sum the passenger is charged after each trip.
Input Specification:
The first line of input contains integer number *n* (1<=≤<=*n*<=≤<=105) — the number of trips made by passenger.
Each of the following *n* lines contains the time of trip *t**i* (0<=≤<=*t**i*<=≤<=109), measured in minutes from the time of starting the system. All *t**i* are different, given in ascending order, i. e. *t**i*<=+<=1<=><=*t**i* holds for all 1<=≤<=*i*<=<<=*n*.
Output Specification:
Output *n* integers. For each trip, print the sum the passenger is charged after it.
Demo Input:
['3\n10\n20\n30\n', '10\n13\n45\n46\n60\n103\n115\n126\n150\n256\n516\n']
Demo Output:
['20\n20\n10\n', '20\n20\n10\n0\n20\n0\n0\n20\n20\n10\n']
Note:
In the first example, the system works as follows: for the first and second trips it is cheaper to pay for two one-trip tickets, so each time 20 rubles is charged, after the third trip the system understands that it would be cheaper to buy a ticket for 90 minutes. This ticket costs 50 rubles, and the passenger had already paid 40 rubles, so it is necessary to charge 10 rubles only. | ```python
from sys import stdin
from bisect import *
rint = lambda: int(stdin.readline())
rint_2d = lambda n: [rint() for _ in range(n)]
n = int(input())
a, mem, out = rint_2d(n), [0] * (n + 1), []
for i in range(n):
vals = [mem[i] + 20]
ix1 = max(bisect_left(a, a[i] - 90 + 1), 0)
vals.append(mem[ix1] + 50)
ix2 = max(bisect_left(a, a[i] - 1440 + 1), 0)
vals.append(mem[ix2] + 120)
mem[i + 1] = min(vals)
out.append(mem[i + 1] - mem[i])
print('\n'.join(map(str, out)))
``` | 3 |
|
92 | A | Chips | PROGRAMMING | 800 | [
"implementation",
"math"
] | A. Chips | 2 | 256 | There are *n* walruses sitting in a circle. All of them are numbered in the clockwise order: the walrus number 2 sits to the left of the walrus number 1, the walrus number 3 sits to the left of the walrus number 2, ..., the walrus number 1 sits to the left of the walrus number *n*.
The presenter has *m* chips. The presenter stands in the middle of the circle and starts giving the chips to the walruses starting from walrus number 1 and moving clockwise. The walrus number *i* gets *i* chips. If the presenter can't give the current walrus the required number of chips, then the presenter takes the remaining chips and the process ends. Determine by the given *n* and *m* how many chips the presenter will get in the end. | The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=50, 1<=≤<=*m*<=≤<=104) — the number of walruses and the number of chips correspondingly. | Print the number of chips the presenter ended up with. | [
"4 11\n",
"17 107\n",
"3 8\n"
] | [
"0\n",
"2\n",
"1\n"
] | In the first sample the presenter gives one chip to the walrus number 1, two chips to the walrus number 2, three chips to the walrus number 3, four chips to the walrus number 4, then again one chip to the walrus number 1. After that the presenter runs out of chips. He can't give anything to the walrus number 2 and the process finishes.
In the third sample the presenter gives one chip to the walrus number 1, two chips to the walrus number 2, three chips to the walrus number 3, then again one chip to the walrus number 1. The presenter has one chip left and he can't give two chips to the walrus number 2, that's why the presenter takes the last chip. | 500 | [
{
"input": "4 11",
"output": "0"
},
{
"input": "17 107",
"output": "2"
},
{
"input": "3 8",
"output": "1"
},
{
"input": "46 7262",
"output": "35"
},
{
"input": "32 6864",
"output": "0"
},
{
"input": "36 6218",
"output": "14"
},
{
"input": "25 9712",
"output": "11"
},
{
"input": "9 7601",
"output": "5"
},
{
"input": "1 9058",
"output": "0"
},
{
"input": "29 7772",
"output": "26"
},
{
"input": "45 9465",
"output": "14"
},
{
"input": "46 866",
"output": "5"
},
{
"input": "29 1241",
"output": "20"
},
{
"input": "17 4248",
"output": "12"
},
{
"input": "20 8082",
"output": "11"
},
{
"input": "50 9555",
"output": "0"
},
{
"input": "4 7455",
"output": "2"
},
{
"input": "36 880",
"output": "4"
},
{
"input": "24 7440",
"output": "9"
},
{
"input": "44 7888",
"output": "12"
},
{
"input": "1 1",
"output": "0"
},
{
"input": "50 10000",
"output": "40"
},
{
"input": "1 10000",
"output": "0"
},
{
"input": "50 1",
"output": "0"
},
{
"input": "50 50",
"output": "5"
}
] | 1,439,355,268 | 2,147,483,647 | Python 3 | OK | TESTS | 25 | 124 | 0 | s = list(map(int, input().split()))
ind = 0
while(s[1] > 0):
if(s[1] < ind+1):
break
s[1] -= ind+1
ind = (ind+1)%s[0]
print(s[1]) | Title: Chips
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
There are *n* walruses sitting in a circle. All of them are numbered in the clockwise order: the walrus number 2 sits to the left of the walrus number 1, the walrus number 3 sits to the left of the walrus number 2, ..., the walrus number 1 sits to the left of the walrus number *n*.
The presenter has *m* chips. The presenter stands in the middle of the circle and starts giving the chips to the walruses starting from walrus number 1 and moving clockwise. The walrus number *i* gets *i* chips. If the presenter can't give the current walrus the required number of chips, then the presenter takes the remaining chips and the process ends. Determine by the given *n* and *m* how many chips the presenter will get in the end.
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=50, 1<=≤<=*m*<=≤<=104) — the number of walruses and the number of chips correspondingly.
Output Specification:
Print the number of chips the presenter ended up with.
Demo Input:
['4 11\n', '17 107\n', '3 8\n']
Demo Output:
['0\n', '2\n', '1\n']
Note:
In the first sample the presenter gives one chip to the walrus number 1, two chips to the walrus number 2, three chips to the walrus number 3, four chips to the walrus number 4, then again one chip to the walrus number 1. After that the presenter runs out of chips. He can't give anything to the walrus number 2 and the process finishes.
In the third sample the presenter gives one chip to the walrus number 1, two chips to the walrus number 2, three chips to the walrus number 3, then again one chip to the walrus number 1. The presenter has one chip left and he can't give two chips to the walrus number 2, that's why the presenter takes the last chip. | ```python
s = list(map(int, input().split()))
ind = 0
while(s[1] > 0):
if(s[1] < ind+1):
break
s[1] -= ind+1
ind = (ind+1)%s[0]
print(s[1])
``` | 3.969 |
567 | A | Lineland Mail | PROGRAMMING | 900 | [
"greedy",
"implementation"
] | null | null | All cities of Lineland are located on the *Ox* coordinate axis. Thus, each city is associated with its position *x**i* — a coordinate on the *Ox* axis. No two cities are located at a single point.
Lineland residents love to send letters to each other. A person may send a letter only if the recipient lives in another city (because if they live in the same city, then it is easier to drop in).
Strange but true, the cost of sending the letter is exactly equal to the distance between the sender's city and the recipient's city.
For each city calculate two values *min**i* and *max**i*, where *min**i* is the minimum cost of sending a letter from the *i*-th city to some other city, and *max**i* is the the maximum cost of sending a letter from the *i*-th city to some other city | The first line of the input contains integer *n* (2<=≤<=*n*<=≤<=105) — the number of cities in Lineland. The second line contains the sequence of *n* distinct integers *x*1,<=*x*2,<=...,<=*x**n* (<=-<=109<=≤<=*x**i*<=≤<=109), where *x**i* is the *x*-coordinate of the *i*-th city. All the *x**i*'s are distinct and follow in ascending order. | Print *n* lines, the *i*-th line must contain two integers *min**i*,<=*max**i*, separated by a space, where *min**i* is the minimum cost of sending a letter from the *i*-th city, and *max**i* is the maximum cost of sending a letter from the *i*-th city. | [
"4\n-5 -2 2 7\n",
"2\n-1 1\n"
] | [
"3 12\n3 9\n4 7\n5 12\n",
"2 2\n2 2\n"
] | none | 500 | [
{
"input": "4\n-5 -2 2 7",
"output": "3 12\n3 9\n4 7\n5 12"
},
{
"input": "2\n-1 1",
"output": "2 2\n2 2"
},
{
"input": "3\n-1 0 1",
"output": "1 2\n1 1\n1 2"
},
{
"input": "4\n-1 0 1 3",
"output": "1 4\n1 3\n1 2\n2 4"
},
{
"input": "3\n-1000000000 0 1000000000",
"output": "1000000000 2000000000\n1000000000 1000000000\n1000000000 2000000000"
},
{
"input": "2\n-1000000000 1000000000",
"output": "2000000000 2000000000\n2000000000 2000000000"
},
{
"input": "10\n1 10 12 15 59 68 130 912 1239 9123",
"output": "9 9122\n2 9113\n2 9111\n3 9108\n9 9064\n9 9055\n62 8993\n327 8211\n327 7884\n7884 9122"
},
{
"input": "5\n-2 -1 0 1 2",
"output": "1 4\n1 3\n1 2\n1 3\n1 4"
},
{
"input": "5\n-2 -1 0 1 3",
"output": "1 5\n1 4\n1 3\n1 3\n2 5"
},
{
"input": "3\n-10000 1 10000",
"output": "10001 20000\n9999 10001\n9999 20000"
},
{
"input": "5\n-1000000000 -999999999 -999999998 -999999997 -999999996",
"output": "1 4\n1 3\n1 2\n1 3\n1 4"
},
{
"input": "10\n-857422304 -529223472 82412729 145077145 188538640 265299215 527377039 588634631 592896147 702473706",
"output": "328198832 1559896010\n328198832 1231697178\n62664416 939835033\n43461495 1002499449\n43461495 1045960944\n76760575 1122721519\n61257592 1384799343\n4261516 1446056935\n4261516 1450318451\n109577559 1559896010"
},
{
"input": "10\n-876779400 -829849659 -781819137 -570920213 18428128 25280705 121178189 219147240 528386329 923854124",
"output": "46929741 1800633524\n46929741 1753703783\n48030522 1705673261\n210898924 1494774337\n6852577 905425996\n6852577 902060105\n95897484 997957589\n97969051 1095926640\n309239089 1405165729\n395467795 1800633524"
},
{
"input": "30\n-15 1 21 25 30 40 59 60 77 81 97 100 103 123 139 141 157 158 173 183 200 215 226 231 244 256 267 279 289 292",
"output": "16 307\n16 291\n4 271\n4 267\n5 262\n10 252\n1 233\n1 232\n4 215\n4 211\n3 195\n3 192\n3 189\n16 169\n2 154\n2 156\n1 172\n1 173\n10 188\n10 198\n15 215\n11 230\n5 241\n5 246\n12 259\n11 271\n11 282\n10 294\n3 304\n3 307"
},
{
"input": "10\n-1000000000 -999999999 -999999997 -999999996 -999999995 -999999994 -999999992 -999999990 -999999988 -999999986",
"output": "1 14\n1 13\n1 11\n1 10\n1 9\n1 8\n2 8\n2 10\n2 12\n2 14"
},
{
"input": "50\n-50000 -49459 -48875 -48456 -48411 -48096 -47901 -47500 -47150 -46808 -46687 -46679 -46337 -45747 -45604 -45194 -44752 -44242 -44231 -44122 -43636 -43274 -42916 -42881 -42386 -42095 -41830 -41618 -41145 -40897 -40534 -40007 -39761 -39389 -39104 -38909 -38630 -38561 -38364 -38255 -38214 -38084 -37959 -37607 -37202 -36890 -36681 -36136 -36123 -35886",
"output": "541 14114\n541 13573\n419 12989\n45 12570\n45 12525\n195 12210\n195 12015\n350 11614\n342 11264\n121 10922\n8 10801\n8 10793\n342 10451\n143 9861\n143 9718\n410 9308\n442 8866\n11 8356\n11 8345\n109 8236\n362 7750\n358 7388\n35 7084\n35 7119\n291 7614\n265 7905\n212 8170\n212 8382\n248 8855\n248 9103\n363 9466\n246 9993\n246 10239\n285 10611\n195 10896\n195 11091\n69 11370\n69 11439\n109 11636\n41 11745\n41 11786\n125 11916\n125 12041\n352 12393\n312 12798\n209 13110\n209 13319\n13 13864\n13 13877\n237 141..."
},
{
"input": "3\n-1000000000 999999999 1000000000",
"output": "1999999999 2000000000\n1 1999999999\n1 2000000000"
}
] | 1,662,773,872 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 28 | 218 | 22,118,400 | t=int(input())
c=list(map(int,input().split()))
l=sorted(c)
m=[]
for j in range(t):
if j==0:
min1 = abs(l[1]-l[0])
max1=abs(l[t-1]-l[0])
m.append(min1)
m.append(max1)
if j!=t-1 and j!=0:
if abs(l[j-1]-l[j])<abs(l[j+1]-l[j]):
min1=abs(l[j-1]-l[j])
if abs(l[j-1]-l[j])>abs(l[j+1]-l[j]):
min1=abs(l[j+1]-l[j])
if abs(l[0]-l[j])>abs(l[t-1]-l[j]):
max1=abs(l[0]-l[j])
else:
max1=abs(l[t-1]-l[j])
m.append(min1)
m.append(max1)
if j==t-1:
min1=abs(l[t-1]-l[t-2])
max1=abs(l[t-1]-l[0])
m.append(min1)
m.append(max1)
k=0
while k<len(m)-1:
print(m[k],m[k+1],end=" ")
print()
k+=2
| Title: Lineland Mail
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
All cities of Lineland are located on the *Ox* coordinate axis. Thus, each city is associated with its position *x**i* — a coordinate on the *Ox* axis. No two cities are located at a single point.
Lineland residents love to send letters to each other. A person may send a letter only if the recipient lives in another city (because if they live in the same city, then it is easier to drop in).
Strange but true, the cost of sending the letter is exactly equal to the distance between the sender's city and the recipient's city.
For each city calculate two values *min**i* and *max**i*, where *min**i* is the minimum cost of sending a letter from the *i*-th city to some other city, and *max**i* is the the maximum cost of sending a letter from the *i*-th city to some other city
Input Specification:
The first line of the input contains integer *n* (2<=≤<=*n*<=≤<=105) — the number of cities in Lineland. The second line contains the sequence of *n* distinct integers *x*1,<=*x*2,<=...,<=*x**n* (<=-<=109<=≤<=*x**i*<=≤<=109), where *x**i* is the *x*-coordinate of the *i*-th city. All the *x**i*'s are distinct and follow in ascending order.
Output Specification:
Print *n* lines, the *i*-th line must contain two integers *min**i*,<=*max**i*, separated by a space, where *min**i* is the minimum cost of sending a letter from the *i*-th city, and *max**i* is the maximum cost of sending a letter from the *i*-th city.
Demo Input:
['4\n-5 -2 2 7\n', '2\n-1 1\n']
Demo Output:
['3 12\n3 9\n4 7\n5 12\n', '2 2\n2 2\n']
Note:
none | ```python
t=int(input())
c=list(map(int,input().split()))
l=sorted(c)
m=[]
for j in range(t):
if j==0:
min1 = abs(l[1]-l[0])
max1=abs(l[t-1]-l[0])
m.append(min1)
m.append(max1)
if j!=t-1 and j!=0:
if abs(l[j-1]-l[j])<abs(l[j+1]-l[j]):
min1=abs(l[j-1]-l[j])
if abs(l[j-1]-l[j])>abs(l[j+1]-l[j]):
min1=abs(l[j+1]-l[j])
if abs(l[0]-l[j])>abs(l[t-1]-l[j]):
max1=abs(l[0]-l[j])
else:
max1=abs(l[t-1]-l[j])
m.append(min1)
m.append(max1)
if j==t-1:
min1=abs(l[t-1]-l[t-2])
max1=abs(l[t-1]-l[0])
m.append(min1)
m.append(max1)
k=0
while k<len(m)-1:
print(m[k],m[k+1],end=" ")
print()
k+=2
``` | 0 |
|
137 | B | Permutation | PROGRAMMING | 1,000 | [
"greedy"
] | null | null | "Hey, it's homework time" — thought Polycarpus and of course he started with his favourite subject, IT. Polycarpus managed to solve all tasks but for the last one in 20 minutes. However, as he failed to solve the last task after some considerable time, the boy asked you to help him.
The sequence of *n* integers is called a permutation if it contains all integers from 1 to *n* exactly once.
You are given an arbitrary sequence *a*1,<=*a*2,<=...,<=*a**n* containing *n* integers. Each integer is not less than 1 and not greater than 5000. Determine what minimum number of elements Polycarpus needs to change to get a permutation (he should not delete or add numbers). In a single change he can modify any single sequence element (i. e. replace it with another integer). | The first line of the input data contains an integer *n* (1<=≤<=*n*<=≤<=5000) which represents how many numbers are in the sequence. The second line contains a sequence of integers *a**i* (1<=≤<=*a**i*<=≤<=5000,<=1<=≤<=*i*<=≤<=*n*). | Print the only number — the minimum number of changes needed to get the permutation. | [
"3\n3 1 2\n",
"2\n2 2\n",
"5\n5 3 3 3 1\n"
] | [
"0\n",
"1\n",
"2\n"
] | The first sample contains the permutation, which is why no replacements are required.
In the second sample it is enough to replace the first element with the number 1 and that will make the sequence the needed permutation.
In the third sample we can replace the second element with number 4 and the fourth element with number 2. | 1,000 | [
{
"input": "3\n3 1 2",
"output": "0"
},
{
"input": "2\n2 2",
"output": "1"
},
{
"input": "5\n5 3 3 3 1",
"output": "2"
},
{
"input": "5\n6 6 6 6 6",
"output": "5"
},
{
"input": "10\n1 1 2 2 8 8 7 7 9 9",
"output": "5"
},
{
"input": "8\n9 8 7 6 5 4 3 2",
"output": "1"
},
{
"input": "15\n1 2 3 4 5 5 4 3 2 1 1 2 3 4 5",
"output": "10"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "1\n5000",
"output": "1"
},
{
"input": "4\n5000 5000 5000 5000",
"output": "4"
},
{
"input": "5\n3366 3461 4 5 4370",
"output": "3"
},
{
"input": "10\n8 2 10 3 4 6 1 7 9 5",
"output": "0"
},
{
"input": "10\n551 3192 3213 2846 3068 1224 3447 1 10 9",
"output": "7"
},
{
"input": "15\n4 1459 12 4281 3241 2748 10 3590 14 845 3518 1721 2 2880 1974",
"output": "10"
},
{
"input": "15\n15 1 8 2 13 11 12 7 3 14 6 10 9 4 5",
"output": "0"
},
{
"input": "15\n2436 2354 4259 1210 2037 2665 700 3578 2880 973 1317 1024 24 3621 4142",
"output": "15"
},
{
"input": "30\n28 1 3449 9 3242 4735 26 3472 15 21 2698 7 4073 3190 10 3 29 1301 4526 22 345 3876 19 12 4562 2535 2 630 18 27",
"output": "14"
},
{
"input": "100\n50 39 95 30 66 78 2169 4326 81 31 74 34 80 40 19 48 97 63 82 6 88 16 21 57 92 77 10 1213 17 93 32 91 38 4375 29 75 44 22 4 45 14 2395 3254 59 3379 2 85 96 8 83 27 94 1512 2960 100 9 73 79 7 25 55 69 90 99 51 87 98 62 18 35 43 4376 4668 28 72 56 4070 61 65 36 54 4106 11 24 15 86 70 71 4087 23 13 76 20 4694 26 4962 4726 37 14 64",
"output": "18"
},
{
"input": "100\n340 14 3275 2283 2673 1107 817 2243 1226 32 2382 3638 4652 418 68 4962 387 764 4647 159 1846 225 2760 4904 3150 403 3 2439 91 4428 92 4705 75 348 1566 1465 69 6 49 4 62 4643 564 1090 3447 1871 2255 139 24 99 2669 969 86 61 4550 158 4537 3993 1589 872 2907 1888 401 80 1825 1483 63 1 2264 4068 4113 2548 41 885 4806 36 67 167 4447 34 1248 2593 82 202 81 1783 1284 4973 16 43 95 7 865 2091 3008 1793 20 947 4912 3604",
"output": "70"
},
{
"input": "1\n2",
"output": "1"
},
{
"input": "2\n5000 5000",
"output": "2"
},
{
"input": "2\n1 2",
"output": "0"
},
{
"input": "2\n1 1",
"output": "1"
},
{
"input": "2\n2 3",
"output": "1"
},
{
"input": "2\n3 4",
"output": "2"
},
{
"input": "10\n1 2 3 4 5 6 7 1000 10 10",
"output": "2"
}
] | 1,566,211,385 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 3 | 218 | 0 | n=int(input())
arr=list(map(int,(input().split())))
l=len(set(arr))
print(n-l) | Title: Permutation
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
"Hey, it's homework time" — thought Polycarpus and of course he started with his favourite subject, IT. Polycarpus managed to solve all tasks but for the last one in 20 minutes. However, as he failed to solve the last task after some considerable time, the boy asked you to help him.
The sequence of *n* integers is called a permutation if it contains all integers from 1 to *n* exactly once.
You are given an arbitrary sequence *a*1,<=*a*2,<=...,<=*a**n* containing *n* integers. Each integer is not less than 1 and not greater than 5000. Determine what minimum number of elements Polycarpus needs to change to get a permutation (he should not delete or add numbers). In a single change he can modify any single sequence element (i. e. replace it with another integer).
Input Specification:
The first line of the input data contains an integer *n* (1<=≤<=*n*<=≤<=5000) which represents how many numbers are in the sequence. The second line contains a sequence of integers *a**i* (1<=≤<=*a**i*<=≤<=5000,<=1<=≤<=*i*<=≤<=*n*).
Output Specification:
Print the only number — the minimum number of changes needed to get the permutation.
Demo Input:
['3\n3 1 2\n', '2\n2 2\n', '5\n5 3 3 3 1\n']
Demo Output:
['0\n', '1\n', '2\n']
Note:
The first sample contains the permutation, which is why no replacements are required.
In the second sample it is enough to replace the first element with the number 1 and that will make the sequence the needed permutation.
In the third sample we can replace the second element with number 4 and the fourth element with number 2. | ```python
n=int(input())
arr=list(map(int,(input().split())))
l=len(set(arr))
print(n-l)
``` | 0 |
|
209 | A | Multicolored Marbles | PROGRAMMING | 1,600 | [
"dp",
"math"
] | null | null | Polycarpus plays with red and blue marbles. He put *n* marbles from the left to the right in a row. As it turned out, the marbles form a zebroid.
A non-empty sequence of red and blue marbles is a zebroid, if the colors of the marbles in this sequence alternate. For example, sequences (red; blue; red) and (blue) are zebroids and sequence (red; red) is not a zebroid.
Now Polycarpus wonders, how many ways there are to pick a zebroid subsequence from this sequence. Help him solve the problem, find the number of ways modulo 1000000007 (109<=+<=7). | The first line contains a single integer *n* (1<=≤<=*n*<=≤<=106) — the number of marbles in Polycarpus's sequence. | Print a single number — the answer to the problem modulo 1000000007 (109<=+<=7). | [
"3\n",
"4\n"
] | [
"6\n",
"11\n"
] | Let's consider the first test sample. Let's assume that Polycarpus initially had sequence (red; blue; red), so there are six ways to pick a zebroid:
- pick the first marble; - pick the second marble; - pick the third marble; - pick the first and second marbles; - pick the second and third marbles; - pick the first, second and third marbles.
It can be proven that if Polycarpus picks (blue; red; blue) as the initial sequence, the number of ways won't change. | 500 | [
{
"input": "3",
"output": "6"
},
{
"input": "4",
"output": "11"
},
{
"input": "1",
"output": "1"
},
{
"input": "2",
"output": "3"
},
{
"input": "5",
"output": "19"
},
{
"input": "6",
"output": "32"
},
{
"input": "7",
"output": "53"
},
{
"input": "8",
"output": "87"
},
{
"input": "9",
"output": "142"
},
{
"input": "10",
"output": "231"
},
{
"input": "11",
"output": "375"
},
{
"input": "12",
"output": "608"
},
{
"input": "13",
"output": "985"
},
{
"input": "14",
"output": "1595"
},
{
"input": "15",
"output": "2582"
},
{
"input": "16",
"output": "4179"
},
{
"input": "17",
"output": "6763"
},
{
"input": "18",
"output": "10944"
},
{
"input": "19",
"output": "17709"
},
{
"input": "20",
"output": "28655"
},
{
"input": "21",
"output": "46366"
},
{
"input": "22",
"output": "75023"
},
{
"input": "30",
"output": "3524576"
},
{
"input": "40",
"output": "433494435"
},
{
"input": "35",
"output": "39088167"
},
{
"input": "747",
"output": "864986803"
},
{
"input": "811",
"output": "121575679"
},
{
"input": "523",
"output": "810594171"
},
{
"input": "939",
"output": "834252191"
},
{
"input": "7218",
"output": "124017976"
},
{
"input": "6578",
"output": "105202924"
},
{
"input": "4994",
"output": "909729063"
},
{
"input": "3410",
"output": "995527079"
},
{
"input": "9122",
"output": "424682518"
},
{
"input": "11500",
"output": "123418864"
},
{
"input": "37212",
"output": "991850249"
},
{
"input": "95628",
"output": "265542306"
},
{
"input": "21341",
"output": "913329411"
},
{
"input": "63405",
"output": "206564501"
},
{
"input": "670293",
"output": "922731630"
},
{
"input": "12357",
"output": "808648710"
},
{
"input": "838069",
"output": "536545985"
},
{
"input": "696485",
"output": "467961960"
},
{
"input": "1000000",
"output": "986892583"
},
{
"input": "999999",
"output": "452491920"
},
{
"input": "999998",
"output": "534400661"
},
{
"input": "999997",
"output": "918091264"
},
{
"input": "999996",
"output": "616309402"
},
{
"input": "999000",
"output": "47031574"
},
{
"input": "100",
"output": "252403354"
}
] | 1,687,893,962 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 50 | 904 | 100,044,800 | import sys
input = lambda: sys.stdin.readline().rstrip()
MOD = 10**9+7
N = int(input())
dp = [[0,0] for _ in range(N+1)]
for i in range(N):
dp[i+1][0]=dp[i][0]
dp[i+1][1]=dp[i][1]
if i%2:
dp[i+1][1]+=dp[i][0]+1
else:
dp[i+1][0]+=dp[i][1]+1
dp[i+1][0]%=MOD
dp[i+1][1]%=MOD
print(sum(dp[-1])%MOD) | Title: Multicolored Marbles
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarpus plays with red and blue marbles. He put *n* marbles from the left to the right in a row. As it turned out, the marbles form a zebroid.
A non-empty sequence of red and blue marbles is a zebroid, if the colors of the marbles in this sequence alternate. For example, sequences (red; blue; red) and (blue) are zebroids and sequence (red; red) is not a zebroid.
Now Polycarpus wonders, how many ways there are to pick a zebroid subsequence from this sequence. Help him solve the problem, find the number of ways modulo 1000000007 (109<=+<=7).
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=106) — the number of marbles in Polycarpus's sequence.
Output Specification:
Print a single number — the answer to the problem modulo 1000000007 (109<=+<=7).
Demo Input:
['3\n', '4\n']
Demo Output:
['6\n', '11\n']
Note:
Let's consider the first test sample. Let's assume that Polycarpus initially had sequence (red; blue; red), so there are six ways to pick a zebroid:
- pick the first marble; - pick the second marble; - pick the third marble; - pick the first and second marbles; - pick the second and third marbles; - pick the first, second and third marbles.
It can be proven that if Polycarpus picks (blue; red; blue) as the initial sequence, the number of ways won't change. | ```python
import sys
input = lambda: sys.stdin.readline().rstrip()
MOD = 10**9+7
N = int(input())
dp = [[0,0] for _ in range(N+1)]
for i in range(N):
dp[i+1][0]=dp[i][0]
dp[i+1][1]=dp[i][1]
if i%2:
dp[i+1][1]+=dp[i][0]+1
else:
dp[i+1][0]+=dp[i][1]+1
dp[i+1][0]%=MOD
dp[i+1][1]%=MOD
print(sum(dp[-1])%MOD)
``` | 3 |
|
224 | A | Parallelepiped | PROGRAMMING | 1,100 | [
"brute force",
"geometry",
"math"
] | null | null | You've got a rectangular parallelepiped with integer edge lengths. You know the areas of its three faces that have a common vertex. Your task is to find the sum of lengths of all 12 edges of this parallelepiped. | The first and the single line contains three space-separated integers — the areas of the parallelepiped's faces. The area's values are positive (<=><=0) and do not exceed 104. It is guaranteed that there exists at least one parallelepiped that satisfies the problem statement. | Print a single number — the sum of all edges of the parallelepiped. | [
"1 1 1\n",
"4 6 6\n"
] | [
"12\n",
"28\n"
] | In the first sample the parallelepiped has sizes 1 × 1 × 1, in the second one — 2 × 2 × 3. | 500 | [
{
"input": "1 1 1",
"output": "12"
},
{
"input": "4 6 6",
"output": "28"
},
{
"input": "20 10 50",
"output": "68"
},
{
"input": "9 4 36",
"output": "56"
},
{
"input": "324 9 36",
"output": "184"
},
{
"input": "1333 93 129",
"output": "308"
},
{
"input": "1022 584 112",
"output": "380"
},
{
"input": "66 174 319",
"output": "184"
},
{
"input": "912 276 1748",
"output": "444"
},
{
"input": "65 156 60",
"output": "120"
},
{
"input": "1 10000 10000",
"output": "40008"
},
{
"input": "1485 55 27",
"output": "332"
},
{
"input": "152 108 4104",
"output": "528"
},
{
"input": "1656 6900 1350",
"output": "740"
},
{
"input": "12 14 42",
"output": "60"
},
{
"input": "615 18 1230",
"output": "856"
},
{
"input": "680 60 408",
"output": "336"
},
{
"input": "644 966 6",
"output": "1308"
},
{
"input": "1 432 432",
"output": "1736"
},
{
"input": "2239 2239 1",
"output": "8964"
},
{
"input": "4106 8212 2",
"output": "16436"
},
{
"input": "10000 10000 10000",
"output": "1200"
},
{
"input": "3623 3623 1",
"output": "14500"
},
{
"input": "9801 9801 9801",
"output": "1188"
},
{
"input": "10000 1 10000",
"output": "40008"
},
{
"input": "9 9 9",
"output": "36"
},
{
"input": "9801 9702 9702",
"output": "1184"
}
] | 1,605,953,178 | 2,147,483,647 | PyPy 3 | OK | TESTS | 27 | 468 | 2,355,200 | import sys,os,io,time,copy
if os.path.exists('input.txt'):
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
import math
def is_int(n):
if n//1==n/1:
return True
else:
return False
def main():
# start=time.time()
a,b,c=map(int,input().split())
for i in range(1,10001):
x=i
y=a/i
z=c/i
if is_int(x) and is_int(y) and is_int(z) and y*z==b:
print(4*int(x+y+z))
break
# end=time.time()
main() | Title: Parallelepiped
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You've got a rectangular parallelepiped with integer edge lengths. You know the areas of its three faces that have a common vertex. Your task is to find the sum of lengths of all 12 edges of this parallelepiped.
Input Specification:
The first and the single line contains three space-separated integers — the areas of the parallelepiped's faces. The area's values are positive (<=><=0) and do not exceed 104. It is guaranteed that there exists at least one parallelepiped that satisfies the problem statement.
Output Specification:
Print a single number — the sum of all edges of the parallelepiped.
Demo Input:
['1 1 1\n', '4 6 6\n']
Demo Output:
['12\n', '28\n']
Note:
In the first sample the parallelepiped has sizes 1 × 1 × 1, in the second one — 2 × 2 × 3. | ```python
import sys,os,io,time,copy
if os.path.exists('input.txt'):
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
import math
def is_int(n):
if n//1==n/1:
return True
else:
return False
def main():
# start=time.time()
a,b,c=map(int,input().split())
for i in range(1,10001):
x=i
y=a/i
z=c/i
if is_int(x) and is_int(y) and is_int(z) and y*z==b:
print(4*int(x+y+z))
break
# end=time.time()
main()
``` | 3 |
|
772 | B | Volatile Kite | PROGRAMMING | 1,800 | [
"geometry"
] | null | null | You are given a convex polygon *P* with *n* distinct vertices *p*1,<=*p*2,<=...,<=*p**n*. Vertex *p**i* has coordinates (*x**i*,<=*y**i*) in the 2D plane. These vertices are listed in clockwise order.
You can choose a real number *D* and move each vertex of the polygon a distance of at most *D* from their original positions.
Find the maximum value of *D* such that no matter how you move the vertices, the polygon does not intersect itself and stays convex. | The first line has one integer *n* (4<=≤<=*n*<=≤<=1<=000) — the number of vertices.
The next *n* lines contain the coordinates of the vertices. Line *i* contains two integers *x**i* and *y**i* (<=-<=109<=≤<=*x**i*,<=*y**i*<=≤<=109) — the coordinates of the *i*-th vertex. These points are guaranteed to be given in clockwise order, and will form a strictly convex polygon (in particular, no three consecutive points lie on the same straight line). | Print one real number *D*, which is the maximum real number such that no matter how you move the vertices, the polygon stays convex.
Your answer will be considered correct if its absolute or relative error does not exceed 10<=-<=6.
Namely, let's assume that your answer is *a* and the answer of the jury is *b*. The checker program will consider your answer correct if . | [
"4\n0 0\n0 1\n1 1\n1 0\n",
"6\n5 0\n10 0\n12 -4\n10 -8\n5 -8\n3 -4\n"
] | [
"0.3535533906\n",
"1.0000000000\n"
] | Here is a picture of the first sample
<img class="tex-graphics" src="https://espresso.codeforces.com/f83aa076d2f437f9bb785cae769c3ae310eff351.png" style="max-width: 100.0%;max-height: 100.0%;"/>
Here is an example of making the polygon non-convex.
<img class="tex-graphics" src="https://espresso.codeforces.com/fbadb81630251ca642bd4ddf9088876ade761630.png" style="max-width: 100.0%;max-height: 100.0%;"/>
This is not an optimal solution, since the maximum distance we moved one point is ≈ 0.4242640687, whereas we can make it non-convex by only moving each point a distance of at most ≈ 0.3535533906. | 1,000 | [
{
"input": "4\n0 0\n0 1\n1 1\n1 0",
"output": "0.3535533906"
},
{
"input": "6\n5 0\n10 0\n12 -4\n10 -8\n5 -8\n3 -4",
"output": "1.0000000000"
},
{
"input": "19\n449447997 711296339\n530233434 692216537\n535464528 613140435\n535533467 100893188\n530498867 -265063956\n519107979 -271820709\n482156929 -287792333\n-303730271 -287970295\n-416935204 -263348201\n-443613873 -249980523\n-453444829 -173903413\n-462102798 -80789280\n-462064673 -13220755\n-461368561 482595837\n-457749751 687048095\n-448625206 709399396\n-145117181 710688825\n159099640 711650577\n400454061 711503381",
"output": "24967.1394973334"
},
{
"input": "4\n0 0\n0 10\n10 10\n6 4",
"output": "0.7071067812"
},
{
"input": "4\n-1000000000 -1000000000\n-999999999 -999999999\n1000000000 999999999\n0 -1",
"output": "0.0000000000"
},
{
"input": "4\n-1000000000 -1000000000\n-1000000000 1000000000\n1000000000 1000000000\n1000000000 -1000000000",
"output": "707106781.1865475000"
},
{
"input": "4\n-100000 -100000\n-99999 -99999\n100000 99999\n0 -100",
"output": "0.0000017678"
},
{
"input": "4\n-10000 -10000\n-9999 -9999\n10000 9999\n0 -1000",
"output": "0.0000176781"
},
{
"input": "5\n0 0\n0 10\n10 10\n20 0\n10 -1",
"output": "0.5000000000"
},
{
"input": "5\n10 -1\n0 0\n0 10\n10 10\n20 0",
"output": "0.5000000000"
},
{
"input": "4\n1000000000 1000000000\n1000000000 -1000000000\n-1000000000 -1000000000\n-1000000000 1000000000",
"output": "707106781.1865475000"
},
{
"input": "4\n2 0\n0 0\n0 14\n8 14",
"output": "0.8682431421"
},
{
"input": "4\n0 0\n1 100\n100 0\n1 -100",
"output": "0.5000000000"
},
{
"input": "4\n-1000000000 1000000000\n1000000000 500000000\n1000000000 -1000000000\n-500000000 -1000000000",
"output": "530330085.8899106400"
}
] | 1,492,381,510 | 2,147,483,647 | Python 3 | OK | TESTS | 36 | 62 | 5,632,000 | from math import inf
def vect(x, y):
return abs(sum([x[i]*(y[(i+1)%3]-y[(i+2)%3]) for i in range(3)]))
def l(x, y):
return ((x[0]-x[2])**2 + (y[0]-y[2])**2)**0.5
def h(x, y):
return vect(x, y) / l(x, y)
n = int(input())
x = []
y = []
for i in range(n):
a, b = [int(x) for x in input().split()]
x.append(a)
y.append(b)
x += x[:2]
y += y[:2]
dmin = inf
for i in range(n):
d = h(x[i:i+3], y[i:i+3])/2
if dmin > d:
dmin = d
print(dmin) | Title: Volatile Kite
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a convex polygon *P* with *n* distinct vertices *p*1,<=*p*2,<=...,<=*p**n*. Vertex *p**i* has coordinates (*x**i*,<=*y**i*) in the 2D plane. These vertices are listed in clockwise order.
You can choose a real number *D* and move each vertex of the polygon a distance of at most *D* from their original positions.
Find the maximum value of *D* such that no matter how you move the vertices, the polygon does not intersect itself and stays convex.
Input Specification:
The first line has one integer *n* (4<=≤<=*n*<=≤<=1<=000) — the number of vertices.
The next *n* lines contain the coordinates of the vertices. Line *i* contains two integers *x**i* and *y**i* (<=-<=109<=≤<=*x**i*,<=*y**i*<=≤<=109) — the coordinates of the *i*-th vertex. These points are guaranteed to be given in clockwise order, and will form a strictly convex polygon (in particular, no three consecutive points lie on the same straight line).
Output Specification:
Print one real number *D*, which is the maximum real number such that no matter how you move the vertices, the polygon stays convex.
Your answer will be considered correct if its absolute or relative error does not exceed 10<=-<=6.
Namely, let's assume that your answer is *a* and the answer of the jury is *b*. The checker program will consider your answer correct if .
Demo Input:
['4\n0 0\n0 1\n1 1\n1 0\n', '6\n5 0\n10 0\n12 -4\n10 -8\n5 -8\n3 -4\n']
Demo Output:
['0.3535533906\n', '1.0000000000\n']
Note:
Here is a picture of the first sample
<img class="tex-graphics" src="https://espresso.codeforces.com/f83aa076d2f437f9bb785cae769c3ae310eff351.png" style="max-width: 100.0%;max-height: 100.0%;"/>
Here is an example of making the polygon non-convex.
<img class="tex-graphics" src="https://espresso.codeforces.com/fbadb81630251ca642bd4ddf9088876ade761630.png" style="max-width: 100.0%;max-height: 100.0%;"/>
This is not an optimal solution, since the maximum distance we moved one point is ≈ 0.4242640687, whereas we can make it non-convex by only moving each point a distance of at most ≈ 0.3535533906. | ```python
from math import inf
def vect(x, y):
return abs(sum([x[i]*(y[(i+1)%3]-y[(i+2)%3]) for i in range(3)]))
def l(x, y):
return ((x[0]-x[2])**2 + (y[0]-y[2])**2)**0.5
def h(x, y):
return vect(x, y) / l(x, y)
n = int(input())
x = []
y = []
for i in range(n):
a, b = [int(x) for x in input().split()]
x.append(a)
y.append(b)
x += x[:2]
y += y[:2]
dmin = inf
for i in range(n):
d = h(x[i:i+3], y[i:i+3])/2
if dmin > d:
dmin = d
print(dmin)
``` | 3 |
|
0 | none | none | none | 0 | [
"none"
] | null | null | Today Pari and Arya are playing a game called Remainders.
Pari chooses two positive integer *x* and *k*, and tells Arya *k* but not *x*. Arya have to find the value . There are *n* ancient numbers *c*1,<=*c*2,<=...,<=*c**n* and Pari has to tell Arya if Arya wants. Given *k* and the ancient values, tell us if Arya has a winning strategy independent of value of *x* or not. Formally, is it true that Arya can understand the value for any positive integer *x*?
Note, that means the remainder of *x* after dividing it by *y*. | The first line of the input contains two integers *n* and *k* (1<=≤<=*n*,<= *k*<=≤<=1<=000<=000) — the number of ancient integers and value *k* that is chosen by Pari.
The second line contains *n* integers *c*1,<=*c*2,<=...,<=*c**n* (1<=≤<=*c**i*<=≤<=1<=000<=000). | Print "Yes" (without quotes) if Arya has a winning strategy independent of value of *x*, or "No" (without quotes) otherwise. | [
"4 5\n2 3 5 12\n",
"2 7\n2 3\n"
] | [
"Yes\n",
"No\n"
] | In the first sample, Arya can understand <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/d170efffcde0907ee6bcf32de21051bce0677a2c.png" style="max-width: 100.0%;max-height: 100.0%;"/> because 5 is one of the ancient numbers.
In the second sample, Arya can't be sure what <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/57b5f6a96f5db073270dd3ed4266c69299ec701d.png" style="max-width: 100.0%;max-height: 100.0%;"/> is. For example 1 and 7 have the same remainders after dividing by 2 and 3, but they differ in remainders after dividing by 7. | 0 | [
{
"input": "4 5\n2 3 5 12",
"output": "Yes"
},
{
"input": "2 7\n2 3",
"output": "No"
},
{
"input": "1 6\n8",
"output": "No"
},
{
"input": "2 3\n9 4",
"output": "Yes"
},
{
"input": "4 16\n19 16 13 9",
"output": "Yes"
},
{
"input": "5 10\n5 16 19 9 17",
"output": "Yes"
},
{
"input": "11 95\n31 49 8 139 169 121 71 17 43 29 125",
"output": "No"
},
{
"input": "17 71\n173 43 139 73 169 199 49 81 11 89 131 107 23 29 125 152 17",
"output": "No"
},
{
"input": "13 86\n41 64 17 31 13 97 19 25 81 47 61 37 71",
"output": "No"
},
{
"input": "15 91\n49 121 83 67 128 125 27 113 41 169 149 19 37 29 71",
"output": "Yes"
},
{
"input": "2 4\n2 2",
"output": "No"
},
{
"input": "14 87\n1619 1619 1619 1619 1619 1619 1619 1619 1619 1619 1619 1619 1619 1619",
"output": "No"
},
{
"input": "12 100\n1766 1766 1766 1766 1766 1766 1766 1766 1766 1766 1766 1766",
"output": "No"
},
{
"input": "1 994619\n216000",
"output": "No"
},
{
"input": "1 651040\n911250",
"output": "No"
},
{
"input": "1 620622\n60060",
"output": "No"
},
{
"input": "1 1\n559872",
"output": "Yes"
},
{
"input": "88 935089\n967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967 967",
"output": "No"
},
{
"input": "93 181476\n426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426 426",
"output": "No"
},
{
"input": "91 4900\n630 630 70 630 910 630 630 630 770 70 770 630 630 770 70 630 70 630 70 630 70 630 630 70 910 630 630 630 770 630 630 630 70 910 70 630 70 630 770 630 630 70 630 770 70 630 70 70 630 630 70 70 70 70 630 70 70 770 910 630 70 630 770 70 910 70 630 910 630 70 770 70 70 630 770 630 70 630 70 70 630 70 630 770 630 70 630 630 70 910 630",
"output": "No"
},
{
"input": "61 531012\n698043 698043 698043 963349 698043 698043 698043 963349 698043 698043 698043 963349 698043 698043 698043 698043 966694 698043 698043 698043 698043 698043 698043 636247 698043 963349 698043 698043 698043 698043 697838 698043 963349 698043 698043 966694 698043 698043 698043 698043 698043 698043 698043 698043 698043 698043 698043 698043 698043 698043 698043 698043 698043 698043 963349 698043 698043 698043 698043 963349 698043",
"output": "No"
},
{
"input": "1 216000\n648000",
"output": "Yes"
},
{
"input": "2 8\n4 4",
"output": "No"
},
{
"input": "3 8\n4 4 4",
"output": "No"
},
{
"input": "2 8\n2 4",
"output": "No"
},
{
"input": "3 12\n2 2 3",
"output": "No"
},
{
"input": "10 4\n2 2 2 2 2 2 2 2 2 2",
"output": "No"
},
{
"input": "10 1024\n1 2 4 8 16 32 64 128 256 512",
"output": "No"
},
{
"input": "3 24\n2 2 3",
"output": "No"
},
{
"input": "1 8\n2",
"output": "No"
},
{
"input": "2 9\n3 3",
"output": "No"
},
{
"input": "3 4\n2 2 2",
"output": "No"
},
{
"input": "3 4\n1 2 2",
"output": "No"
},
{
"input": "1 4\n2",
"output": "No"
},
{
"input": "1 100003\n2",
"output": "No"
},
{
"input": "1 2\n12",
"output": "Yes"
},
{
"input": "2 988027\n989018 995006",
"output": "Yes"
},
{
"input": "3 9\n3 3 3",
"output": "No"
},
{
"input": "1 49\n7",
"output": "No"
},
{
"input": "2 600000\n200000 300000",
"output": "Yes"
},
{
"input": "3 8\n2 2 2",
"output": "No"
},
{
"input": "7 510510\n524288 531441 390625 823543 161051 371293 83521",
"output": "Yes"
},
{
"input": "2 30\n6 10",
"output": "Yes"
},
{
"input": "2 27000\n5400 4500",
"output": "Yes"
},
{
"input": "3 8\n1 2 4",
"output": "No"
},
{
"input": "4 16\n2 2 2 2",
"output": "No"
},
{
"input": "2 16\n4 8",
"output": "No"
},
{
"input": "2 8\n4 2",
"output": "No"
},
{
"input": "3 4\n2 2 3",
"output": "No"
},
{
"input": "1 8\n4",
"output": "No"
},
{
"input": "1 999983\n2",
"output": "No"
},
{
"input": "3 16\n2 4 8",
"output": "No"
},
{
"input": "2 216\n12 18",
"output": "No"
},
{
"input": "2 16\n8 8",
"output": "No"
},
{
"input": "2 36\n18 12",
"output": "Yes"
},
{
"input": "2 36\n12 18",
"output": "Yes"
},
{
"input": "2 1000000\n1000000 1000000",
"output": "Yes"
},
{
"input": "3 20\n2 2 5",
"output": "No"
},
{
"input": "1 2\n6",
"output": "Yes"
},
{
"input": "4 4\n2 3 6 5",
"output": "No"
},
{
"input": "1 2\n1",
"output": "No"
},
{
"input": "1 6\n6",
"output": "Yes"
},
{
"input": "2 16\n4 4",
"output": "No"
},
{
"input": "2 3779\n1 2",
"output": "No"
},
{
"input": "2 8\n4 12",
"output": "No"
},
{
"input": "2 24\n4 6",
"output": "No"
},
{
"input": "1 1\n5",
"output": "Yes"
},
{
"input": "10 255255\n1000000 700000 300000 110000 130000 170000 190000 230000 290000 310000",
"output": "Yes"
},
{
"input": "2 1000\n500 2",
"output": "No"
},
{
"input": "4 8\n2 2 2 2",
"output": "No"
},
{
"input": "1 666013\n1",
"output": "No"
},
{
"input": "1 999983\n1",
"output": "No"
},
{
"input": "1 125\n5",
"output": "No"
},
{
"input": "2 32\n4 8",
"output": "No"
},
{
"input": "3 32\n2 4 8",
"output": "No"
},
{
"input": "3 8\n2 4 11",
"output": "No"
},
{
"input": "1 32\n2",
"output": "No"
},
{
"input": "4 4\n2 2 2 2",
"output": "No"
},
{
"input": "8 32\n2 2 2 2 2 2 2 2",
"output": "No"
},
{
"input": "3 24\n2 4 3",
"output": "No"
},
{
"input": "2 4\n2 6",
"output": "No"
},
{
"input": "2 24\n6 4",
"output": "No"
},
{
"input": "1 20998\n2",
"output": "No"
},
{
"input": "1 1\n1",
"output": "Yes"
},
{
"input": "2 20998\n2 10499",
"output": "Yes"
},
{
"input": "1 5\n1",
"output": "No"
},
{
"input": "2 25\n5 5",
"output": "No"
},
{
"input": "5 2\n2 2 2 2 2",
"output": "Yes"
},
{
"input": "3 12\n2 3 6",
"output": "No"
},
{
"input": "1 1\n3",
"output": "Yes"
},
{
"input": "2 16\n8 4",
"output": "No"
},
{
"input": "40 10\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2",
"output": "No"
}
] | 1,697,404,079 | 2,147,483,647 | PyPy 3-64 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | #include <iostream>
using namespace std;
int main() {
int a = 12;
int b = 34;
int times = 1;
// Find the appropriate multiplier for 'a'
while (times <= b)
times *= 10;
// Merge 'a' and 'b'
int merged = a * times + b;
cout << "Merged number: " << merged << endl;
return 0;
}
| Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Today Pari and Arya are playing a game called Remainders.
Pari chooses two positive integer *x* and *k*, and tells Arya *k* but not *x*. Arya have to find the value . There are *n* ancient numbers *c*1,<=*c*2,<=...,<=*c**n* and Pari has to tell Arya if Arya wants. Given *k* and the ancient values, tell us if Arya has a winning strategy independent of value of *x* or not. Formally, is it true that Arya can understand the value for any positive integer *x*?
Note, that means the remainder of *x* after dividing it by *y*.
Input Specification:
The first line of the input contains two integers *n* and *k* (1<=≤<=*n*,<= *k*<=≤<=1<=000<=000) — the number of ancient integers and value *k* that is chosen by Pari.
The second line contains *n* integers *c*1,<=*c*2,<=...,<=*c**n* (1<=≤<=*c**i*<=≤<=1<=000<=000).
Output Specification:
Print "Yes" (without quotes) if Arya has a winning strategy independent of value of *x*, or "No" (without quotes) otherwise.
Demo Input:
['4 5\n2 3 5 12\n', '2 7\n2 3\n']
Demo Output:
['Yes\n', 'No\n']
Note:
In the first sample, Arya can understand <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/d170efffcde0907ee6bcf32de21051bce0677a2c.png" style="max-width: 100.0%;max-height: 100.0%;"/> because 5 is one of the ancient numbers.
In the second sample, Arya can't be sure what <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/57b5f6a96f5db073270dd3ed4266c69299ec701d.png" style="max-width: 100.0%;max-height: 100.0%;"/> is. For example 1 and 7 have the same remainders after dividing by 2 and 3, but they differ in remainders after dividing by 7. | ```python
#include <iostream>
using namespace std;
int main() {
int a = 12;
int b = 34;
int times = 1;
// Find the appropriate multiplier for 'a'
while (times <= b)
times *= 10;
// Merge 'a' and 'b'
int merged = a * times + b;
cout << "Merged number: " << merged << endl;
return 0;
}
``` | -1 |
|
369 | A | Valera and Plates | PROGRAMMING | 900 | [
"greedy",
"implementation"
] | null | null | Valera is a lazy student. He has *m* clean bowls and *k* clean plates.
Valera has made an eating plan for the next *n* days. As Valera is lazy, he will eat exactly one dish per day. At that, in order to eat a dish, he needs exactly one clean plate or bowl. We know that Valera can cook only two types of dishes. He can eat dishes of the first type from bowls and dishes of the second type from either bowls or plates.
When Valera finishes eating, he leaves a dirty plate/bowl behind. His life philosophy doesn't let him eat from dirty kitchenware. So sometimes he needs to wash his plate/bowl before eating. Find the minimum number of times Valera will need to wash a plate/bowl, if he acts optimally. | The first line of the input contains three integers *n*, *m*, *k* (1<=≤<=*n*,<=*m*,<=*k*<=≤<=1000) — the number of the planned days, the number of clean bowls and the number of clean plates.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=2). If *a**i* equals one, then on day *i* Valera will eat a first type dish. If *a**i* equals two, then on day *i* Valera will eat a second type dish. | Print a single integer — the minimum number of times Valera will need to wash a plate/bowl. | [
"3 1 1\n1 2 1\n",
"4 3 1\n1 1 1 1\n",
"3 1 2\n2 2 2\n",
"8 2 2\n1 2 1 2 1 2 1 2\n"
] | [
"1\n",
"1\n",
"0\n",
"4\n"
] | In the first sample Valera will wash a bowl only on the third day, so the answer is one.
In the second sample, Valera will have the first type of the dish during all four days, and since there are only three bowls, he will wash a bowl exactly once.
In the third sample, Valera will have the second type of dish for all three days, and as they can be eaten from either a plate or a bowl, he will never need to wash a plate/bowl. | 500 | [
{
"input": "3 1 1\n1 2 1",
"output": "1"
},
{
"input": "4 3 1\n1 1 1 1",
"output": "1"
},
{
"input": "3 1 2\n2 2 2",
"output": "0"
},
{
"input": "8 2 2\n1 2 1 2 1 2 1 2",
"output": "4"
},
{
"input": "2 100 100\n2 2",
"output": "0"
},
{
"input": "1 1 1\n2",
"output": "0"
},
{
"input": "233 100 1\n2 2 1 1 1 2 2 2 2 1 1 2 2 2 1 2 2 1 1 1 2 2 1 1 1 1 2 1 2 2 1 1 2 2 1 2 2 1 2 1 2 1 2 2 2 1 1 1 1 2 1 2 1 1 2 1 1 2 2 1 2 1 2 1 1 1 1 1 1 1 1 1 2 1 2 2 2 1 1 2 2 1 1 1 1 2 1 1 2 1 2 2 2 1 1 1 2 2 2 1 1 1 1 2 1 2 1 1 1 1 2 2 2 1 1 2 1 2 1 1 1 1 1 2 1 1 1 1 1 2 1 1 2 2 1 2 1 1 2 2 1 1 2 2 1 1 1 2 2 1 1 2 1 2 1 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 2 1 2 2 1 1 1 2 2 1 1 2 2 1 1 2 1 1 2 2 1 2 2 2 2 2 2 1 2 2 2 2 2 1 1 2 2 2 2 2 2 1 1 1 2 1 2 2 2 2 2 2 2 2 1 1 2 1 2 1 2 2",
"output": "132"
},
{
"input": "123 100 1\n2 2 2 1 1 2 2 2 2 1 1 2 2 2 1 2 2 2 2 1 2 2 2 1 1 1 2 2 2 2 1 2 2 2 2 2 2 1 2 1 2 1 2 2 2 1 2 1 2 2 1 2 2 1 2 2 1 2 2 1 2 2 2 1 1 1 1 1 1 1 1 1 2 2 2 2 2 1 1 2 2 1 1 1 1 2 1 2 2 1 2 2 2 1 1 1 2 2 2 1 2 2 2 2 1 2 2 2 2 1 2 2 2 1 1 2 1 2 1 2 1 1 1",
"output": "22"
},
{
"input": "188 100 1\n2 2 1 1 1 2 2 2 2 1 1 2 2 2 1 2 2 1 1 1 2 2 1 1 1 1 2 1 2 2 1 1 2 2 1 2 2 1 2 1 2 1 2 2 2 1 1 1 1 2 1 2 1 1 2 1 1 2 2 1 2 1 2 1 1 1 1 1 1 1 1 1 2 1 2 2 2 1 1 2 2 1 1 1 1 2 1 1 2 1 2 2 2 1 1 1 2 2 2 1 1 1 1 2 1 2 1 1 1 1 2 2 2 1 1 2 1 2 1 1 1 1 1 2 1 1 1 1 1 2 1 1 2 2 1 2 1 1 2 2 1 1 2 2 1 1 1 2 2 1 1 2 1 2 1 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 2 1 2 2 1 1 1 2 2 1 1 2 2 1 1 2 1",
"output": "87"
},
{
"input": "3 1 2\n1 1 1",
"output": "2"
},
{
"input": "3 2 2\n1 1 1",
"output": "1"
},
{
"input": "3 2 1\n1 1 1",
"output": "1"
},
{
"input": "3 1 1\n1 1 1",
"output": "2"
},
{
"input": "5 1 2\n2 2 2 2 2",
"output": "2"
},
{
"input": "5 2 2\n2 2 2 2 2",
"output": "1"
},
{
"input": "5 2 1\n2 2 2 2 2",
"output": "2"
},
{
"input": "5 1 1\n2 2 2 2 2",
"output": "3"
},
{
"input": "1 1 2\n2",
"output": "0"
},
{
"input": "1 2 2\n2",
"output": "0"
},
{
"input": "1 2 1\n2",
"output": "0"
},
{
"input": "1 1 1\n2",
"output": "0"
},
{
"input": "6 3 1\n1 1 2 2 2 2",
"output": "2"
},
{
"input": "100 40 20\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "40"
},
{
"input": "7 5 2\n2 2 1 1 1 1 1",
"output": "0"
},
{
"input": "10 4 4\n2 2 2 2 2 2 1 1 1 1",
"output": "2"
},
{
"input": "3 2 1\n2 1 1",
"output": "0"
},
{
"input": "7 6 1\n2 1 1 1 1 1 1",
"output": "0"
},
{
"input": "7 5 1\n1 1 1 2 2 2 2",
"output": "1"
},
{
"input": "5 3 1\n1 1 2 2 2",
"output": "1"
},
{
"input": "3 1 1\n2 2 2",
"output": "1"
},
{
"input": "5 2 2\n2 2 2 2 2",
"output": "1"
},
{
"input": "3 1 3\n1 1 1",
"output": "2"
},
{
"input": "5 2 1\n1 1 2 2 2",
"output": "2"
},
{
"input": "4 3 2\n2 1 1 1",
"output": "0"
},
{
"input": "4 2 1\n1 2 2 2",
"output": "1"
},
{
"input": "14 4 7\n1 1 1 2 2 2 2 2 2 2 2 2 2 2",
"output": "3"
},
{
"input": "12 10 4\n2 2 2 2 2 2 1 1 1 1 1 1",
"output": "0"
},
{
"input": "5 3 2\n2 2 1 1 1",
"output": "0"
}
] | 1,642,666,918 | 2,147,483,647 | Python 3 | OK | TESTS | 63 | 46 | 0 | n, m, k = list(map(int, input().split()))
a = list(map(int, input().split()))
def solve(n, m, k, a):
cnt = 0
for dish in a:
if dish == 1:
if m > 0:
m -= 1
else:
cnt += 1
else:
if m == 0 and k == 0:
cnt += 1
elif k > 0:
k -= 1
else:
m -= 1
return cnt
print(solve(n, m, k, a))
| Title: Valera and Plates
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Valera is a lazy student. He has *m* clean bowls and *k* clean plates.
Valera has made an eating plan for the next *n* days. As Valera is lazy, he will eat exactly one dish per day. At that, in order to eat a dish, he needs exactly one clean plate or bowl. We know that Valera can cook only two types of dishes. He can eat dishes of the first type from bowls and dishes of the second type from either bowls or plates.
When Valera finishes eating, he leaves a dirty plate/bowl behind. His life philosophy doesn't let him eat from dirty kitchenware. So sometimes he needs to wash his plate/bowl before eating. Find the minimum number of times Valera will need to wash a plate/bowl, if he acts optimally.
Input Specification:
The first line of the input contains three integers *n*, *m*, *k* (1<=≤<=*n*,<=*m*,<=*k*<=≤<=1000) — the number of the planned days, the number of clean bowls and the number of clean plates.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=2). If *a**i* equals one, then on day *i* Valera will eat a first type dish. If *a**i* equals two, then on day *i* Valera will eat a second type dish.
Output Specification:
Print a single integer — the minimum number of times Valera will need to wash a plate/bowl.
Demo Input:
['3 1 1\n1 2 1\n', '4 3 1\n1 1 1 1\n', '3 1 2\n2 2 2\n', '8 2 2\n1 2 1 2 1 2 1 2\n']
Demo Output:
['1\n', '1\n', '0\n', '4\n']
Note:
In the first sample Valera will wash a bowl only on the third day, so the answer is one.
In the second sample, Valera will have the first type of the dish during all four days, and since there are only three bowls, he will wash a bowl exactly once.
In the third sample, Valera will have the second type of dish for all three days, and as they can be eaten from either a plate or a bowl, he will never need to wash a plate/bowl. | ```python
n, m, k = list(map(int, input().split()))
a = list(map(int, input().split()))
def solve(n, m, k, a):
cnt = 0
for dish in a:
if dish == 1:
if m > 0:
m -= 1
else:
cnt += 1
else:
if m == 0 and k == 0:
cnt += 1
elif k > 0:
k -= 1
else:
m -= 1
return cnt
print(solve(n, m, k, a))
``` | 3 |
|
650 | A | Watchmen | PROGRAMMING | 1,400 | [
"data structures",
"geometry",
"math"
] | null | null | Watchmen are in a danger and Doctor Manhattan together with his friend Daniel Dreiberg should warn them as soon as possible. There are *n* watchmen on a plane, the *i*-th watchman is located at point (*x**i*,<=*y**i*).
They need to arrange a plan, but there are some difficulties on their way. As you know, Doctor Manhattan considers the distance between watchmen *i* and *j* to be |*x**i*<=-<=*x**j*|<=+<=|*y**i*<=-<=*y**j*|. Daniel, as an ordinary person, calculates the distance using the formula .
The success of the operation relies on the number of pairs (*i*,<=*j*) (1<=≤<=*i*<=<<=*j*<=≤<=*n*), such that the distance between watchman *i* and watchmen *j* calculated by Doctor Manhattan is equal to the distance between them calculated by Daniel. You were asked to compute the number of such pairs. | The first line of the input contains the single integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of watchmen.
Each of the following *n* lines contains two integers *x**i* and *y**i* (|*x**i*|,<=|*y**i*|<=≤<=109).
Some positions may coincide. | Print the number of pairs of watchmen such that the distance between them calculated by Doctor Manhattan is equal to the distance calculated by Daniel. | [
"3\n1 1\n7 5\n1 5\n",
"6\n0 0\n0 1\n0 2\n-1 1\n0 1\n1 1\n"
] | [
"2\n",
"11\n"
] | In the first sample, the distance between watchman 1 and watchman 2 is equal to |1 - 7| + |1 - 5| = 10 for Doctor Manhattan and <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/bcb5b7064b5f02088da0fdcf677e6fda495dd0df.png" style="max-width: 100.0%;max-height: 100.0%;"/> for Daniel. For pairs (1, 1), (1, 5) and (7, 5), (1, 5) Doctor Manhattan and Daniel will calculate the same distances. | 500 | [
{
"input": "3\n1 1\n7 5\n1 5",
"output": "2"
},
{
"input": "6\n0 0\n0 1\n0 2\n-1 1\n0 1\n1 1",
"output": "11"
},
{
"input": "10\n46 -55\n46 45\n46 45\n83 -55\n46 45\n83 -55\n46 45\n83 45\n83 45\n46 -55",
"output": "33"
},
{
"input": "1\n-5 -90",
"output": "0"
},
{
"input": "2\n315 845\n-669 -762",
"output": "0"
},
{
"input": "3\n8911 7861\n-6888 7861\n8911 7861",
"output": "3"
},
{
"input": "2\n-1 1000000000\n0 -1",
"output": "0"
},
{
"input": "2\n1000000000 0\n-7 1",
"output": "0"
},
{
"input": "2\n1 4\n2 1",
"output": "0"
},
{
"input": "2\n1 0\n0 2333333",
"output": "0"
},
{
"input": "2\n2 1\n1 2",
"output": "0"
},
{
"input": "2\n1 1000000000\n2 -1000000000",
"output": "0"
},
{
"input": "2\n0 1000000000\n1 -7",
"output": "0"
},
{
"input": "2\n1 0\n0 19990213",
"output": "0"
}
] | 1,565,118,719 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 2 | 124 | 0 | import sys
input = sys.stdin.readline
n = int(input())
N = 0
x, y = {}, {}
pp = []
for i in range(n):
xi, yi = map(int, input().split())
if xi not in x:
x[xi] = []
x[xi].append(yi)
if yi not in y:
y[yi] = []
y[yi].append(xi)
p = (xi, yi)
if p not in pp:
pp.append(p)
##print(x)
##print(y)
##print(pp)
s = 0
for xi in x:
l = len(x[xi])
s += l*(l-1)//2
for yi in y:
l = len(y[yi])
s += l*(l-1)//2
s -= (n - len(pp))
print(s)
| Title: Watchmen
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Watchmen are in a danger and Doctor Manhattan together with his friend Daniel Dreiberg should warn them as soon as possible. There are *n* watchmen on a plane, the *i*-th watchman is located at point (*x**i*,<=*y**i*).
They need to arrange a plan, but there are some difficulties on their way. As you know, Doctor Manhattan considers the distance between watchmen *i* and *j* to be |*x**i*<=-<=*x**j*|<=+<=|*y**i*<=-<=*y**j*|. Daniel, as an ordinary person, calculates the distance using the formula .
The success of the operation relies on the number of pairs (*i*,<=*j*) (1<=≤<=*i*<=<<=*j*<=≤<=*n*), such that the distance between watchman *i* and watchmen *j* calculated by Doctor Manhattan is equal to the distance between them calculated by Daniel. You were asked to compute the number of such pairs.
Input Specification:
The first line of the input contains the single integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of watchmen.
Each of the following *n* lines contains two integers *x**i* and *y**i* (|*x**i*|,<=|*y**i*|<=≤<=109).
Some positions may coincide.
Output Specification:
Print the number of pairs of watchmen such that the distance between them calculated by Doctor Manhattan is equal to the distance calculated by Daniel.
Demo Input:
['3\n1 1\n7 5\n1 5\n', '6\n0 0\n0 1\n0 2\n-1 1\n0 1\n1 1\n']
Demo Output:
['2\n', '11\n']
Note:
In the first sample, the distance between watchman 1 and watchman 2 is equal to |1 - 7| + |1 - 5| = 10 for Doctor Manhattan and <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/bcb5b7064b5f02088da0fdcf677e6fda495dd0df.png" style="max-width: 100.0%;max-height: 100.0%;"/> for Daniel. For pairs (1, 1), (1, 5) and (7, 5), (1, 5) Doctor Manhattan and Daniel will calculate the same distances. | ```python
import sys
input = sys.stdin.readline
n = int(input())
N = 0
x, y = {}, {}
pp = []
for i in range(n):
xi, yi = map(int, input().split())
if xi not in x:
x[xi] = []
x[xi].append(yi)
if yi not in y:
y[yi] = []
y[yi].append(xi)
p = (xi, yi)
if p not in pp:
pp.append(p)
##print(x)
##print(y)
##print(pp)
s = 0
for xi in x:
l = len(x[xi])
s += l*(l-1)//2
for yi in y:
l = len(y[yi])
s += l*(l-1)//2
s -= (n - len(pp))
print(s)
``` | 0 |
|
96 | A | Football | PROGRAMMING | 900 | [
"implementation",
"strings"
] | A. Football | 2 | 256 | Petya loves football very much. One day, as he was watching a football match, he was writing the players' current positions on a piece of paper. To simplify the situation he depicted it as a string consisting of zeroes and ones. A zero corresponds to players of one team; a one corresponds to players of another team. If there are at least 7 players of some team standing one after another, then the situation is considered dangerous. For example, the situation 00100110111111101 is dangerous and 11110111011101 is not. You are given the current situation. Determine whether it is dangerous or not. | The first input line contains a non-empty string consisting of characters "0" and "1", which represents players. The length of the string does not exceed 100 characters. There's at least one player from each team present on the field. | Print "YES" if the situation is dangerous. Otherwise, print "NO". | [
"001001\n",
"1000000001\n"
] | [
"NO\n",
"YES\n"
] | none | 500 | [
{
"input": "001001",
"output": "NO"
},
{
"input": "1000000001",
"output": "YES"
},
{
"input": "00100110111111101",
"output": "YES"
},
{
"input": "11110111111111111",
"output": "YES"
},
{
"input": "01",
"output": "NO"
},
{
"input": "10100101",
"output": "NO"
},
{
"input": "1010010100000000010",
"output": "YES"
},
{
"input": "101010101",
"output": "NO"
},
{
"input": "000000000100000000000110101100000",
"output": "YES"
},
{
"input": "100001000000110101100000",
"output": "NO"
},
{
"input": "100001000011010110000",
"output": "NO"
},
{
"input": "010",
"output": "NO"
},
{
"input": "10101011111111111111111111111100",
"output": "YES"
},
{
"input": "1001101100",
"output": "NO"
},
{
"input": "1001101010",
"output": "NO"
},
{
"input": "1111100111",
"output": "NO"
},
{
"input": "00110110001110001111",
"output": "NO"
},
{
"input": "11110001001111110001",
"output": "NO"
},
{
"input": "10001111001011111101",
"output": "NO"
},
{
"input": "10000010100000001000110001010100001001001010011",
"output": "YES"
},
{
"input": "01111011111010111100101100001011001010111110000010",
"output": "NO"
},
{
"input": "00100000100100101110011001011011101110110110010100",
"output": "NO"
},
{
"input": "10110100110001001011110101110010100010000000000100101010111110111110100011",
"output": "YES"
},
{
"input": "00011101010101111001011011001101101011111101000010100000111000011100101011",
"output": "NO"
},
{
"input": "01110000110100110101110100111000101101011101011110110100100111100001110111",
"output": "NO"
},
{
"input": "11110110011000100111100111101101011111110100010101011011111101110110110111",
"output": "YES"
},
{
"input": "100100010101110010001011001110100011100010011110100101100011010001001010001001101111001100",
"output": "NO"
},
{
"input": "111110010001011010010011111100110110001111000010100011011100111101111101110010101111011110000001010",
"output": "NO"
},
{
"input": "111110111100010100000100001010111011101011000111011011011010110010100010000101011111000011010011110",
"output": "NO"
},
{
"input": "1011110110111010110111111010010010100011111011110000011000110010011110111010110100011010100010111000",
"output": "NO"
},
{
"input": "0010100111100010110110000011100111110100111110001010000100111111111010111100101101010101001011010110",
"output": "YES"
},
{
"input": "1111010100010100101011101100101101110011000010100010000001111100010011100101010001101111000001011000",
"output": "NO"
},
{
"input": "0101100011001110001110100111100011010101011000000000110110010010111100101111010111100011101100100101",
"output": "YES"
},
{
"input": "0001101110011101110000000010011111101001101111100001001010110000110001100000010001111011011110001101",
"output": "YES"
},
{
"input": "1000010000100000100010000100001000010000100001000010000100001000010000100001000010000100001000010000",
"output": "NO"
},
{
"input": "1000000000100000000010000000001000000000100000000010000000001000000000100000000010000000001000000000",
"output": "YES"
},
{
"input": "0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001",
"output": "YES"
},
{
"input": "1111111111111111111111111111111111111111011111111111111111111111111111111111111111111111111111111111",
"output": "YES"
},
{
"input": "10100101000",
"output": "NO"
},
{
"input": "11110111011101",
"output": "NO"
},
{
"input": "10000000",
"output": "YES"
},
{
"input": "00000001",
"output": "YES"
},
{
"input": "01111111",
"output": "YES"
},
{
"input": "11111110",
"output": "YES"
}
] | 1,688,282,538 | 2,147,483,647 | Python 3 | OK | TESTS | 44 | 92 | 0 | s = input()
dangerous = False
if '0000000' in s or '1111111' in s:
dangerous = True
if dangerous:
print("YES")
else:
print("NO")
| Title: Football
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Petya loves football very much. One day, as he was watching a football match, he was writing the players' current positions on a piece of paper. To simplify the situation he depicted it as a string consisting of zeroes and ones. A zero corresponds to players of one team; a one corresponds to players of another team. If there are at least 7 players of some team standing one after another, then the situation is considered dangerous. For example, the situation 00100110111111101 is dangerous and 11110111011101 is not. You are given the current situation. Determine whether it is dangerous or not.
Input Specification:
The first input line contains a non-empty string consisting of characters "0" and "1", which represents players. The length of the string does not exceed 100 characters. There's at least one player from each team present on the field.
Output Specification:
Print "YES" if the situation is dangerous. Otherwise, print "NO".
Demo Input:
['001001\n', '1000000001\n']
Demo Output:
['NO\n', 'YES\n']
Note:
none | ```python
s = input()
dangerous = False
if '0000000' in s or '1111111' in s:
dangerous = True
if dangerous:
print("YES")
else:
print("NO")
``` | 3.977 |
4 | A | Watermelon | PROGRAMMING | 800 | [
"brute force",
"math"
] | A. Watermelon | 1 | 64 | One hot summer day Pete and his friend Billy decided to buy a watermelon. They chose the biggest and the ripest one, in their opinion. After that the watermelon was weighed, and the scales showed *w* kilos. They rushed home, dying of thirst, and decided to divide the berry, however they faced a hard problem.
Pete and Billy are great fans of even numbers, that's why they want to divide the watermelon in such a way that each of the two parts weighs even number of kilos, at the same time it is not obligatory that the parts are equal. The boys are extremely tired and want to start their meal as soon as possible, that's why you should help them and find out, if they can divide the watermelon in the way they want. For sure, each of them should get a part of positive weight. | The first (and the only) input line contains integer number *w* (1<=≤<=*w*<=≤<=100) — the weight of the watermelon bought by the boys. | Print YES, if the boys can divide the watermelon into two parts, each of them weighing even number of kilos; and NO in the opposite case. | [
"8\n"
] | [
"YES\n"
] | For example, the boys can divide the watermelon into two parts of 2 and 6 kilos respectively (another variant — two parts of 4 and 4 kilos). | 0 | [
{
"input": "8",
"output": "YES"
},
{
"input": "5",
"output": "NO"
},
{
"input": "4",
"output": "YES"
},
{
"input": "3",
"output": "NO"
},
{
"input": "2",
"output": "NO"
},
{
"input": "1",
"output": "NO"
},
{
"input": "7",
"output": "NO"
},
{
"input": "6",
"output": "YES"
},
{
"input": "10",
"output": "YES"
},
{
"input": "9",
"output": "NO"
},
{
"input": "53",
"output": "NO"
},
{
"input": "77",
"output": "NO"
},
{
"input": "32",
"output": "YES"
},
{
"input": "44",
"output": "YES"
},
{
"input": "98",
"output": "YES"
},
{
"input": "99",
"output": "NO"
},
{
"input": "90",
"output": "YES"
},
{
"input": "67",
"output": "NO"
},
{
"input": "100",
"output": "YES"
},
{
"input": "88",
"output": "YES"
}
] | 1,692,929,643 | 2,147,483,647 | Python 3 | OK | TESTS | 20 | 92 | 0 | def can_divide_watermelon(w):
# If the weight is less than 4, it's not possible to divide it into even parts
if w < 4:
return "NO"
# If the weight is even, it can be divided into two even parts
if w % 2 == 0:
return "YES"
return "NO"
# Read the input
w = int(input())
# Call the function and print the result
print(can_divide_watermelon(w))
| Title: Watermelon
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
One hot summer day Pete and his friend Billy decided to buy a watermelon. They chose the biggest and the ripest one, in their opinion. After that the watermelon was weighed, and the scales showed *w* kilos. They rushed home, dying of thirst, and decided to divide the berry, however they faced a hard problem.
Pete and Billy are great fans of even numbers, that's why they want to divide the watermelon in such a way that each of the two parts weighs even number of kilos, at the same time it is not obligatory that the parts are equal. The boys are extremely tired and want to start their meal as soon as possible, that's why you should help them and find out, if they can divide the watermelon in the way they want. For sure, each of them should get a part of positive weight.
Input Specification:
The first (and the only) input line contains integer number *w* (1<=≤<=*w*<=≤<=100) — the weight of the watermelon bought by the boys.
Output Specification:
Print YES, if the boys can divide the watermelon into two parts, each of them weighing even number of kilos; and NO in the opposite case.
Demo Input:
['8\n']
Demo Output:
['YES\n']
Note:
For example, the boys can divide the watermelon into two parts of 2 and 6 kilos respectively (another variant — two parts of 4 and 4 kilos). | ```python
def can_divide_watermelon(w):
# If the weight is less than 4, it's not possible to divide it into even parts
if w < 4:
return "NO"
# If the weight is even, it can be divided into two even parts
if w % 2 == 0:
return "YES"
return "NO"
# Read the input
w = int(input())
# Call the function and print the result
print(can_divide_watermelon(w))
``` | 3.954 |
573 | A | Bear and Poker | PROGRAMMING | 1,300 | [
"implementation",
"math",
"number theory"
] | null | null | Limak is an old brown bear. He often plays poker with his friends. Today they went to a casino. There are *n* players (including Limak himself) and right now all of them have bids on the table. *i*-th of them has bid with size *a**i* dollars.
Each player can double his bid any number of times and triple his bid any number of times. The casino has a great jackpot for making all bids equal. Is it possible that Limak and his friends will win a jackpot? | First line of input contains an integer *n* (2<=≤<=*n*<=≤<=105), the number of players.
The second line contains *n* integer numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the bids of players. | Print "Yes" (without the quotes) if players can make their bids become equal, or "No" otherwise. | [
"4\n75 150 75 50\n",
"3\n100 150 250\n"
] | [
"Yes\n",
"No\n"
] | In the first sample test first and third players should double their bids twice, second player should double his bid once and fourth player should both double and triple his bid.
It can be shown that in the second sample test there is no way to make all bids equal. | 500 | [
{
"input": "4\n75 150 75 50",
"output": "Yes"
},
{
"input": "3\n100 150 250",
"output": "No"
},
{
"input": "7\n34 34 68 34 34 68 34",
"output": "Yes"
},
{
"input": "10\n72 96 12 18 81 20 6 2 54 1",
"output": "No"
},
{
"input": "20\n958692492 954966768 77387000 724664764 101294996 614007760 202904092 555293973 707655552 108023967 73123445 612562357 552908390 914853758 915004122 466129205 122853497 814592742 373389439 818473058",
"output": "No"
},
{
"input": "2\n1 1",
"output": "Yes"
},
{
"input": "2\n72 72",
"output": "Yes"
},
{
"input": "2\n49 42",
"output": "No"
},
{
"input": "3\n1000000000 1000000000 1000000000",
"output": "Yes"
},
{
"input": "6\n162000 96000 648000 1000 864000 432000",
"output": "Yes"
},
{
"input": "8\n600000 100000 100000 100000 900000 600000 900000 600000",
"output": "Yes"
},
{
"input": "12\n2048 1024 6144 1024 3072 3072 6144 1024 4096 2048 6144 3072",
"output": "Yes"
},
{
"input": "20\n246 246 246 246 246 246 246 246 246 246 246 246 246 246 246 246 246 246 246 246",
"output": "Yes"
},
{
"input": "50\n840868705 387420489 387420489 795385082 634350497 206851546 536870912 536870912 414927754 387420489 387420489 536870912 387420489 149011306 373106005 536870912 700746206 387420489 777952883 847215247 176645254 576664386 387420489 230876513 536870912 536870912 536870912 387420489 387420489 536870912 460495524 528643722 387420489 536870912 470369206 899619085 387420489 631148352 387420489 387420489 536870912 414666674 521349938 776784669 387420489 102428009 536870912 387420489 536870912 718311009",
"output": "No"
},
{
"input": "2\n5 6",
"output": "No"
},
{
"input": "3\n536870912 387420489 257407169",
"output": "No"
},
{
"input": "4\n2 2 5 2",
"output": "No"
},
{
"input": "2\n33554432 59049",
"output": "Yes"
},
{
"input": "3\n536870912 387420489 387420489",
"output": "Yes"
},
{
"input": "2\n1 5",
"output": "No"
},
{
"input": "18\n2 3 5 7 11 13 17 19 23 29 31 37 43 47 53 59 67 71",
"output": "No"
},
{
"input": "2\n1 30",
"output": "No"
},
{
"input": "3\n335544320 71744535 71744535",
"output": "Yes"
},
{
"input": "5\n1000000000 999999999 999999998 999999997 999999996",
"output": "No"
},
{
"input": "2\n25 5",
"output": "No"
},
{
"input": "4\n75 150 75 5",
"output": "No"
},
{
"input": "3\n536870912 387420489 362797056",
"output": "Yes"
},
{
"input": "3\n536870912 387420489 89",
"output": "No"
},
{
"input": "4\n547 2606459 222763549 143466789",
"output": "No"
},
{
"input": "3\n129140163 33554432 1",
"output": "Yes"
},
{
"input": "10\n244140625 244140625 244140625 244140625 244140625 244140625 244140625 244140625 536870912 387420489",
"output": "No"
},
{
"input": "3\n5 5 1",
"output": "No"
},
{
"input": "5\n3 7 29 36760123 823996703",
"output": "No"
}
] | 1,668,173,679 | 2,147,483,647 | Python 3 | OK | TESTS | 70 | 639 | 8,806,400 | n = int(input())
l = list(map(int, input().split()))
for i in range(n):
while(l[i]%2 == 0):
l[i] //= 2
for i in range(n):
while(l[i]%3 == 0):
l[i] //= 3
c = l.count(l[0])
if(c == n):
print("YES")
else:
print("NO") | Title: Bear and Poker
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Limak is an old brown bear. He often plays poker with his friends. Today they went to a casino. There are *n* players (including Limak himself) and right now all of them have bids on the table. *i*-th of them has bid with size *a**i* dollars.
Each player can double his bid any number of times and triple his bid any number of times. The casino has a great jackpot for making all bids equal. Is it possible that Limak and his friends will win a jackpot?
Input Specification:
First line of input contains an integer *n* (2<=≤<=*n*<=≤<=105), the number of players.
The second line contains *n* integer numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the bids of players.
Output Specification:
Print "Yes" (without the quotes) if players can make their bids become equal, or "No" otherwise.
Demo Input:
['4\n75 150 75 50\n', '3\n100 150 250\n']
Demo Output:
['Yes\n', 'No\n']
Note:
In the first sample test first and third players should double their bids twice, second player should double his bid once and fourth player should both double and triple his bid.
It can be shown that in the second sample test there is no way to make all bids equal. | ```python
n = int(input())
l = list(map(int, input().split()))
for i in range(n):
while(l[i]%2 == 0):
l[i] //= 2
for i in range(n):
while(l[i]%3 == 0):
l[i] //= 3
c = l.count(l[0])
if(c == n):
print("YES")
else:
print("NO")
``` | 3 |
|
863 | B | Kayaking | PROGRAMMING | 1,500 | [
"brute force",
"greedy",
"sortings"
] | null | null | Vadim is really keen on travelling. Recently he heard about kayaking activity near his town and became very excited about it, so he joined a party of kayakers.
Now the party is ready to start its journey, but firstly they have to choose kayaks. There are 2·*n* people in the group (including Vadim), and they have exactly *n*<=-<=1 tandem kayaks (each of which, obviously, can carry two people) and 2 single kayaks. *i*-th person's weight is *w**i*, and weight is an important matter in kayaking — if the difference between the weights of two people that sit in the same tandem kayak is too large, then it can crash. And, of course, people want to distribute their seats in kayaks in order to minimize the chances that kayaks will crash.
Formally, the instability of a single kayak is always 0, and the instability of a tandem kayak is the absolute difference between weights of the people that are in this kayak. Instability of the whole journey is the total instability of all kayaks.
Help the party to determine minimum possible total instability! | The first line contains one number *n* (2<=≤<=*n*<=≤<=50).
The second line contains 2·*n* integer numbers *w*1, *w*2, ..., *w*2*n*, where *w**i* is weight of person *i* (1<=≤<=*w**i*<=≤<=1000). | Print minimum possible total instability. | [
"2\n1 2 3 4\n",
"4\n1 3 4 6 3 4 100 200\n"
] | [
"1\n",
"5\n"
] | none | 0 | [
{
"input": "2\n1 2 3 4",
"output": "1"
},
{
"input": "4\n1 3 4 6 3 4 100 200",
"output": "5"
},
{
"input": "3\n305 139 205 406 530 206",
"output": "102"
},
{
"input": "3\n610 750 778 6 361 407",
"output": "74"
},
{
"input": "5\n97 166 126 164 154 98 221 7 51 47",
"output": "35"
},
{
"input": "50\n1 1 2 2 1 3 2 2 1 1 1 1 2 3 3 1 2 1 3 3 2 1 2 3 1 1 2 1 3 1 3 1 3 3 3 1 1 1 3 3 2 2 2 2 3 2 2 2 2 3 1 3 3 3 3 1 3 3 1 3 3 3 3 2 3 1 3 3 1 1 1 3 1 2 2 2 1 1 1 3 1 2 3 2 1 3 3 2 2 1 3 1 3 1 2 2 1 2 3 2",
"output": "0"
},
{
"input": "50\n5 5 5 5 4 2 2 3 2 2 4 1 5 5 1 2 4 2 4 2 5 2 2 2 2 3 2 4 2 5 5 4 3 1 2 3 3 5 4 2 2 5 2 4 5 5 4 4 1 5 5 3 2 2 5 1 3 3 2 4 4 5 1 2 3 4 4 1 3 3 3 5 1 2 4 4 4 4 2 5 2 5 3 2 4 5 5 2 1 1 2 4 5 3 2 1 2 4 4 4",
"output": "1"
},
{
"input": "50\n499 780 837 984 481 526 944 482 862 136 265 605 5 631 974 967 574 293 969 467 573 845 102 224 17 873 648 120 694 996 244 313 404 129 899 583 541 314 525 496 443 857 297 78 575 2 430 137 387 319 382 651 594 411 845 746 18 232 6 289 889 81 174 175 805 1000 799 950 475 713 951 685 729 925 262 447 139 217 788 514 658 572 784 185 112 636 10 251 621 218 210 89 597 553 430 532 264 11 160 476",
"output": "368"
},
{
"input": "50\n873 838 288 87 889 364 720 410 565 651 577 356 740 99 549 592 994 385 777 435 486 118 887 440 749 533 356 790 413 681 267 496 475 317 88 660 374 186 61 437 729 860 880 538 277 301 667 180 60 393 955 540 896 241 362 146 74 680 734 767 851 337 751 860 542 735 444 793 340 259 495 903 743 961 964 966 87 275 22 776 368 701 835 732 810 735 267 988 352 647 924 183 1 924 217 944 322 252 758 597",
"output": "393"
},
{
"input": "50\n297 787 34 268 439 629 600 398 425 833 721 908 830 636 64 509 420 647 499 675 427 599 396 119 798 742 577 355 22 847 389 574 766 453 196 772 808 261 106 844 726 975 173 992 874 89 775 616 678 52 69 591 181 573 258 381 665 301 589 379 362 146 790 842 765 100 229 916 938 97 340 793 758 177 736 396 247 562 571 92 923 861 165 748 345 703 431 930 101 761 862 595 505 393 126 846 431 103 596 21",
"output": "387"
},
{
"input": "50\n721 631 587 746 692 406 583 90 388 16 161 948 921 70 387 426 39 398 517 724 879 377 906 502 359 950 798 408 846 718 911 845 57 886 9 668 537 632 344 762 19 193 658 447 870 173 98 156 592 519 183 539 274 393 962 615 551 626 148 183 769 763 829 120 796 761 14 744 537 231 696 284 581 688 611 826 703 145 224 600 965 613 791 275 984 375 402 281 851 580 992 8 816 454 35 532 347 250 242 637",
"output": "376"
},
{
"input": "50\n849 475 37 120 754 183 758 374 543 198 896 691 11 607 198 343 761 660 239 669 628 259 223 182 216 158 20 565 454 884 137 923 156 22 310 77 267 707 582 169 120 308 439 309 59 152 206 696 210 177 296 887 559 22 154 553 142 247 491 692 473 572 461 206 532 319 503 164 328 365 541 366 300 392 486 257 863 432 877 404 520 69 418 99 519 239 374 927 601 103 226 316 423 219 240 26 455 101 184 61",
"output": "351"
},
{
"input": "3\n1 2 10 11 100 100",
"output": "1"
},
{
"input": "17\n814 744 145 886 751 1000 272 914 270 529 467 164 410 369 123 424 991 12 702 582 561 858 746 950 598 393 606 498 648 686 455 873 728 858",
"output": "318"
},
{
"input": "45\n476 103 187 696 463 457 588 632 763 77 391 721 95 124 378 812 980 193 694 898 859 572 721 274 605 264 929 615 257 918 42 493 1 3 697 349 990 800 82 535 382 816 943 735 11 272 562 323 653 370 766 332 666 130 704 604 645 717 267 255 37 470 925 941 376 611 332 758 504 40 477 263 708 434 38 596 650 990 714 662 572 467 949 799 648 581 545 828 508 636",
"output": "355"
},
{
"input": "2\n55 5 25 51",
"output": "4"
},
{
"input": "25\n89 50 640 463 858 301 522 241 923 378 892 822 550 17 42 66 706 779 657 840 273 222 444 459 94 925 437 159 182 727 92 851 742 215 653 891 782 533 29 128 133 883 317 475 165 994 802 434 744 973",
"output": "348"
},
{
"input": "4\n35 48 71 44 78 79 57 48",
"output": "10"
},
{
"input": "3\n58 89 73 15 5 47",
"output": "21"
},
{
"input": "2\n1 20 99 100",
"output": "1"
}
] | 1,671,909,382 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 30 | 0 | n = int(input())
weights_string = input()
weights_set = weights_string.split()
weights = []
for i in range(len(weights_set)):
weights.append(int(weights_set[i]))
weights_sort = sorted(weights)
runmax = 0
runtotal = 0
for i in range(0, 2*n, 2):
if runmax <= weights_sort[i+1] - weights_sort[i]:
runmax = weights_sort[i+1] - weights_sort[i]
runtotal = runtotal + weights_sort[i + 1] - weights_sort[i]
print(runtotal-runmax) | Title: Kayaking
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vadim is really keen on travelling. Recently he heard about kayaking activity near his town and became very excited about it, so he joined a party of kayakers.
Now the party is ready to start its journey, but firstly they have to choose kayaks. There are 2·*n* people in the group (including Vadim), and they have exactly *n*<=-<=1 tandem kayaks (each of which, obviously, can carry two people) and 2 single kayaks. *i*-th person's weight is *w**i*, and weight is an important matter in kayaking — if the difference between the weights of two people that sit in the same tandem kayak is too large, then it can crash. And, of course, people want to distribute their seats in kayaks in order to minimize the chances that kayaks will crash.
Formally, the instability of a single kayak is always 0, and the instability of a tandem kayak is the absolute difference between weights of the people that are in this kayak. Instability of the whole journey is the total instability of all kayaks.
Help the party to determine minimum possible total instability!
Input Specification:
The first line contains one number *n* (2<=≤<=*n*<=≤<=50).
The second line contains 2·*n* integer numbers *w*1, *w*2, ..., *w*2*n*, where *w**i* is weight of person *i* (1<=≤<=*w**i*<=≤<=1000).
Output Specification:
Print minimum possible total instability.
Demo Input:
['2\n1 2 3 4\n', '4\n1 3 4 6 3 4 100 200\n']
Demo Output:
['1\n', '5\n']
Note:
none | ```python
n = int(input())
weights_string = input()
weights_set = weights_string.split()
weights = []
for i in range(len(weights_set)):
weights.append(int(weights_set[i]))
weights_sort = sorted(weights)
runmax = 0
runtotal = 0
for i in range(0, 2*n, 2):
if runmax <= weights_sort[i+1] - weights_sort[i]:
runmax = weights_sort[i+1] - weights_sort[i]
runtotal = runtotal + weights_sort[i + 1] - weights_sort[i]
print(runtotal-runmax)
``` | 0 |
|
43 | A | Football | PROGRAMMING | 1,000 | [
"strings"
] | A. Football | 2 | 256 | One day Vasya decided to have a look at the results of Berland 1910 Football Championship’s finals. Unfortunately he didn't find the overall score of the match; however, he got hold of a profound description of the match's process. On the whole there are *n* lines in that description each of which described one goal. Every goal was marked with the name of the team that had scored it. Help Vasya, learn the name of the team that won the finals. It is guaranteed that the match did not end in a tie. | The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of lines in the description. Then follow *n* lines — for each goal the names of the teams that scored it. The names are non-empty lines consisting of uppercase Latin letters whose lengths do not exceed 10 symbols. It is guaranteed that the match did not end in a tie and the description contains no more than two different teams. | Print the name of the winning team. We remind you that in football the team that scores more goals is considered the winner. | [
"1\nABC\n",
"5\nA\nABA\nABA\nA\nA\n"
] | [
"ABC\n",
"A\n"
] | none | 500 | [
{
"input": "1\nABC",
"output": "ABC"
},
{
"input": "5\nA\nABA\nABA\nA\nA",
"output": "A"
},
{
"input": "2\nXTSJEP\nXTSJEP",
"output": "XTSJEP"
},
{
"input": "3\nXZYDJAEDZ\nXZYDJAEDZ\nXZYDJAEDZ",
"output": "XZYDJAEDZ"
},
{
"input": "3\nQCCYXL\nQCCYXL\nAXGLFQDD",
"output": "QCCYXL"
},
{
"input": "3\nAZID\nEERWBC\nEERWBC",
"output": "EERWBC"
},
{
"input": "3\nHNCGYL\nHNCGYL\nHNCGYL",
"output": "HNCGYL"
},
{
"input": "4\nZZWZTG\nZZWZTG\nZZWZTG\nZZWZTG",
"output": "ZZWZTG"
},
{
"input": "4\nA\nA\nKUDLJMXCSE\nA",
"output": "A"
},
{
"input": "5\nPHBTW\nPHBTW\nPHBTW\nPHBTW\nPHBTW",
"output": "PHBTW"
},
{
"input": "5\nPKUZYTFYWN\nPKUZYTFYWN\nSTC\nPKUZYTFYWN\nPKUZYTFYWN",
"output": "PKUZYTFYWN"
},
{
"input": "5\nHH\nHH\nNTQWPA\nNTQWPA\nHH",
"output": "HH"
},
{
"input": "10\nW\nW\nW\nW\nW\nD\nW\nD\nD\nW",
"output": "W"
},
{
"input": "19\nXBCP\nTGACNIH\nXBCP\nXBCP\nXBCP\nXBCP\nXBCP\nTGACNIH\nXBCP\nXBCP\nXBCP\nXBCP\nXBCP\nTGACNIH\nXBCP\nXBCP\nTGACNIH\nTGACNIH\nXBCP",
"output": "XBCP"
},
{
"input": "33\nOWQWCKLLF\nOWQWCKLLF\nOWQWCKLLF\nPYPAS\nPYPAS\nPYPAS\nOWQWCKLLF\nPYPAS\nOWQWCKLLF\nPYPAS\nPYPAS\nOWQWCKLLF\nOWQWCKLLF\nOWQWCKLLF\nPYPAS\nOWQWCKLLF\nPYPAS\nPYPAS\nPYPAS\nPYPAS\nOWQWCKLLF\nPYPAS\nPYPAS\nOWQWCKLLF\nOWQWCKLLF\nPYPAS\nOWQWCKLLF\nOWQWCKLLF\nPYPAS\nPYPAS\nOWQWCKLLF\nPYPAS\nPYPAS",
"output": "PYPAS"
},
{
"input": "51\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC",
"output": "NC"
},
{
"input": "89\nH\nVOCI\nVOCI\nH\nVOCI\nH\nH\nVOCI\nVOCI\nVOCI\nH\nH\nH\nVOCI\nVOCI\nVOCI\nH\nVOCI\nVOCI\nH\nVOCI\nVOCI\nVOCI\nH\nVOCI\nH\nVOCI\nH\nVOCI\nH\nVOCI\nVOCI\nH\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nH\nVOCI\nVOCI\nVOCI\nVOCI\nH\nVOCI\nH\nH\nVOCI\nH\nVOCI\nH\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nH\nH\nVOCI\nH\nH\nVOCI\nH\nVOCI\nH\nVOCI\nVOCI\nH\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nH\nH\nH\nH\nH\nVOCI\nH\nVOCI\nH\nVOCI\nVOCI",
"output": "VOCI"
},
{
"input": "100\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nM\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA",
"output": "HA"
},
{
"input": "100\nG\nG\nS\nS\nG\nG\nS\nS\nG\nS\nS\nS\nG\nS\nG\nG\nS\nG\nS\nS\nG\nS\nS\nS\nS\nS\nG\nS\nG\nS\nS\nG\nG\nG\nS\nS\nS\nS\nG\nS\nS\nG\nG\nG\nG\nG\nS\nG\nG\nS\nS\nS\nS\nS\nG\nG\nS\nG\nG\nG\nG\nG\nS\nS\nG\nS\nS\nS\nS\nG\nS\nS\nG\nS\nG\nG\nG\nG\nG\nG\nG\nG\nG\nG\nG\nS\nS\nG\nS\nS\nS\nS\nG\nG\nG\nS\nG\nG\nG\nS",
"output": "G"
},
{
"input": "100\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nOBH\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL",
"output": "WL"
}
] | 1,662,417,972 | 2,147,483,647 | Python 3 | OK | TESTS | 34 | 92 | 0 | '''
One day Vasya decided to have a look at the results of Berland 1910 Football Championship’s finals. Unfortunately he didn't find the overall score of the match; however, he got hold of a profound description of the match's process. On the whole there are n lines in that description each of which described one goal. Every goal was marked with the name of the team that had scored it. Help Vasya, learn the name of the team that won the finals. It is guaranteed that the match did not end in a tie.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the number of lines in the description. Then follow n lines — for each goal the names of the teams that scored it. The names are non-empty lines consisting of uppercase Latin letters whose lengths do not exceed 10 symbols. It is guaranteed that the match did not end in a tie and the description contains no more than two different teams.
Output
Print the name of the winning team. We remind you that in football the team that scores more goals is considered the winner.
'''
games = int(input())
win_list = []
for game in range(games):
winner = input().strip()
win_list.append(winner)
print(max(win_list,key=win_list.count)) | Title: Football
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
One day Vasya decided to have a look at the results of Berland 1910 Football Championship’s finals. Unfortunately he didn't find the overall score of the match; however, he got hold of a profound description of the match's process. On the whole there are *n* lines in that description each of which described one goal. Every goal was marked with the name of the team that had scored it. Help Vasya, learn the name of the team that won the finals. It is guaranteed that the match did not end in a tie.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of lines in the description. Then follow *n* lines — for each goal the names of the teams that scored it. The names are non-empty lines consisting of uppercase Latin letters whose lengths do not exceed 10 symbols. It is guaranteed that the match did not end in a tie and the description contains no more than two different teams.
Output Specification:
Print the name of the winning team. We remind you that in football the team that scores more goals is considered the winner.
Demo Input:
['1\nABC\n', '5\nA\nABA\nABA\nA\nA\n']
Demo Output:
['ABC\n', 'A\n']
Note:
none | ```python
'''
One day Vasya decided to have a look at the results of Berland 1910 Football Championship’s finals. Unfortunately he didn't find the overall score of the match; however, he got hold of a profound description of the match's process. On the whole there are n lines in that description each of which described one goal. Every goal was marked with the name of the team that had scored it. Help Vasya, learn the name of the team that won the finals. It is guaranteed that the match did not end in a tie.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the number of lines in the description. Then follow n lines — for each goal the names of the teams that scored it. The names are non-empty lines consisting of uppercase Latin letters whose lengths do not exceed 10 symbols. It is guaranteed that the match did not end in a tie and the description contains no more than two different teams.
Output
Print the name of the winning team. We remind you that in football the team that scores more goals is considered the winner.
'''
games = int(input())
win_list = []
for game in range(games):
winner = input().strip()
win_list.append(winner)
print(max(win_list,key=win_list.count))
``` | 3.977 |
919 | B | Perfect Number | PROGRAMMING | 1,100 | [
"binary search",
"brute force",
"dp",
"implementation",
"number theory"
] | null | null | We consider a positive integer perfect, if and only if the sum of its digits is exactly $10$. Given a positive integer $k$, your task is to find the $k$-th smallest perfect positive integer. | A single line with a positive integer $k$ ($1 \leq k \leq 10\,000$). | A single number, denoting the $k$-th smallest perfect integer. | [
"1\n",
"2\n"
] | [
"19\n",
"28\n"
] | The first perfect integer is $19$ and the second one is $28$. | 750 | [
{
"input": "1",
"output": "19"
},
{
"input": "2",
"output": "28"
},
{
"input": "13",
"output": "136"
},
{
"input": "101",
"output": "1432"
},
{
"input": "1023",
"output": "100270"
},
{
"input": "9999",
"output": "10800010"
},
{
"input": "10000",
"output": "10800100"
},
{
"input": "2333",
"output": "310060"
},
{
"input": "9139",
"output": "10134010"
},
{
"input": "9859",
"output": "10422001"
},
{
"input": "5987",
"output": "2221201"
},
{
"input": "3",
"output": "37"
},
{
"input": "4",
"output": "46"
},
{
"input": "5",
"output": "55"
},
{
"input": "7",
"output": "73"
},
{
"input": "9",
"output": "91"
},
{
"input": "10",
"output": "109"
},
{
"input": "15",
"output": "154"
},
{
"input": "51",
"output": "613"
},
{
"input": "99",
"output": "1414"
},
{
"input": "138",
"output": "2224"
},
{
"input": "233",
"output": "5050"
},
{
"input": "666",
"output": "27100"
},
{
"input": "1234",
"output": "110206"
},
{
"input": "3567",
"output": "1033003"
},
{
"input": "4445",
"output": "1221301"
},
{
"input": "5689",
"output": "2114002"
},
{
"input": "6666",
"output": "3102004"
},
{
"input": "7777",
"output": "5300200"
},
{
"input": "8888",
"output": "10110061"
},
{
"input": "9411",
"output": "10214200"
},
{
"input": "5539",
"output": "2101114"
},
{
"input": "6259",
"output": "2511100"
},
{
"input": "2387",
"output": "312220"
},
{
"input": "8515",
"output": "10030114"
}
] | 1,687,935,628 | 2,147,483,647 | PyPy 3-64 | TIME_LIMIT_EXCEEDED | TESTS | 5 | 2,000 | 8,806,400 | n=int(input())
i=0
k=0
def s(x):
x=list(str(x))
x=list(map(int,x))
return sum(x)
while(s(i)!=10 or k!=n):
i=i+1
if(s(i)==10):
k=k+1
print(i)
| Title: Perfect Number
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
We consider a positive integer perfect, if and only if the sum of its digits is exactly $10$. Given a positive integer $k$, your task is to find the $k$-th smallest perfect positive integer.
Input Specification:
A single line with a positive integer $k$ ($1 \leq k \leq 10\,000$).
Output Specification:
A single number, denoting the $k$-th smallest perfect integer.
Demo Input:
['1\n', '2\n']
Demo Output:
['19\n', '28\n']
Note:
The first perfect integer is $19$ and the second one is $28$. | ```python
n=int(input())
i=0
k=0
def s(x):
x=list(str(x))
x=list(map(int,x))
return sum(x)
while(s(i)!=10 or k!=n):
i=i+1
if(s(i)==10):
k=k+1
print(i)
``` | 0 |
|
387 | B | George and Round | PROGRAMMING | 1,200 | [
"brute force",
"greedy",
"two pointers"
] | null | null | George decided to prepare a Codesecrof round, so he has prepared *m* problems for the round. Let's number the problems with integers 1 through *m*. George estimates the *i*-th problem's complexity by integer *b**i*.
To make the round good, he needs to put at least *n* problems there. Besides, he needs to have at least one problem with complexity exactly *a*1, at least one with complexity exactly *a*2, ..., and at least one with complexity exactly *a**n*. Of course, the round can also have problems with other complexities.
George has a poor imagination. It's easier for him to make some already prepared problem simpler than to come up with a new one and prepare it. George is magnificent at simplifying problems. He can simplify any already prepared problem with complexity *c* to any positive integer complexity *d* (*c*<=≥<=*d*), by changing limits on the input data.
However, nothing is so simple. George understood that even if he simplifies some problems, he can run out of problems for a good round. That's why he decided to find out the minimum number of problems he needs to come up with in addition to the *m* he's prepared in order to make a good round. Note that George can come up with a new problem of any complexity. | The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=3000) — the minimal number of problems in a good round and the number of problems George's prepared. The second line contains space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a*1<=<<=*a*2<=<<=...<=<<=*a**n*<=≤<=106) — the requirements for the complexity of the problems in a good round. The third line contains space-separated integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b*1<=≤<=*b*2...<=≤<=*b**m*<=≤<=106) — the complexities of the problems prepared by George. | Print a single integer — the answer to the problem. | [
"3 5\n1 2 3\n1 2 2 3 3\n",
"3 5\n1 2 3\n1 1 1 1 1\n",
"3 1\n2 3 4\n1\n"
] | [
"0\n",
"2\n",
"3\n"
] | In the first sample the set of the prepared problems meets the requirements for a good round.
In the second sample, it is enough to come up with and prepare two problems with complexities 2 and 3 to get a good round.
In the third sample it is very easy to get a good round if come up with and prepare extra problems with complexities: 2, 3, 4. | 1,000 | [
{
"input": "3 5\n1 2 3\n1 2 2 3 3",
"output": "0"
},
{
"input": "3 5\n1 2 3\n1 1 1 1 1",
"output": "2"
},
{
"input": "3 1\n2 3 4\n1",
"output": "3"
},
{
"input": "29 100\n20 32 41 67 72 155 331 382 399 412 465 470 484 511 515 529 616 637 679 715 733 763 826 843 862 903 925 979 989\n15 15 15 17 18 19 19 20 21 21 22 24 25 26 26 27 28 31 32 32 37 38 38 39 39 40 41 42 43 43 45 45 46 47 49 49 50 50 50 51 52 53 53 55 56 57 59 59 59 60 60 62 62 63 63 64 64 64 66 67 69 69 70 70 72 72 73 74 75 76 77 78 80 80 81 81 83 83 83 84 86 86 86 86 87 88 89 91 91 91 92 93 94 94 96 97 97 97 98 98",
"output": "24"
}
] | 1,573,355,547 | 2,147,483,647 | Python 3 | OK | TESTS | 41 | 124 | 819,200 | def main():
nm=input().split(' ')
s1=input().split(' ')
s2=input().split(' ')
n=int(nm[0])
m=int(nm[1])
j=0
for i in range(len(s1)):
while j<m and int(s2[j])<int(s1[i]):
j+=1
if j<m:
n-=1
j+=1
else:
break
print(n)
if __name__ == "__main__":
main() | Title: George and Round
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
George decided to prepare a Codesecrof round, so he has prepared *m* problems for the round. Let's number the problems with integers 1 through *m*. George estimates the *i*-th problem's complexity by integer *b**i*.
To make the round good, he needs to put at least *n* problems there. Besides, he needs to have at least one problem with complexity exactly *a*1, at least one with complexity exactly *a*2, ..., and at least one with complexity exactly *a**n*. Of course, the round can also have problems with other complexities.
George has a poor imagination. It's easier for him to make some already prepared problem simpler than to come up with a new one and prepare it. George is magnificent at simplifying problems. He can simplify any already prepared problem with complexity *c* to any positive integer complexity *d* (*c*<=≥<=*d*), by changing limits on the input data.
However, nothing is so simple. George understood that even if he simplifies some problems, he can run out of problems for a good round. That's why he decided to find out the minimum number of problems he needs to come up with in addition to the *m* he's prepared in order to make a good round. Note that George can come up with a new problem of any complexity.
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=3000) — the minimal number of problems in a good round and the number of problems George's prepared. The second line contains space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a*1<=<<=*a*2<=<<=...<=<<=*a**n*<=≤<=106) — the requirements for the complexity of the problems in a good round. The third line contains space-separated integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b*1<=≤<=*b*2...<=≤<=*b**m*<=≤<=106) — the complexities of the problems prepared by George.
Output Specification:
Print a single integer — the answer to the problem.
Demo Input:
['3 5\n1 2 3\n1 2 2 3 3\n', '3 5\n1 2 3\n1 1 1 1 1\n', '3 1\n2 3 4\n1\n']
Demo Output:
['0\n', '2\n', '3\n']
Note:
In the first sample the set of the prepared problems meets the requirements for a good round.
In the second sample, it is enough to come up with and prepare two problems with complexities 2 and 3 to get a good round.
In the third sample it is very easy to get a good round if come up with and prepare extra problems with complexities: 2, 3, 4. | ```python
def main():
nm=input().split(' ')
s1=input().split(' ')
s2=input().split(' ')
n=int(nm[0])
m=int(nm[1])
j=0
for i in range(len(s1)):
while j<m and int(s2[j])<int(s1[i]):
j+=1
if j<m:
n-=1
j+=1
else:
break
print(n)
if __name__ == "__main__":
main()
``` | 3 |
|
59 | A | Word | PROGRAMMING | 800 | [
"implementation",
"strings"
] | A. Word | 2 | 256 | Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word. | The first line contains a word *s* — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100. | Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one. | [
"HoUse\n",
"ViP\n",
"maTRIx\n"
] | [
"house\n",
"VIP\n",
"matrix\n"
] | none | 500 | [
{
"input": "HoUse",
"output": "house"
},
{
"input": "ViP",
"output": "VIP"
},
{
"input": "maTRIx",
"output": "matrix"
},
{
"input": "BNHWpnpawg",
"output": "bnhwpnpawg"
},
{
"input": "VTYGP",
"output": "VTYGP"
},
{
"input": "CHNenu",
"output": "chnenu"
},
{
"input": "ERPZGrodyu",
"output": "erpzgrodyu"
},
{
"input": "KSXBXWpebh",
"output": "KSXBXWPEBH"
},
{
"input": "qvxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaiv",
"output": "qvxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaiv"
},
{
"input": "Amnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfd",
"output": "amnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfd"
},
{
"input": "ISAGFJFARYFBLOPQDSHWGMCNKMFTLVFUGNJEWGWNBLXUIATXEkqiettmmjgydwcpafqrppdsrrrtguinqbgmzzfqwonkpgpcwenv",
"output": "isagfjfaryfblopqdshwgmcnkmftlvfugnjewgwnblxuiatxekqiettmmjgydwcpafqrppdsrrrtguinqbgmzzfqwonkpgpcwenv"
},
{
"input": "XHRPXZEGHSOCJPICUIXSKFUZUPYTSGJSDIYBCMNMNBPNDBXLXBzhbfnqvwcffvrdhtickyqhupmcehlsyvncqmfhautvxudqdhgg",
"output": "xhrpxzeghsocjpicuixskfuzupytsgjsdiybcmnmnbpndbxlxbzhbfnqvwcffvrdhtickyqhupmcehlsyvncqmfhautvxudqdhgg"
},
{
"input": "RJIQZMJCIMSNDBOHBRAWIENODSALETAKGKPYUFGVEFGCBRENZGAdkcetqjljtmttlonpekcovdzebzdkzggwfsxhapmjkdbuceak",
"output": "RJIQZMJCIMSNDBOHBRAWIENODSALETAKGKPYUFGVEFGCBRENZGADKCETQJLJTMTTLONPEKCOVDZEBZDKZGGWFSXHAPMJKDBUCEAK"
},
{
"input": "DWLWOBHNMMGTFOLFAECKBRNNGLYLYDXTGTVRLMEESZOIUATZZZXUFUZDLSJXMEVRTESSFBWLNZZCLCQWEVNNUCXYVHNGNXHCBDFw",
"output": "DWLWOBHNMMGTFOLFAECKBRNNGLYLYDXTGTVRLMEESZOIUATZZZXUFUZDLSJXMEVRTESSFBWLNZZCLCQWEVNNUCXYVHNGNXHCBDFW"
},
{
"input": "NYCNHJWGBOCOTSPETKKHVWFGAQYNHOVJWJHCIEFOUQZXOYUIEQDZALFKTEHTVDBVJMEUBJUBCMNVPWGDPNCHQHZJRCHYRFPVIGUB",
"output": "NYCNHJWGBOCOTSPETKKHVWFGAQYNHOVJWJHCIEFOUQZXOYUIEQDZALFKTEHTVDBVJMEUBJUBCMNVPWGDPNCHQHZJRCHYRFPVIGUB"
},
{
"input": "igxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwge",
"output": "igxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwge"
},
{
"input": "Ykkekrsqolzryiwsmdlnbmfautxxxauoojrddvwklgnlyrfcvhorrzbmtcrvpaypqhcffdqhwziipyyskcmztjprjqvmzzqhqnw",
"output": "ykkekrsqolzryiwsmdlnbmfautxxxauoojrddvwklgnlyrfcvhorrzbmtcrvpaypqhcffdqhwziipyyskcmztjprjqvmzzqhqnw"
},
{
"input": "YQOMLKYAORUQQUCQZCDYMIVDHGWZFFRMUVTAWCHERFPMNRYRIkgqrciokgajamehmcxgerpudvsqyonjonsxgbnefftzmygncks",
"output": "yqomlkyaoruqqucqzcdymivdhgwzffrmuvtawcherfpmnryrikgqrciokgajamehmcxgerpudvsqyonjonsxgbnefftzmygncks"
},
{
"input": "CDOZDPBVVVHNBJVBYHEOXWFLJKRWJCAJMIFCOZWWYFKVWOGTVJcuusigdqfkumewjtdyitveeiaybwrhomrwmpdipjwiuxfnwuz",
"output": "CDOZDPBVVVHNBJVBYHEOXWFLJKRWJCAJMIFCOZWWYFKVWOGTVJCUUSIGDQFKUMEWJTDYITVEEIAYBWRHOMRWMPDIPJWIUXFNWUZ"
},
{
"input": "WHIUVEXHVOOIJIDVJVPQUBJMEVPMPDKQWJKFBZSGSKUXMIPPMJWuckzcpxosodcjaaakvlxpbiigsiauviilylnnqlyucziihqg",
"output": "WHIUVEXHVOOIJIDVJVPQUBJMEVPMPDKQWJKFBZSGSKUXMIPPMJWUCKZCPXOSODCJAAAKVLXPBIIGSIAUVIILYLNNQLYUCZIIHQG"
},
{
"input": "VGHUNFOXKETUYMZDJNGTAOIOANYXSGYNFOGOFFLDAWEUKYFOZXCJTCAFXZYLQZERYZLRSQXYQGAPCSUDPMEYTNCTTTMFAGVDWBO",
"output": "VGHUNFOXKETUYMZDJNGTAOIOANYXSGYNFOGOFFLDAWEUKYFOZXCJTCAFXZYLQZERYZLRSQXYQGAPCSUDPMEYTNCTTTMFAGVDWBO"
},
{
"input": "EDUANHCQDOCJHFONTTSWBUJSTTIXBIXMAIUFSGFLXWAYIURTVAVZPYQDLAWIGCLMPHXCEFCJPFAAHXVNGQUFNHADAIUAZIDMHDd",
"output": "EDUANHCQDOCJHFONTTSWBUJSTTIXBIXMAIUFSGFLXWAYIURTVAVZPYQDLAWIGCLMPHXCEFCJPFAAHXVNGQUFNHADAIUAZIDMHDD"
},
{
"input": "EBWOVSNLYTWWXrnovgugogtjahnmatomubebyxakas",
"output": "ebwovsnlytwwxrnovgugogtjahnmatomubebyxakas"
},
{
"input": "AORNNDKTRLRVGDPXJKXFTPFpopuzrlqumrxssskvbm",
"output": "AORNNDKTRLRVGDPXJKXFTPFPOPUZRLQUMRXSSSKVBM"
},
{
"input": "DBBZJCWQEVGYHEEUHJAWRHBDPsnplijbwtdibqrrheebfxqcvn",
"output": "dbbzjcwqevgyheeuhjawrhbdpsnplijbwtdibqrrheebfxqcvn"
},
{
"input": "t",
"output": "t"
},
{
"input": "N",
"output": "N"
},
{
"input": "kv",
"output": "kv"
},
{
"input": "Ur",
"output": "ur"
},
{
"input": "CN",
"output": "CN"
}
] | 1,617,198,805 | 2,147,483,647 | Python 3 | OK | TESTS | 30 | 154 | 0 | #word codeforces
# if lower > upper = use lower
# if equal use lower
n =input()
up = 0 ; low = 0
for i in n:
if i.isupper():
up+=1
else:
low+=1
if up > low:
print(n.upper())
elif up < low:
print(n.lower())
else: print(n.lower())
| Title: Word
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word.
Input Specification:
The first line contains a word *s* — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100.
Output Specification:
Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one.
Demo Input:
['HoUse\n', 'ViP\n', 'maTRIx\n']
Demo Output:
['house\n', 'VIP\n', 'matrix\n']
Note:
none | ```python
#word codeforces
# if lower > upper = use lower
# if equal use lower
n =input()
up = 0 ; low = 0
for i in n:
if i.isupper():
up+=1
else:
low+=1
if up > low:
print(n.upper())
elif up < low:
print(n.lower())
else: print(n.lower())
``` | 3.9615 |
317 | A | Perfect Pair | PROGRAMMING | 1,600 | [
"brute force"
] | null | null | Let us call a pair of integer numbers *m*-perfect, if at least one number in the pair is greater than or equal to *m*. Thus, the pairs (3, 3) and (0, 2) are 2-perfect while the pair (-1, 1) is not.
Two integers *x*, *y* are written on the blackboard. It is allowed to erase one of them and replace it with the sum of the numbers, (*x*<=+<=*y*).
What is the minimum number of such operations one has to perform in order to make the given pair of integers *m*-perfect? | Single line of the input contains three integers *x*, *y* and *m* (<=-<=1018<=≤<=*x*, *y*, *m*<=≤<=1018).
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preffered to use the cin, cout streams or the %I64d specifier. | Print the minimum number of operations or "-1" (without quotes), if it is impossible to transform the given pair to the *m*-perfect one. | [
"1 2 5\n",
"-1 4 15\n",
"0 -1 5\n"
] | [
"2\n",
"4\n",
"-1\n"
] | In the first sample the following sequence of operations is suitable: (1, 2) <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> (3, 2) <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> (5, 2).
In the second sample: (-1, 4) <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> (3, 4) <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> (7, 4) <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> (11, 4) <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> (15, 4).
Finally, in the third sample *x*, *y* cannot be made positive, hence there is no proper sequence of operations. | 500 | [
{
"input": "1 2 5",
"output": "2"
},
{
"input": "-1 4 15",
"output": "4"
},
{
"input": "0 -1 5",
"output": "-1"
},
{
"input": "0 1 8",
"output": "5"
},
{
"input": "-134 -345 -134",
"output": "0"
},
{
"input": "-134 -345 -133",
"output": "-1"
},
{
"input": "999999999 -1000000000 1000000000",
"output": "3"
},
{
"input": "0 0 0",
"output": "0"
},
{
"input": "0 0 1",
"output": "-1"
},
{
"input": "-1000000000000000000 1 1000000000000000000",
"output": "1000000000000000087"
},
{
"input": "-3 26 -1",
"output": "0"
},
{
"input": "-25 4 -8",
"output": "0"
},
{
"input": "12 30 -8",
"output": "0"
},
{
"input": "-12 17 3",
"output": "0"
},
{
"input": "4 -11 28",
"output": "8"
},
{
"input": "38 174 957147453",
"output": "33"
},
{
"input": "154 30 763391461",
"output": "33"
},
{
"input": "3 193 648520002",
"output": "32"
},
{
"input": "139 82 923851170",
"output": "33"
},
{
"input": "171 185 534908267",
"output": "31"
},
{
"input": "-868993006 -389009632 -766659629",
"output": "0"
},
{
"input": "-429468031 69656014 39767881",
"output": "0"
},
{
"input": "185212727 871828769 828159476",
"output": "0"
},
{
"input": "140457446 731228634 -75123935",
"output": "0"
},
{
"input": "223567628 -731033737 352248633",
"output": "5"
},
{
"input": "-187818082 -372699371 -301077133",
"output": "0"
},
{
"input": "-552043292 -693546115 415527936",
"output": "-1"
},
{
"input": "-29007970 -344600631 62206369",
"output": "-1"
},
{
"input": "101292660 -305927896 648565756",
"output": "8"
},
{
"input": "702748103 -278432024 995244274",
"output": "2"
},
{
"input": "0 0 -1",
"output": "0"
},
{
"input": "0 0 0",
"output": "0"
},
{
"input": "0 0 1000000000",
"output": "-1"
},
{
"input": "0 0 1",
"output": "-1"
},
{
"input": "1 -999999999 239239239",
"output": "1000000040"
},
{
"input": "-1 -1 0",
"output": "-1"
},
{
"input": "-1 0 0",
"output": "0"
},
{
"input": "-1 0 1",
"output": "-1"
},
{
"input": "-1000000000 -1000000000 -1000000000",
"output": "0"
},
{
"input": "-1000000000 -1000000000 1000000000",
"output": "-1"
},
{
"input": "999999999 999999999 1000000000",
"output": "1"
},
{
"input": "-1 1 609276626",
"output": "44"
},
{
"input": "-1 2 926098525",
"output": "43"
},
{
"input": "0 0 698431198",
"output": "-1"
},
{
"input": "0 -3 702455284",
"output": "-1"
},
{
"input": "0 0 648749804",
"output": "-1"
},
{
"input": "-1 0 861856808",
"output": "-1"
},
{
"input": "2 2 -213745374",
"output": "0"
},
{
"input": "-1 1 21065410",
"output": "37"
},
{
"input": "3 -3 607820420",
"output": "42"
},
{
"input": "36 -58 428518679",
"output": "37"
},
{
"input": "8 55 931239629",
"output": "36"
},
{
"input": "-99 -91 523666385",
"output": "-1"
},
{
"input": "-56 -11 631827324",
"output": "-1"
},
{
"input": "-4 6 447215792",
"output": "39"
},
{
"input": "-47 -31 -669106932",
"output": "0"
},
{
"input": "12 -76 -999189104",
"output": "0"
},
{
"input": "39 15 -960040787",
"output": "0"
},
{
"input": "-96 26 -210129802",
"output": "0"
},
{
"input": "93 86 -850132431",
"output": "0"
},
{
"input": "1 -1000000000000000000 1000000000000000000",
"output": "1000000000000000087"
},
{
"input": "-2348349823443 234234545453 1000000000000000000",
"output": "43"
},
{
"input": "0 1 679891637638612258",
"output": "86"
},
{
"input": "-1000000000000000000 -1000000000000000000 -1000000000000000000",
"output": "0"
},
{
"input": "-1000000000000000000 -1000000000000000000 -999999999999999999",
"output": "-1"
},
{
"input": "-100000000000000 1 233",
"output": "100000000000012"
},
{
"input": "-1000000000000 2 1000000000000",
"output": "500000000057"
},
{
"input": "-1000000000000 3 1000000000",
"output": "333333333375"
},
{
"input": "10 -10 0",
"output": "0"
},
{
"input": "-1000000000000000 2 444",
"output": "500000000000012"
}
] | 1,489,573,620 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 3 | 122 | 4,608,000 | x,y,m = input().strip().split(' ')
x = int(x)
y = int(y)
m = int(m)
val = (m-x)/y
if(val<0): print (-1)
elif((m-x)%y!=0): print (-1)
else: print(int(val))
| Title: Perfect Pair
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Let us call a pair of integer numbers *m*-perfect, if at least one number in the pair is greater than or equal to *m*. Thus, the pairs (3, 3) and (0, 2) are 2-perfect while the pair (-1, 1) is not.
Two integers *x*, *y* are written on the blackboard. It is allowed to erase one of them and replace it with the sum of the numbers, (*x*<=+<=*y*).
What is the minimum number of such operations one has to perform in order to make the given pair of integers *m*-perfect?
Input Specification:
Single line of the input contains three integers *x*, *y* and *m* (<=-<=1018<=≤<=*x*, *y*, *m*<=≤<=1018).
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preffered to use the cin, cout streams or the %I64d specifier.
Output Specification:
Print the minimum number of operations or "-1" (without quotes), if it is impossible to transform the given pair to the *m*-perfect one.
Demo Input:
['1 2 5\n', '-1 4 15\n', '0 -1 5\n']
Demo Output:
['2\n', '4\n', '-1\n']
Note:
In the first sample the following sequence of operations is suitable: (1, 2) <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> (3, 2) <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> (5, 2).
In the second sample: (-1, 4) <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> (3, 4) <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> (7, 4) <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> (11, 4) <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> (15, 4).
Finally, in the third sample *x*, *y* cannot be made positive, hence there is no proper sequence of operations. | ```python
x,y,m = input().strip().split(' ')
x = int(x)
y = int(y)
m = int(m)
val = (m-x)/y
if(val<0): print (-1)
elif((m-x)%y!=0): print (-1)
else: print(int(val))
``` | 0 |
|
404 | A | Valera and X | PROGRAMMING | 1,000 | [
"implementation"
] | null | null | Valera is a little boy. Yesterday he got a huge Math hometask at school, so Valera didn't have enough time to properly learn the English alphabet for his English lesson. Unfortunately, the English teacher decided to have a test on alphabet today. At the test Valera got a square piece of squared paper. The length of the side equals *n* squares (*n* is an odd number) and each unit square contains some small letter of the English alphabet.
Valera needs to know if the letters written on the square piece of paper form letter "X". Valera's teacher thinks that the letters on the piece of paper form an "X", if:
- on both diagonals of the square paper all letters are the same; - all other squares of the paper (they are not on the diagonals) contain the same letter that is different from the letters on the diagonals.
Help Valera, write the program that completes the described task for him. | The first line contains integer *n* (3<=≤<=*n*<=<<=300; *n* is odd). Each of the next *n* lines contains *n* small English letters — the description of Valera's paper. | Print string "YES", if the letters on the paper form letter "X". Otherwise, print string "NO". Print the strings without quotes. | [
"5\nxooox\noxoxo\nsoxoo\noxoxo\nxooox\n",
"3\nwsw\nsws\nwsw\n",
"3\nxpx\npxp\nxpe\n"
] | [
"NO\n",
"YES\n",
"NO\n"
] | none | 500 | [
{
"input": "5\nxooox\noxoxo\nsoxoo\noxoxo\nxooox",
"output": "NO"
},
{
"input": "3\nwsw\nsws\nwsw",
"output": "YES"
},
{
"input": "3\nxpx\npxp\nxpe",
"output": "NO"
},
{
"input": "5\nliiil\nilili\niilii\nilili\nliiil",
"output": "YES"
},
{
"input": "7\nbwccccb\nckcccbj\nccbcbcc\ncccbccc\nccbcbcc\ncbcccbc\nbccccdt",
"output": "NO"
},
{
"input": "13\nsooooooooooos\nosoooooooooso\noosooooooosoo\nooosooooosooo\noooosooosoooo\nooooososooooo\noooooosoooooo\nooooososooooo\noooosooosoooo\nooosooooosooo\noosooooooosoo\nosoooooooooso\nsooooooooooos",
"output": "YES"
},
{
"input": "3\naaa\naaa\naaa",
"output": "NO"
},
{
"input": "3\naca\noec\nzba",
"output": "NO"
},
{
"input": "15\nrxeeeeeeeeeeeer\nereeeeeeeeeeere\needeeeeeeeeeoee\neeereeeeeeeewee\neeeereeeeebeeee\nqeeeereeejedyee\neeeeeerereeeeee\neeeeeeereeeeeee\neeeeeerereeeeze\neeeeereeereeeee\neeeereeeeegeeee\neeereeeeeeereee\neereeeeeeqeeved\ncreeeeeeceeeere\nreeerneeeeeeeer",
"output": "NO"
},
{
"input": "5\nxxxxx\nxxxxx\nxxxxx\nxxxxx\nxxxxx",
"output": "NO"
},
{
"input": "5\nxxxxx\nxxxxx\nxoxxx\nxxxxx\nxxxxx",
"output": "NO"
},
{
"input": "5\noxxxo\nxoxox\nxxxxx\nxoxox\noxxxo",
"output": "NO"
},
{
"input": "5\noxxxo\nxoxox\nxxoox\nxoxox\noxxxo",
"output": "NO"
},
{
"input": "5\noxxxo\nxoxox\nxxaxx\nxoxox\noxxxo",
"output": "NO"
},
{
"input": "5\noxxxo\nxoxox\noxoxx\nxoxox\noxxxo",
"output": "NO"
},
{
"input": "3\nxxx\naxa\nxax",
"output": "NO"
},
{
"input": "3\nxax\naxx\nxax",
"output": "NO"
},
{
"input": "3\nxax\naxa\nxxx",
"output": "NO"
},
{
"input": "3\nxax\nxxa\nxax",
"output": "NO"
},
{
"input": "3\nxax\naaa\nxax",
"output": "NO"
},
{
"input": "3\naax\naxa\nxax",
"output": "NO"
},
{
"input": "3\nxaa\naxa\nxax",
"output": "NO"
},
{
"input": "3\nxax\naxa\naax",
"output": "NO"
},
{
"input": "3\nxax\naxa\nxaa",
"output": "NO"
},
{
"input": "3\nxfx\naxa\nxax",
"output": "NO"
},
{
"input": "3\nxax\nafa\nxax",
"output": "NO"
},
{
"input": "3\nxax\naxa\nxaf",
"output": "NO"
},
{
"input": "3\nxox\nxxx\nxxx",
"output": "NO"
},
{
"input": "3\naxa\naax\nxxa",
"output": "NO"
},
{
"input": "3\nxox\noxx\nxox",
"output": "NO"
},
{
"input": "3\nxox\nooo\nxox",
"output": "NO"
},
{
"input": "3\naaa\naab\nbbb",
"output": "NO"
},
{
"input": "3\nxxx\nsxs\nxsx",
"output": "NO"
},
{
"input": "5\nabbba\nbabab\nbbbbb\nbaaab\nabbba",
"output": "NO"
},
{
"input": "5\nabaaa\nbbbbb\nbbabb\nbabab\nabbba",
"output": "NO"
},
{
"input": "5\nxoxox\noxoxo\nooxoo\noxoxo\nxooox",
"output": "NO"
},
{
"input": "3\nxox\noxx\nxxx",
"output": "NO"
},
{
"input": "5\nxoooo\noxooo\nooxoo\noooxo\noooox",
"output": "NO"
},
{
"input": "5\nxoooo\noxoxx\nooxoo\noxoxo\noxoox",
"output": "NO"
},
{
"input": "3\naaa\nbab\naba",
"output": "NO"
}
] | 1,599,312,524 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 6 | 93 | 0 | n,arr1,arr2=int(input()),[],[]
for i in range(n):
s=input()
for j in range(len(s)):
if i==j or j==n-1-i:arr1.append(s[j])
else:arr2.append(s[j])
if len(set(arr1))==1==len(set(arr2)):print('YES')
else:print('NO') | Title: Valera and X
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Valera is a little boy. Yesterday he got a huge Math hometask at school, so Valera didn't have enough time to properly learn the English alphabet for his English lesson. Unfortunately, the English teacher decided to have a test on alphabet today. At the test Valera got a square piece of squared paper. The length of the side equals *n* squares (*n* is an odd number) and each unit square contains some small letter of the English alphabet.
Valera needs to know if the letters written on the square piece of paper form letter "X". Valera's teacher thinks that the letters on the piece of paper form an "X", if:
- on both diagonals of the square paper all letters are the same; - all other squares of the paper (they are not on the diagonals) contain the same letter that is different from the letters on the diagonals.
Help Valera, write the program that completes the described task for him.
Input Specification:
The first line contains integer *n* (3<=≤<=*n*<=<<=300; *n* is odd). Each of the next *n* lines contains *n* small English letters — the description of Valera's paper.
Output Specification:
Print string "YES", if the letters on the paper form letter "X". Otherwise, print string "NO". Print the strings without quotes.
Demo Input:
['5\nxooox\noxoxo\nsoxoo\noxoxo\nxooox\n', '3\nwsw\nsws\nwsw\n', '3\nxpx\npxp\nxpe\n']
Demo Output:
['NO\n', 'YES\n', 'NO\n']
Note:
none | ```python
n,arr1,arr2=int(input()),[],[]
for i in range(n):
s=input()
for j in range(len(s)):
if i==j or j==n-1-i:arr1.append(s[j])
else:arr2.append(s[j])
if len(set(arr1))==1==len(set(arr2)):print('YES')
else:print('NO')
``` | 0 |
|
887 | A | Div. 64 | PROGRAMMING | 1,000 | [
"implementation"
] | null | null | Top-model Izabella participates in the competition. She wants to impress judges and show her mathematical skills.
Her problem is following: for given string, consisting of only 0 and 1, tell if it's possible to remove some digits in such a way, that remaining number is a representation of some positive integer, divisible by 64, in the binary numerical system. | In the only line given a non-empty binary string *s* with length up to 100. | Print «yes» (without quotes) if it's possible to remove digits required way and «no» otherwise. | [
"100010001\n",
"100\n"
] | [
"yes",
"no"
] | In the first test case, you can get string 1 000 000 after removing two ones which is a representation of number 64 in the binary numerical system.
You can read more about binary numeral system representation here: [https://en.wikipedia.org/wiki/Binary_system](https://en.wikipedia.org/wiki/Binary_system) | 500 | [
{
"input": "100010001",
"output": "yes"
},
{
"input": "100",
"output": "no"
},
{
"input": "0000001000000",
"output": "yes"
},
{
"input": "1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111",
"output": "no"
},
{
"input": "1111111111111111111111111111111111111111111111111111111111111111111111110111111111111111111111111111",
"output": "no"
},
{
"input": "0111111101111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111",
"output": "no"
},
{
"input": "1111011111111111111111111111110111110111111111111111111111011111111111111110111111111111111111111111",
"output": "no"
},
{
"input": "1111111111101111111111111111111111111011111111111111111111111101111011111101111111111101111111111111",
"output": "yes"
},
{
"input": "0110111111111111111111011111111110110111110111111111111111111111111111111111111110111111111111111111",
"output": "yes"
},
{
"input": "1100110001111011001101101000001110111110011110111110010100011000100101000010010111100000010001001101",
"output": "yes"
},
{
"input": "000000",
"output": "no"
},
{
"input": "0001000",
"output": "no"
},
{
"input": "0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000",
"output": "no"
},
{
"input": "1000000",
"output": "yes"
},
{
"input": "0",
"output": "no"
},
{
"input": "1",
"output": "no"
},
{
"input": "10000000000",
"output": "yes"
},
{
"input": "0000000000",
"output": "no"
},
{
"input": "0010000",
"output": "no"
},
{
"input": "000000011",
"output": "no"
},
{
"input": "000000000",
"output": "no"
},
{
"input": "00000000",
"output": "no"
},
{
"input": "000000000011",
"output": "no"
},
{
"input": "0000000",
"output": "no"
},
{
"input": "00000000011",
"output": "no"
},
{
"input": "000000001",
"output": "no"
},
{
"input": "000000000000000000000000000",
"output": "no"
},
{
"input": "0000001",
"output": "no"
},
{
"input": "00000001",
"output": "no"
},
{
"input": "00000000100",
"output": "no"
},
{
"input": "00000000000000000000",
"output": "no"
},
{
"input": "0000000000000000000",
"output": "no"
},
{
"input": "00001000",
"output": "no"
},
{
"input": "0000000000010",
"output": "no"
},
{
"input": "000000000010",
"output": "no"
},
{
"input": "000000000000010",
"output": "no"
},
{
"input": "0100000",
"output": "no"
},
{
"input": "00010000",
"output": "no"
},
{
"input": "00000000000000000",
"output": "no"
},
{
"input": "00000000000",
"output": "no"
},
{
"input": "000001000",
"output": "no"
},
{
"input": "000000000000",
"output": "no"
},
{
"input": "100000000000000",
"output": "yes"
},
{
"input": "000010000",
"output": "no"
},
{
"input": "00000100",
"output": "no"
},
{
"input": "0001100000",
"output": "no"
},
{
"input": "000000000000000000000000001",
"output": "no"
},
{
"input": "000000100",
"output": "no"
},
{
"input": "0000000000001111111111",
"output": "no"
},
{
"input": "00000010",
"output": "no"
},
{
"input": "0001110000",
"output": "no"
},
{
"input": "0000000000000000000000",
"output": "no"
},
{
"input": "000000010010",
"output": "no"
},
{
"input": "0000100",
"output": "no"
},
{
"input": "0000000001",
"output": "no"
},
{
"input": "000000111",
"output": "no"
},
{
"input": "0000000000000",
"output": "no"
},
{
"input": "000000000000000000",
"output": "no"
},
{
"input": "0000000000000000000000000",
"output": "no"
},
{
"input": "000000000000000",
"output": "no"
},
{
"input": "0010000000000100",
"output": "yes"
},
{
"input": "0000001000",
"output": "no"
},
{
"input": "00000000000000000001",
"output": "no"
},
{
"input": "100000000",
"output": "yes"
},
{
"input": "000000000001",
"output": "no"
},
{
"input": "0000011001",
"output": "no"
},
{
"input": "000",
"output": "no"
},
{
"input": "000000000000000000000",
"output": "no"
},
{
"input": "0000000000011",
"output": "no"
},
{
"input": "0000000000000000",
"output": "no"
},
{
"input": "00000000000000001",
"output": "no"
},
{
"input": "00000000000000",
"output": "no"
},
{
"input": "0000000000000000010",
"output": "no"
},
{
"input": "00000000000000000000000000000000000000000000000000000000",
"output": "no"
},
{
"input": "000011000",
"output": "no"
},
{
"input": "00000011",
"output": "no"
},
{
"input": "0000000000001100",
"output": "no"
},
{
"input": "00000",
"output": "no"
},
{
"input": "000000000000000000000000000111111111111111",
"output": "no"
},
{
"input": "000000010",
"output": "no"
},
{
"input": "00000000111",
"output": "no"
},
{
"input": "000000000000001",
"output": "no"
},
{
"input": "0000000000000011111111111111111",
"output": "no"
},
{
"input": "0000000010",
"output": "no"
},
{
"input": "0000000000000000000000000000000000000000000000000",
"output": "no"
},
{
"input": "00000000010",
"output": "no"
},
{
"input": "101000000000",
"output": "yes"
},
{
"input": "00100000",
"output": "no"
},
{
"input": "00000000000001",
"output": "no"
},
{
"input": "0000000000100",
"output": "no"
},
{
"input": "0000",
"output": "no"
},
{
"input": "00000000000111",
"output": "no"
},
{
"input": "0000000000000011",
"output": "no"
},
{
"input": "0000000000000000000000000000000000000000",
"output": "no"
},
{
"input": "0000000000000010",
"output": "no"
},
{
"input": "0010101010",
"output": "no"
},
{
"input": "0000000000000001",
"output": "no"
},
{
"input": "1010101",
"output": "no"
}
] | 1,509,810,005 | 905 | Python 3 | WRONG_ANSWER | TESTS | 0 | 31 | 0 | __author__ = 'Matias'
def onkoBinaari(inp):
valid = True
count = 0
for a in inp:
if a == '0' or a == '1':
count += 1
else:
valid = False
break
return [valid, count]
def onko1ja6nollaa(inp):
onefound = 0
countToSix = 0
for a in inp:
if a == "1":
onefound = 1
if a == "0" and onefound == 1:
countToSix += 1
if countToSix > 5:
return True
break
return False
inp = input("Input: ")
valid, count = onkoBinaari(inp)
if valid:
if count > 6:
ans = onko1ja6nollaa(inp)
if ans:
print("yes")
else:
print("no")
else:
print("no")
else:
print("no") | Title: Div. 64
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Top-model Izabella participates in the competition. She wants to impress judges and show her mathematical skills.
Her problem is following: for given string, consisting of only 0 and 1, tell if it's possible to remove some digits in such a way, that remaining number is a representation of some positive integer, divisible by 64, in the binary numerical system.
Input Specification:
In the only line given a non-empty binary string *s* with length up to 100.
Output Specification:
Print «yes» (without quotes) if it's possible to remove digits required way and «no» otherwise.
Demo Input:
['100010001\n', '100\n']
Demo Output:
['yes', 'no']
Note:
In the first test case, you can get string 1 000 000 after removing two ones which is a representation of number 64 in the binary numerical system.
You can read more about binary numeral system representation here: [https://en.wikipedia.org/wiki/Binary_system](https://en.wikipedia.org/wiki/Binary_system) | ```python
__author__ = 'Matias'
def onkoBinaari(inp):
valid = True
count = 0
for a in inp:
if a == '0' or a == '1':
count += 1
else:
valid = False
break
return [valid, count]
def onko1ja6nollaa(inp):
onefound = 0
countToSix = 0
for a in inp:
if a == "1":
onefound = 1
if a == "0" and onefound == 1:
countToSix += 1
if countToSix > 5:
return True
break
return False
inp = input("Input: ")
valid, count = onkoBinaari(inp)
if valid:
if count > 6:
ans = onko1ja6nollaa(inp)
if ans:
print("yes")
else:
print("no")
else:
print("no")
else:
print("no")
``` | 0 |
|
231 | A | Team | PROGRAMMING | 800 | [
"brute force",
"greedy"
] | null | null | One day three best friends Petya, Vasya and Tonya decided to form a team and take part in programming contests. Participants are usually offered several problems during programming contests. Long before the start the friends decided that they will implement a problem if at least two of them are sure about the solution. Otherwise, the friends won't write the problem's solution.
This contest offers *n* problems to the participants. For each problem we know, which friend is sure about the solution. Help the friends find the number of problems for which they will write a solution. | The first input line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of problems in the contest. Then *n* lines contain three integers each, each integer is either 0 or 1. If the first number in the line equals 1, then Petya is sure about the problem's solution, otherwise he isn't sure. The second number shows Vasya's view on the solution, the third number shows Tonya's view. The numbers on the lines are separated by spaces. | Print a single integer — the number of problems the friends will implement on the contest. | [
"3\n1 1 0\n1 1 1\n1 0 0\n",
"2\n1 0 0\n0 1 1\n"
] | [
"2\n",
"1\n"
] | In the first sample Petya and Vasya are sure that they know how to solve the first problem and all three of them know how to solve the second problem. That means that they will write solutions for these problems. Only Petya is sure about the solution for the third problem, but that isn't enough, so the friends won't take it.
In the second sample the friends will only implement the second problem, as Vasya and Tonya are sure about the solution. | 500 | [
{
"input": "3\n1 1 0\n1 1 1\n1 0 0",
"output": "2"
},
{
"input": "2\n1 0 0\n0 1 1",
"output": "1"
},
{
"input": "1\n1 0 0",
"output": "0"
},
{
"input": "2\n1 0 0\n1 1 1",
"output": "1"
},
{
"input": "5\n1 0 0\n0 1 0\n1 1 1\n0 0 1\n0 0 0",
"output": "1"
},
{
"input": "10\n0 1 0\n0 1 0\n1 1 0\n1 0 0\n0 0 1\n0 1 1\n1 1 1\n1 1 0\n0 0 0\n0 0 0",
"output": "4"
},
{
"input": "15\n0 1 0\n1 0 0\n1 1 0\n1 1 1\n0 1 0\n0 0 1\n1 0 1\n1 0 1\n1 0 1\n0 0 0\n1 1 1\n1 1 0\n0 1 1\n1 1 0\n1 1 1",
"output": "10"
},
{
"input": "50\n0 0 0\n0 1 1\n1 1 1\n0 1 0\n1 0 1\n1 1 1\n0 0 1\n1 0 0\n1 1 0\n1 0 1\n0 1 0\n0 0 1\n1 1 0\n0 1 0\n1 1 0\n0 0 0\n1 1 1\n1 0 1\n0 0 1\n1 1 0\n1 1 1\n0 1 1\n1 1 0\n0 0 0\n0 0 0\n1 1 1\n0 0 0\n1 1 1\n0 1 1\n0 0 1\n0 0 0\n0 0 0\n1 1 0\n1 1 0\n1 0 1\n1 0 0\n1 0 1\n1 0 1\n0 1 1\n1 1 0\n1 1 0\n0 1 0\n1 0 1\n0 0 0\n0 0 0\n0 0 0\n0 0 1\n1 1 1\n0 1 1\n1 0 1",
"output": "29"
},
{
"input": "1\n1 1 1",
"output": "1"
},
{
"input": "8\n0 0 0\n0 0 1\n0 0 0\n0 1 1\n1 0 0\n1 0 1\n1 1 0\n1 1 1",
"output": "4"
},
{
"input": "16\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1",
"output": "16"
}
] | 1,696,587,043 | 2,147,483,647 | Python 3 | OK | TESTS | 21 | 92 | 102,400 | n=int(input())
L=[]
for i in range(n):
ss=input()
W=ss.split()
L.append(W)
maincount=0
for i in L:
count=0
for j in i:
if int(j)==1:
count+=1
if count>=2:
maincount+=1
print(maincount) | Title: Team
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day three best friends Petya, Vasya and Tonya decided to form a team and take part in programming contests. Participants are usually offered several problems during programming contests. Long before the start the friends decided that they will implement a problem if at least two of them are sure about the solution. Otherwise, the friends won't write the problem's solution.
This contest offers *n* problems to the participants. For each problem we know, which friend is sure about the solution. Help the friends find the number of problems for which they will write a solution.
Input Specification:
The first input line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of problems in the contest. Then *n* lines contain three integers each, each integer is either 0 or 1. If the first number in the line equals 1, then Petya is sure about the problem's solution, otherwise he isn't sure. The second number shows Vasya's view on the solution, the third number shows Tonya's view. The numbers on the lines are separated by spaces.
Output Specification:
Print a single integer — the number of problems the friends will implement on the contest.
Demo Input:
['3\n1 1 0\n1 1 1\n1 0 0\n', '2\n1 0 0\n0 1 1\n']
Demo Output:
['2\n', '1\n']
Note:
In the first sample Petya and Vasya are sure that they know how to solve the first problem and all three of them know how to solve the second problem. That means that they will write solutions for these problems. Only Petya is sure about the solution for the third problem, but that isn't enough, so the friends won't take it.
In the second sample the friends will only implement the second problem, as Vasya and Tonya are sure about the solution. | ```python
n=int(input())
L=[]
for i in range(n):
ss=input()
W=ss.split()
L.append(W)
maincount=0
for i in L:
count=0
for j in i:
if int(j)==1:
count+=1
if count>=2:
maincount+=1
print(maincount)
``` | 3 |
|
964 | A | Splits | PROGRAMMING | 800 | [
"math"
] | null | null | Let's define a split of $n$ as a nonincreasing sequence of positive integers, the sum of which is $n$.
For example, the following sequences are splits of $8$: $[4, 4]$, $[3, 3, 2]$, $[2, 2, 1, 1, 1, 1]$, $[5, 2, 1]$.
The following sequences aren't splits of $8$: $[1, 7]$, $[5, 4]$, $[11, -3]$, $[1, 1, 4, 1, 1]$.
The weight of a split is the number of elements in the split that are equal to the first element. For example, the weight of the split $[1, 1, 1, 1, 1]$ is $5$, the weight of the split $[5, 5, 3, 3, 3]$ is $2$ and the weight of the split $[9]$ equals $1$.
For a given $n$, find out the number of different weights of its splits. | The first line contains one integer $n$ ($1 \leq n \leq 10^9$). | Output one integer — the answer to the problem. | [
"7\n",
"8\n",
"9\n"
] | [
"4\n",
"5\n",
"5\n"
] | In the first sample, there are following possible weights of splits of $7$:
Weight 1: [$\textbf 7$]
Weight 2: [$\textbf 3$, $\textbf 3$, 1]
Weight 3: [$\textbf 2$, $\textbf 2$, $\textbf 2$, 1]
Weight 7: [$\textbf 1$, $\textbf 1$, $\textbf 1$, $\textbf 1$, $\textbf 1$, $\textbf 1$, $\textbf 1$] | 500 | [
{
"input": "7",
"output": "4"
},
{
"input": "8",
"output": "5"
},
{
"input": "9",
"output": "5"
},
{
"input": "1",
"output": "1"
},
{
"input": "286",
"output": "144"
},
{
"input": "48",
"output": "25"
},
{
"input": "941",
"output": "471"
},
{
"input": "45154",
"output": "22578"
},
{
"input": "60324",
"output": "30163"
},
{
"input": "91840",
"output": "45921"
},
{
"input": "41909",
"output": "20955"
},
{
"input": "58288",
"output": "29145"
},
{
"input": "91641",
"output": "45821"
},
{
"input": "62258",
"output": "31130"
},
{
"input": "79811",
"output": "39906"
},
{
"input": "88740",
"output": "44371"
},
{
"input": "12351",
"output": "6176"
},
{
"input": "1960",
"output": "981"
},
{
"input": "29239",
"output": "14620"
},
{
"input": "85801",
"output": "42901"
},
{
"input": "43255",
"output": "21628"
},
{
"input": "13439",
"output": "6720"
},
{
"input": "35668",
"output": "17835"
},
{
"input": "19122",
"output": "9562"
},
{
"input": "60169",
"output": "30085"
},
{
"input": "50588",
"output": "25295"
},
{
"input": "2467",
"output": "1234"
},
{
"input": "39315",
"output": "19658"
},
{
"input": "29950",
"output": "14976"
},
{
"input": "17286",
"output": "8644"
},
{
"input": "7359066",
"output": "3679534"
},
{
"input": "1016391",
"output": "508196"
},
{
"input": "7928871",
"output": "3964436"
},
{
"input": "3968891",
"output": "1984446"
},
{
"input": "2636452",
"output": "1318227"
},
{
"input": "5076901",
"output": "2538451"
},
{
"input": "9870265",
"output": "4935133"
},
{
"input": "2453786",
"output": "1226894"
},
{
"input": "7263670",
"output": "3631836"
},
{
"input": "1890845",
"output": "945423"
},
{
"input": "574128507",
"output": "287064254"
},
{
"input": "648476655",
"output": "324238328"
},
{
"input": "97349542",
"output": "48674772"
},
{
"input": "716489761",
"output": "358244881"
},
{
"input": "858771038",
"output": "429385520"
},
{
"input": "520778784",
"output": "260389393"
},
{
"input": "439004204",
"output": "219502103"
},
{
"input": "589992198",
"output": "294996100"
},
{
"input": "371106544",
"output": "185553273"
},
{
"input": "894241590",
"output": "447120796"
},
{
"input": "123957268",
"output": "61978635"
},
{
"input": "234149297",
"output": "117074649"
},
{
"input": "789954052",
"output": "394977027"
},
{
"input": "667978920",
"output": "333989461"
},
{
"input": "154647261",
"output": "77323631"
},
{
"input": "751453521",
"output": "375726761"
},
{
"input": "848862308",
"output": "424431155"
},
{
"input": "323926781",
"output": "161963391"
},
{
"input": "576768825",
"output": "288384413"
},
{
"input": "31293802",
"output": "15646902"
},
{
"input": "2",
"output": "2"
},
{
"input": "1000000000",
"output": "500000001"
},
{
"input": "3",
"output": "2"
}
] | 1,524,242,036 | 2,147,483,647 | Python 3 | OK | TESTS | 63 | 93 | 7,065,600 | # from dust i have come, dust i will be
n=int(input())
print(n//2+1) | Title: Splits
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Let's define a split of $n$ as a nonincreasing sequence of positive integers, the sum of which is $n$.
For example, the following sequences are splits of $8$: $[4, 4]$, $[3, 3, 2]$, $[2, 2, 1, 1, 1, 1]$, $[5, 2, 1]$.
The following sequences aren't splits of $8$: $[1, 7]$, $[5, 4]$, $[11, -3]$, $[1, 1, 4, 1, 1]$.
The weight of a split is the number of elements in the split that are equal to the first element. For example, the weight of the split $[1, 1, 1, 1, 1]$ is $5$, the weight of the split $[5, 5, 3, 3, 3]$ is $2$ and the weight of the split $[9]$ equals $1$.
For a given $n$, find out the number of different weights of its splits.
Input Specification:
The first line contains one integer $n$ ($1 \leq n \leq 10^9$).
Output Specification:
Output one integer — the answer to the problem.
Demo Input:
['7\n', '8\n', '9\n']
Demo Output:
['4\n', '5\n', '5\n']
Note:
In the first sample, there are following possible weights of splits of $7$:
Weight 1: [$\textbf 7$]
Weight 2: [$\textbf 3$, $\textbf 3$, 1]
Weight 3: [$\textbf 2$, $\textbf 2$, $\textbf 2$, 1]
Weight 7: [$\textbf 1$, $\textbf 1$, $\textbf 1$, $\textbf 1$, $\textbf 1$, $\textbf 1$, $\textbf 1$] | ```python
# from dust i have come, dust i will be
n=int(input())
print(n//2+1)
``` | 3 |
|
0 | none | none | none | 0 | [
"none"
] | null | null | Squirrel Liss lived in a forest peacefully, but unexpected trouble happens. Stones fall from a mountain. Initially Squirrel Liss occupies an interval [0,<=1]. Next, *n* stones will fall and Liss will escape from the stones. The stones are numbered from 1 to *n* in order.
The stones always fall to the center of Liss's interval. When Liss occupies the interval [*k*<=-<=*d*,<=*k*<=+<=*d*] and a stone falls to *k*, she will escape to the left or to the right. If she escapes to the left, her new interval will be [*k*<=-<=*d*,<=*k*]. If she escapes to the right, her new interval will be [*k*,<=*k*<=+<=*d*].
You are given a string *s* of length *n*. If the *i*-th character of *s* is "l" or "r", when the *i*-th stone falls Liss will escape to the left or to the right, respectively. Find the sequence of stones' numbers from left to right after all the *n* stones falls. | The input consists of only one line. The only line contains the string *s* (1<=≤<=|*s*|<=≤<=106). Each character in *s* will be either "l" or "r". | Output *n* lines — on the *i*-th line you should print the *i*-th stone's number from the left. | [
"llrlr\n",
"rrlll\n",
"lrlrr\n"
] | [
"3\n5\n4\n2\n1\n",
"1\n2\n5\n4\n3\n",
"2\n4\n5\n3\n1\n"
] | In the first example, the positions of stones 1, 2, 3, 4, 5 will be <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/58fdb5684df807bfcb705a9da9ce175613362b7d.png" style="max-width: 100.0%;max-height: 100.0%;"/>, respectively. So you should print the sequence: 3, 5, 4, 2, 1. | 0 | [
{
"input": "llrlr",
"output": "3\n5\n4\n2\n1"
},
{
"input": "rrlll",
"output": "1\n2\n5\n4\n3"
},
{
"input": "lrlrr",
"output": "2\n4\n5\n3\n1"
},
{
"input": "lllrlrllrl",
"output": "4\n6\n9\n10\n8\n7\n5\n3\n2\n1"
},
{
"input": "llrlrrrlrr",
"output": "3\n5\n6\n7\n9\n10\n8\n4\n2\n1"
},
{
"input": "rlrrrllrrr",
"output": "1\n3\n4\n5\n8\n9\n10\n7\n6\n2"
},
{
"input": "lrrlrrllrrrrllllllrr",
"output": "2\n3\n5\n6\n9\n10\n11\n12\n19\n20\n18\n17\n16\n15\n14\n13\n8\n7\n4\n1"
},
{
"input": "rlrrrlrrrllrrllrlrll",
"output": "1\n3\n4\n5\n7\n8\n9\n12\n13\n16\n18\n20\n19\n17\n15\n14\n11\n10\n6\n2"
},
{
"input": "lllrrlrlrllrrrrrllrl",
"output": "4\n5\n7\n9\n12\n13\n14\n15\n16\n19\n20\n18\n17\n11\n10\n8\n6\n3\n2\n1"
},
{
"input": "rrrllrrrlllrlllrlrrr",
"output": "1\n2\n3\n6\n7\n8\n12\n16\n18\n19\n20\n17\n15\n14\n13\n11\n10\n9\n5\n4"
},
{
"input": "rrlllrrrlrrlrrrlllrlrlrrrlllrllrrllrllrrlrlrrllllrlrrrrlrlllrlrrrlrlrllrlrlrrlrrllrrrlrlrlllrrllllrl",
"output": "1\n2\n6\n7\n8\n10\n11\n13\n14\n15\n19\n21\n23\n24\n25\n29\n32\n33\n36\n39\n40\n42\n44\n45\n50\n52\n53\n54\n55\n57\n61\n63\n64\n65\n67\n69\n72\n74\n76\n77\n79\n80\n83\n84\n85\n87\n89\n93\n94\n99\n100\n98\n97\n96\n95\n92\n91\n90\n88\n86\n82\n81\n78\n75\n73\n71\n70\n68\n66\n62\n60\n59\n58\n56\n51\n49\n48\n47\n46\n43\n41\n38\n37\n35\n34\n31\n30\n28\n27\n26\n22\n20\n18\n17\n16\n12\n9\n5\n4\n3"
},
{
"input": "llrlrlllrrllrllllrlrrlrlrrllrlrlrrlrrrrrrlllrrlrrrrrlrrrlrlrlrrlllllrrrrllrrlrlrrrllllrlrrlrrlrlrrll",
"output": "3\n5\n9\n10\n13\n18\n20\n21\n23\n25\n26\n29\n31\n33\n34\n36\n37\n38\n39\n40\n41\n45\n46\n48\n49\n50\n51\n52\n54\n55\n56\n58\n60\n62\n63\n69\n70\n71\n72\n75\n76\n78\n80\n81\n82\n87\n89\n90\n92\n93\n95\n97\n98\n100\n99\n96\n94\n91\n88\n86\n85\n84\n83\n79\n77\n74\n73\n68\n67\n66\n65\n64\n61\n59\n57\n53\n47\n44\n43\n42\n35\n32\n30\n28\n27\n24\n22\n19\n17\n16\n15\n14\n12\n11\n8\n7\n6\n4\n2\n1"
},
{
"input": "llrrrrllrrlllrlrllrlrllllllrrrrrrrrllrrrrrrllrlrrrlllrrrrrrlllllllrrlrrllrrrllllrrlllrrrlrlrrlrlrllr",
"output": "3\n4\n5\n6\n9\n10\n14\n16\n19\n21\n28\n29\n30\n31\n32\n33\n34\n35\n38\n39\n40\n41\n42\n43\n46\n48\n49\n50\n54\n55\n56\n57\n58\n59\n67\n68\n70\n71\n74\n75\n76\n81\n82\n86\n87\n88\n90\n92\n93\n95\n97\n100\n99\n98\n96\n94\n91\n89\n85\n84\n83\n80\n79\n78\n77\n73\n72\n69\n66\n65\n64\n63\n62\n61\n60\n53\n52\n51\n47\n45\n44\n37\n36\n27\n26\n25\n24\n23\n22\n20\n18\n17\n15\n13\n12\n11\n8\n7\n2\n1"
},
{
"input": "lllllrllrrlllrrrllrrrrlrrlrllllrrrrrllrlrllllllrrlrllrlrllrlrrlrlrrlrrrlrrrrllrlrrrrrrrllrllrrlrllrl",
"output": "6\n9\n10\n14\n15\n16\n19\n20\n21\n22\n24\n25\n27\n32\n33\n34\n35\n36\n39\n41\n48\n49\n51\n54\n56\n59\n61\n62\n64\n66\n67\n69\n70\n71\n73\n74\n75\n76\n79\n81\n82\n83\n84\n85\n86\n87\n90\n93\n94\n96\n99\n100\n98\n97\n95\n92\n91\n89\n88\n80\n78\n77\n72\n68\n65\n63\n60\n58\n57\n55\n53\n52\n50\n47\n46\n45\n44\n43\n42\n40\n38\n37\n31\n30\n29\n28\n26\n23\n18\n17\n13\n12\n11\n8\n7\n5\n4\n3\n2\n1"
},
{
"input": "llrlrlrlrlrlrrlllllllrllllrllrrrlllrrllrllrrlllrrlllrlrrllllrrlllrrllrrllllrrlllrlllrrrllrrrrrrllrrl",
"output": "3\n5\n7\n9\n11\n13\n14\n22\n27\n30\n31\n32\n36\n37\n40\n43\n44\n48\n49\n53\n55\n56\n61\n62\n66\n67\n70\n71\n76\n77\n81\n85\n86\n87\n90\n91\n92\n93\n94\n95\n98\n99\n100\n97\n96\n89\n88\n84\n83\n82\n80\n79\n78\n75\n74\n73\n72\n69\n68\n65\n64\n63\n60\n59\n58\n57\n54\n52\n51\n50\n47\n46\n45\n42\n41\n39\n38\n35\n34\n33\n29\n28\n26\n25\n24\n23\n21\n20\n19\n18\n17\n16\n15\n12\n10\n8\n6\n4\n2\n1"
},
{
"input": "l",
"output": "1"
},
{
"input": "r",
"output": "1"
}
] | 1,642,745,741 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 57 | 1,107 | 29,388,800 | s=str(input())
n=len(s)
lstl=[]
lstr=[]
for i in range(0,n):
if(s[i]=='l'):
lstl.append(i+1)
else:
lstr.append(i+1)
lstl.reverse()
for i in lstr:
print(i)
for i in lstl:
print(i)
| Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Squirrel Liss lived in a forest peacefully, but unexpected trouble happens. Stones fall from a mountain. Initially Squirrel Liss occupies an interval [0,<=1]. Next, *n* stones will fall and Liss will escape from the stones. The stones are numbered from 1 to *n* in order.
The stones always fall to the center of Liss's interval. When Liss occupies the interval [*k*<=-<=*d*,<=*k*<=+<=*d*] and a stone falls to *k*, she will escape to the left or to the right. If she escapes to the left, her new interval will be [*k*<=-<=*d*,<=*k*]. If she escapes to the right, her new interval will be [*k*,<=*k*<=+<=*d*].
You are given a string *s* of length *n*. If the *i*-th character of *s* is "l" or "r", when the *i*-th stone falls Liss will escape to the left or to the right, respectively. Find the sequence of stones' numbers from left to right after all the *n* stones falls.
Input Specification:
The input consists of only one line. The only line contains the string *s* (1<=≤<=|*s*|<=≤<=106). Each character in *s* will be either "l" or "r".
Output Specification:
Output *n* lines — on the *i*-th line you should print the *i*-th stone's number from the left.
Demo Input:
['llrlr\n', 'rrlll\n', 'lrlrr\n']
Demo Output:
['3\n5\n4\n2\n1\n', '1\n2\n5\n4\n3\n', '2\n4\n5\n3\n1\n']
Note:
In the first example, the positions of stones 1, 2, 3, 4, 5 will be <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/58fdb5684df807bfcb705a9da9ce175613362b7d.png" style="max-width: 100.0%;max-height: 100.0%;"/>, respectively. So you should print the sequence: 3, 5, 4, 2, 1. | ```python
s=str(input())
n=len(s)
lstl=[]
lstr=[]
for i in range(0,n):
if(s[i]=='l'):
lstl.append(i+1)
else:
lstr.append(i+1)
lstl.reverse()
for i in lstr:
print(i)
for i in lstl:
print(i)
``` | 3 |
|
567 | A | Lineland Mail | PROGRAMMING | 900 | [
"greedy",
"implementation"
] | null | null | All cities of Lineland are located on the *Ox* coordinate axis. Thus, each city is associated with its position *x**i* — a coordinate on the *Ox* axis. No two cities are located at a single point.
Lineland residents love to send letters to each other. A person may send a letter only if the recipient lives in another city (because if they live in the same city, then it is easier to drop in).
Strange but true, the cost of sending the letter is exactly equal to the distance between the sender's city and the recipient's city.
For each city calculate two values *min**i* and *max**i*, where *min**i* is the minimum cost of sending a letter from the *i*-th city to some other city, and *max**i* is the the maximum cost of sending a letter from the *i*-th city to some other city | The first line of the input contains integer *n* (2<=≤<=*n*<=≤<=105) — the number of cities in Lineland. The second line contains the sequence of *n* distinct integers *x*1,<=*x*2,<=...,<=*x**n* (<=-<=109<=≤<=*x**i*<=≤<=109), where *x**i* is the *x*-coordinate of the *i*-th city. All the *x**i*'s are distinct and follow in ascending order. | Print *n* lines, the *i*-th line must contain two integers *min**i*,<=*max**i*, separated by a space, where *min**i* is the minimum cost of sending a letter from the *i*-th city, and *max**i* is the maximum cost of sending a letter from the *i*-th city. | [
"4\n-5 -2 2 7\n",
"2\n-1 1\n"
] | [
"3 12\n3 9\n4 7\n5 12\n",
"2 2\n2 2\n"
] | none | 500 | [
{
"input": "4\n-5 -2 2 7",
"output": "3 12\n3 9\n4 7\n5 12"
},
{
"input": "2\n-1 1",
"output": "2 2\n2 2"
},
{
"input": "3\n-1 0 1",
"output": "1 2\n1 1\n1 2"
},
{
"input": "4\n-1 0 1 3",
"output": "1 4\n1 3\n1 2\n2 4"
},
{
"input": "3\n-1000000000 0 1000000000",
"output": "1000000000 2000000000\n1000000000 1000000000\n1000000000 2000000000"
},
{
"input": "2\n-1000000000 1000000000",
"output": "2000000000 2000000000\n2000000000 2000000000"
},
{
"input": "10\n1 10 12 15 59 68 130 912 1239 9123",
"output": "9 9122\n2 9113\n2 9111\n3 9108\n9 9064\n9 9055\n62 8993\n327 8211\n327 7884\n7884 9122"
},
{
"input": "5\n-2 -1 0 1 2",
"output": "1 4\n1 3\n1 2\n1 3\n1 4"
},
{
"input": "5\n-2 -1 0 1 3",
"output": "1 5\n1 4\n1 3\n1 3\n2 5"
},
{
"input": "3\n-10000 1 10000",
"output": "10001 20000\n9999 10001\n9999 20000"
},
{
"input": "5\n-1000000000 -999999999 -999999998 -999999997 -999999996",
"output": "1 4\n1 3\n1 2\n1 3\n1 4"
},
{
"input": "10\n-857422304 -529223472 82412729 145077145 188538640 265299215 527377039 588634631 592896147 702473706",
"output": "328198832 1559896010\n328198832 1231697178\n62664416 939835033\n43461495 1002499449\n43461495 1045960944\n76760575 1122721519\n61257592 1384799343\n4261516 1446056935\n4261516 1450318451\n109577559 1559896010"
},
{
"input": "10\n-876779400 -829849659 -781819137 -570920213 18428128 25280705 121178189 219147240 528386329 923854124",
"output": "46929741 1800633524\n46929741 1753703783\n48030522 1705673261\n210898924 1494774337\n6852577 905425996\n6852577 902060105\n95897484 997957589\n97969051 1095926640\n309239089 1405165729\n395467795 1800633524"
},
{
"input": "30\n-15 1 21 25 30 40 59 60 77 81 97 100 103 123 139 141 157 158 173 183 200 215 226 231 244 256 267 279 289 292",
"output": "16 307\n16 291\n4 271\n4 267\n5 262\n10 252\n1 233\n1 232\n4 215\n4 211\n3 195\n3 192\n3 189\n16 169\n2 154\n2 156\n1 172\n1 173\n10 188\n10 198\n15 215\n11 230\n5 241\n5 246\n12 259\n11 271\n11 282\n10 294\n3 304\n3 307"
},
{
"input": "10\n-1000000000 -999999999 -999999997 -999999996 -999999995 -999999994 -999999992 -999999990 -999999988 -999999986",
"output": "1 14\n1 13\n1 11\n1 10\n1 9\n1 8\n2 8\n2 10\n2 12\n2 14"
},
{
"input": "50\n-50000 -49459 -48875 -48456 -48411 -48096 -47901 -47500 -47150 -46808 -46687 -46679 -46337 -45747 -45604 -45194 -44752 -44242 -44231 -44122 -43636 -43274 -42916 -42881 -42386 -42095 -41830 -41618 -41145 -40897 -40534 -40007 -39761 -39389 -39104 -38909 -38630 -38561 -38364 -38255 -38214 -38084 -37959 -37607 -37202 -36890 -36681 -36136 -36123 -35886",
"output": "541 14114\n541 13573\n419 12989\n45 12570\n45 12525\n195 12210\n195 12015\n350 11614\n342 11264\n121 10922\n8 10801\n8 10793\n342 10451\n143 9861\n143 9718\n410 9308\n442 8866\n11 8356\n11 8345\n109 8236\n362 7750\n358 7388\n35 7084\n35 7119\n291 7614\n265 7905\n212 8170\n212 8382\n248 8855\n248 9103\n363 9466\n246 9993\n246 10239\n285 10611\n195 10896\n195 11091\n69 11370\n69 11439\n109 11636\n41 11745\n41 11786\n125 11916\n125 12041\n352 12393\n312 12798\n209 13110\n209 13319\n13 13864\n13 13877\n237 141..."
},
{
"input": "3\n-1000000000 999999999 1000000000",
"output": "1999999999 2000000000\n1 1999999999\n1 2000000000"
}
] | 1,605,716,903 | 2,147,483,647 | Python 3 | OK | TESTS | 54 | 639 | 9,523,200 | n = int(input())
cities = list(map(int, input().split()))
mins = []
maxs = []
for i in range(n):
comm = 0
if i == 0:
var = abs(cities[i] - cities[i+1])
mins.append(var)
comm = abs(cities[i]-cities[n-1])
maxs.append(comm)
elif i == n-1:
comm = abs(cities[0]-cities[n-1])
maxs.append(comm)
var = abs(cities[i]-cities[i-1])
mins.append(var)
else:
var = abs(cities[i]-cities[i-1])
par = abs(cities[i]-cities[i+1])
mins.append(min(var, par))
var = abs(cities[i]-cities[0])
par = abs(cities[i]-cities[n-1])
maxs.append(max(var, par))
[print(mins[i], maxs[i]) for i in range(n)]
| Title: Lineland Mail
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
All cities of Lineland are located on the *Ox* coordinate axis. Thus, each city is associated with its position *x**i* — a coordinate on the *Ox* axis. No two cities are located at a single point.
Lineland residents love to send letters to each other. A person may send a letter only if the recipient lives in another city (because if they live in the same city, then it is easier to drop in).
Strange but true, the cost of sending the letter is exactly equal to the distance between the sender's city and the recipient's city.
For each city calculate two values *min**i* and *max**i*, where *min**i* is the minimum cost of sending a letter from the *i*-th city to some other city, and *max**i* is the the maximum cost of sending a letter from the *i*-th city to some other city
Input Specification:
The first line of the input contains integer *n* (2<=≤<=*n*<=≤<=105) — the number of cities in Lineland. The second line contains the sequence of *n* distinct integers *x*1,<=*x*2,<=...,<=*x**n* (<=-<=109<=≤<=*x**i*<=≤<=109), where *x**i* is the *x*-coordinate of the *i*-th city. All the *x**i*'s are distinct and follow in ascending order.
Output Specification:
Print *n* lines, the *i*-th line must contain two integers *min**i*,<=*max**i*, separated by a space, where *min**i* is the minimum cost of sending a letter from the *i*-th city, and *max**i* is the maximum cost of sending a letter from the *i*-th city.
Demo Input:
['4\n-5 -2 2 7\n', '2\n-1 1\n']
Demo Output:
['3 12\n3 9\n4 7\n5 12\n', '2 2\n2 2\n']
Note:
none | ```python
n = int(input())
cities = list(map(int, input().split()))
mins = []
maxs = []
for i in range(n):
comm = 0
if i == 0:
var = abs(cities[i] - cities[i+1])
mins.append(var)
comm = abs(cities[i]-cities[n-1])
maxs.append(comm)
elif i == n-1:
comm = abs(cities[0]-cities[n-1])
maxs.append(comm)
var = abs(cities[i]-cities[i-1])
mins.append(var)
else:
var = abs(cities[i]-cities[i-1])
par = abs(cities[i]-cities[i+1])
mins.append(min(var, par))
var = abs(cities[i]-cities[0])
par = abs(cities[i]-cities[n-1])
maxs.append(max(var, par))
[print(mins[i], maxs[i]) for i in range(n)]
``` | 3 |
|
811 | A | Vladik and Courtesy | PROGRAMMING | 800 | [
"brute force",
"implementation"
] | null | null | At regular competition Vladik and Valera won *a* and *b* candies respectively. Vladik offered 1 his candy to Valera. After that Valera gave Vladik 2 his candies, so that no one thought that he was less generous. Vladik for same reason gave 3 candies to Valera in next turn.
More formally, the guys take turns giving each other one candy more than they received in the previous turn.
This continued until the moment when one of them couldn’t give the right amount of candy. Candies, which guys got from each other, they don’t consider as their own. You need to know, who is the first who can’t give the right amount of candy. | Single line of input data contains two space-separated integers *a*, *b* (1<=≤<=*a*,<=*b*<=≤<=109) — number of Vladik and Valera candies respectively. | Pring a single line "Vladik’’ in case, if Vladik first who can’t give right amount of candy, or "Valera’’ otherwise. | [
"1 1\n",
"7 6\n"
] | [
"Valera\n",
"Vladik\n"
] | Illustration for first test case:
<img class="tex-graphics" src="https://espresso.codeforces.com/ad9b7d0e481208de8e3a585aa1d96b9e1dda4fd7.png" style="max-width: 100.0%;max-height: 100.0%;"/>
Illustration for second test case:
<img class="tex-graphics" src="https://espresso.codeforces.com/9f4836d2ccdffaee5a63898e5d4e6caf2ed4678c.png" style="max-width: 100.0%;max-height: 100.0%;"/> | 500 | [
{
"input": "1 1",
"output": "Valera"
},
{
"input": "7 6",
"output": "Vladik"
},
{
"input": "25 38",
"output": "Vladik"
},
{
"input": "8311 2468",
"output": "Valera"
},
{
"input": "250708 857756",
"output": "Vladik"
},
{
"input": "957985574 24997558",
"output": "Valera"
},
{
"input": "999963734 999994456",
"output": "Vladik"
},
{
"input": "1000000000 1000000000",
"output": "Vladik"
},
{
"input": "946 879",
"output": "Valera"
},
{
"input": "10819 45238",
"output": "Vladik"
},
{
"input": "101357 236928",
"output": "Vladik"
},
{
"input": "1033090 7376359",
"output": "Vladik"
},
{
"input": "9754309 9525494",
"output": "Valera"
},
{
"input": "90706344 99960537",
"output": "Vladik"
},
{
"input": "965161805 908862070",
"output": "Valera"
},
{
"input": "9 11",
"output": "Valera"
},
{
"input": "3 2",
"output": "Vladik"
},
{
"input": "6 6",
"output": "Vladik"
},
{
"input": "4 4",
"output": "Valera"
},
{
"input": "5 5",
"output": "Valera"
},
{
"input": "5 4",
"output": "Valera"
},
{
"input": "12345680 1",
"output": "Valera"
},
{
"input": "9 10",
"output": "Valera"
},
{
"input": "678 76687",
"output": "Vladik"
},
{
"input": "1 678",
"output": "Vladik"
},
{
"input": "45 1678",
"output": "Vladik"
},
{
"input": "3 3",
"output": "Vladik"
},
{
"input": "10 11",
"output": "Valera"
},
{
"input": "2 1",
"output": "Valera"
},
{
"input": "1 2",
"output": "Vladik"
},
{
"input": "2 2",
"output": "Vladik"
},
{
"input": "4 5",
"output": "Valera"
},
{
"input": "9 6",
"output": "Valera"
},
{
"input": "1 5",
"output": "Vladik"
},
{
"input": "7 8",
"output": "Vladik"
},
{
"input": "1000000000 999982505",
"output": "Valera"
},
{
"input": "12 12",
"output": "Vladik"
},
{
"input": "1000 950",
"output": "Valera"
},
{
"input": "10 9",
"output": "Valera"
},
{
"input": "100 9",
"output": "Valera"
},
{
"input": "1000 996",
"output": "Vladik"
},
{
"input": "9 5",
"output": "Valera"
}
] | 1,657,724,458 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 42 | 62 | 1,536,000 | def main():
a , b = map(int,input().split())
i = 1
while 1:
if(i&1):
if(a >= i):
a -= i
else :
print("Vladik")
break
else :
if( b >= i):
b -= i
else :
print("Valera")
break
i +=1
if __name__ == '__main__':
main() | Title: Vladik and Courtesy
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
At regular competition Vladik and Valera won *a* and *b* candies respectively. Vladik offered 1 his candy to Valera. After that Valera gave Vladik 2 his candies, so that no one thought that he was less generous. Vladik for same reason gave 3 candies to Valera in next turn.
More formally, the guys take turns giving each other one candy more than they received in the previous turn.
This continued until the moment when one of them couldn’t give the right amount of candy. Candies, which guys got from each other, they don’t consider as their own. You need to know, who is the first who can’t give the right amount of candy.
Input Specification:
Single line of input data contains two space-separated integers *a*, *b* (1<=≤<=*a*,<=*b*<=≤<=109) — number of Vladik and Valera candies respectively.
Output Specification:
Pring a single line "Vladik’’ in case, if Vladik first who can’t give right amount of candy, or "Valera’’ otherwise.
Demo Input:
['1 1\n', '7 6\n']
Demo Output:
['Valera\n', 'Vladik\n']
Note:
Illustration for first test case:
<img class="tex-graphics" src="https://espresso.codeforces.com/ad9b7d0e481208de8e3a585aa1d96b9e1dda4fd7.png" style="max-width: 100.0%;max-height: 100.0%;"/>
Illustration for second test case:
<img class="tex-graphics" src="https://espresso.codeforces.com/9f4836d2ccdffaee5a63898e5d4e6caf2ed4678c.png" style="max-width: 100.0%;max-height: 100.0%;"/> | ```python
def main():
a , b = map(int,input().split())
i = 1
while 1:
if(i&1):
if(a >= i):
a -= i
else :
print("Vladik")
break
else :
if( b >= i):
b -= i
else :
print("Valera")
break
i +=1
if __name__ == '__main__':
main()
``` | 3 |
|
892 | A | Greed | PROGRAMMING | 900 | [
"greedy",
"implementation"
] | null | null | Jafar has *n* cans of cola. Each can is described by two integers: remaining volume of cola *a**i* and can's capacity *b**i* (*a**i* <=≤<= *b**i*).
Jafar has decided to pour all remaining cola into just 2 cans, determine if he can do this or not! | The first line of the input contains one integer *n* (2<=≤<=*n*<=≤<=100<=000) — number of cola cans.
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=109) — volume of remaining cola in cans.
The third line contains *n* space-separated integers that *b*1,<=*b*2,<=...,<=*b**n* (*a**i*<=≤<=*b**i*<=≤<=109) — capacities of the cans. | Print "YES" (without quotes) if it is possible to pour all remaining cola in 2 cans. Otherwise print "NO" (without quotes).
You can print each letter in any case (upper or lower). | [
"2\n3 5\n3 6\n",
"3\n6 8 9\n6 10 12\n",
"5\n0 0 5 0 0\n1 1 8 10 5\n",
"4\n4 1 0 3\n5 2 2 3\n"
] | [
"YES\n",
"NO\n",
"YES\n",
"YES\n"
] | In the first sample, there are already 2 cans, so the answer is "YES". | 500 | [
{
"input": "2\n3 5\n3 6",
"output": "YES"
},
{
"input": "3\n6 8 9\n6 10 12",
"output": "NO"
},
{
"input": "5\n0 0 5 0 0\n1 1 8 10 5",
"output": "YES"
},
{
"input": "4\n4 1 0 3\n5 2 2 3",
"output": "YES"
},
{
"input": "10\n9 10 24 11 1 7 8 3 28 14\n86 20 34 11 22 94 8 16 73 85",
"output": "YES"
},
{
"input": "4\n25 35 7 31\n70 37 43 35",
"output": "YES"
},
{
"input": "10\n15 26 15 14 14 39 40 4 25 39\n27 72 16 44 69 48 53 17 63 42",
"output": "NO"
},
{
"input": "5\n22 5 19 16 32\n26 10 43 38 37",
"output": "NO"
},
{
"input": "5\n32 4 22 40 26\n39 20 36 98 44",
"output": "YES"
},
{
"input": "6\n18 25 3 10 13 37\n38 73 19 35 24 37",
"output": "YES"
},
{
"input": "2\n2 2\n2 2",
"output": "YES"
},
{
"input": "2\n2 5\n2 5",
"output": "YES"
},
{
"input": "2\n1000 1008\n10000 2352",
"output": "YES"
},
{
"input": "5\n1 2 3 4 5\n1 2 3 4 11",
"output": "YES"
},
{
"input": "4\n1 0 0 0\n2 0 0 0",
"output": "YES"
},
{
"input": "2\n0 0\n1 2",
"output": "YES"
},
{
"input": "3\n9 13 4\n10 14 5",
"output": "NO"
},
{
"input": "2\n0 0\n1 1",
"output": "YES"
},
{
"input": "5\n1 1 2 3 1\n1 1 2 3 4",
"output": "NO"
},
{
"input": "2\n0 0\n0 0",
"output": "YES"
},
{
"input": "3\n5 1 1\n5 5 5",
"output": "YES"
}
] | 1,644,507,796 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 39 | 249 | 17,305,600 | n=int(input())
left=[int(x) for x in input().split()]
cap=[int(x) for x in input().split()]
m1=max(cap)
l=cap;l.remove(m1);
m2=max(l)
if m1+m2>=sum(left):
print("YES")
else:
print('NO')
| Title: Greed
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Jafar has *n* cans of cola. Each can is described by two integers: remaining volume of cola *a**i* and can's capacity *b**i* (*a**i* <=≤<= *b**i*).
Jafar has decided to pour all remaining cola into just 2 cans, determine if he can do this or not!
Input Specification:
The first line of the input contains one integer *n* (2<=≤<=*n*<=≤<=100<=000) — number of cola cans.
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=109) — volume of remaining cola in cans.
The third line contains *n* space-separated integers that *b*1,<=*b*2,<=...,<=*b**n* (*a**i*<=≤<=*b**i*<=≤<=109) — capacities of the cans.
Output Specification:
Print "YES" (without quotes) if it is possible to pour all remaining cola in 2 cans. Otherwise print "NO" (without quotes).
You can print each letter in any case (upper or lower).
Demo Input:
['2\n3 5\n3 6\n', '3\n6 8 9\n6 10 12\n', '5\n0 0 5 0 0\n1 1 8 10 5\n', '4\n4 1 0 3\n5 2 2 3\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n', 'YES\n']
Note:
In the first sample, there are already 2 cans, so the answer is "YES". | ```python
n=int(input())
left=[int(x) for x in input().split()]
cap=[int(x) for x in input().split()]
m1=max(cap)
l=cap;l.remove(m1);
m2=max(l)
if m1+m2>=sum(left):
print("YES")
else:
print('NO')
``` | 3 |
|
375 | B | Maximum Submatrix 2 | PROGRAMMING | 1,600 | [
"data structures",
"dp",
"implementation",
"sortings"
] | null | null | You are given a matrix consisting of digits zero and one, its size is *n*<=×<=*m*. You are allowed to rearrange its rows. What is the maximum area of the submatrix that only consists of ones and can be obtained in the given problem by the described operations?
Let's assume that the rows of matrix *a* are numbered from 1 to *n* from top to bottom and the columns are numbered from 1 to *m* from left to right. A matrix cell on the intersection of the *i*-th row and the *j*-th column can be represented as (*i*,<=*j*). Formally, a submatrix of matrix *a* is a group of four integers *d*,<=*u*,<=*l*,<=*r* (1<=≤<=*d*<=≤<=*u*<=≤<=*n*; 1<=≤<=*l*<=≤<=*r*<=≤<=*m*). We will assume that the submatrix contains cells (*i*,<=*j*) (*d*<=≤<=*i*<=≤<=*u*; *l*<=≤<=*j*<=≤<=*r*). The area of the submatrix is the number of cells it contains. | The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=5000). Next *n* lines contain *m* characters each — matrix *a*. Matrix *a* only contains characters: "0" and "1". Note that the elements of the matrix follow without any spaces in the lines. | Print a single integer — the area of the maximum obtained submatrix. If we cannot obtain a matrix of numbers one, print 0. | [
"1 1\n1\n",
"2 2\n10\n11\n",
"4 3\n100\n011\n000\n101\n"
] | [
"1\n",
"2\n",
"2\n"
] | none | 1,000 | [
{
"input": "1 1\n1",
"output": "1"
},
{
"input": "2 2\n10\n11",
"output": "2"
},
{
"input": "4 3\n100\n011\n000\n101",
"output": "2"
},
{
"input": "11 16\n0111110101100011\n1000101100010000\n0010110110010101\n0110110010110010\n0011101101110000\n1001100011010111\n0010011111111000\n0100100100111110\n1001000000100111\n0110000011001000\n1011111011010000",
"output": "9"
},
{
"input": "19 12\n110001100110\n100100000000\n101011001111\n010111110001\n011000100100\n011111010000\n010011101100\n011010011110\n011001111110\n010111110001\n010000010111\n001111110100\n100100110001\n100110000000\n110000010010\n111101011101\n010111100000\n100000011010\n000100100101",
"output": "16"
},
{
"input": "13 19\n0000111111111111011\n0111000001110001101\n1110100110111011101\n0001101011100001110\n1101100100010000101\n1010100011110011010\n1010011101010000001\n1011101000001111000\n1101110001101011110\n0110101010001111100\n0001011010100111001\n1111101000110001000\n0010010000011100010",
"output": "14"
},
{
"input": "8 5\n00000\n00000\n00000\n00000\n00000\n00000\n00000\n00000",
"output": "0"
},
{
"input": "15 18\n111111111111111111\n111111111111111111\n111111111111111111\n111111111111111111\n111111111111111111\n111111111111111111\n111111111111111111\n111111111111111111\n111111111111111111\n111111111111111111\n111111111111111111\n111111111111111111\n111111111111111111\n111111111111111111\n111111111111111111",
"output": "270"
},
{
"input": "1 1\n0",
"output": "0"
}
] | 1,560,456,430 | 2,147,483,647 | PyPy 3 | TIME_LIMIT_EXCEEDED | TESTS | 17 | 2,000 | 268,185,600 | '''input
1 1
1
'''
from sys import stdin
import math
from collections import defaultdict, deque
def pre_calculation(matrix, n, m):
dp = [[0 for x in range(m)] for y in range(n)]
for i in range(n):
for j in range(m - 1, -1, -1):
if j + 1 < m:
if matrix[i][j] == '1':
dp[i][j] = 1 + dp[i][j + 1]
else:
if matrix[i][j] == '1':
dp[i][j] = 1
return dp
def brute(aux):
ans = -float('inf')
for i in range(len(aux)):
ans = max(ans, aux[i] * (i + 1))
return ans
def myfunction(dp, n, m, index):
aux = []
for i in range(n):
aux.append(dp[i][index])
aux.sort(reverse = True)
return brute(aux)
# main starts
n, m = list(map(int, stdin.readline().split()))
matrix = []
for _ in range(n):
matrix.append(list(stdin.readline().strip()))
dp = pre_calculation(matrix, n, m)
mx = -float('inf')
for i in range(m):
mx = max(mx, myfunction(dp, n, m, i))
print(mx) | Title: Maximum Submatrix 2
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a matrix consisting of digits zero and one, its size is *n*<=×<=*m*. You are allowed to rearrange its rows. What is the maximum area of the submatrix that only consists of ones and can be obtained in the given problem by the described operations?
Let's assume that the rows of matrix *a* are numbered from 1 to *n* from top to bottom and the columns are numbered from 1 to *m* from left to right. A matrix cell on the intersection of the *i*-th row and the *j*-th column can be represented as (*i*,<=*j*). Formally, a submatrix of matrix *a* is a group of four integers *d*,<=*u*,<=*l*,<=*r* (1<=≤<=*d*<=≤<=*u*<=≤<=*n*; 1<=≤<=*l*<=≤<=*r*<=≤<=*m*). We will assume that the submatrix contains cells (*i*,<=*j*) (*d*<=≤<=*i*<=≤<=*u*; *l*<=≤<=*j*<=≤<=*r*). The area of the submatrix is the number of cells it contains.
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=5000). Next *n* lines contain *m* characters each — matrix *a*. Matrix *a* only contains characters: "0" and "1". Note that the elements of the matrix follow without any spaces in the lines.
Output Specification:
Print a single integer — the area of the maximum obtained submatrix. If we cannot obtain a matrix of numbers one, print 0.
Demo Input:
['1 1\n1\n', '2 2\n10\n11\n', '4 3\n100\n011\n000\n101\n']
Demo Output:
['1\n', '2\n', '2\n']
Note:
none | ```python
'''input
1 1
1
'''
from sys import stdin
import math
from collections import defaultdict, deque
def pre_calculation(matrix, n, m):
dp = [[0 for x in range(m)] for y in range(n)]
for i in range(n):
for j in range(m - 1, -1, -1):
if j + 1 < m:
if matrix[i][j] == '1':
dp[i][j] = 1 + dp[i][j + 1]
else:
if matrix[i][j] == '1':
dp[i][j] = 1
return dp
def brute(aux):
ans = -float('inf')
for i in range(len(aux)):
ans = max(ans, aux[i] * (i + 1))
return ans
def myfunction(dp, n, m, index):
aux = []
for i in range(n):
aux.append(dp[i][index])
aux.sort(reverse = True)
return brute(aux)
# main starts
n, m = list(map(int, stdin.readline().split()))
matrix = []
for _ in range(n):
matrix.append(list(stdin.readline().strip()))
dp = pre_calculation(matrix, n, m)
mx = -float('inf')
for i in range(m):
mx = max(mx, myfunction(dp, n, m, i))
print(mx)
``` | 0 |
|
991 | B | Getting an A | PROGRAMMING | 900 | [
"greedy",
"sortings"
] | null | null | Translator's note: in Russia's most widespread grading system, there are four grades: 5, 4, 3, 2, the higher the better, roughly corresponding to A, B, C and F respectively in American grading system.
The term is coming to an end and students start thinking about their grades. Today, a professor told his students that the grades for his course would be given out automatically — he would calculate the simple average (arithmetic mean) of all grades given out for lab works this term and round to the nearest integer. The rounding would be done in favour of the student — $4.5$ would be rounded up to $5$ (as in example 3), but $4.4$ would be rounded down to $4$.
This does not bode well for Vasya who didn't think those lab works would influence anything, so he may receive a grade worse than $5$ (maybe even the dreaded $2$). However, the professor allowed him to redo some of his works of Vasya's choosing to increase his average grade. Vasya wants to redo as as few lab works as possible in order to get $5$ for the course. Of course, Vasya will get $5$ for the lab works he chooses to redo.
Help Vasya — calculate the minimum amount of lab works Vasya has to redo. | The first line contains a single integer $n$ — the number of Vasya's grades ($1 \leq n \leq 100$).
The second line contains $n$ integers from $2$ to $5$ — Vasya's grades for his lab works. | Output a single integer — the minimum amount of lab works that Vasya has to redo. It can be shown that Vasya can always redo enough lab works to get a $5$. | [
"3\n4 4 4\n",
"4\n5 4 5 5\n",
"4\n5 3 3 5\n"
] | [
"2\n",
"0\n",
"1\n"
] | In the first sample, it is enough to redo two lab works to make two $4$s into $5$s.
In the second sample, Vasya's average is already $4.75$ so he doesn't have to redo anything to get a $5$.
In the second sample Vasya has to redo one lab work to get rid of one of the $3$s, that will make the average exactly $4.5$ so the final grade would be $5$. | 1,000 | [
{
"input": "3\n4 4 4",
"output": "2"
},
{
"input": "4\n5 4 5 5",
"output": "0"
},
{
"input": "4\n5 3 3 5",
"output": "1"
},
{
"input": "1\n5",
"output": "0"
},
{
"input": "4\n3 2 5 4",
"output": "2"
},
{
"input": "5\n5 4 3 2 5",
"output": "2"
},
{
"input": "8\n5 4 2 5 5 2 5 5",
"output": "1"
},
{
"input": "5\n5 5 2 5 5",
"output": "1"
},
{
"input": "6\n5 5 5 5 5 2",
"output": "0"
},
{
"input": "6\n2 2 2 2 2 2",
"output": "5"
},
{
"input": "100\n3 2 4 3 3 3 4 2 3 5 5 2 5 2 3 2 4 4 4 5 5 4 2 5 4 3 2 5 3 4 3 4 2 4 5 4 2 4 3 4 5 2 5 3 3 4 2 2 4 4 4 5 4 3 3 3 2 5 2 2 2 3 5 4 3 2 4 5 5 5 2 2 4 2 3 3 3 5 3 2 2 4 5 5 4 5 5 4 2 3 2 2 2 2 5 3 5 2 3 4",
"output": "40"
},
{
"input": "1\n2",
"output": "1"
},
{
"input": "1\n3",
"output": "1"
},
{
"input": "1\n4",
"output": "1"
},
{
"input": "4\n3 2 5 5",
"output": "1"
},
{
"input": "6\n4 3 3 3 3 4",
"output": "4"
},
{
"input": "8\n3 3 5 3 3 3 5 5",
"output": "3"
},
{
"input": "10\n2 4 5 5 5 5 2 3 3 2",
"output": "3"
},
{
"input": "20\n5 2 5 2 2 2 2 2 5 2 2 5 2 5 5 2 2 5 2 2",
"output": "10"
},
{
"input": "25\n4 4 4 4 3 4 3 3 3 3 3 4 4 3 4 4 4 4 4 3 3 3 4 3 4",
"output": "13"
},
{
"input": "30\n4 2 4 2 4 2 2 4 4 4 4 2 4 4 4 2 2 2 2 4 2 4 4 4 2 4 2 4 2 2",
"output": "15"
},
{
"input": "52\n5 3 4 4 4 3 5 3 4 5 3 4 4 3 5 5 4 3 3 3 4 5 4 4 5 3 5 3 5 4 5 5 4 3 4 5 3 4 3 3 4 4 4 3 5 3 4 5 3 5 4 5",
"output": "14"
},
{
"input": "77\n5 3 2 3 2 3 2 3 5 2 2 3 3 3 3 5 3 3 2 2 2 5 5 5 5 3 2 2 5 2 3 2 2 5 2 5 3 3 2 2 5 5 2 3 3 2 3 3 3 2 5 5 2 2 3 3 5 5 2 2 5 5 3 3 5 5 2 2 5 2 2 5 5 5 2 5 2",
"output": "33"
},
{
"input": "55\n3 4 2 3 3 2 4 4 3 3 4 2 4 4 3 3 2 3 2 2 3 3 2 3 2 3 2 4 4 3 2 3 2 3 3 2 2 4 2 4 4 3 4 3 2 4 3 2 4 2 2 3 2 3 4",
"output": "34"
},
{
"input": "66\n5 4 5 5 4 4 4 4 4 2 5 5 2 4 2 2 2 5 4 4 4 4 5 2 2 5 5 2 2 4 4 2 4 2 2 5 2 5 4 5 4 5 4 4 2 5 2 4 4 4 2 2 5 5 5 5 4 4 4 4 4 2 4 5 5 5",
"output": "16"
},
{
"input": "99\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2",
"output": "83"
},
{
"input": "100\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2",
"output": "84"
},
{
"input": "99\n3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3",
"output": "75"
},
{
"input": "100\n3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3",
"output": "75"
},
{
"input": "99\n2 2 3 3 3 3 3 2 2 3 2 3 2 3 2 2 3 2 3 2 3 3 3 3 2 2 2 2 3 2 3 3 3 3 3 2 3 3 3 3 2 3 2 3 3 3 2 3 2 3 3 3 3 2 2 3 2 3 2 3 2 3 2 2 2 3 3 2 3 2 2 2 2 2 2 2 2 3 3 3 3 2 3 2 3 3 2 3 2 3 2 3 3 2 2 2 3 2 3",
"output": "75"
},
{
"input": "100\n3 2 3 3 2 2 3 2 2 3 3 2 3 2 2 2 2 2 3 2 2 2 3 2 3 3 2 2 3 2 2 2 2 3 2 3 3 2 2 3 2 2 3 2 3 2 2 3 2 3 2 2 3 2 2 3 3 3 3 3 2 2 3 2 3 3 2 2 3 2 2 2 3 2 2 3 3 2 2 3 3 3 3 2 3 2 2 2 3 3 2 2 3 2 2 2 2 3 2 2",
"output": "75"
},
{
"input": "99\n4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4",
"output": "50"
},
{
"input": "100\n4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4",
"output": "50"
},
{
"input": "99\n2 2 2 2 4 2 2 2 2 4 4 4 4 2 4 4 2 2 4 4 2 2 2 4 4 2 4 4 2 4 4 2 2 2 4 4 2 2 2 2 4 4 4 2 2 2 4 4 2 4 2 4 2 2 4 2 4 4 4 4 4 2 2 4 4 4 2 2 2 2 4 2 4 2 2 2 2 2 2 4 4 2 4 2 2 4 2 2 2 2 2 4 2 4 2 2 4 4 4",
"output": "54"
},
{
"input": "100\n4 2 4 4 2 4 2 2 4 4 4 4 4 4 4 4 4 2 4 4 2 2 4 4 2 2 4 4 2 2 2 4 4 2 4 4 2 4 2 2 4 4 2 4 2 4 4 4 2 2 2 2 2 2 2 4 2 2 2 4 4 4 2 2 2 2 4 2 2 2 2 2 2 2 4 4 4 4 4 4 4 4 4 2 2 2 2 2 2 2 2 4 4 4 4 2 4 2 2 4",
"output": "50"
},
{
"input": "99\n4 3 4 4 4 4 4 3 4 3 3 4 3 3 4 4 3 3 3 4 3 4 3 3 4 3 3 3 3 4 3 4 4 3 4 4 3 3 4 4 4 3 3 3 4 4 3 3 4 3 4 3 4 3 4 3 3 3 3 4 3 4 4 4 4 4 4 3 4 4 3 3 3 3 3 3 3 3 4 3 3 3 4 4 4 4 4 4 3 3 3 3 4 4 4 3 3 4 3",
"output": "51"
},
{
"input": "100\n3 3 4 4 4 4 4 3 4 4 3 3 3 3 4 4 4 4 4 4 3 3 3 4 3 4 3 4 3 3 4 3 3 3 3 3 3 3 3 4 3 4 3 3 4 3 3 3 4 4 3 4 4 3 3 4 4 4 4 4 4 3 4 4 3 4 3 3 3 4 4 3 3 4 4 3 4 4 4 3 3 4 3 3 4 3 4 3 4 3 3 4 4 4 3 3 4 3 3 4",
"output": "51"
},
{
"input": "99\n3 3 4 4 4 2 4 4 3 2 3 4 4 4 2 2 2 3 2 4 4 2 4 3 2 2 2 4 2 3 4 3 4 2 3 3 4 2 3 3 2 3 4 4 3 2 4 3 4 3 3 3 3 3 4 4 3 3 4 4 2 4 3 4 3 2 3 3 3 4 4 2 4 4 2 3 4 2 3 3 3 4 2 2 3 2 4 3 2 3 3 2 3 4 2 3 3 2 3",
"output": "58"
},
{
"input": "100\n2 2 4 2 2 3 2 3 4 4 3 3 4 4 4 2 3 2 2 3 4 2 3 2 4 3 4 2 3 3 3 2 4 3 3 2 2 3 2 4 4 2 4 3 4 4 3 3 3 2 4 2 2 2 2 2 2 3 2 3 2 3 4 4 4 2 2 3 4 4 3 4 3 3 2 3 3 3 4 3 2 3 3 2 4 2 3 3 4 4 3 3 4 3 4 3 3 4 3 3",
"output": "61"
},
{
"input": "99\n5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5",
"output": "0"
},
{
"input": "100\n5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5",
"output": "0"
},
{
"input": "99\n2 2 2 2 2 5 2 2 5 2 5 2 5 2 2 2 2 2 5 2 2 2 5 2 2 5 2 2 2 5 5 2 5 2 2 5 2 5 2 2 5 5 2 2 2 2 5 5 2 2 2 5 2 2 5 2 2 2 2 2 5 5 5 5 2 2 5 2 5 2 2 2 2 2 5 2 2 5 5 2 2 2 2 2 5 5 2 2 5 5 2 2 2 2 5 5 5 2 5",
"output": "48"
},
{
"input": "100\n5 5 2 2 2 2 2 2 5 5 2 5 2 2 2 2 5 2 5 2 5 5 2 5 5 2 2 2 2 2 2 5 2 2 2 5 2 2 5 2 2 5 5 5 2 5 5 5 5 5 5 2 2 5 2 2 5 5 5 5 5 2 5 2 5 2 2 2 5 2 5 2 5 5 2 5 5 2 2 5 2 5 5 2 5 2 2 5 2 2 2 5 2 2 2 2 5 5 2 5",
"output": "38"
},
{
"input": "99\n5 3 3 3 5 3 3 3 3 3 3 3 3 5 3 3 3 3 3 3 3 3 5 3 3 3 5 5 3 5 5 3 3 5 5 5 3 5 3 3 3 3 5 3 3 5 5 3 5 5 5 3 5 3 5 3 5 5 5 5 3 3 3 5 3 5 3 3 3 5 5 5 5 5 3 5 5 3 3 5 5 3 5 5 3 5 5 3 3 5 5 5 3 3 3 5 3 3 3",
"output": "32"
},
{
"input": "100\n3 3 3 5 3 3 3 3 3 3 5 5 5 5 3 3 3 3 5 3 3 3 3 3 5 3 5 3 3 5 5 5 5 5 5 3 3 5 3 3 5 3 5 5 5 3 5 3 3 3 3 3 3 3 3 3 3 3 5 5 3 5 3 5 5 3 5 3 3 5 3 5 5 5 5 3 5 3 3 3 5 5 5 3 3 3 5 3 5 5 5 3 3 3 5 3 5 5 3 5",
"output": "32"
},
{
"input": "99\n5 3 5 5 3 3 3 2 2 5 2 5 3 2 5 2 5 2 3 5 3 2 3 2 5 5 2 2 3 3 5 5 3 5 5 2 3 3 5 2 2 5 3 2 5 2 3 5 5 2 5 2 2 5 3 3 5 3 3 5 3 2 3 5 3 2 3 2 3 2 2 2 2 5 2 2 3 2 5 5 5 3 3 2 5 3 5 5 5 2 3 2 5 5 2 5 2 5 3",
"output": "39"
},
{
"input": "100\n3 5 3 3 5 5 3 3 2 5 5 3 3 3 2 2 3 2 5 3 2 2 3 3 3 3 2 5 3 2 3 3 5 2 2 2 3 2 3 5 5 3 2 5 2 2 5 5 3 5 5 5 2 2 5 5 3 3 2 2 2 5 3 3 2 2 3 5 3 2 3 5 5 3 2 3 5 5 3 3 2 3 5 2 5 5 5 5 5 5 3 5 3 2 3 3 2 5 2 2",
"output": "42"
},
{
"input": "99\n4 4 4 5 4 4 5 5 4 4 5 5 5 4 5 4 5 5 5 4 4 5 5 5 5 4 5 5 5 4 4 5 5 4 5 4 4 4 5 5 5 5 4 4 5 4 4 5 4 4 4 4 5 5 5 4 5 4 5 5 5 5 5 4 5 4 5 4 4 4 4 5 5 5 4 5 5 4 4 5 5 5 4 5 4 4 5 5 4 5 5 5 5 4 5 5 4 4 4",
"output": "0"
},
{
"input": "100\n4 4 5 5 5 5 5 5 4 4 5 5 4 4 5 5 4 5 4 4 4 4 4 4 4 4 5 5 5 5 5 4 4 4 4 4 5 4 4 5 4 4 4 5 5 5 4 5 5 5 5 5 5 4 4 4 4 4 4 5 5 4 5 4 4 5 4 4 4 4 5 5 4 5 5 4 4 4 5 5 5 5 4 5 5 5 4 4 5 5 5 4 5 4 5 4 4 5 5 4",
"output": "1"
},
{
"input": "99\n2 2 2 5 2 2 2 2 2 4 4 5 5 2 2 4 2 5 2 2 2 5 2 2 5 5 5 4 5 5 4 4 2 2 5 2 2 2 2 5 5 2 2 4 4 4 2 2 2 5 2 4 4 2 4 2 4 2 5 4 2 2 5 2 4 4 4 2 5 2 2 5 4 2 2 5 5 5 2 4 5 4 5 5 4 4 4 5 4 5 4 5 4 2 5 2 2 2 4",
"output": "37"
},
{
"input": "100\n4 4 5 2 2 5 4 5 2 2 2 4 2 5 4 4 2 2 4 5 2 4 2 5 5 4 2 4 4 2 2 5 4 2 5 4 5 2 5 2 4 2 5 4 5 2 2 2 5 2 5 2 5 2 2 4 4 5 5 5 5 5 5 5 4 2 2 2 4 2 2 4 5 5 4 5 4 2 2 2 2 4 2 2 5 5 4 2 2 5 4 5 5 5 4 5 5 5 2 2",
"output": "31"
},
{
"input": "99\n5 3 4 4 5 4 4 4 3 5 4 3 3 4 3 5 5 5 5 4 3 3 5 3 4 5 3 5 4 4 3 5 5 4 4 4 4 3 5 3 3 5 5 5 5 5 4 3 4 4 3 5 5 3 3 4 4 4 5 4 4 5 4 4 4 4 5 5 4 3 3 4 3 5 3 3 3 3 4 4 4 4 3 4 5 4 4 5 5 5 3 4 5 3 4 5 4 3 3",
"output": "24"
},
{
"input": "100\n5 4 4 4 5 5 5 4 5 4 4 3 3 4 4 4 5 4 5 5 3 5 5 4 5 5 5 4 4 5 3 5 3 5 3 3 5 4 4 5 5 4 5 5 3 4 5 4 4 3 4 4 3 3 5 4 5 4 5 3 4 5 3 4 5 4 3 5 4 5 4 4 4 3 4 5 3 4 3 5 3 4 4 4 3 4 4 5 3 3 4 4 5 5 4 3 4 4 3 5",
"output": "19"
},
{
"input": "99\n2 2 5 2 5 3 4 2 3 5 4 3 4 2 5 3 2 2 4 2 4 4 5 4 4 5 2 5 5 3 2 3 2 2 3 4 5 3 5 2 5 4 4 5 4 2 2 3 2 3 3 3 4 4 3 2 2 4 4 2 5 3 5 3 5 4 4 4 5 4 5 2 2 5 4 4 4 3 3 2 5 2 5 2 3 2 5 2 2 5 5 3 4 5 3 4 4 4 4",
"output": "37"
},
{
"input": "2\n5 2",
"output": "1"
},
{
"input": "5\n2 2 2 2 2",
"output": "5"
},
{
"input": "100\n2 3 2 2 2 3 2 3 3 3 3 3 2 3 3 2 2 3 3 2 3 2 3 2 3 4 4 4 3 3 3 3 3 4 4 3 3 4 3 2 3 4 3 3 3 3 2 3 4 3 4 3 3 2 4 4 2 4 4 3 3 3 3 4 3 2 3 4 3 4 4 4 4 4 3 2 2 3 4 2 4 4 4 2 2 4 2 2 3 2 2 4 4 3 4 2 3 3 2 2",
"output": "61"
},
{
"input": "100\n5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4",
"output": "1"
},
{
"input": "100\n5 4 3 5 3 5 4 2 3 3 4 5 4 5 5 4 2 4 2 2 5 2 5 3 4 4 4 5 5 5 3 4 4 4 3 5 3 2 5 4 3 3 3 5 2 3 4 2 5 4 3 4 5 2 2 3 4 4 2 3 3 3 2 5 2 3 4 3 3 3 2 5 4 3 4 5 4 2 5 4 5 2 2 4 2 2 5 5 4 5 2 2 2 2 5 2 4 4 4 5",
"output": "35"
},
{
"input": "2\n2 2",
"output": "2"
},
{
"input": "20\n4 4 4 4 4 4 4 4 4 4 4 5 5 5 5 5 5 5 5 5",
"output": "1"
}
] | 1,601,285,028 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 2 | 109 | 0 | import math
n=int(input())
li=list(map(int,input().split()))
k=sum(li)
count=0
while True:
if round(k/n)==5:
break
else:
k+=1
count+=1
print(count)
| Title: Getting an A
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Translator's note: in Russia's most widespread grading system, there are four grades: 5, 4, 3, 2, the higher the better, roughly corresponding to A, B, C and F respectively in American grading system.
The term is coming to an end and students start thinking about their grades. Today, a professor told his students that the grades for his course would be given out automatically — he would calculate the simple average (arithmetic mean) of all grades given out for lab works this term and round to the nearest integer. The rounding would be done in favour of the student — $4.5$ would be rounded up to $5$ (as in example 3), but $4.4$ would be rounded down to $4$.
This does not bode well for Vasya who didn't think those lab works would influence anything, so he may receive a grade worse than $5$ (maybe even the dreaded $2$). However, the professor allowed him to redo some of his works of Vasya's choosing to increase his average grade. Vasya wants to redo as as few lab works as possible in order to get $5$ for the course. Of course, Vasya will get $5$ for the lab works he chooses to redo.
Help Vasya — calculate the minimum amount of lab works Vasya has to redo.
Input Specification:
The first line contains a single integer $n$ — the number of Vasya's grades ($1 \leq n \leq 100$).
The second line contains $n$ integers from $2$ to $5$ — Vasya's grades for his lab works.
Output Specification:
Output a single integer — the minimum amount of lab works that Vasya has to redo. It can be shown that Vasya can always redo enough lab works to get a $5$.
Demo Input:
['3\n4 4 4\n', '4\n5 4 5 5\n', '4\n5 3 3 5\n']
Demo Output:
['2\n', '0\n', '1\n']
Note:
In the first sample, it is enough to redo two lab works to make two $4$s into $5$s.
In the second sample, Vasya's average is already $4.75$ so he doesn't have to redo anything to get a $5$.
In the second sample Vasya has to redo one lab work to get rid of one of the $3$s, that will make the average exactly $4.5$ so the final grade would be $5$. | ```python
import math
n=int(input())
li=list(map(int,input().split()))
k=sum(li)
count=0
while True:
if round(k/n)==5:
break
else:
k+=1
count+=1
print(count)
``` | 0 |
|
701 | C | They Are Everywhere | PROGRAMMING | 1,500 | [
"binary search",
"strings",
"two pointers"
] | null | null | Sergei B., the young coach of Pokemons, has found the big house which consists of *n* flats ordered in a row from left to right. It is possible to enter each flat from the street. It is possible to go out from each flat. Also, each flat is connected with the flat to the left and the flat to the right. Flat number 1 is only connected with the flat number 2 and the flat number *n* is only connected with the flat number *n*<=-<=1.
There is exactly one Pokemon of some type in each of these flats. Sergei B. asked residents of the house to let him enter their flats in order to catch Pokemons. After consulting the residents of the house decided to let Sergei B. enter one flat from the street, visit several flats and then go out from some flat. But they won't let him visit the same flat more than once.
Sergei B. was very pleased, and now he wants to visit as few flats as possible in order to collect Pokemons of all types that appear in this house. Your task is to help him and determine this minimum number of flats he has to visit. | The first line contains the integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of flats in the house.
The second line contains the row *s* with the length *n*, it consists of uppercase and lowercase letters of English alphabet, the *i*-th letter equals the type of Pokemon, which is in the flat number *i*. | Print the minimum number of flats which Sergei B. should visit in order to catch Pokemons of all types which there are in the house. | [
"3\nAaA\n",
"7\nbcAAcbc\n",
"6\naaBCCe\n"
] | [
"2\n",
"3\n",
"5\n"
] | In the first test Sergei B. can begin, for example, from the flat number 1 and end in the flat number 2.
In the second test Sergei B. can begin, for example, from the flat number 4 and end in the flat number 6.
In the third test Sergei B. must begin from the flat number 2 and end in the flat number 6. | 1,000 | [
{
"input": "3\nAaA",
"output": "2"
},
{
"input": "7\nbcAAcbc",
"output": "3"
},
{
"input": "6\naaBCCe",
"output": "5"
},
{
"input": "1\nA",
"output": "1"
},
{
"input": "1\ng",
"output": "1"
},
{
"input": "52\nabcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ",
"output": "52"
},
{
"input": "2\nAA",
"output": "1"
},
{
"input": "4\nqqqE",
"output": "2"
},
{
"input": "10\nrrrrroooro",
"output": "2"
},
{
"input": "15\nOCOCCCCiCOCCCOi",
"output": "3"
},
{
"input": "20\nVEVnVVnWnVEVVnEVBEWn",
"output": "5"
},
{
"input": "25\ncpcyPPjPPcPPPPcppPcPpppcP",
"output": "6"
},
{
"input": "30\nsssssAsesssssssssssssessssssss",
"output": "3"
},
{
"input": "35\ngdXdddgddddddddggggXdbgdggdgddddddb",
"output": "4"
},
{
"input": "40\nIgsggIiIggzgigIIiiIIIiIgIggIzgIiiiggggIi",
"output": "9"
},
{
"input": "45\neteeeeeteaattaeetaetteeettoetettteyeteeeotaae",
"output": "9"
},
{
"input": "50\nlUlUllUlUllllUllllUllllUlUlllUlllUlllllUUlllUUlkUl",
"output": "3"
},
{
"input": "55\nAAAAASAAAASAASAAAAAAAAAAAAASAAAAAAAAAAAAAAAASAAAAAAAAAA",
"output": "2"
},
{
"input": "60\nRRRrSRRRRRRRRRSSRRRSRRRRRRRRrRSRRRRRRRRRRRRRRSRRRRRSSRSRrRRR",
"output": "3"
},
{
"input": "65\nhhMhMhhhhhhhhhhhMhhMMMhhhhBhhhhMhhhhMhhhhhMhhhBhhhhhhhhhhBhhhhhhh",
"output": "5"
},
{
"input": "70\nwAwwwAwwwwwwwwwwwwwwAwAAwwAwwwwwwwwAwAAAwAAwwwwwwwwwAwwwwwwwwwwwwAAwww",
"output": "2"
},
{
"input": "75\niiiXXiiyiiiXyXiiyXiiXiiiiiiXXyiiiiXXiiXiiXifiXiXXiifiiiiiiXfXiyiXXiXiiiiXiX",
"output": "4"
},
{
"input": "80\nSrSrrrrrrrrrrrrrrSSSrrrrrrSrrrrSrrrrrrrrrrSSrrrrrrrrrrrSrrrSrrrrSrrrrSrrrrSSrSSr",
"output": "2"
},
{
"input": "85\nwkMMMwMMkMMMMMMMkkkkMMMMzkkMMwMMkkwMkMwkMMkMMwwMzMMMkkMwwMMMMMMkMMkMzMMMkMMkwMkMMMkMM",
"output": "6"
},
{
"input": "90\nZllZZZyZlZlllZlylllZlllZZllllllllZZllllllllllyylZZyvZvZlllZZlZllZlZlllZyllZyZlllZlllllllZl",
"output": "5"
},
{
"input": "95\nEmuBuEBmmEBBElBlElmmBEmmmEmmEuBEEmummmEmBBBBEWBBBmEEBmmummBBmmlluBBmElmEBEmBmBBmBmuLmEBBmlEBmBu",
"output": "39"
},
{
"input": "100\nfAfhfNNNhfffAfNffNANhffNffffffNfffffhNfffNfffNfffNfNNNffNfffNfhNNNffffNfNfffAffNffNffNNfhfffNhNffNNN",
"output": "4"
},
{
"input": "7\nabcbbab",
"output": "3"
},
{
"input": "65\nkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkjjjjkkkkkkkkkkkkkkkkkkkllllllljk",
"output": "3"
},
{
"input": "9\nbcdddbddc",
"output": "3"
},
{
"input": "71\nlqcrdopdpfagcilpcmheqhjbmnciqmeoabflrenbkrsaabaanpehdmpqcdqnljcrsplnlco",
"output": "36"
},
{
"input": "8\nabbbbbbc",
"output": "8"
}
] | 1,698,091,810 | 2,147,483,647 | PyPy 3-64 | TIME_LIMIT_EXCEEDED | TESTS | 10 | 2,000 | 13,107,200 | # love thy neighbour
from sys import stdin,stdout
def ArrayIn():return list(map(int,input().split()))
#def print(x):stdout.write(f'{x}\n')
n=int(input())
flats=input()
def Valid(row,k):
target=set(row)
for i in range(len(row)-k+1):
if set(row[i:i+k])==target:
return True
return False
l, r=0, n
ans=int(10e6)
while l<=r:
mid = l + (r-l)//2
if Valid(flats,mid):
ans=min(ans,mid)
r=mid-1
else:l=mid+1
print(ans)
| Title: They Are Everywhere
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Sergei B., the young coach of Pokemons, has found the big house which consists of *n* flats ordered in a row from left to right. It is possible to enter each flat from the street. It is possible to go out from each flat. Also, each flat is connected with the flat to the left and the flat to the right. Flat number 1 is only connected with the flat number 2 and the flat number *n* is only connected with the flat number *n*<=-<=1.
There is exactly one Pokemon of some type in each of these flats. Sergei B. asked residents of the house to let him enter their flats in order to catch Pokemons. After consulting the residents of the house decided to let Sergei B. enter one flat from the street, visit several flats and then go out from some flat. But they won't let him visit the same flat more than once.
Sergei B. was very pleased, and now he wants to visit as few flats as possible in order to collect Pokemons of all types that appear in this house. Your task is to help him and determine this minimum number of flats he has to visit.
Input Specification:
The first line contains the integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of flats in the house.
The second line contains the row *s* with the length *n*, it consists of uppercase and lowercase letters of English alphabet, the *i*-th letter equals the type of Pokemon, which is in the flat number *i*.
Output Specification:
Print the minimum number of flats which Sergei B. should visit in order to catch Pokemons of all types which there are in the house.
Demo Input:
['3\nAaA\n', '7\nbcAAcbc\n', '6\naaBCCe\n']
Demo Output:
['2\n', '3\n', '5\n']
Note:
In the first test Sergei B. can begin, for example, from the flat number 1 and end in the flat number 2.
In the second test Sergei B. can begin, for example, from the flat number 4 and end in the flat number 6.
In the third test Sergei B. must begin from the flat number 2 and end in the flat number 6. | ```python
# love thy neighbour
from sys import stdin,stdout
def ArrayIn():return list(map(int,input().split()))
#def print(x):stdout.write(f'{x}\n')
n=int(input())
flats=input()
def Valid(row,k):
target=set(row)
for i in range(len(row)-k+1):
if set(row[i:i+k])==target:
return True
return False
l, r=0, n
ans=int(10e6)
while l<=r:
mid = l + (r-l)//2
if Valid(flats,mid):
ans=min(ans,mid)
r=mid-1
else:l=mid+1
print(ans)
``` | 0 |
|
71 | A | Way Too Long Words | PROGRAMMING | 800 | [
"strings"
] | A. Way Too Long Words | 1 | 256 | Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes. | The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters. | Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data. | [
"4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n"
] | [
"word\nl10n\ni18n\np43s\n"
] | none | 500 | [
{
"input": "4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis",
"output": "word\nl10n\ni18n\np43s"
},
{
"input": "5\nabcdefgh\nabcdefghi\nabcdefghij\nabcdefghijk\nabcdefghijklm",
"output": "abcdefgh\nabcdefghi\nabcdefghij\na9k\na11m"
},
{
"input": "3\nnjfngnrurunrgunrunvurn\njfvnjfdnvjdbfvsbdubruvbubvkdb\nksdnvidnviudbvibd",
"output": "n20n\nj27b\nk15d"
},
{
"input": "1\ntcyctkktcctrcyvbyiuhihhhgyvyvyvyvjvytchjckt",
"output": "t41t"
},
{
"input": "24\nyou\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nunofficially\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings",
"output": "you\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nu10y\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings"
},
{
"input": "1\na",
"output": "a"
},
{
"input": "26\na\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz",
"output": "a\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz"
},
{
"input": "1\nabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghij",
"output": "a98j"
},
{
"input": "10\ngyartjdxxlcl\nfzsck\nuidwu\nxbymclornemdmtj\nilppyoapitawgje\ncibzc\ndrgbeu\nhezplmsdekhhbo\nfeuzlrimbqbytdu\nkgdco",
"output": "g10l\nfzsck\nuidwu\nx13j\ni13e\ncibzc\ndrgbeu\nh12o\nf13u\nkgdco"
},
{
"input": "20\nlkpmx\nkovxmxorlgwaomlswjxlpnbvltfv\nhykasjxqyjrmybejnmeumzha\ntuevlumpqbbhbww\nqgqsphvrmupxxc\ntrissbaf\nqfgrlinkzvzqdryckaizutd\nzzqtoaxkvwoscyx\noswytrlnhpjvvnwookx\nlpuzqgec\ngyzqfwxggtvpjhzmzmdw\nrlxjgmvdftvrmvbdwudra\nvsntnjpepnvdaxiporggmglhagv\nxlvcqkqgcrbgtgglj\nlyxwxbiszyhlsrgzeedzprbmcpduvq\nyrmqqvrkqskqukzqrwukpsifgtdc\nxpuohcsjhhuhvr\nvvlfrlxpvqejngwrbfbpmqeirxlw\nsvmasocxdvadmaxtrpakysmeaympy\nyuflqboqfdt",
"output": "lkpmx\nk26v\nh22a\nt13w\nq12c\ntrissbaf\nq21d\nz13x\no17x\nlpuzqgec\ng18w\nr19a\nv25v\nx15j\nl28q\ny26c\nx12r\nv26w\ns27y\ny9t"
},
{
"input": "100\nm\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr",
"output": "m\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr"
}
] | 1,686,159,136 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 15 | 0 | def wtl():
string1 = input("Enter")
if len(string1) > 10:
string2 = ""
string2+=string1[0]
string2+=str(len(string1)-2)
string2+=string1[-1]
print(string2)
else:
print(string1)
test = int(input())
for i in range(test):
wtl() | Title: Way Too Long Words
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
Output Specification:
Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data.
Demo Input:
['4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n']
Demo Output:
['word\nl10n\ni18n\np43s\n']
Note:
none | ```python
def wtl():
string1 = input("Enter")
if len(string1) > 10:
string2 = ""
string2+=string1[0]
string2+=str(len(string1)-2)
string2+=string1[-1]
print(string2)
else:
print(string1)
test = int(input())
for i in range(test):
wtl()
``` | 0 |
820 | A | Mister B and Book Reading | PROGRAMMING | 900 | [
"implementation"
] | null | null | Mister B once received a gift: it was a book about aliens, which he started read immediately. This book had *c* pages.
At first day Mister B read *v*0 pages, but after that he started to speed up. Every day, starting from the second, he read *a* pages more than on the previous day (at first day he read *v*0 pages, at second — *v*0<=+<=*a* pages, at third — *v*0<=+<=2*a* pages, and so on). But Mister B is just a human, so he physically wasn't able to read more than *v*1 pages per day.
Also, to refresh his memory, every day, starting from the second, Mister B had to reread last *l* pages he read on the previous day. Mister B finished the book when he read the last page for the first time.
Help Mister B to calculate how many days he needed to finish the book. | First and only line contains five space-separated integers: *c*, *v*0, *v*1, *a* and *l* (1<=≤<=*c*<=≤<=1000, 0<=≤<=*l*<=<<=*v*0<=≤<=*v*1<=≤<=1000, 0<=≤<=*a*<=≤<=1000) — the length of the book in pages, the initial reading speed, the maximum reading speed, the acceleration in reading speed and the number of pages for rereading. | Print one integer — the number of days Mister B needed to finish the book. | [
"5 5 10 5 4\n",
"12 4 12 4 1\n",
"15 1 100 0 0\n"
] | [
"1\n",
"3\n",
"15\n"
] | In the first sample test the book contains 5 pages, so Mister B read it right at the first day.
In the second sample test at first day Mister B read pages number 1 - 4, at second day — 4 - 11, at third day — 11 - 12 and finished the book.
In third sample test every day Mister B read 1 page of the book, so he finished in 15 days. | 500 | [
{
"input": "5 5 10 5 4",
"output": "1"
},
{
"input": "12 4 12 4 1",
"output": "3"
},
{
"input": "15 1 100 0 0",
"output": "15"
},
{
"input": "1 1 1 0 0",
"output": "1"
},
{
"input": "1000 999 1000 1000 998",
"output": "2"
},
{
"input": "1000 2 2 5 1",
"output": "999"
},
{
"input": "1000 1 1 1000 0",
"output": "1000"
},
{
"input": "737 41 74 12 11",
"output": "13"
},
{
"input": "1000 1000 1000 0 999",
"output": "1"
},
{
"input": "765 12 105 5 7",
"output": "17"
},
{
"input": "15 2 2 1000 0",
"output": "8"
},
{
"input": "1000 1 1000 1000 0",
"output": "2"
},
{
"input": "20 3 7 1 2",
"output": "6"
},
{
"input": "1000 500 500 1000 499",
"output": "501"
},
{
"input": "1 1000 1000 1000 0",
"output": "1"
},
{
"input": "1000 2 1000 56 0",
"output": "7"
},
{
"input": "1000 2 1000 802 0",
"output": "3"
},
{
"input": "16 1 8 2 0",
"output": "4"
},
{
"input": "20 6 10 2 2",
"output": "3"
},
{
"input": "8 2 12 4 1",
"output": "3"
},
{
"input": "8 6 13 2 5",
"output": "2"
},
{
"input": "70 4 20 87 0",
"output": "5"
},
{
"input": "97 8 13 234 5",
"output": "13"
},
{
"input": "16 4 23 8 3",
"output": "3"
},
{
"input": "65 7 22 7 4",
"output": "5"
},
{
"input": "93 10 18 11 7",
"output": "9"
},
{
"input": "86 13 19 15 9",
"output": "9"
},
{
"input": "333 17 50 10 16",
"output": "12"
},
{
"input": "881 16 55 10 12",
"output": "23"
},
{
"input": "528 11 84 3 9",
"output": "19"
},
{
"input": "896 2 184 8 1",
"output": "16"
},
{
"input": "236 10 930 9 8",
"output": "8"
},
{
"input": "784 1 550 14 0",
"output": "12"
},
{
"input": "506 1 10 4 0",
"output": "53"
},
{
"input": "460 1 3 2 0",
"output": "154"
},
{
"input": "701 1 3 1 0",
"output": "235"
},
{
"input": "100 49 50 1000 2",
"output": "3"
},
{
"input": "100 1 100 100 0",
"output": "2"
},
{
"input": "12 1 4 2 0",
"output": "4"
},
{
"input": "22 10 12 0 0",
"output": "3"
},
{
"input": "20 10 15 1 4",
"output": "3"
},
{
"input": "1000 5 10 1 4",
"output": "169"
},
{
"input": "1000 1 1000 1 0",
"output": "45"
},
{
"input": "4 1 2 2 0",
"output": "3"
},
{
"input": "1 5 5 1 1",
"output": "1"
},
{
"input": "19 10 11 0 2",
"output": "3"
},
{
"input": "1 2 3 0 0",
"output": "1"
},
{
"input": "10 1 4 10 0",
"output": "4"
},
{
"input": "20 3 100 1 1",
"output": "5"
},
{
"input": "1000 5 9 5 0",
"output": "112"
},
{
"input": "1 11 12 0 10",
"output": "1"
},
{
"input": "1 1 1 1 0",
"output": "1"
},
{
"input": "1000 1 20 1 0",
"output": "60"
},
{
"input": "9 1 4 2 0",
"output": "4"
},
{
"input": "129 2 3 4 0",
"output": "44"
},
{
"input": "4 2 2 0 1",
"output": "3"
},
{
"input": "1000 1 10 100 0",
"output": "101"
},
{
"input": "100 1 100 1 0",
"output": "14"
},
{
"input": "8 3 4 2 0",
"output": "3"
},
{
"input": "20 1 6 4 0",
"output": "5"
},
{
"input": "8 2 4 2 0",
"output": "3"
},
{
"input": "11 5 6 7 2",
"output": "3"
},
{
"input": "100 120 130 120 0",
"output": "1"
},
{
"input": "7 1 4 1 0",
"output": "4"
},
{
"input": "5 3 10 0 2",
"output": "3"
},
{
"input": "5 2 2 0 0",
"output": "3"
},
{
"input": "1000 10 1000 10 0",
"output": "14"
},
{
"input": "25 3 50 4 2",
"output": "4"
},
{
"input": "9 10 10 10 9",
"output": "1"
},
{
"input": "17 10 12 6 5",
"output": "2"
},
{
"input": "15 5 10 3 0",
"output": "3"
},
{
"input": "8 3 5 1 0",
"output": "3"
},
{
"input": "19 1 12 5 0",
"output": "4"
},
{
"input": "1000 10 1000 1 0",
"output": "37"
},
{
"input": "100 1 2 1000 0",
"output": "51"
},
{
"input": "20 10 11 1000 9",
"output": "6"
},
{
"input": "16 2 100 1 1",
"output": "5"
},
{
"input": "18 10 13 2 5",
"output": "3"
},
{
"input": "12 3 5 3 1",
"output": "4"
},
{
"input": "17 3 11 2 0",
"output": "4"
},
{
"input": "4 2 100 1 1",
"output": "2"
},
{
"input": "7 4 5 2 3",
"output": "3"
},
{
"input": "100 1 2 2 0",
"output": "51"
},
{
"input": "50 4 5 5 0",
"output": "11"
},
{
"input": "1 2 2 0 1",
"output": "1"
},
{
"input": "1000 2 3 10 1",
"output": "500"
},
{
"input": "500 10 500 1000 0",
"output": "2"
},
{
"input": "1000 4 12 1 0",
"output": "87"
},
{
"input": "18 10 13 1 5",
"output": "3"
},
{
"input": "7 3 6 2 2",
"output": "3"
},
{
"input": "15 5 100 1 2",
"output": "4"
},
{
"input": "100 1 10 1 0",
"output": "15"
},
{
"input": "8 2 7 5 1",
"output": "2"
},
{
"input": "11 2 4 1 1",
"output": "5"
},
{
"input": "1000 500 900 100 300",
"output": "3"
},
{
"input": "7 1 2 5 0",
"output": "4"
},
{
"input": "7 3 5 3 2",
"output": "3"
},
{
"input": "7 3 10 2 1",
"output": "2"
},
{
"input": "1000 501 510 1 499",
"output": "50"
},
{
"input": "1000 1 1000 2 0",
"output": "32"
},
{
"input": "1 5 5 0 0",
"output": "1"
},
{
"input": "18 10 15 1 5",
"output": "3"
},
{
"input": "100 4 1000 1 2",
"output": "13"
},
{
"input": "20 2 40 1 1",
"output": "6"
},
{
"input": "1 11 1000 100 1",
"output": "1"
},
{
"input": "6 4 4 1 2",
"output": "2"
},
{
"input": "8 3 5 3 1",
"output": "3"
},
{
"input": "10 5 7 1 2",
"output": "3"
},
{
"input": "400 100 198 1 99",
"output": "25"
},
{
"input": "3 1 2 5 0",
"output": "2"
}
] | 1,498,575,686 | 1,586 | Python 3 | WRONG_ANSWER | PRETESTS | 5 | 46 | 5,529,600 | import math
s = input()
arr = s.split()
totalPages = int(arr[0])
firstDayPages = int(arr[1])
maxPages = int(arr[2])
acceleration = int(arr[3])
reRead = int(arr[4])
def getPagesOnDay(day):
global firstDayPages, maxPages, acceleration
pages = firstDayPages + (day - 1) * acceleration
if day > 1:
pages = pages - reRead
if pages > maxPages:
return maxPages
else:
return pages
day = 0
readedPages = 0
while True:
day = day + 1
readedPages = readedPages + getPagesOnDay(day)
if readedPages >= totalPages:
break
print(day)
| Title: Mister B and Book Reading
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mister B once received a gift: it was a book about aliens, which he started read immediately. This book had *c* pages.
At first day Mister B read *v*0 pages, but after that he started to speed up. Every day, starting from the second, he read *a* pages more than on the previous day (at first day he read *v*0 pages, at second — *v*0<=+<=*a* pages, at third — *v*0<=+<=2*a* pages, and so on). But Mister B is just a human, so he physically wasn't able to read more than *v*1 pages per day.
Also, to refresh his memory, every day, starting from the second, Mister B had to reread last *l* pages he read on the previous day. Mister B finished the book when he read the last page for the first time.
Help Mister B to calculate how many days he needed to finish the book.
Input Specification:
First and only line contains five space-separated integers: *c*, *v*0, *v*1, *a* and *l* (1<=≤<=*c*<=≤<=1000, 0<=≤<=*l*<=<<=*v*0<=≤<=*v*1<=≤<=1000, 0<=≤<=*a*<=≤<=1000) — the length of the book in pages, the initial reading speed, the maximum reading speed, the acceleration in reading speed and the number of pages for rereading.
Output Specification:
Print one integer — the number of days Mister B needed to finish the book.
Demo Input:
['5 5 10 5 4\n', '12 4 12 4 1\n', '15 1 100 0 0\n']
Demo Output:
['1\n', '3\n', '15\n']
Note:
In the first sample test the book contains 5 pages, so Mister B read it right at the first day.
In the second sample test at first day Mister B read pages number 1 - 4, at second day — 4 - 11, at third day — 11 - 12 and finished the book.
In third sample test every day Mister B read 1 page of the book, so he finished in 15 days. | ```python
import math
s = input()
arr = s.split()
totalPages = int(arr[0])
firstDayPages = int(arr[1])
maxPages = int(arr[2])
acceleration = int(arr[3])
reRead = int(arr[4])
def getPagesOnDay(day):
global firstDayPages, maxPages, acceleration
pages = firstDayPages + (day - 1) * acceleration
if day > 1:
pages = pages - reRead
if pages > maxPages:
return maxPages
else:
return pages
day = 0
readedPages = 0
while True:
day = day + 1
readedPages = readedPages + getPagesOnDay(day)
if readedPages >= totalPages:
break
print(day)
``` | 0 |
|
375 | B | Maximum Submatrix 2 | PROGRAMMING | 1,600 | [
"data structures",
"dp",
"implementation",
"sortings"
] | null | null | You are given a matrix consisting of digits zero and one, its size is *n*<=×<=*m*. You are allowed to rearrange its rows. What is the maximum area of the submatrix that only consists of ones and can be obtained in the given problem by the described operations?
Let's assume that the rows of matrix *a* are numbered from 1 to *n* from top to bottom and the columns are numbered from 1 to *m* from left to right. A matrix cell on the intersection of the *i*-th row and the *j*-th column can be represented as (*i*,<=*j*). Formally, a submatrix of matrix *a* is a group of four integers *d*,<=*u*,<=*l*,<=*r* (1<=≤<=*d*<=≤<=*u*<=≤<=*n*; 1<=≤<=*l*<=≤<=*r*<=≤<=*m*). We will assume that the submatrix contains cells (*i*,<=*j*) (*d*<=≤<=*i*<=≤<=*u*; *l*<=≤<=*j*<=≤<=*r*). The area of the submatrix is the number of cells it contains. | The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=5000). Next *n* lines contain *m* characters each — matrix *a*. Matrix *a* only contains characters: "0" and "1". Note that the elements of the matrix follow without any spaces in the lines. | Print a single integer — the area of the maximum obtained submatrix. If we cannot obtain a matrix of numbers one, print 0. | [
"1 1\n1\n",
"2 2\n10\n11\n",
"4 3\n100\n011\n000\n101\n"
] | [
"1\n",
"2\n",
"2\n"
] | none | 1,000 | [
{
"input": "1 1\n1",
"output": "1"
},
{
"input": "2 2\n10\n11",
"output": "2"
},
{
"input": "4 3\n100\n011\n000\n101",
"output": "2"
},
{
"input": "11 16\n0111110101100011\n1000101100010000\n0010110110010101\n0110110010110010\n0011101101110000\n1001100011010111\n0010011111111000\n0100100100111110\n1001000000100111\n0110000011001000\n1011111011010000",
"output": "9"
},
{
"input": "19 12\n110001100110\n100100000000\n101011001111\n010111110001\n011000100100\n011111010000\n010011101100\n011010011110\n011001111110\n010111110001\n010000010111\n001111110100\n100100110001\n100110000000\n110000010010\n111101011101\n010111100000\n100000011010\n000100100101",
"output": "16"
},
{
"input": "13 19\n0000111111111111011\n0111000001110001101\n1110100110111011101\n0001101011100001110\n1101100100010000101\n1010100011110011010\n1010011101010000001\n1011101000001111000\n1101110001101011110\n0110101010001111100\n0001011010100111001\n1111101000110001000\n0010010000011100010",
"output": "14"
},
{
"input": "8 5\n00000\n00000\n00000\n00000\n00000\n00000\n00000\n00000",
"output": "0"
},
{
"input": "15 18\n111111111111111111\n111111111111111111\n111111111111111111\n111111111111111111\n111111111111111111\n111111111111111111\n111111111111111111\n111111111111111111\n111111111111111111\n111111111111111111\n111111111111111111\n111111111111111111\n111111111111111111\n111111111111111111\n111111111111111111",
"output": "270"
},
{
"input": "1 1\n0",
"output": "0"
}
] | 1,579,145,946 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 3 | 249 | 5,632,000 | # -*- coding:utf-8 -*-
"""
created by shuangquan.huang at 1/15/20
"""
import collections
import time
import os
import sys
import bisect
import heapq
from typing import List
def solve(N, M, A):
# for each index i of each row, we count how many consecutive '1' right at i,
# this is the max width if we include this row into the target submatrix which start from col i
for r in range(N):
for c in range(1, M):
if A[r][c] != 0:
A[r][c] = A[r][c-1] + 1
ans = 0
# we fix the right index of the sub-matrix, and scrub it to right
heights = [0 for _ in range(M+1)]
for right in range(M):
for row in A:
heights[row[right]] += 1
count = 0
for w in range(M+1):
if heights[w] > 0:
ans = max(ans, (N-count) * w)
count += heights[w]
print(ans)
N, M = map(int, input().split())
A = []
for i in range(N):
row = [int(x) for x in list(input())]
A.append(row)
solve(N, M, A) | Title: Maximum Submatrix 2
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a matrix consisting of digits zero and one, its size is *n*<=×<=*m*. You are allowed to rearrange its rows. What is the maximum area of the submatrix that only consists of ones and can be obtained in the given problem by the described operations?
Let's assume that the rows of matrix *a* are numbered from 1 to *n* from top to bottom and the columns are numbered from 1 to *m* from left to right. A matrix cell on the intersection of the *i*-th row and the *j*-th column can be represented as (*i*,<=*j*). Formally, a submatrix of matrix *a* is a group of four integers *d*,<=*u*,<=*l*,<=*r* (1<=≤<=*d*<=≤<=*u*<=≤<=*n*; 1<=≤<=*l*<=≤<=*r*<=≤<=*m*). We will assume that the submatrix contains cells (*i*,<=*j*) (*d*<=≤<=*i*<=≤<=*u*; *l*<=≤<=*j*<=≤<=*r*). The area of the submatrix is the number of cells it contains.
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=5000). Next *n* lines contain *m* characters each — matrix *a*. Matrix *a* only contains characters: "0" and "1". Note that the elements of the matrix follow without any spaces in the lines.
Output Specification:
Print a single integer — the area of the maximum obtained submatrix. If we cannot obtain a matrix of numbers one, print 0.
Demo Input:
['1 1\n1\n', '2 2\n10\n11\n', '4 3\n100\n011\n000\n101\n']
Demo Output:
['1\n', '2\n', '2\n']
Note:
none | ```python
# -*- coding:utf-8 -*-
"""
created by shuangquan.huang at 1/15/20
"""
import collections
import time
import os
import sys
import bisect
import heapq
from typing import List
def solve(N, M, A):
# for each index i of each row, we count how many consecutive '1' right at i,
# this is the max width if we include this row into the target submatrix which start from col i
for r in range(N):
for c in range(1, M):
if A[r][c] != 0:
A[r][c] = A[r][c-1] + 1
ans = 0
# we fix the right index of the sub-matrix, and scrub it to right
heights = [0 for _ in range(M+1)]
for right in range(M):
for row in A:
heights[row[right]] += 1
count = 0
for w in range(M+1):
if heights[w] > 0:
ans = max(ans, (N-count) * w)
count += heights[w]
print(ans)
N, M = map(int, input().split())
A = []
for i in range(N):
row = [int(x) for x in list(input())]
A.append(row)
solve(N, M, A)
``` | 0 |
|
875 | A | Classroom Watch | PROGRAMMING | 1,200 | [
"brute force",
"math"
] | null | null | Eighth-grader Vova is on duty today in the class. After classes, he went into the office to wash the board, and found on it the number *n*. He asked what is this number and the teacher of mathematics Inna Petrovna answered Vova that *n* is the answer to the arithmetic task for first-graders. In the textbook, a certain positive integer *x* was given. The task was to add *x* to the sum of the digits of the number *x* written in decimal numeral system.
Since the number *n* on the board was small, Vova quickly guessed which *x* could be in the textbook. Now he wants to get a program which will search for arbitrary values of the number *n* for all suitable values of *x* or determine that such *x* does not exist. Write such a program for Vova. | The first line contains integer *n* (1<=≤<=*n*<=≤<=109). | In the first line print one integer *k* — number of different values of *x* satisfying the condition.
In next *k* lines print these values in ascending order. | [
"21\n",
"20\n"
] | [
"1\n15\n",
"0\n"
] | In the first test case *x* = 15 there is only one variant: 15 + 1 + 5 = 21.
In the second test case there are no such *x*. | 500 | [
{
"input": "21",
"output": "1\n15"
},
{
"input": "20",
"output": "0"
},
{
"input": "1",
"output": "0"
},
{
"input": "2",
"output": "1\n1"
},
{
"input": "3",
"output": "0"
},
{
"input": "100000001",
"output": "2\n99999937\n100000000"
},
{
"input": "1000000000",
"output": "1\n999999932"
},
{
"input": "999999979",
"output": "2\n999999899\n999999908"
},
{
"input": "9",
"output": "0"
},
{
"input": "10",
"output": "1\n5"
},
{
"input": "11",
"output": "1\n10"
},
{
"input": "39",
"output": "1\n33"
},
{
"input": "66",
"output": "1\n60"
},
{
"input": "75",
"output": "0"
},
{
"input": "100",
"output": "1\n86"
},
{
"input": "101",
"output": "2\n91\n100"
},
{
"input": "2014",
"output": "2\n1988\n2006"
},
{
"input": "999999994",
"output": "0"
}
] | 1,643,959,123 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 5 | 1,000 | 0 | n = input()
total = 0;
numPossiveis = [];
for num in range(int(n), -1, -1):
totalDigitos = sum(int(i) for i in str(num));
if((num + totalDigitos) == int(n)):
numPossiveis.append(num);
print(len(numPossiveis))
if(len(numPossiveis) > 0):
sorted(numPossiveis);
for key in numPossiveis:
print(key)
| Title: Classroom Watch
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Eighth-grader Vova is on duty today in the class. After classes, he went into the office to wash the board, and found on it the number *n*. He asked what is this number and the teacher of mathematics Inna Petrovna answered Vova that *n* is the answer to the arithmetic task for first-graders. In the textbook, a certain positive integer *x* was given. The task was to add *x* to the sum of the digits of the number *x* written in decimal numeral system.
Since the number *n* on the board was small, Vova quickly guessed which *x* could be in the textbook. Now he wants to get a program which will search for arbitrary values of the number *n* for all suitable values of *x* or determine that such *x* does not exist. Write such a program for Vova.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=109).
Output Specification:
In the first line print one integer *k* — number of different values of *x* satisfying the condition.
In next *k* lines print these values in ascending order.
Demo Input:
['21\n', '20\n']
Demo Output:
['1\n15\n', '0\n']
Note:
In the first test case *x* = 15 there is only one variant: 15 + 1 + 5 = 21.
In the second test case there are no such *x*. | ```python
n = input()
total = 0;
numPossiveis = [];
for num in range(int(n), -1, -1):
totalDigitos = sum(int(i) for i in str(num));
if((num + totalDigitos) == int(n)):
numPossiveis.append(num);
print(len(numPossiveis))
if(len(numPossiveis) > 0):
sorted(numPossiveis);
for key in numPossiveis:
print(key)
``` | 0 |
|
522 | B | Photo to Remember | PROGRAMMING | 1,100 | [
"*special",
"data structures",
"dp",
"implementation"
] | null | null | One day *n* friends met at a party, they hadn't seen each other for a long time and so they decided to make a group photo together.
Simply speaking, the process of taking photos can be described as follows. On the photo, each photographed friend occupies a rectangle of pixels: the *i*-th of them occupies the rectangle of width *w**i* pixels and height *h**i* pixels. On the group photo everybody stands in a line, thus the minimum pixel size of the photo including all the photographed friends, is *W*<=×<=*H*, where *W* is the total sum of all widths and *H* is the maximum height of all the photographed friends.
As is usually the case, the friends made *n* photos — the *j*-th (1<=≤<=*j*<=≤<=*n*) photo had everybody except for the *j*-th friend as he was the photographer.
Print the minimum size of each made photo in pixels. | The first line contains integer *n* (2<=≤<=*n*<=≤<=200<=000) — the number of friends.
Then *n* lines follow: the *i*-th line contains information about the *i*-th friend. The line contains a pair of integers *w**i*,<=*h**i* (1<=≤<=*w**i*<=≤<=10,<=1<=≤<=*h**i*<=≤<=1000) — the width and height in pixels of the corresponding rectangle. | Print *n* space-separated numbers *b*1,<=*b*2,<=...,<=*b**n*, where *b**i* — the total number of pixels on the minimum photo containing all friends expect for the *i*-th one. | [
"3\n1 10\n5 5\n10 1\n",
"3\n2 1\n1 2\n2 1\n"
] | [
"75 110 60 ",
"6 4 6 "
] | none | 1,000 | [
{
"input": "3\n1 10\n5 5\n10 1",
"output": "75 110 60 "
},
{
"input": "3\n2 1\n1 2\n2 1",
"output": "6 4 6 "
},
{
"input": "2\n1 5\n2 3",
"output": "6 5 "
},
{
"input": "2\n2 3\n1 1",
"output": "1 6 "
},
{
"input": "3\n1 10\n2 10\n3 10",
"output": "50 40 30 "
},
{
"input": "3\n2 10\n1 9\n3 7",
"output": "36 50 30 "
},
{
"input": "3\n1 1\n3 2\n2 3",
"output": "15 9 8 "
},
{
"input": "3\n3 123\n1 456\n2 789",
"output": "2367 3945 1824 "
},
{
"input": "3\n2 987\n3 654\n1 321",
"output": "2616 2961 4935 "
},
{
"input": "3\n3 143\n2 543\n1 893",
"output": "2679 3572 2715 "
},
{
"input": "2\n1 1\n1 2",
"output": "2 1 "
},
{
"input": "3\n2 22\n1 11\n2 22",
"output": "66 88 66 "
},
{
"input": "3\n1 11\n1 12\n1 13",
"output": "26 26 24 "
},
{
"input": "3\n1 11\n1 12\n2 10",
"output": "36 33 24 "
},
{
"input": "10\n6 20\n1 175\n1 758\n1 169\n2 490\n2 600\n4 463\n7 377\n9 40\n4 961",
"output": "29791 34596 34596 34596 33635 33635 31713 28830 26908 25014 "
},
{
"input": "10\n8 158\n1 709\n6 766\n4 335\n5 356\n2 972\n1 108\n4 235\n3 631\n1 414",
"output": "26244 33048 28188 30132 29160 25278 33048 30132 31104 33048 "
},
{
"input": "10\n7 549\n9 115\n8 141\n3 650\n5 730\n3 841\n7 18\n9 170\n2 217\n1 155",
"output": "39527 37845 38686 42891 41209 37230 39527 37845 43732 44573 "
},
{
"input": "10\n6 386\n9 816\n9 268\n9 481\n8 284\n10 715\n9 351\n7 580\n4 327\n7 392",
"output": "58752 49335 56304 56304 57120 55488 56304 57936 60384 57936 "
},
{
"input": "10\n9 292\n4 6\n6 638\n8 461\n10 970\n10 488\n9 769\n10 644\n8 280\n5 334",
"output": "67900 72750 70810 68870 53061 66930 67900 66930 68870 71780 "
},
{
"input": "10\n10 1000\n10 1000\n10 1000\n10 1000\n10 1000\n10 1000\n10 1000\n10 1000\n10 1000\n10 1000",
"output": "90000 90000 90000 90000 90000 90000 90000 90000 90000 90000 "
}
] | 1,645,944,680 | 2,147,483,647 | PyPy 3 | TIME_LIMIT_EXCEEDED | TESTS | 14 | 2,000 | 38,707,200 | n = int(input())
wd = {}
hd = {}
for i in range(1,n+1):
wd[i] = 0
hd[i] = 0
for i in range(1,n+1):
w,h = map(int,input().split(" "))
for j in range(1,n+1):
if j != i:
wd[j] += w
hd[j] = max(hd[j], h)
for i in range(1,n+1):
print(wd[i]*hd[i], end=" ") | Title: Photo to Remember
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day *n* friends met at a party, they hadn't seen each other for a long time and so they decided to make a group photo together.
Simply speaking, the process of taking photos can be described as follows. On the photo, each photographed friend occupies a rectangle of pixels: the *i*-th of them occupies the rectangle of width *w**i* pixels and height *h**i* pixels. On the group photo everybody stands in a line, thus the minimum pixel size of the photo including all the photographed friends, is *W*<=×<=*H*, where *W* is the total sum of all widths and *H* is the maximum height of all the photographed friends.
As is usually the case, the friends made *n* photos — the *j*-th (1<=≤<=*j*<=≤<=*n*) photo had everybody except for the *j*-th friend as he was the photographer.
Print the minimum size of each made photo in pixels.
Input Specification:
The first line contains integer *n* (2<=≤<=*n*<=≤<=200<=000) — the number of friends.
Then *n* lines follow: the *i*-th line contains information about the *i*-th friend. The line contains a pair of integers *w**i*,<=*h**i* (1<=≤<=*w**i*<=≤<=10,<=1<=≤<=*h**i*<=≤<=1000) — the width and height in pixels of the corresponding rectangle.
Output Specification:
Print *n* space-separated numbers *b*1,<=*b*2,<=...,<=*b**n*, where *b**i* — the total number of pixels on the minimum photo containing all friends expect for the *i*-th one.
Demo Input:
['3\n1 10\n5 5\n10 1\n', '3\n2 1\n1 2\n2 1\n']
Demo Output:
['75 110 60 ', '6 4 6 ']
Note:
none | ```python
n = int(input())
wd = {}
hd = {}
for i in range(1,n+1):
wd[i] = 0
hd[i] = 0
for i in range(1,n+1):
w,h = map(int,input().split(" "))
for j in range(1,n+1):
if j != i:
wd[j] += w
hd[j] = max(hd[j], h)
for i in range(1,n+1):
print(wd[i]*hd[i], end=" ")
``` | 0 |
|
615 | A | Bulbs | PROGRAMMING | 800 | [
"implementation"
] | null | null | Vasya wants to turn on Christmas lights consisting of *m* bulbs. Initially, all bulbs are turned off. There are *n* buttons, each of them is connected to some set of bulbs. Vasya can press any of these buttons. When the button is pressed, it turns on all the bulbs it's connected to. Can Vasya light up all the bulbs?
If Vasya presses the button such that some bulbs connected to it are already turned on, they do not change their state, i.e. remain turned on. | The first line of the input contains integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of buttons and the number of bulbs respectively.
Each of the next *n* lines contains *x**i* (0<=≤<=*x**i*<=≤<=*m*) — the number of bulbs that are turned on by the *i*-th button, and then *x**i* numbers *y**ij* (1<=≤<=*y**ij*<=≤<=*m*) — the numbers of these bulbs. | If it's possible to turn on all *m* bulbs print "YES", otherwise print "NO". | [
"3 4\n2 1 4\n3 1 3 1\n1 2\n",
"3 3\n1 1\n1 2\n1 1\n"
] | [
"YES\n",
"NO\n"
] | In the first sample you can press each button once and turn on all the bulbs. In the 2 sample it is impossible to turn on the 3-rd lamp. | 500 | [
{
"input": "3 4\n2 1 4\n3 1 3 1\n1 2",
"output": "YES"
},
{
"input": "3 3\n1 1\n1 2\n1 1",
"output": "NO"
},
{
"input": "3 4\n1 1\n1 2\n1 3",
"output": "NO"
},
{
"input": "1 5\n5 1 2 3 4 5",
"output": "YES"
},
{
"input": "1 5\n5 4 4 1 2 3",
"output": "NO"
},
{
"input": "1 5\n5 1 1 1 1 5",
"output": "NO"
},
{
"input": "2 5\n4 3 1 4 2\n4 2 3 4 5",
"output": "YES"
},
{
"input": "5 7\n2 6 7\n5 1 1 1 1 1\n3 6 5 4\n0\n4 4 3 2 1",
"output": "YES"
},
{
"input": "100 100\n0\n0\n0\n1 53\n0\n0\n1 34\n1 54\n0\n1 14\n0\n1 33\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n1 82\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n1 34\n0\n0\n1 26\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n1 34\n0\n0\n0\n0\n0\n1 3\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n1 40\n0\n0\n0\n1 26\n0\n0\n0\n0\n0\n1 97\n0\n1 5\n0\n0\n0\n0\n0",
"output": "NO"
},
{
"input": "100 100\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0",
"output": "NO"
},
{
"input": "5 6\n3 1 2 6\n3 1 2 6\n1 1\n2 3 4\n3 1 5 6",
"output": "YES"
},
{
"input": "5 2\n1 1\n1 1\n1 1\n1 1\n1 1",
"output": "NO"
},
{
"input": "1 4\n3 1 2 3",
"output": "NO"
},
{
"input": "1 4\n3 2 3 4",
"output": "NO"
},
{
"input": "2 4\n3 2 3 4\n1 1",
"output": "YES"
},
{
"input": "2 4\n3 1 2 3\n1 4",
"output": "YES"
},
{
"input": "5 1\n0\n0\n0\n0\n0",
"output": "NO"
},
{
"input": "1 1\n0",
"output": "NO"
},
{
"input": "1 10\n10 1 2 3 4 5 6 7 8 9 10",
"output": "YES"
},
{
"input": "1 1\n1 1",
"output": "YES"
},
{
"input": "1 100\n99 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99",
"output": "NO"
},
{
"input": "1 3\n3 1 2 1",
"output": "NO"
},
{
"input": "1 100\n100 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100",
"output": "YES"
}
] | 1,665,845,310 | 2,147,483,647 | Python 3 | OK | TESTS | 45 | 46 | 0 | n,m=map(int,input().split())
p=[]
po=[]
for i in range(m):
p.append('0')
po.append('1')
for i in range(1,n+1):
q=input().split()
um=q[0]
for ui in range(1,int(um)+1):
p[int(q[ui])-1]='1'
if p==po:
print('YES')
else:print('NO') | Title: Bulbs
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya wants to turn on Christmas lights consisting of *m* bulbs. Initially, all bulbs are turned off. There are *n* buttons, each of them is connected to some set of bulbs. Vasya can press any of these buttons. When the button is pressed, it turns on all the bulbs it's connected to. Can Vasya light up all the bulbs?
If Vasya presses the button such that some bulbs connected to it are already turned on, they do not change their state, i.e. remain turned on.
Input Specification:
The first line of the input contains integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of buttons and the number of bulbs respectively.
Each of the next *n* lines contains *x**i* (0<=≤<=*x**i*<=≤<=*m*) — the number of bulbs that are turned on by the *i*-th button, and then *x**i* numbers *y**ij* (1<=≤<=*y**ij*<=≤<=*m*) — the numbers of these bulbs.
Output Specification:
If it's possible to turn on all *m* bulbs print "YES", otherwise print "NO".
Demo Input:
['3 4\n2 1 4\n3 1 3 1\n1 2\n', '3 3\n1 1\n1 2\n1 1\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first sample you can press each button once and turn on all the bulbs. In the 2 sample it is impossible to turn on the 3-rd lamp. | ```python
n,m=map(int,input().split())
p=[]
po=[]
for i in range(m):
p.append('0')
po.append('1')
for i in range(1,n+1):
q=input().split()
um=q[0]
for ui in range(1,int(um)+1):
p[int(q[ui])-1]='1'
if p==po:
print('YES')
else:print('NO')
``` | 3 |
|
1,003 | A | Polycarp's Pockets | PROGRAMMING | 800 | [
"implementation"
] | null | null | Polycarp has $n$ coins, the value of the $i$-th coin is $a_i$. Polycarp wants to distribute all the coins between his pockets, but he cannot put two coins with the same value into the same pocket.
For example, if Polycarp has got six coins represented as an array $a = [1, 2, 4, 3, 3, 2]$, he can distribute the coins into two pockets as follows: $[1, 2, 3], [2, 3, 4]$.
Polycarp wants to distribute all the coins with the minimum number of used pockets. Help him to do that. | The first line of the input contains one integer $n$ ($1 \le n \le 100$) — the number of coins.
The second line of the input contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 100$) — values of coins. | Print only one integer — the minimum number of pockets Polycarp needs to distribute all the coins so no two coins with the same value are put into the same pocket. | [
"6\n1 2 4 3 3 2\n",
"1\n100\n"
] | [
"2\n",
"1\n"
] | none | 0 | [
{
"input": "6\n1 2 4 3 3 2",
"output": "2"
},
{
"input": "1\n100",
"output": "1"
},
{
"input": "100\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100",
"output": "100"
},
{
"input": "100\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "100"
},
{
"input": "100\n59 47 39 47 47 71 47 28 58 47 35 79 58 47 38 47 47 47 47 27 47 43 29 95 47 49 46 71 47 74 79 47 47 32 45 67 47 47 30 37 47 47 16 67 22 76 47 86 84 10 5 47 47 47 47 47 1 51 47 54 47 8 47 47 9 47 47 47 47 28 47 47 26 47 47 47 47 47 47 92 47 47 77 47 47 24 45 47 10 47 47 89 47 27 47 89 47 67 24 71",
"output": "51"
},
{
"input": "100\n45 99 10 27 16 85 39 38 17 32 15 23 67 48 50 97 42 70 62 30 44 81 64 73 34 22 46 5 83 52 58 60 33 74 47 88 18 61 78 53 25 95 94 31 3 75 1 57 20 54 59 9 68 7 77 43 21 87 86 24 4 80 11 49 2 72 36 84 71 8 65 55 79 100 41 14 35 89 66 69 93 37 56 82 90 91 51 19 26 92 6 96 13 98 12 28 76 40 63 29",
"output": "1"
},
{
"input": "100\n45 29 5 2 6 50 22 36 14 15 9 48 46 20 8 37 7 47 12 50 21 38 18 27 33 19 40 10 5 49 38 42 34 37 27 30 35 24 10 3 40 49 41 3 4 44 13 25 28 31 46 36 23 1 1 23 7 22 35 26 21 16 48 42 32 8 11 16 34 11 39 32 47 28 43 41 39 4 14 19 26 45 13 18 15 25 2 44 17 29 17 33 43 6 12 30 9 20 31 24",
"output": "2"
},
{
"input": "50\n7 7 3 3 7 4 5 6 4 3 7 5 6 4 5 4 4 5 6 7 7 7 4 5 5 5 3 7 6 3 4 6 3 6 4 4 5 4 6 6 3 5 6 3 5 3 3 7 7 6",
"output": "10"
},
{
"input": "100\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 99 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100",
"output": "99"
},
{
"input": "7\n1 2 3 3 3 1 2",
"output": "3"
},
{
"input": "5\n1 2 3 4 5",
"output": "1"
},
{
"input": "7\n1 2 3 4 5 6 7",
"output": "1"
},
{
"input": "8\n1 2 3 4 5 6 7 8",
"output": "1"
},
{
"input": "9\n1 2 3 4 5 6 7 8 9",
"output": "1"
},
{
"input": "10\n1 2 3 4 5 6 7 8 9 10",
"output": "1"
},
{
"input": "3\n2 1 1",
"output": "2"
},
{
"input": "11\n1 2 3 4 5 6 7 8 9 1 1",
"output": "3"
},
{
"input": "12\n1 2 1 1 1 1 1 1 1 1 1 1",
"output": "11"
},
{
"input": "13\n1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "13"
},
{
"input": "14\n1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "14"
},
{
"input": "15\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "15"
},
{
"input": "16\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "16"
},
{
"input": "3\n1 1 1",
"output": "3"
},
{
"input": "3\n1 2 3",
"output": "1"
},
{
"input": "10\n1 1 1 1 2 2 1 1 9 10",
"output": "6"
},
{
"input": "2\n1 1",
"output": "2"
},
{
"input": "56\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "56"
},
{
"input": "99\n35 96 73 72 70 83 22 93 98 75 45 32 81 82 45 54 25 7 53 72 29 2 94 19 21 98 34 28 39 99 55 85 44 23 6 47 98 2 33 34 19 57 49 35 67 4 60 4 4 23 55 6 57 66 16 68 34 45 84 79 48 63 4 9 46 88 98 13 19 27 83 12 4 63 57 22 44 77 44 62 28 52 44 64 9 24 55 22 48 4 2 9 80 76 45 1 56 22 92",
"output": "6"
},
{
"input": "10\n1 2 2 3 3 3 4 4 4 4",
"output": "4"
},
{
"input": "99\n97 44 33 56 42 10 61 85 64 26 40 39 82 34 75 9 51 51 39 73 58 38 74 31 13 99 58 1 28 89 76 19 52 7 40 56 12 27 72 72 67 75 62 46 22 55 35 16 18 39 60 63 92 42 85 69 34 61 73 50 57 95 30 4 45 63 76 58 32 35 48 81 10 78 95 79 55 97 21 21 22 94 30 17 78 57 89 93 100 44 16 89 68 55 19 46 42 73 21",
"output": "3"
},
{
"input": "5\n5 5 5 5 1",
"output": "4"
},
{
"input": "6\n2 3 2 5 2 6",
"output": "3"
},
{
"input": "3\n58 59 58",
"output": "2"
},
{
"input": "9\n1 2 3 4 5 6 7 8 8",
"output": "2"
},
{
"input": "97\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "97"
},
{
"input": "3\n95 95 4",
"output": "2"
},
{
"input": "3\n2 2 5",
"output": "2"
}
] | 1,621,411,997 | 2,147,483,647 | Python 3 | OK | TESTS | 37 | 77 | 102,400 | import collections
n = int(input())
c = collections.Counter(int(x) for x in input().split())
print(max(c.values())) | Title: Polycarp's Pockets
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp has $n$ coins, the value of the $i$-th coin is $a_i$. Polycarp wants to distribute all the coins between his pockets, but he cannot put two coins with the same value into the same pocket.
For example, if Polycarp has got six coins represented as an array $a = [1, 2, 4, 3, 3, 2]$, he can distribute the coins into two pockets as follows: $[1, 2, 3], [2, 3, 4]$.
Polycarp wants to distribute all the coins with the minimum number of used pockets. Help him to do that.
Input Specification:
The first line of the input contains one integer $n$ ($1 \le n \le 100$) — the number of coins.
The second line of the input contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 100$) — values of coins.
Output Specification:
Print only one integer — the minimum number of pockets Polycarp needs to distribute all the coins so no two coins with the same value are put into the same pocket.
Demo Input:
['6\n1 2 4 3 3 2\n', '1\n100\n']
Demo Output:
['2\n', '1\n']
Note:
none | ```python
import collections
n = int(input())
c = collections.Counter(int(x) for x in input().split())
print(max(c.values()))
``` | 3 |
|
432 | A | Choosing Teams | PROGRAMMING | 800 | [
"greedy",
"implementation",
"sortings"
] | null | null | The Saratov State University Olympiad Programmers Training Center (SSU OPTC) has *n* students. For each student you know the number of times he/she has participated in the ACM ICPC world programming championship. According to the ACM ICPC rules, each person can participate in the world championship at most 5 times.
The head of the SSU OPTC is recently gathering teams to participate in the world championship. Each team must consist of exactly three people, at that, any person cannot be a member of two or more teams. What maximum number of teams can the head make if he wants each team to participate in the world championship with the same members at least *k* times? | The first line contains two integers, *n* and *k* (1<=≤<=*n*<=≤<=2000; 1<=≤<=*k*<=≤<=5). The next line contains *n* integers: *y*1,<=*y*2,<=...,<=*y**n* (0<=≤<=*y**i*<=≤<=5), where *y**i* shows the number of times the *i*-th person participated in the ACM ICPC world championship. | Print a single number — the answer to the problem. | [
"5 2\n0 4 5 1 0\n",
"6 4\n0 1 2 3 4 5\n",
"6 5\n0 0 0 0 0 0\n"
] | [
"1\n",
"0\n",
"2\n"
] | In the first sample only one team could be made: the first, the fourth and the fifth participants.
In the second sample no teams could be created.
In the third sample two teams could be created. Any partition into two teams fits. | 500 | [
{
"input": "5 2\n0 4 5 1 0",
"output": "1"
},
{
"input": "6 4\n0 1 2 3 4 5",
"output": "0"
},
{
"input": "6 5\n0 0 0 0 0 0",
"output": "2"
},
{
"input": "3 4\n0 1 0",
"output": "1"
},
{
"input": "3 4\n0 2 0",
"output": "0"
},
{
"input": "6 5\n0 0 0 0 0 0",
"output": "2"
},
{
"input": "12 2\n0 1 2 3 4 0 1 2 3 4 0 1",
"output": "3"
},
{
"input": "15 2\n0 1 2 3 4 0 1 2 3 4 0 1 2 3 4",
"output": "4"
},
{
"input": "13 1\n5 0 5 0 1 5 0 4 1 1 1 4 1",
"output": "3"
},
{
"input": "20 1\n5 0 4 2 2 3 2 1 2 4 1 3 5 5 5 4 4 1 3 0",
"output": "5"
},
{
"input": "20 2\n1 4 5 1 3 4 0 5 0 3 3 0 4 0 0 1 4 2 0 4",
"output": "4"
},
{
"input": "20 2\n1 4 5 1 3 4 0 5 0 3 3 0 4 0 0 1 4 2 0 4",
"output": "4"
},
{
"input": "20 2\n1 4 5 1 3 4 0 5 0 3 3 0 4 0 0 1 4 2 0 4",
"output": "4"
},
{
"input": "20 2\n1 4 5 1 3 4 0 5 0 3 3 0 4 0 0 1 4 2 0 4",
"output": "4"
},
{
"input": "6 3\n4 4 4 4 4 4",
"output": "0"
},
{
"input": "1 4\n1",
"output": "0"
},
{
"input": "1 1\n0",
"output": "0"
},
{
"input": "2 3\n0 0",
"output": "0"
}
] | 1,684,596,500 | 2,147,483,647 | Python 3 | OK | TESTS | 35 | 46 | 0 | n,k=map(int,input().split())
l=list(map(int,input().split()))
re=5-k
c=0
for i in l:
if(i <= re):
c+=1
print(c//3) | Title: Choosing Teams
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Saratov State University Olympiad Programmers Training Center (SSU OPTC) has *n* students. For each student you know the number of times he/she has participated in the ACM ICPC world programming championship. According to the ACM ICPC rules, each person can participate in the world championship at most 5 times.
The head of the SSU OPTC is recently gathering teams to participate in the world championship. Each team must consist of exactly three people, at that, any person cannot be a member of two or more teams. What maximum number of teams can the head make if he wants each team to participate in the world championship with the same members at least *k* times?
Input Specification:
The first line contains two integers, *n* and *k* (1<=≤<=*n*<=≤<=2000; 1<=≤<=*k*<=≤<=5). The next line contains *n* integers: *y*1,<=*y*2,<=...,<=*y**n* (0<=≤<=*y**i*<=≤<=5), where *y**i* shows the number of times the *i*-th person participated in the ACM ICPC world championship.
Output Specification:
Print a single number — the answer to the problem.
Demo Input:
['5 2\n0 4 5 1 0\n', '6 4\n0 1 2 3 4 5\n', '6 5\n0 0 0 0 0 0\n']
Demo Output:
['1\n', '0\n', '2\n']
Note:
In the first sample only one team could be made: the first, the fourth and the fifth participants.
In the second sample no teams could be created.
In the third sample two teams could be created. Any partition into two teams fits. | ```python
n,k=map(int,input().split())
l=list(map(int,input().split()))
re=5-k
c=0
for i in l:
if(i <= re):
c+=1
print(c//3)
``` | 3 |
|
982 | A | Row | PROGRAMMING | 1,200 | [
"brute force",
"constructive algorithms"
] | null | null | You're given a row with $n$ chairs. We call a seating of people "maximal" if the two following conditions hold:
1. There are no neighbors adjacent to anyone seated. 1. It's impossible to seat one more person without violating the first rule.
The seating is given as a string consisting of zeros and ones ($0$ means that the corresponding seat is empty, $1$ — occupied). The goal is to determine whether this seating is "maximal".
Note that the first and last seats are not adjacent (if $n \ne 2$). | The first line contains a single integer $n$ ($1 \leq n \leq 1000$) — the number of chairs.
The next line contains a string of $n$ characters, each of them is either zero or one, describing the seating. | Output "Yes" (without quotation marks) if the seating is "maximal". Otherwise print "No".
You are allowed to print letters in whatever case you'd like (uppercase or lowercase). | [
"3\n101\n",
"4\n1011\n",
"5\n10001\n"
] | [
"Yes\n",
"No\n",
"No\n"
] | In sample case one the given seating is maximal.
In sample case two the person at chair three has a neighbour to the right.
In sample case three it is possible to seat yet another person into chair three. | 500 | [
{
"input": "3\n101",
"output": "Yes"
},
{
"input": "4\n1011",
"output": "No"
},
{
"input": "5\n10001",
"output": "No"
},
{
"input": "1\n0",
"output": "No"
},
{
"input": "1\n1",
"output": "Yes"
},
{
"input": "100\n0101001010101001010010010101001010100101001001001010010101010010101001001010101001001001010100101010",
"output": "Yes"
},
{
"input": "4\n0100",
"output": "No"
},
{
"input": "42\n011000100101001001101011011010100010011010",
"output": "No"
},
{
"input": "3\n001",
"output": "No"
},
{
"input": "64\n1001001010010010100101010010010100100101001001001001010100101001",
"output": "Yes"
},
{
"input": "3\n111",
"output": "No"
},
{
"input": "4\n0000",
"output": "No"
},
{
"input": "4\n0001",
"output": "No"
},
{
"input": "4\n0010",
"output": "No"
},
{
"input": "4\n0011",
"output": "No"
},
{
"input": "4\n0101",
"output": "Yes"
},
{
"input": "4\n0110",
"output": "No"
},
{
"input": "4\n0111",
"output": "No"
},
{
"input": "4\n1000",
"output": "No"
},
{
"input": "4\n1001",
"output": "Yes"
},
{
"input": "4\n1010",
"output": "Yes"
},
{
"input": "4\n1100",
"output": "No"
},
{
"input": "4\n1101",
"output": "No"
},
{
"input": "4\n1110",
"output": "No"
},
{
"input": "4\n1111",
"output": "No"
},
{
"input": "2\n00",
"output": "No"
},
{
"input": "2\n01",
"output": "Yes"
},
{
"input": "2\n10",
"output": "Yes"
},
{
"input": "2\n11",
"output": "No"
},
{
"input": "3\n000",
"output": "No"
},
{
"input": "3\n010",
"output": "Yes"
},
{
"input": "3\n011",
"output": "No"
},
{
"input": "3\n100",
"output": "No"
},
{
"input": "3\n110",
"output": "No"
},
{
"input": "100\n0111001010101110001100000010011000100101110010001100111110101110001110101010111000010010011000000110",
"output": "No"
},
{
"input": "357\n100101010010010010010100101001001010101010100100100100101001010101001010010100101001010100101001010010100100101001010101010101001001010100101010010100101001010100100100101010010010010100101010010010101001010010010101001001010010010101010100100101010010100100101001010100101001010100101001010010010010100101001010100100100100100100100101010101010010010100101",
"output": "Yes"
},
{
"input": "459\n000111000101010000100001001010111110011011010001100101111010111011101110111101111101100101100011011001100110001001111001101000111001011100110100011111011111000010000110010011100110011011111110011100001101001111000100111011001000001011111100110100001001001100101011100001110110100101011011110100100111101011000101110000100110100100010000000100001001111111000011101010010011001111010111001100000100111001010111011010000011000011100101101011101000011011000110011",
"output": "No"
},
{
"input": "3\n001",
"output": "No"
},
{
"input": "4\n0010",
"output": "No"
},
{
"input": "3\n111",
"output": "No"
},
{
"input": "4\n0110",
"output": "No"
},
{
"input": "2\n11",
"output": "No"
},
{
"input": "2\n10",
"output": "Yes"
},
{
"input": "2\n01",
"output": "Yes"
},
{
"input": "4\n1101",
"output": "No"
},
{
"input": "2\n00",
"output": "No"
},
{
"input": "4\n1111",
"output": "No"
},
{
"input": "15\n000010101010000",
"output": "No"
},
{
"input": "8\n10010100",
"output": "No"
},
{
"input": "5\n00100",
"output": "No"
},
{
"input": "6\n100100",
"output": "No"
},
{
"input": "7\n1000001",
"output": "No"
},
{
"input": "98\n10101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010",
"output": "Yes"
}
] | 1,609,840,364 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | #include<bits/stdc++.h>
using namespace std;
int n;
string s;
int main()
{
scanf ("%d", &n);
cin>>s;
for (int i=0;i<n-1;i++)
if (s[i]==s[i+1])
{
printf("NO\n");
return 0;
}
printf("Yes\n");
return 0;
}
| Title: Row
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You're given a row with $n$ chairs. We call a seating of people "maximal" if the two following conditions hold:
1. There are no neighbors adjacent to anyone seated. 1. It's impossible to seat one more person without violating the first rule.
The seating is given as a string consisting of zeros and ones ($0$ means that the corresponding seat is empty, $1$ — occupied). The goal is to determine whether this seating is "maximal".
Note that the first and last seats are not adjacent (if $n \ne 2$).
Input Specification:
The first line contains a single integer $n$ ($1 \leq n \leq 1000$) — the number of chairs.
The next line contains a string of $n$ characters, each of them is either zero or one, describing the seating.
Output Specification:
Output "Yes" (without quotation marks) if the seating is "maximal". Otherwise print "No".
You are allowed to print letters in whatever case you'd like (uppercase or lowercase).
Demo Input:
['3\n101\n', '4\n1011\n', '5\n10001\n']
Demo Output:
['Yes\n', 'No\n', 'No\n']
Note:
In sample case one the given seating is maximal.
In sample case two the person at chair three has a neighbour to the right.
In sample case three it is possible to seat yet another person into chair three. | ```python
#include<bits/stdc++.h>
using namespace std;
int n;
string s;
int main()
{
scanf ("%d", &n);
cin>>s;
for (int i=0;i<n-1;i++)
if (s[i]==s[i+1])
{
printf("NO\n");
return 0;
}
printf("Yes\n");
return 0;
}
``` | -1 |
|
615 | A | Bulbs | PROGRAMMING | 800 | [
"implementation"
] | null | null | Vasya wants to turn on Christmas lights consisting of *m* bulbs. Initially, all bulbs are turned off. There are *n* buttons, each of them is connected to some set of bulbs. Vasya can press any of these buttons. When the button is pressed, it turns on all the bulbs it's connected to. Can Vasya light up all the bulbs?
If Vasya presses the button such that some bulbs connected to it are already turned on, they do not change their state, i.e. remain turned on. | The first line of the input contains integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of buttons and the number of bulbs respectively.
Each of the next *n* lines contains *x**i* (0<=≤<=*x**i*<=≤<=*m*) — the number of bulbs that are turned on by the *i*-th button, and then *x**i* numbers *y**ij* (1<=≤<=*y**ij*<=≤<=*m*) — the numbers of these bulbs. | If it's possible to turn on all *m* bulbs print "YES", otherwise print "NO". | [
"3 4\n2 1 4\n3 1 3 1\n1 2\n",
"3 3\n1 1\n1 2\n1 1\n"
] | [
"YES\n",
"NO\n"
] | In the first sample you can press each button once and turn on all the bulbs. In the 2 sample it is impossible to turn on the 3-rd lamp. | 500 | [
{
"input": "3 4\n2 1 4\n3 1 3 1\n1 2",
"output": "YES"
},
{
"input": "3 3\n1 1\n1 2\n1 1",
"output": "NO"
},
{
"input": "3 4\n1 1\n1 2\n1 3",
"output": "NO"
},
{
"input": "1 5\n5 1 2 3 4 5",
"output": "YES"
},
{
"input": "1 5\n5 4 4 1 2 3",
"output": "NO"
},
{
"input": "1 5\n5 1 1 1 1 5",
"output": "NO"
},
{
"input": "2 5\n4 3 1 4 2\n4 2 3 4 5",
"output": "YES"
},
{
"input": "5 7\n2 6 7\n5 1 1 1 1 1\n3 6 5 4\n0\n4 4 3 2 1",
"output": "YES"
},
{
"input": "100 100\n0\n0\n0\n1 53\n0\n0\n1 34\n1 54\n0\n1 14\n0\n1 33\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n1 82\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n1 34\n0\n0\n1 26\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n1 34\n0\n0\n0\n0\n0\n1 3\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n1 40\n0\n0\n0\n1 26\n0\n0\n0\n0\n0\n1 97\n0\n1 5\n0\n0\n0\n0\n0",
"output": "NO"
},
{
"input": "100 100\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0",
"output": "NO"
},
{
"input": "5 6\n3 1 2 6\n3 1 2 6\n1 1\n2 3 4\n3 1 5 6",
"output": "YES"
},
{
"input": "5 2\n1 1\n1 1\n1 1\n1 1\n1 1",
"output": "NO"
},
{
"input": "1 4\n3 1 2 3",
"output": "NO"
},
{
"input": "1 4\n3 2 3 4",
"output": "NO"
},
{
"input": "2 4\n3 2 3 4\n1 1",
"output": "YES"
},
{
"input": "2 4\n3 1 2 3\n1 4",
"output": "YES"
},
{
"input": "5 1\n0\n0\n0\n0\n0",
"output": "NO"
},
{
"input": "1 1\n0",
"output": "NO"
},
{
"input": "1 10\n10 1 2 3 4 5 6 7 8 9 10",
"output": "YES"
},
{
"input": "1 1\n1 1",
"output": "YES"
},
{
"input": "1 100\n99 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99",
"output": "NO"
},
{
"input": "1 3\n3 1 2 1",
"output": "NO"
},
{
"input": "1 100\n100 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100",
"output": "YES"
}
] | 1,678,458,110 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 30 | 0 | Error = 1
while Error:
Number_Switches_And_Bulbs =input()
Str = Number_Switches_And_Bulbs.split(' ')
Number_Switches , Number_Bulbs = int(Str[0]) , int(Str[1])
Bulbs_to_work = [(i+1) for i in range(Number_Bulbs)]
bulbs_to_corss = set()
for i in range(Number_Switches):
Working_logic =input()
Str = Working_logic.split(' ')
Num_of_bulbsPerswitch = int(Str[0])
for j in range(Num_of_bulbsPerswitch):
try:
bulbs_to_corss.add(int(Str[j+1]))
except:
print("incorrect entry, please try again")
Error = 1
Error = 0
if(list(bulbs_to_corss) == Bulbs_to_work):
print("TRUE")
else:
print("FALSE")
| Title: Bulbs
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya wants to turn on Christmas lights consisting of *m* bulbs. Initially, all bulbs are turned off. There are *n* buttons, each of them is connected to some set of bulbs. Vasya can press any of these buttons. When the button is pressed, it turns on all the bulbs it's connected to. Can Vasya light up all the bulbs?
If Vasya presses the button such that some bulbs connected to it are already turned on, they do not change their state, i.e. remain turned on.
Input Specification:
The first line of the input contains integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of buttons and the number of bulbs respectively.
Each of the next *n* lines contains *x**i* (0<=≤<=*x**i*<=≤<=*m*) — the number of bulbs that are turned on by the *i*-th button, and then *x**i* numbers *y**ij* (1<=≤<=*y**ij*<=≤<=*m*) — the numbers of these bulbs.
Output Specification:
If it's possible to turn on all *m* bulbs print "YES", otherwise print "NO".
Demo Input:
['3 4\n2 1 4\n3 1 3 1\n1 2\n', '3 3\n1 1\n1 2\n1 1\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first sample you can press each button once and turn on all the bulbs. In the 2 sample it is impossible to turn on the 3-rd lamp. | ```python
Error = 1
while Error:
Number_Switches_And_Bulbs =input()
Str = Number_Switches_And_Bulbs.split(' ')
Number_Switches , Number_Bulbs = int(Str[0]) , int(Str[1])
Bulbs_to_work = [(i+1) for i in range(Number_Bulbs)]
bulbs_to_corss = set()
for i in range(Number_Switches):
Working_logic =input()
Str = Working_logic.split(' ')
Num_of_bulbsPerswitch = int(Str[0])
for j in range(Num_of_bulbsPerswitch):
try:
bulbs_to_corss.add(int(Str[j+1]))
except:
print("incorrect entry, please try again")
Error = 1
Error = 0
if(list(bulbs_to_corss) == Bulbs_to_work):
print("TRUE")
else:
print("FALSE")
``` | 0 |
|
22 | A | Second Order Statistics | PROGRAMMING | 800 | [
"brute force"
] | A. Second Order Statistics | 2 | 256 | Once Bob needed to find the second order statistics of a sequence of integer numbers. Lets choose each number from the sequence exactly once and sort them. The value on the second position is the second order statistics of the given sequence. In other words it is the smallest element strictly greater than the minimum. Help Bob solve this problem. | The first input line contains integer *n* (1<=≤<=*n*<=≤<=100) — amount of numbers in the sequence. The second line contains *n* space-separated integer numbers — elements of the sequence. These numbers don't exceed 100 in absolute value. | If the given sequence has the second order statistics, output this order statistics, otherwise output NO. | [
"4\n1 2 2 -4\n",
"5\n1 2 3 1 1\n"
] | [
"1\n",
"2\n"
] | none | 0 | [
{
"input": "4\n1 2 2 -4",
"output": "1"
},
{
"input": "5\n1 2 3 1 1",
"output": "2"
},
{
"input": "1\n28",
"output": "NO"
},
{
"input": "2\n-28 12",
"output": "12"
},
{
"input": "3\n-83 40 -80",
"output": "-80"
},
{
"input": "8\n93 77 -92 26 21 -48 53 91",
"output": "-48"
},
{
"input": "20\n-72 -9 -86 80 7 -10 40 -27 -94 92 96 56 28 -19 79 36 -3 -73 -63 -49",
"output": "-86"
},
{
"input": "49\n-74 -100 -80 23 -8 -83 -41 -20 48 17 46 -73 -55 67 85 4 40 -60 -69 -75 56 -74 -42 93 74 -95 64 -46 97 -47 55 0 -78 -34 -31 40 -63 -49 -76 48 21 -1 -49 -29 -98 -11 76 26 94",
"output": "-98"
},
{
"input": "88\n63 48 1 -53 -89 -49 64 -70 -49 71 -17 -16 76 81 -26 -50 67 -59 -56 97 2 100 14 18 -91 -80 42 92 -25 -88 59 8 -56 38 48 -71 -78 24 -14 48 -1 69 73 -76 54 16 -92 44 47 33 -34 -17 -81 21 -59 -61 53 26 10 -76 67 35 -29 70 65 -13 -29 81 80 32 74 -6 34 46 57 1 -45 -55 69 79 -58 11 -2 22 -18 -16 -89 -46",
"output": "-91"
},
{
"input": "100\n34 32 88 20 76 53 -71 -39 -98 -10 57 37 63 -3 -54 -64 -78 -82 73 20 -30 -4 22 75 51 -64 -91 29 -52 -48 83 19 18 -47 46 57 -44 95 89 89 -30 84 -83 67 58 -99 -90 -53 92 -60 -5 -56 -61 27 68 -48 52 -95 64 -48 -30 -67 66 89 14 -33 -31 -91 39 7 -94 -54 92 -96 -99 -83 -16 91 -28 -66 81 44 14 -85 -21 18 40 16 -13 -82 -33 47 -10 -40 -19 10 25 60 -34 -89",
"output": "-98"
},
{
"input": "2\n-1 -1",
"output": "NO"
},
{
"input": "3\n-2 -2 -2",
"output": "NO"
},
{
"input": "100\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0",
"output": "NO"
},
{
"input": "100\n100 100 100 100 100 100 100 100 100 100 100 100 -100 100 100 100 100 100 100 100 100 100 100 100 -100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 -100 100 100 100 100 100 100 100 100 100 100 -100 100 100 100 100 -100 100 100 100 100 100 100 100 100 100 100 100",
"output": "100"
},
{
"input": "10\n40 71 -85 -85 40 -85 -85 64 -85 47",
"output": "40"
},
{
"input": "23\n-90 -90 -41 -64 -64 -90 -15 10 -43 -90 -64 -64 89 -64 36 47 38 -90 -64 -90 -90 68 -90",
"output": "-64"
},
{
"input": "39\n-97 -93 -42 -93 -97 -93 56 -97 -97 -97 76 -33 -60 91 7 82 17 47 -97 -97 -93 73 -97 12 -97 -97 -97 -97 56 -92 -83 -93 -93 49 -93 -97 -97 -17 -93",
"output": "-93"
},
{
"input": "51\n-21 6 -35 -98 -86 -98 -86 -43 -65 32 -98 -40 96 -98 -98 -98 -98 -86 -86 -98 56 -86 -98 -98 -30 -98 -86 -31 -98 -86 -86 -86 -86 -30 96 -86 -86 -86 -60 25 88 -86 -86 58 31 -47 57 -86 37 44 -83",
"output": "-86"
},
{
"input": "66\n-14 -95 65 -95 -95 -97 -90 -71 -97 -97 70 -95 -95 -97 -95 -27 35 -87 -95 -5 -97 -97 87 34 -49 -95 -97 -95 -97 -95 -30 -95 -97 47 -95 -17 -97 -95 -97 -69 51 -97 -97 -95 -75 87 59 21 63 56 76 -91 98 -97 6 -97 -95 -95 -97 -73 11 -97 -35 -95 -95 -43",
"output": "-95"
},
{
"input": "77\n-67 -93 -93 -92 97 29 93 -93 -93 -5 -93 -7 60 -92 -93 44 -84 68 -92 -93 69 -92 -37 56 43 -93 35 -92 -93 19 -79 18 -92 -93 -93 -37 -93 -47 -93 -92 -92 74 67 19 40 -92 -92 -92 -92 -93 -93 -41 -93 -92 -93 -93 -92 -93 51 -80 6 -42 -92 -92 -66 -12 -92 -92 -3 93 -92 -49 -93 40 62 -92 -92",
"output": "-92"
},
{
"input": "89\n-98 40 16 -87 -98 63 -100 55 -96 -98 -21 -100 -93 26 -98 -98 -100 -89 -98 -5 -65 -28 -100 -6 -66 67 -100 -98 -98 10 -98 -98 -70 7 -98 2 -100 -100 -98 25 -100 -100 -98 23 -68 -100 -98 3 98 -100 -98 -98 -98 -98 -24 -100 -100 -9 -98 35 -100 99 -5 -98 -100 -100 37 -100 -84 57 -98 40 -47 -100 -1 -92 -76 -98 -98 -100 -100 -100 -63 30 21 -100 -100 -100 -12",
"output": "-98"
},
{
"input": "99\n10 -84 -100 -100 73 -64 -100 -94 33 -100 -100 -100 -100 71 64 24 7 -100 -32 -100 -100 77 -100 62 -12 55 45 -100 -100 -80 -100 -100 -100 -100 -100 -100 -100 -100 -100 -39 -48 -100 -34 47 -100 -100 -100 -100 -100 -77 -100 -100 -100 -100 -100 -100 -52 40 -55 -100 -44 -100 72 33 70 -100 -100 -78 -100 -3 100 -77 22 -100 95 -30 -100 10 -69 -100 -100 -100 -100 52 -39 -100 -100 -100 7 -100 -98 -66 95 -17 -100 52 -100 68 -100",
"output": "-98"
},
{
"input": "100\n-99 -98 -64 89 53 57 -99 29 -78 18 -3 -54 76 -98 -99 -98 37 -98 19 -47 89 73 -98 -91 -99 -99 -98 -48 -99 22 -99 -98 -99 -99 -98 -60 84 67 -99 -98 20 -98 88 -98 46 -98 -99 -98 -99 -71 -99 -98 -98 -39 83 95 -98 63 -98 -99 32 -98 -99 -64 57 -30 -53 -83 -4 -99 58 20 -98 -10 -99 -44 -99 -99 -99 -99 75 34 -98 -52 -98 -30 -98 -99 -98 -98 51 -99 -99 -99 -98 -99 -99 -82 -90 92",
"output": "-98"
},
{
"input": "3\n1 2 3",
"output": "2"
},
{
"input": "3\n1 3 2",
"output": "2"
},
{
"input": "3\n2 1 3",
"output": "2"
},
{
"input": "3\n2 3 1",
"output": "2"
},
{
"input": "3\n3 1 2",
"output": "2"
},
{
"input": "3\n3 2 1",
"output": "2"
},
{
"input": "9\n99 99 99 99 100 100 100 100 100",
"output": "100"
},
{
"input": "5\n-100 -100 -100 -100 -100",
"output": "NO"
}
] | 1,618,557,570 | 2,147,483,647 | Python 3 | OK | TESTS | 31 | 154 | 0 | n=int(input())
a=[int(x) for x in input().split()]
a=list(set(a))
a.sort()
if(len(a)==1):
print("NO")
else:
print(a[1])
| Title: Second Order Statistics
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Once Bob needed to find the second order statistics of a sequence of integer numbers. Lets choose each number from the sequence exactly once and sort them. The value on the second position is the second order statistics of the given sequence. In other words it is the smallest element strictly greater than the minimum. Help Bob solve this problem.
Input Specification:
The first input line contains integer *n* (1<=≤<=*n*<=≤<=100) — amount of numbers in the sequence. The second line contains *n* space-separated integer numbers — elements of the sequence. These numbers don't exceed 100 in absolute value.
Output Specification:
If the given sequence has the second order statistics, output this order statistics, otherwise output NO.
Demo Input:
['4\n1 2 2 -4\n', '5\n1 2 3 1 1\n']
Demo Output:
['1\n', '2\n']
Note:
none | ```python
n=int(input())
a=[int(x) for x in input().split()]
a=list(set(a))
a.sort()
if(len(a)==1):
print("NO")
else:
print(a[1])
``` | 3.9615 |
471 | A | MUH and Sticks | PROGRAMMING | 1,100 | [
"implementation"
] | null | null | Two polar bears Menshykov and Uslada from the St.Petersburg zoo and elephant Horace from the Kiev zoo got six sticks to play with and assess the animals' creativity. Menshykov, Uslada and Horace decided to make either an elephant or a bear from those sticks. They can make an animal from sticks in the following way:
- Four sticks represent the animal's legs, these sticks should have the same length. - Two remaining sticks represent the animal's head and body. The bear's head stick must be shorter than the body stick. The elephant, however, has a long trunk, so his head stick must be as long as the body stick. Note that there are no limits on the relations between the leg sticks and the head and body sticks.
Your task is to find out which animal can be made from the given stick set. The zoo keeper wants the sticks back after the game, so they must never be broken, even bears understand it. | The single line contains six space-separated integers *l**i* (1<=≤<=*l**i*<=≤<=9) — the lengths of the six sticks. It is guaranteed that the input is such that you cannot make both animals from the sticks. | If you can make a bear from the given set, print string "Bear" (without the quotes). If you can make an elephant, print string "Elephant" (wıthout the quotes). If you can make neither a bear nor an elephant, print string "Alien" (without the quotes). | [
"4 2 5 4 4 4\n",
"4 4 5 4 4 5\n",
"1 2 3 4 5 6\n"
] | [
"Bear",
"Elephant",
"Alien"
] | If you're out of creative ideas, see instructions below which show how to make a bear and an elephant in the first two samples. The stick of length 2 is in red, the sticks of length 4 are in green, the sticks of length 5 are in blue. | 500 | [
{
"input": "4 2 5 4 4 4",
"output": "Bear"
},
{
"input": "4 4 5 4 4 5",
"output": "Elephant"
},
{
"input": "1 2 3 4 5 6",
"output": "Alien"
},
{
"input": "5 5 5 5 5 5",
"output": "Elephant"
},
{
"input": "1 1 1 2 3 5",
"output": "Alien"
},
{
"input": "1 1 1 1 1 1",
"output": "Elephant"
},
{
"input": "9 9 9 9 9 9",
"output": "Elephant"
},
{
"input": "1 8 9 1 1 1",
"output": "Bear"
},
{
"input": "9 9 9 1 9 9",
"output": "Bear"
},
{
"input": "1 2 3 8 9 7",
"output": "Alien"
},
{
"input": "5 5 5 6 6 6",
"output": "Alien"
},
{
"input": "1 1 2 2 3 4",
"output": "Alien"
},
{
"input": "4 4 4 4 4 2",
"output": "Bear"
},
{
"input": "2 2 3 3 4 4",
"output": "Alien"
},
{
"input": "4 4 4 4 4 5",
"output": "Bear"
},
{
"input": "1 1 2 2 2 2",
"output": "Elephant"
},
{
"input": "1 2 5 5 5 5",
"output": "Bear"
},
{
"input": "4 4 2 2 2 2",
"output": "Elephant"
},
{
"input": "1 1 1 1 1 2",
"output": "Bear"
},
{
"input": "2 2 4 4 4 4",
"output": "Elephant"
},
{
"input": "4 4 4 4 4 3",
"output": "Bear"
},
{
"input": "4 4 5 6 7 8",
"output": "Alien"
},
{
"input": "4 4 4 4 2 2",
"output": "Elephant"
},
{
"input": "1 1 1 1 2 2",
"output": "Elephant"
},
{
"input": "1 1 3 3 3 5",
"output": "Alien"
},
{
"input": "1 2 2 3 3 3",
"output": "Alien"
},
{
"input": "1 2 2 2 2 2",
"output": "Bear"
},
{
"input": "1 3 3 3 4 5",
"output": "Alien"
},
{
"input": "5 1 1 1 1 1",
"output": "Bear"
}
] | 1,584,956,751 | 2,147,483,647 | Python 3 | OK | TESTS | 29 | 109 | 204,800 | a = list(map(int, input().split()))
h = {}
for n in a:
if n not in h:
h[n] = 1
else:
h[n] += 1
k = 4
if len(h)==1:
print('Elephant')
elif k+1 in h.values():
print('Bear')
elif k in h.values():
if len(h)==2:
print('Elephant')
elif len(h)==3:
print('Bear')
else:
print('Alien')
| Title: MUH and Sticks
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Two polar bears Menshykov and Uslada from the St.Petersburg zoo and elephant Horace from the Kiev zoo got six sticks to play with and assess the animals' creativity. Menshykov, Uslada and Horace decided to make either an elephant or a bear from those sticks. They can make an animal from sticks in the following way:
- Four sticks represent the animal's legs, these sticks should have the same length. - Two remaining sticks represent the animal's head and body. The bear's head stick must be shorter than the body stick. The elephant, however, has a long trunk, so his head stick must be as long as the body stick. Note that there are no limits on the relations between the leg sticks and the head and body sticks.
Your task is to find out which animal can be made from the given stick set. The zoo keeper wants the sticks back after the game, so they must never be broken, even bears understand it.
Input Specification:
The single line contains six space-separated integers *l**i* (1<=≤<=*l**i*<=≤<=9) — the lengths of the six sticks. It is guaranteed that the input is such that you cannot make both animals from the sticks.
Output Specification:
If you can make a bear from the given set, print string "Bear" (without the quotes). If you can make an elephant, print string "Elephant" (wıthout the quotes). If you can make neither a bear nor an elephant, print string "Alien" (without the quotes).
Demo Input:
['4 2 5 4 4 4\n', '4 4 5 4 4 5\n', '1 2 3 4 5 6\n']
Demo Output:
['Bear', 'Elephant', 'Alien']
Note:
If you're out of creative ideas, see instructions below which show how to make a bear and an elephant in the first two samples. The stick of length 2 is in red, the sticks of length 4 are in green, the sticks of length 5 are in blue. | ```python
a = list(map(int, input().split()))
h = {}
for n in a:
if n not in h:
h[n] = 1
else:
h[n] += 1
k = 4
if len(h)==1:
print('Elephant')
elif k+1 in h.values():
print('Bear')
elif k in h.values():
if len(h)==2:
print('Elephant')
elif len(h)==3:
print('Bear')
else:
print('Alien')
``` | 3 |
|
50 | A | Domino piling | PROGRAMMING | 800 | [
"greedy",
"math"
] | A. Domino piling | 2 | 256 | You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions. | In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16). | Output one number — the maximal number of dominoes, which can be placed. | [
"2 4\n",
"3 3\n"
] | [
"4\n",
"4\n"
] | none | 500 | [
{
"input": "2 4",
"output": "4"
},
{
"input": "3 3",
"output": "4"
},
{
"input": "1 5",
"output": "2"
},
{
"input": "1 6",
"output": "3"
},
{
"input": "1 15",
"output": "7"
},
{
"input": "1 16",
"output": "8"
},
{
"input": "2 5",
"output": "5"
},
{
"input": "2 6",
"output": "6"
},
{
"input": "2 7",
"output": "7"
},
{
"input": "2 14",
"output": "14"
},
{
"input": "2 15",
"output": "15"
},
{
"input": "1 4",
"output": "2"
},
{
"input": "2 16",
"output": "16"
},
{
"input": "3 5",
"output": "7"
},
{
"input": "3 6",
"output": "9"
},
{
"input": "3 10",
"output": "15"
},
{
"input": "3 14",
"output": "21"
},
{
"input": "3 15",
"output": "22"
},
{
"input": "3 16",
"output": "24"
},
{
"input": "5 7",
"output": "17"
},
{
"input": "16 16",
"output": "128"
},
{
"input": "15 16",
"output": "120"
},
{
"input": "2 3",
"output": "3"
},
{
"input": "15 15",
"output": "112"
},
{
"input": "14 16",
"output": "112"
},
{
"input": "11 13",
"output": "71"
},
{
"input": "5 16",
"output": "40"
},
{
"input": "8 15",
"output": "60"
},
{
"input": "2 2",
"output": "2"
},
{
"input": "3 4",
"output": "6"
},
{
"input": "4 4",
"output": "8"
},
{
"input": "1 1",
"output": "0"
},
{
"input": "1 2",
"output": "1"
},
{
"input": "1 3",
"output": "1"
},
{
"input": "14 15",
"output": "105"
}
] | 1,694,264,193 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 60 | 0 | str_list = input().split()
m = int(str_list[0])
n = int(str_list[1])
if m % 2 == 0 or n % 2 == 0:
result = m * n / 2
else:
result = (m * n - 1)/2
res = int(result)
print(result) | Title: Domino piling
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions.
Input Specification:
In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16).
Output Specification:
Output one number — the maximal number of dominoes, which can be placed.
Demo Input:
['2 4\n', '3 3\n']
Demo Output:
['4\n', '4\n']
Note:
none | ```python
str_list = input().split()
m = int(str_list[0])
n = int(str_list[1])
if m % 2 == 0 or n % 2 == 0:
result = m * n / 2
else:
result = (m * n - 1)/2
res = int(result)
print(result)
``` | 0 |
78 | B | Easter Eggs | PROGRAMMING | 1,200 | [
"constructive algorithms",
"implementation"
] | B. Easter Eggs | 2 | 256 | The Easter Rabbit laid *n* eggs in a circle and is about to paint them.
Each egg should be painted one color out of 7: red, orange, yellow, green, blue, indigo or violet. Also, the following conditions should be satisfied:
- Each of the seven colors should be used to paint at least one egg. - Any four eggs lying sequentially should be painted different colors.
Help the Easter Rabbit paint the eggs in the required manner. We know that it is always possible. | The only line contains an integer *n* — the amount of eggs (7<=≤<=*n*<=≤<=100). | Print one line consisting of *n* characters. The *i*-th character should describe the color of the *i*-th egg in the order they lie in the circle. The colors should be represented as follows: "R" stands for red, "O" stands for orange, "Y" stands for yellow, "G" stands for green, "B" stands for blue, "I" stands for indigo, "V" stands for violet.
If there are several answers, print any of them. | [
"8\n",
"13\n"
] | [
"ROYGRBIV\n",
"ROYGBIVGBIVYG\n"
] | The way the eggs will be painted in the first sample is shown on the picture: | 1,000 | [
{
"input": "8",
"output": "ROYGBIVG"
},
{
"input": "13",
"output": "ROYGBIVOYGBIV"
},
{
"input": "7",
"output": "ROYGBIV"
},
{
"input": "10",
"output": "ROYGBIVYGB"
},
{
"input": "14",
"output": "ROYGBIVROYGBIV"
},
{
"input": "50",
"output": "ROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVG"
},
{
"input": "9",
"output": "ROYGBIVGB"
},
{
"input": "11",
"output": "ROYGBIVYGBI"
},
{
"input": "12",
"output": "ROYGBIVOYGBI"
},
{
"input": "15",
"output": "ROYGBIVROYGBIVG"
},
{
"input": "16",
"output": "ROYGBIVROYGBIVGB"
},
{
"input": "17",
"output": "ROYGBIVROYGBIVYGB"
},
{
"input": "18",
"output": "ROYGBIVROYGBIVYGBI"
},
{
"input": "19",
"output": "ROYGBIVROYGBIVOYGBI"
},
{
"input": "20",
"output": "ROYGBIVROYGBIVOYGBIV"
},
{
"input": "21",
"output": "ROYGBIVROYGBIVROYGBIV"
},
{
"input": "22",
"output": "ROYGBIVROYGBIVROYGBIVG"
},
{
"input": "23",
"output": "ROYGBIVROYGBIVROYGBIVGB"
},
{
"input": "24",
"output": "ROYGBIVROYGBIVROYGBIVYGB"
},
{
"input": "25",
"output": "ROYGBIVROYGBIVROYGBIVYGBI"
},
{
"input": "28",
"output": "ROYGBIVROYGBIVROYGBIVROYGBIV"
},
{
"input": "29",
"output": "ROYGBIVROYGBIVROYGBIVROYGBIVG"
},
{
"input": "34",
"output": "ROYGBIVROYGBIVROYGBIVROYGBIVOYGBIV"
},
{
"input": "43",
"output": "ROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVG"
},
{
"input": "61",
"output": "ROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVOYGBI"
},
{
"input": "79",
"output": "ROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVGB"
},
{
"input": "81",
"output": "ROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVYGBI"
},
{
"input": "92",
"output": "ROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVG"
},
{
"input": "95",
"output": "ROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVYGBI"
},
{
"input": "96",
"output": "ROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVOYGBI"
},
{
"input": "97",
"output": "ROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVOYGBIV"
},
{
"input": "98",
"output": "ROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIV"
},
{
"input": "99",
"output": "ROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVG"
},
{
"input": "100",
"output": "ROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVROYGBIVGB"
}
] | 1,667,849,509 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 30 | 0 | n = int(input(""))
repeat = "ROYRBIV"
if n < len(repeat):
print(repeat[:n])
exit()
elif n == len(repeat):
print(repeat)
exit()
else:
rep = n // len(repeat)
#print(len(repeat)%n)
repeat = (repeat * rep) + repeat[len(repeat) - (n % len(repeat)):]
print(repeat)
exit()
| Title: Easter Eggs
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
The Easter Rabbit laid *n* eggs in a circle and is about to paint them.
Each egg should be painted one color out of 7: red, orange, yellow, green, blue, indigo or violet. Also, the following conditions should be satisfied:
- Each of the seven colors should be used to paint at least one egg. - Any four eggs lying sequentially should be painted different colors.
Help the Easter Rabbit paint the eggs in the required manner. We know that it is always possible.
Input Specification:
The only line contains an integer *n* — the amount of eggs (7<=≤<=*n*<=≤<=100).
Output Specification:
Print one line consisting of *n* characters. The *i*-th character should describe the color of the *i*-th egg in the order they lie in the circle. The colors should be represented as follows: "R" stands for red, "O" stands for orange, "Y" stands for yellow, "G" stands for green, "B" stands for blue, "I" stands for indigo, "V" stands for violet.
If there are several answers, print any of them.
Demo Input:
['8\n', '13\n']
Demo Output:
['ROYGRBIV\n', 'ROYGBIVGBIVYG\n']
Note:
The way the eggs will be painted in the first sample is shown on the picture: | ```python
n = int(input(""))
repeat = "ROYRBIV"
if n < len(repeat):
print(repeat[:n])
exit()
elif n == len(repeat):
print(repeat)
exit()
else:
rep = n // len(repeat)
#print(len(repeat)%n)
repeat = (repeat * rep) + repeat[len(repeat) - (n % len(repeat)):]
print(repeat)
exit()
``` | 0 |
651 | A | Joysticks | PROGRAMMING | 1,100 | [
"dp",
"greedy",
"implementation",
"math"
] | null | null | Friends are going to play console. They have two joysticks and only one charger for them. Initially first joystick is charged at *a*1 percent and second one is charged at *a*2 percent. You can connect charger to a joystick only at the beginning of each minute. In one minute joystick either discharges by 2 percent (if not connected to a charger) or charges by 1 percent (if connected to a charger).
Game continues while both joysticks have a positive charge. Hence, if at the beginning of minute some joystick is charged by 1 percent, it has to be connected to a charger, otherwise the game stops. If some joystick completely discharges (its charge turns to 0), the game also stops.
Determine the maximum number of minutes that game can last. It is prohibited to pause the game, i. e. at each moment both joysticks should be enabled. It is allowed for joystick to be charged by more than 100 percent. | The first line of the input contains two positive integers *a*1 and *a*2 (1<=≤<=*a*1,<=*a*2<=≤<=100), the initial charge level of first and second joystick respectively. | Output the only integer, the maximum number of minutes that the game can last. Game continues until some joystick is discharged. | [
"3 5\n",
"4 4\n"
] | [
"6\n",
"5\n"
] | In the first sample game lasts for 6 minute by using the following algorithm:
- at the beginning of the first minute connect first joystick to the charger, by the end of this minute first joystick is at 4%, second is at 3%; - continue the game without changing charger, by the end of the second minute the first joystick is at 5%, second is at 1%; - at the beginning of the third minute connect second joystick to the charger, after this minute the first joystick is at 3%, the second one is at 2%; - continue the game without changing charger, by the end of the fourth minute first joystick is at 1%, second one is at 3%; - at the beginning of the fifth minute connect first joystick to the charger, after this minute the first joystick is at 2%, the second one is at 1%; - at the beginning of the sixth minute connect second joystick to the charger, after this minute the first joystick is at 0%, the second one is at 2%.
After that the first joystick is completely discharged and the game is stopped. | 500 | [
{
"input": "3 5",
"output": "6"
},
{
"input": "4 4",
"output": "5"
},
{
"input": "100 100",
"output": "197"
},
{
"input": "1 100",
"output": "98"
},
{
"input": "100 1",
"output": "98"
},
{
"input": "1 4",
"output": "2"
},
{
"input": "1 1",
"output": "0"
},
{
"input": "8 8",
"output": "13"
},
{
"input": "7 2",
"output": "7"
},
{
"input": "24 15",
"output": "36"
},
{
"input": "19 30",
"output": "47"
},
{
"input": "15 31",
"output": "44"
},
{
"input": "14 15",
"output": "27"
},
{
"input": "58 33",
"output": "89"
},
{
"input": "15 25",
"output": "38"
},
{
"input": "59 45",
"output": "102"
},
{
"input": "3 73",
"output": "74"
},
{
"input": "48 1",
"output": "47"
},
{
"input": "100 25",
"output": "122"
},
{
"input": "40 49",
"output": "86"
},
{
"input": "85 73",
"output": "155"
},
{
"input": "29 1",
"output": "28"
},
{
"input": "74 25",
"output": "97"
},
{
"input": "24 57",
"output": "78"
},
{
"input": "23 12",
"output": "33"
},
{
"input": "2 99",
"output": "99"
},
{
"input": "98 2",
"output": "97"
},
{
"input": "2 97",
"output": "97"
},
{
"input": "30 54",
"output": "81"
},
{
"input": "32 53",
"output": "82"
},
{
"input": "32 54",
"output": "84"
},
{
"input": "1 2",
"output": "1"
},
{
"input": "2 1",
"output": "1"
},
{
"input": "2 2",
"output": "1"
},
{
"input": "1 3",
"output": "2"
},
{
"input": "3 1",
"output": "2"
},
{
"input": "1 4",
"output": "2"
},
{
"input": "2 3",
"output": "3"
},
{
"input": "3 2",
"output": "3"
}
] | 1,690,455,363 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 6 | 46 | 0 | a1,a2=list( map( int, input().split(" ")))
i=0
while a1 and a2:
if a1 >= a2:
a1=a1-2
a2=a2+1
else:
a2=a2-2
a1=a1+1
i = i + 1
print(i) | Title: Joysticks
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Friends are going to play console. They have two joysticks and only one charger for them. Initially first joystick is charged at *a*1 percent and second one is charged at *a*2 percent. You can connect charger to a joystick only at the beginning of each minute. In one minute joystick either discharges by 2 percent (if not connected to a charger) or charges by 1 percent (if connected to a charger).
Game continues while both joysticks have a positive charge. Hence, if at the beginning of minute some joystick is charged by 1 percent, it has to be connected to a charger, otherwise the game stops. If some joystick completely discharges (its charge turns to 0), the game also stops.
Determine the maximum number of minutes that game can last. It is prohibited to pause the game, i. e. at each moment both joysticks should be enabled. It is allowed for joystick to be charged by more than 100 percent.
Input Specification:
The first line of the input contains two positive integers *a*1 and *a*2 (1<=≤<=*a*1,<=*a*2<=≤<=100), the initial charge level of first and second joystick respectively.
Output Specification:
Output the only integer, the maximum number of minutes that the game can last. Game continues until some joystick is discharged.
Demo Input:
['3 5\n', '4 4\n']
Demo Output:
['6\n', '5\n']
Note:
In the first sample game lasts for 6 minute by using the following algorithm:
- at the beginning of the first minute connect first joystick to the charger, by the end of this minute first joystick is at 4%, second is at 3%; - continue the game without changing charger, by the end of the second minute the first joystick is at 5%, second is at 1%; - at the beginning of the third minute connect second joystick to the charger, after this minute the first joystick is at 3%, the second one is at 2%; - continue the game without changing charger, by the end of the fourth minute first joystick is at 1%, second one is at 3%; - at the beginning of the fifth minute connect first joystick to the charger, after this minute the first joystick is at 2%, the second one is at 1%; - at the beginning of the sixth minute connect second joystick to the charger, after this minute the first joystick is at 0%, the second one is at 2%.
After that the first joystick is completely discharged and the game is stopped. | ```python
a1,a2=list( map( int, input().split(" ")))
i=0
while a1 and a2:
if a1 >= a2:
a1=a1-2
a2=a2+1
else:
a2=a2-2
a1=a1+1
i = i + 1
print(i)
``` | 0 |
|
1,006 | E | Military Problem | PROGRAMMING | 1,600 | [
"dfs and similar",
"graphs",
"trees"
] | null | null | In this problem you will have to help Berland army with organizing their command delivery system.
There are $n$ officers in Berland army. The first officer is the commander of the army, and he does not have any superiors. Every other officer has exactly one direct superior. If officer $a$ is the direct superior of officer $b$, then we also can say that officer $b$ is a direct subordinate of officer $a$.
Officer $x$ is considered to be a subordinate (direct or indirect) of officer $y$ if one of the following conditions holds:
- officer $y$ is the direct superior of officer $x$; - the direct superior of officer $x$ is a subordinate of officer $y$.
For example, on the picture below the subordinates of the officer $3$ are: $5, 6, 7, 8, 9$.
The structure of Berland army is organized in such a way that every officer, except for the commander, is a subordinate of the commander of the army.
Formally, let's represent Berland army as a tree consisting of $n$ vertices, in which vertex $u$ corresponds to officer $u$. The parent of vertex $u$ corresponds to the direct superior of officer $u$. The root (which has index $1$) corresponds to the commander of the army.
Berland War Ministry has ordered you to give answers on $q$ queries, the $i$-th query is given as $(u_i, k_i)$, where $u_i$ is some officer, and $k_i$ is a positive integer.
To process the $i$-th query imagine how a command from $u_i$ spreads to the subordinates of $u_i$. Typical DFS (depth first search) algorithm is used here.
Suppose the current officer is $a$ and he spreads a command. Officer $a$ chooses $b$ — one of his direct subordinates (i.e. a child in the tree) who has not received this command yet. If there are many such direct subordinates, then $a$ chooses the one having minimal index. Officer $a$ gives a command to officer $b$. Afterwards, $b$ uses exactly the same algorithm to spread the command to its subtree. After $b$ finishes spreading the command, officer $a$ chooses the next direct subordinate again (using the same strategy). When officer $a$ cannot choose any direct subordinate who still hasn't received this command, officer $a$ finishes spreading the command.
Let's look at the following example:
If officer $1$ spreads a command, officers receive it in the following order: $[1, 2, 3, 5 ,6, 8, 7, 9, 4]$.
If officer $3$ spreads a command, officers receive it in the following order: $[3, 5, 6, 8, 7, 9]$.
If officer $7$ spreads a command, officers receive it in the following order: $[7, 9]$.
If officer $9$ spreads a command, officers receive it in the following order: $[9]$.
To answer the $i$-th query $(u_i, k_i)$, construct a sequence which describes the order in which officers will receive the command if the $u_i$-th officer spreads it. Return the $k_i$-th element of the constructed list or -1 if there are fewer than $k_i$ elements in it.
You should process queries independently. A query doesn't affect the following queries. | The first line of the input contains two integers $n$ and $q$ ($2 \le n \le 2 \cdot 10^5, 1 \le q \le 2 \cdot 10^5$) — the number of officers in Berland army and the number of queries.
The second line of the input contains $n - 1$ integers $p_2, p_3, \dots, p_n$ ($1 \le p_i < i$), where $p_i$ is the index of the direct superior of the officer having the index $i$. The commander has index $1$ and doesn't have any superiors.
The next $q$ lines describe the queries. The $i$-th query is given as a pair ($u_i, k_i$) ($1 \le u_i, k_i \le n$), where $u_i$ is the index of the officer which starts spreading a command, and $k_i$ is the index of the required officer in the command spreading sequence. | Print $q$ numbers, where the $i$-th number is the officer at the position $k_i$ in the list which describes the order in which officers will receive the command if it starts spreading from officer $u_i$. Print "-1" if the number of officers which receive the command is less than $k_i$.
You should process queries independently. They do not affect each other. | [
"9 6\n1 1 1 3 5 3 5 7\n3 1\n1 5\n3 4\n7 3\n1 8\n1 9\n"
] | [
"3\n6\n8\n-1\n9\n4\n"
] | none | 0 | [
{
"input": "9 6\n1 1 1 3 5 3 5 7\n3 1\n1 5\n3 4\n7 3\n1 8\n1 9",
"output": "3\n6\n8\n-1\n9\n4"
},
{
"input": "2 1\n1\n1 1",
"output": "1"
},
{
"input": "13 12\n1 1 1 1 1 1 1 1 1 1 1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1",
"output": "1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1"
}
] | 1,602,676,616 | 2,147,483,647 | PyPy 3 | RUNTIME_ERROR | TESTS | 1 | 342 | 26,316,800 | from collections import defaultdict
dfs=[]
class Graph:
def __init__(self):
self.graph = defaultdict(list)
def addEdge(self, u, v):
# self.graph[u].append(v)
self.graph[v].append(u)
def DFSUtil(self, v):
visited[v] = True
dfs.append(v)
for i in self.graph[v]:
if visited[i] == False:
self.DFSUtil(i)
def DFS(self, v):
# visited = [False] * (max(self.graph) + 1)
self.DFSUtil(v)
def numberofnodes(self,s,e):
count1[s]=1
for u in self.graph[s]:
if u == e:
continue
self.numberofnodes(u, s)
count1[s] += count1[u]
g = Graph()
n,q=map(int,input().split())
count1=[0]*(n+1)
pi=list(map(int,input().split()))
visited=[0]*(n+2)
for i in range(0,n-1):
g.addEdge(i+2,pi[i])
g.DFS(1)
g.numberofnodes(1,0)
index=defaultdict(int)
for i in range(len(dfs)):
index[dfs[i]]=i
for qq in range(q):
ui,vi=map(int,input().split())
if count1[ui]<vi:
print(-1)
continue
else:
ind=index[ui]+vi
print(dfs[ind-1])
| Title: Military Problem
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In this problem you will have to help Berland army with organizing their command delivery system.
There are $n$ officers in Berland army. The first officer is the commander of the army, and he does not have any superiors. Every other officer has exactly one direct superior. If officer $a$ is the direct superior of officer $b$, then we also can say that officer $b$ is a direct subordinate of officer $a$.
Officer $x$ is considered to be a subordinate (direct or indirect) of officer $y$ if one of the following conditions holds:
- officer $y$ is the direct superior of officer $x$; - the direct superior of officer $x$ is a subordinate of officer $y$.
For example, on the picture below the subordinates of the officer $3$ are: $5, 6, 7, 8, 9$.
The structure of Berland army is organized in such a way that every officer, except for the commander, is a subordinate of the commander of the army.
Formally, let's represent Berland army as a tree consisting of $n$ vertices, in which vertex $u$ corresponds to officer $u$. The parent of vertex $u$ corresponds to the direct superior of officer $u$. The root (which has index $1$) corresponds to the commander of the army.
Berland War Ministry has ordered you to give answers on $q$ queries, the $i$-th query is given as $(u_i, k_i)$, where $u_i$ is some officer, and $k_i$ is a positive integer.
To process the $i$-th query imagine how a command from $u_i$ spreads to the subordinates of $u_i$. Typical DFS (depth first search) algorithm is used here.
Suppose the current officer is $a$ and he spreads a command. Officer $a$ chooses $b$ — one of his direct subordinates (i.e. a child in the tree) who has not received this command yet. If there are many such direct subordinates, then $a$ chooses the one having minimal index. Officer $a$ gives a command to officer $b$. Afterwards, $b$ uses exactly the same algorithm to spread the command to its subtree. After $b$ finishes spreading the command, officer $a$ chooses the next direct subordinate again (using the same strategy). When officer $a$ cannot choose any direct subordinate who still hasn't received this command, officer $a$ finishes spreading the command.
Let's look at the following example:
If officer $1$ spreads a command, officers receive it in the following order: $[1, 2, 3, 5 ,6, 8, 7, 9, 4]$.
If officer $3$ spreads a command, officers receive it in the following order: $[3, 5, 6, 8, 7, 9]$.
If officer $7$ spreads a command, officers receive it in the following order: $[7, 9]$.
If officer $9$ spreads a command, officers receive it in the following order: $[9]$.
To answer the $i$-th query $(u_i, k_i)$, construct a sequence which describes the order in which officers will receive the command if the $u_i$-th officer spreads it. Return the $k_i$-th element of the constructed list or -1 if there are fewer than $k_i$ elements in it.
You should process queries independently. A query doesn't affect the following queries.
Input Specification:
The first line of the input contains two integers $n$ and $q$ ($2 \le n \le 2 \cdot 10^5, 1 \le q \le 2 \cdot 10^5$) — the number of officers in Berland army and the number of queries.
The second line of the input contains $n - 1$ integers $p_2, p_3, \dots, p_n$ ($1 \le p_i < i$), where $p_i$ is the index of the direct superior of the officer having the index $i$. The commander has index $1$ and doesn't have any superiors.
The next $q$ lines describe the queries. The $i$-th query is given as a pair ($u_i, k_i$) ($1 \le u_i, k_i \le n$), where $u_i$ is the index of the officer which starts spreading a command, and $k_i$ is the index of the required officer in the command spreading sequence.
Output Specification:
Print $q$ numbers, where the $i$-th number is the officer at the position $k_i$ in the list which describes the order in which officers will receive the command if it starts spreading from officer $u_i$. Print "-1" if the number of officers which receive the command is less than $k_i$.
You should process queries independently. They do not affect each other.
Demo Input:
['9 6\n1 1 1 3 5 3 5 7\n3 1\n1 5\n3 4\n7 3\n1 8\n1 9\n']
Demo Output:
['3\n6\n8\n-1\n9\n4\n']
Note:
none | ```python
from collections import defaultdict
dfs=[]
class Graph:
def __init__(self):
self.graph = defaultdict(list)
def addEdge(self, u, v):
# self.graph[u].append(v)
self.graph[v].append(u)
def DFSUtil(self, v):
visited[v] = True
dfs.append(v)
for i in self.graph[v]:
if visited[i] == False:
self.DFSUtil(i)
def DFS(self, v):
# visited = [False] * (max(self.graph) + 1)
self.DFSUtil(v)
def numberofnodes(self,s,e):
count1[s]=1
for u in self.graph[s]:
if u == e:
continue
self.numberofnodes(u, s)
count1[s] += count1[u]
g = Graph()
n,q=map(int,input().split())
count1=[0]*(n+1)
pi=list(map(int,input().split()))
visited=[0]*(n+2)
for i in range(0,n-1):
g.addEdge(i+2,pi[i])
g.DFS(1)
g.numberofnodes(1,0)
index=defaultdict(int)
for i in range(len(dfs)):
index[dfs[i]]=i
for qq in range(q):
ui,vi=map(int,input().split())
if count1[ui]<vi:
print(-1)
continue
else:
ind=index[ui]+vi
print(dfs[ind-1])
``` | -1 |
|
285 | C | Building Permutation | PROGRAMMING | 1,200 | [
"greedy",
"implementation",
"sortings"
] | null | null | Permutation *p* is an ordered set of integers *p*1,<=<=*p*2,<=<=...,<=<=*p**n*, consisting of *n* distinct positive integers, each of them doesn't exceed *n*. We'll denote the *i*-th element of permutation *p* as *p**i*. We'll call number *n* the size or the length of permutation *p*1,<=<=*p*2,<=<=...,<=<=*p**n*.
You have a sequence of integers *a*1,<=*a*2,<=...,<=*a**n*. In one move, you are allowed to decrease or increase any number by one. Count the minimum number of moves, needed to build a permutation from this sequence. | The first line contains integer *n* (1<=≤<=*n*<=≤<=3·105) — the size of the sought permutation. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=109<=≤<=*a**i*<=≤<=109). | Print a single number — the minimum number of moves.
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier. | [
"2\n3 0\n",
"3\n-1 -1 2\n"
] | [
"2\n",
"6\n"
] | In the first sample you should decrease the first number by one and then increase the second number by one. The resulting permutation is (2, 1).
In the second sample you need 6 moves to build permutation (1, 3, 2). | 1,500 | [
{
"input": "2\n3 0",
"output": "2"
},
{
"input": "3\n-1 -1 2",
"output": "6"
},
{
"input": "5\n-3 5 -3 3 3",
"output": "10"
},
{
"input": "10\n9 6 -2 4 1 1 1 9 6 2",
"output": "18"
},
{
"input": "9\n2 0 0 6 5 4 1 9 3",
"output": "15"
},
{
"input": "100\n-77 57 -95 -23 53 -28 82 -83 38 -73 85 28 25 6 -43 4 -10 -30 -9 -92 14 34 -93 61 36 -100 90 -68 28 16 100 -3 97 30 36 -55 62 -62 53 74 -50 -23 67 11 22 -30 -19 83 7 84 43 90 -65 -75 -15 97 90 15 66 2 13 -91 91 -44 46 51 51 -58 95 77 20 30 76 79 91 60 76 2 82 42 -93 94 -57 88 65 -95 -66 100 -9 33 -67 54 -99 97 53 13 54 66 60 -48",
"output": "3459"
}
] | 1,621,090,760 | 2,147,483,647 | PyPy 3 | OK | TESTS | 33 | 436 | 25,907,200 | n = int(input())
lista = [int(item) for item in input().split()]
lista.sort()
count = 0
for i in range(n):
count += abs(lista[i]-(i+1))
print(count)
| Title: Building Permutation
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Permutation *p* is an ordered set of integers *p*1,<=<=*p*2,<=<=...,<=<=*p**n*, consisting of *n* distinct positive integers, each of them doesn't exceed *n*. We'll denote the *i*-th element of permutation *p* as *p**i*. We'll call number *n* the size or the length of permutation *p*1,<=<=*p*2,<=<=...,<=<=*p**n*.
You have a sequence of integers *a*1,<=*a*2,<=...,<=*a**n*. In one move, you are allowed to decrease or increase any number by one. Count the minimum number of moves, needed to build a permutation from this sequence.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=3·105) — the size of the sought permutation. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=109<=≤<=*a**i*<=≤<=109).
Output Specification:
Print a single number — the minimum number of moves.
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.
Demo Input:
['2\n3 0\n', '3\n-1 -1 2\n']
Demo Output:
['2\n', '6\n']
Note:
In the first sample you should decrease the first number by one and then increase the second number by one. The resulting permutation is (2, 1).
In the second sample you need 6 moves to build permutation (1, 3, 2). | ```python
n = int(input())
lista = [int(item) for item in input().split()]
lista.sort()
count = 0
for i in range(n):
count += abs(lista[i]-(i+1))
print(count)
``` | 3 |
|
25 | B | Phone numbers | PROGRAMMING | 1,100 | [
"implementation"
] | B. Phone numbers | 2 | 256 | Phone number in Berland is a sequence of *n* digits. Often, to make it easier to memorize the number, it is divided into groups of two or three digits. For example, the phone number 1198733 is easier to remember as 11-987-33. Your task is to find for a given phone number any of its divisions into groups of two or three digits. | The first line contains integer *n* (2<=≤<=*n*<=≤<=100) — amount of digits in the phone number. The second line contains *n* digits — the phone number to divide into groups. | Output any of divisions of the given phone number into groups of two or three digits. Separate groups by single character -. If the answer is not unique, output any. | [
"6\n549871\n",
"7\n1198733\n"
] | [
"54-98-71",
"11-987-33\n"
] | none | 0 | [
{
"input": "6\n549871",
"output": "54-98-71"
},
{
"input": "7\n1198733",
"output": "119-87-33"
},
{
"input": "2\n74",
"output": "74"
},
{
"input": "2\n33",
"output": "33"
},
{
"input": "3\n074",
"output": "074"
},
{
"input": "3\n081",
"output": "081"
},
{
"input": "4\n3811",
"output": "38-11"
},
{
"input": "5\n21583",
"output": "215-83"
},
{
"input": "8\n33408349",
"output": "33-40-83-49"
},
{
"input": "9\n988808426",
"output": "988-80-84-26"
},
{
"input": "10\n0180990956",
"output": "01-80-99-09-56"
},
{
"input": "15\n433488906230138",
"output": "433-48-89-06-23-01-38"
},
{
"input": "22\n7135498415686025907059",
"output": "71-35-49-84-15-68-60-25-90-70-59"
},
{
"input": "49\n2429965524999668169991253653390090510755018570235",
"output": "242-99-65-52-49-99-66-81-69-99-12-53-65-33-90-09-05-10-75-50-18-57-02-35"
},
{
"input": "72\n491925337784111770500147619881727525570039735507439360627744863794794290",
"output": "49-19-25-33-77-84-11-17-70-50-01-47-61-98-81-72-75-25-57-00-39-73-55-07-43-93-60-62-77-44-86-37-94-79-42-90"
},
{
"input": "95\n32543414456047900690980198395035321172843693417425457554204776648220562494524275489599199209210",
"output": "325-43-41-44-56-04-79-00-69-09-80-19-83-95-03-53-21-17-28-43-69-34-17-42-54-57-55-42-04-77-66-48-22-05-62-49-45-24-27-54-89-59-91-99-20-92-10"
},
{
"input": "97\n9362344595153688016434451101547661156123505108492010669557671355055642365998461003851354321478898",
"output": "936-23-44-59-51-53-68-80-16-43-44-51-10-15-47-66-11-56-12-35-05-10-84-92-01-06-69-55-76-71-35-50-55-64-23-65-99-84-61-00-38-51-35-43-21-47-88-98"
},
{
"input": "98\n65521815795893886057122984634320900545031770769333931308009346017867969790810907868670369236928568",
"output": "65-52-18-15-79-58-93-88-60-57-12-29-84-63-43-20-90-05-45-03-17-70-76-93-33-93-13-08-00-93-46-01-78-67-96-97-90-81-09-07-86-86-70-36-92-36-92-85-68"
},
{
"input": "99\n455213856470326729480192345541970106407563996625458559297407682539801838244443866898560852503660390",
"output": "455-21-38-56-47-03-26-72-94-80-19-23-45-54-19-70-10-64-07-56-39-96-62-54-58-55-92-97-40-76-82-53-98-01-83-82-44-44-38-66-89-85-60-85-25-03-66-03-90"
},
{
"input": "100\n4004223124942730640235383244438257614581534320356060987241659784249551110165034719443327659510644224",
"output": "40-04-22-31-24-94-27-30-64-02-35-38-32-44-43-82-57-61-45-81-53-43-20-35-60-60-98-72-41-65-97-84-24-95-51-11-01-65-03-47-19-44-33-27-65-95-10-64-42-24"
}
] | 1,550,318,510 | 2,147,483,647 | Python 3 | OK | TESTS | 20 | 342 | 0 | n = int(input())
s = input()
l = []
if n > 3:
if n % 2 == 0:
for i in range(0, n, 2):
l.append(s[i:i + 2])
else:
l.append(s[:3])
for i in range(3, n, 2):
l.append(s[i:i + 2])
print(*l, sep='-')
else:
print(s)
| Title: Phone numbers
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Phone number in Berland is a sequence of *n* digits. Often, to make it easier to memorize the number, it is divided into groups of two or three digits. For example, the phone number 1198733 is easier to remember as 11-987-33. Your task is to find for a given phone number any of its divisions into groups of two or three digits.
Input Specification:
The first line contains integer *n* (2<=≤<=*n*<=≤<=100) — amount of digits in the phone number. The second line contains *n* digits — the phone number to divide into groups.
Output Specification:
Output any of divisions of the given phone number into groups of two or three digits. Separate groups by single character -. If the answer is not unique, output any.
Demo Input:
['6\n549871\n', '7\n1198733\n']
Demo Output:
['54-98-71', '11-987-33\n']
Note:
none | ```python
n = int(input())
s = input()
l = []
if n > 3:
if n % 2 == 0:
for i in range(0, n, 2):
l.append(s[i:i + 2])
else:
l.append(s[:3])
for i in range(3, n, 2):
l.append(s[i:i + 2])
print(*l, sep='-')
else:
print(s)
``` | 3.9145 |
747 | A | Display Size | PROGRAMMING | 800 | [
"brute force",
"math"
] | null | null | A big company decided to launch a new series of rectangular displays, and decided that the display must have exactly *n* pixels.
Your task is to determine the size of the rectangular display — the number of lines (rows) of pixels *a* and the number of columns of pixels *b*, so that:
- there are exactly *n* pixels on the display; - the number of rows does not exceed the number of columns, it means *a*<=≤<=*b*; - the difference *b*<=-<=*a* is as small as possible. | The first line contains the positive integer *n* (1<=≤<=*n*<=≤<=106) — the number of pixels display should have. | Print two integers — the number of rows and columns on the display. | [
"8\n",
"64\n",
"5\n",
"999999\n"
] | [
"2 4\n",
"8 8\n",
"1 5\n",
"999 1001\n"
] | In the first example the minimum possible difference equals 2, so on the display should be 2 rows of 4 pixels.
In the second example the minimum possible difference equals 0, so on the display should be 8 rows of 8 pixels.
In the third example the minimum possible difference equals 4, so on the display should be 1 row of 5 pixels. | 500 | [
{
"input": "8",
"output": "2 4"
},
{
"input": "64",
"output": "8 8"
},
{
"input": "5",
"output": "1 5"
},
{
"input": "999999",
"output": "999 1001"
},
{
"input": "716539",
"output": "97 7387"
},
{
"input": "1",
"output": "1 1"
},
{
"input": "2",
"output": "1 2"
},
{
"input": "3",
"output": "1 3"
},
{
"input": "4",
"output": "2 2"
},
{
"input": "6",
"output": "2 3"
},
{
"input": "7",
"output": "1 7"
},
{
"input": "9",
"output": "3 3"
},
{
"input": "10",
"output": "2 5"
},
{
"input": "11",
"output": "1 11"
},
{
"input": "12",
"output": "3 4"
},
{
"input": "15",
"output": "3 5"
},
{
"input": "100",
"output": "10 10"
},
{
"input": "101",
"output": "1 101"
},
{
"input": "169",
"output": "13 13"
},
{
"input": "179",
"output": "1 179"
},
{
"input": "190",
"output": "10 19"
},
{
"input": "1000",
"output": "25 40"
},
{
"input": "10000",
"output": "100 100"
},
{
"input": "10001",
"output": "73 137"
},
{
"input": "100000",
"output": "250 400"
},
{
"input": "100001",
"output": "11 9091"
},
{
"input": "1000000",
"output": "1000 1000"
},
{
"input": "999983",
"output": "1 999983"
},
{
"input": "524288",
"output": "512 1024"
},
{
"input": "954493",
"output": "971 983"
},
{
"input": "966289",
"output": "983 983"
},
{
"input": "944663",
"output": "961 983"
},
{
"input": "912673",
"output": "97 9409"
},
{
"input": "732641",
"output": "679 1079"
},
{
"input": "232897",
"output": "343 679"
},
{
"input": "16807",
"output": "49 343"
},
{
"input": "999958",
"output": "2 499979"
},
{
"input": "990151",
"output": "1 990151"
},
{
"input": "997002",
"output": "998 999"
},
{
"input": "20",
"output": "4 5"
},
{
"input": "20261",
"output": "1 20261"
},
{
"input": "999123",
"output": "3 333041"
},
{
"input": "901841",
"output": "1 901841"
}
] | 1,516,528,274 | 2,147,483,647 | Python 3 | OK | TESTS | 43 | 62 | 5,632,000 | from math import sqrt
n = int(input().strip())
sq = int(sqrt(n))
for i in range(sq, 0, -1):
if not n%i:
print(i, n//i)
break
| Title: Display Size
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A big company decided to launch a new series of rectangular displays, and decided that the display must have exactly *n* pixels.
Your task is to determine the size of the rectangular display — the number of lines (rows) of pixels *a* and the number of columns of pixels *b*, so that:
- there are exactly *n* pixels on the display; - the number of rows does not exceed the number of columns, it means *a*<=≤<=*b*; - the difference *b*<=-<=*a* is as small as possible.
Input Specification:
The first line contains the positive integer *n* (1<=≤<=*n*<=≤<=106) — the number of pixels display should have.
Output Specification:
Print two integers — the number of rows and columns on the display.
Demo Input:
['8\n', '64\n', '5\n', '999999\n']
Demo Output:
['2 4\n', '8 8\n', '1 5\n', '999 1001\n']
Note:
In the first example the minimum possible difference equals 2, so on the display should be 2 rows of 4 pixels.
In the second example the minimum possible difference equals 0, so on the display should be 8 rows of 8 pixels.
In the third example the minimum possible difference equals 4, so on the display should be 1 row of 5 pixels. | ```python
from math import sqrt
n = int(input().strip())
sq = int(sqrt(n))
for i in range(sq, 0, -1):
if not n%i:
print(i, n//i)
break
``` | 3 |
|
876 | A | Trip For Meal | PROGRAMMING | 900 | [
"math"
] | null | null | Winnie-the-Pooh likes honey very much! That is why he decided to visit his friends. Winnie has got three best friends: Rabbit, Owl and Eeyore, each of them lives in his own house. There are winding paths between each pair of houses. The length of a path between Rabbit's and Owl's houses is *a* meters, between Rabbit's and Eeyore's house is *b* meters, between Owl's and Eeyore's house is *c* meters.
For enjoying his life and singing merry songs Winnie-the-Pooh should have a meal *n* times a day. Now he is in the Rabbit's house and has a meal for the first time. Each time when in the friend's house where Winnie is now the supply of honey is about to end, Winnie leaves that house. If Winnie has not had a meal the required amount of times, he comes out from the house and goes to someone else of his two friends. For this he chooses one of two adjacent paths, arrives to the house on the other end and visits his friend. You may assume that when Winnie is eating in one of his friend's house, the supply of honey in other friend's houses recover (most probably, they go to the supply store).
Winnie-the-Pooh does not like physical activity. He wants to have a meal *n* times, traveling minimum possible distance. Help him to find this distance. | First line contains an integer *n* (1<=≤<=*n*<=≤<=100) — number of visits.
Second line contains an integer *a* (1<=≤<=*a*<=≤<=100) — distance between Rabbit's and Owl's houses.
Third line contains an integer *b* (1<=≤<=*b*<=≤<=100) — distance between Rabbit's and Eeyore's houses.
Fourth line contains an integer *c* (1<=≤<=*c*<=≤<=100) — distance between Owl's and Eeyore's houses. | Output one number — minimum distance in meters Winnie must go through to have a meal *n* times. | [
"3\n2\n3\n1\n",
"1\n2\n3\n5\n"
] | [
"3\n",
"0\n"
] | In the first test case the optimal path for Winnie is the following: first have a meal in Rabbit's house, then in Owl's house, then in Eeyore's house. Thus he will pass the distance 2 + 1 = 3.
In the second test case Winnie has a meal in Rabbit's house and that is for him. So he doesn't have to walk anywhere at all. | 500 | [
{
"input": "3\n2\n3\n1",
"output": "3"
},
{
"input": "1\n2\n3\n5",
"output": "0"
},
{
"input": "10\n1\n8\n3",
"output": "9"
},
{
"input": "7\n10\n5\n6",
"output": "30"
},
{
"input": "9\n9\n7\n5",
"output": "42"
},
{
"input": "9\n37\n85\n76",
"output": "296"
},
{
"input": "76\n46\n77\n11",
"output": "860"
},
{
"input": "80\n42\n1\n37",
"output": "79"
},
{
"input": "8\n80\n55\n1",
"output": "61"
},
{
"input": "10\n13\n72\n17",
"output": "117"
},
{
"input": "9\n24\n1\n63",
"output": "8"
},
{
"input": "65\n5\n8\n7",
"output": "320"
},
{
"input": "56\n8\n9\n3",
"output": "170"
},
{
"input": "59\n8\n1\n2",
"output": "58"
},
{
"input": "75\n50\n50\n5",
"output": "415"
},
{
"input": "75\n54\n76\n66",
"output": "3996"
},
{
"input": "73\n71\n69\n66",
"output": "4755"
},
{
"input": "83\n58\n88\n16",
"output": "1354"
},
{
"input": "74\n31\n11\n79",
"output": "803"
},
{
"input": "62\n27\n16\n72",
"output": "976"
},
{
"input": "72\n95\n27\n9",
"output": "657"
},
{
"input": "1\n2\n2\n1",
"output": "0"
},
{
"input": "1\n1\n1\n1",
"output": "0"
},
{
"input": "1\n1\n1\n99",
"output": "0"
},
{
"input": "100\n100\n100\n100",
"output": "9900"
},
{
"input": "2\n1\n1\n3",
"output": "1"
},
{
"input": "1\n3\n2\n1",
"output": "0"
},
{
"input": "1\n5\n6\n1",
"output": "0"
},
{
"input": "1\n2\n6\n1",
"output": "0"
},
{
"input": "1\n30\n20\n1",
"output": "0"
},
{
"input": "1\n15\n13\n11",
"output": "0"
},
{
"input": "1\n100\n50\n1",
"output": "0"
},
{
"input": "1\n2\n3\n1",
"output": "0"
},
{
"input": "5\n5\n5\n5",
"output": "20"
},
{
"input": "1\n100\n50\n5",
"output": "0"
},
{
"input": "1\n5\n7\n3",
"output": "0"
},
{
"input": "1\n3\n4\n2",
"output": "0"
},
{
"input": "3\n3\n3\n3",
"output": "6"
},
{
"input": "1\n99\n98\n1",
"output": "0"
},
{
"input": "5\n3\n3\n1",
"output": "6"
},
{
"input": "2\n1\n1\n1",
"output": "1"
},
{
"input": "2\n1\n7\n8",
"output": "1"
},
{
"input": "2\n2\n2\n1",
"output": "2"
},
{
"input": "2\n2\n3\n1",
"output": "2"
},
{
"input": "1\n5\n6\n2",
"output": "0"
},
{
"input": "100\n1\n1\n100",
"output": "99"
},
{
"input": "2\n3\n2\n1",
"output": "2"
},
{
"input": "1\n99\n98\n97",
"output": "0"
}
] | 1,508,242,338 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 62 | 5,529,600 | n = int(input())
a = int(input())
b = int(input())
c = int(input())
lista = []
lista.append(a)
lista.append(b)
lista.append(c)
minimalni = 101
nekaj = 0
for i in lista:
if i < minimalni:
minimalni = i
if minimalni < n:
nekaj += minimalni
else:
break
print(nekaj)
| Title: Trip For Meal
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Winnie-the-Pooh likes honey very much! That is why he decided to visit his friends. Winnie has got three best friends: Rabbit, Owl and Eeyore, each of them lives in his own house. There are winding paths between each pair of houses. The length of a path between Rabbit's and Owl's houses is *a* meters, between Rabbit's and Eeyore's house is *b* meters, between Owl's and Eeyore's house is *c* meters.
For enjoying his life and singing merry songs Winnie-the-Pooh should have a meal *n* times a day. Now he is in the Rabbit's house and has a meal for the first time. Each time when in the friend's house where Winnie is now the supply of honey is about to end, Winnie leaves that house. If Winnie has not had a meal the required amount of times, he comes out from the house and goes to someone else of his two friends. For this he chooses one of two adjacent paths, arrives to the house on the other end and visits his friend. You may assume that when Winnie is eating in one of his friend's house, the supply of honey in other friend's houses recover (most probably, they go to the supply store).
Winnie-the-Pooh does not like physical activity. He wants to have a meal *n* times, traveling minimum possible distance. Help him to find this distance.
Input Specification:
First line contains an integer *n* (1<=≤<=*n*<=≤<=100) — number of visits.
Second line contains an integer *a* (1<=≤<=*a*<=≤<=100) — distance between Rabbit's and Owl's houses.
Third line contains an integer *b* (1<=≤<=*b*<=≤<=100) — distance between Rabbit's and Eeyore's houses.
Fourth line contains an integer *c* (1<=≤<=*c*<=≤<=100) — distance between Owl's and Eeyore's houses.
Output Specification:
Output one number — minimum distance in meters Winnie must go through to have a meal *n* times.
Demo Input:
['3\n2\n3\n1\n', '1\n2\n3\n5\n']
Demo Output:
['3\n', '0\n']
Note:
In the first test case the optimal path for Winnie is the following: first have a meal in Rabbit's house, then in Owl's house, then in Eeyore's house. Thus he will pass the distance 2 + 1 = 3.
In the second test case Winnie has a meal in Rabbit's house and that is for him. So he doesn't have to walk anywhere at all. | ```python
n = int(input())
a = int(input())
b = int(input())
c = int(input())
lista = []
lista.append(a)
lista.append(b)
lista.append(c)
minimalni = 101
nekaj = 0
for i in lista:
if i < minimalni:
minimalni = i
if minimalni < n:
nekaj += minimalni
else:
break
print(nekaj)
``` | 0 |
|
298 | A | Snow Footprints | PROGRAMMING | 1,300 | [
"greedy",
"implementation"
] | null | null | There is a straight snowy road, divided into *n* blocks. The blocks are numbered from 1 to *n* from left to right. If one moves from the *i*-th block to the (*i*<=+<=1)-th block, he will leave a right footprint on the *i*-th block. Similarly, if one moves from the *i*-th block to the (*i*<=-<=1)-th block, he will leave a left footprint on the *i*-th block. If there already is a footprint on the *i*-th block, the new footprint will cover the old one.
At the beginning, there were no footprints. Then polar bear Alice starts from the *s*-th block, makes a sequence of moves and ends in the *t*-th block. It is known that Alice never moves outside of the road.
You are given the description of Alice's footprints. Your task is to find a pair of possible values of *s*,<=*t* by looking at the footprints. | The first line of the input contains integer *n* (3<=≤<=*n*<=≤<=1000).
The second line contains the description of the road — the string that consists of *n* characters. Each character will be either "." (a block without footprint), or "L" (a block with a left footprint), "R" (a block with a right footprint).
It's guaranteed that the given string contains at least one character not equal to ".". Also, the first and the last character will always be ".". It's guaranteed that a solution exists. | Print two space-separated integers — the values of *s* and *t*. If there are several possible solutions you can print any of them. | [
"9\n..RRLL...\n",
"11\n.RRRLLLLL..\n"
] | [
"3 4\n",
"7 5\n"
] | The first test sample is the one in the picture. | 500 | [
{
"input": "11\n.RRRLLLLL..",
"output": "7 5"
},
{
"input": "4\n.RL.",
"output": "3 2"
},
{
"input": "3\n.L.",
"output": "2 1"
},
{
"input": "3\n.R.",
"output": "2 3"
}
] | 1,545,207,909 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 3 | 186 | 0 | n = int(input())
a = list(input())
start = ends = counter = 0
for i in range(n-1):
if a[i] == '.':
continue
elif a[i] == 'R':
start = i + 1
if a[i+1] == 'R':
continue
elif a[i+1] == 'L' or a[i+1] == '.':
ends = i + 2
break
else:
start = i + 1
if a[i+1] == 'L':
continue
elif a[i+1] == 'R' or a[i+1] == '.':
ends = i + 2
break
print(start, ends) | Title: Snow Footprints
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There is a straight snowy road, divided into *n* blocks. The blocks are numbered from 1 to *n* from left to right. If one moves from the *i*-th block to the (*i*<=+<=1)-th block, he will leave a right footprint on the *i*-th block. Similarly, if one moves from the *i*-th block to the (*i*<=-<=1)-th block, he will leave a left footprint on the *i*-th block. If there already is a footprint on the *i*-th block, the new footprint will cover the old one.
At the beginning, there were no footprints. Then polar bear Alice starts from the *s*-th block, makes a sequence of moves and ends in the *t*-th block. It is known that Alice never moves outside of the road.
You are given the description of Alice's footprints. Your task is to find a pair of possible values of *s*,<=*t* by looking at the footprints.
Input Specification:
The first line of the input contains integer *n* (3<=≤<=*n*<=≤<=1000).
The second line contains the description of the road — the string that consists of *n* characters. Each character will be either "." (a block without footprint), or "L" (a block with a left footprint), "R" (a block with a right footprint).
It's guaranteed that the given string contains at least one character not equal to ".". Also, the first and the last character will always be ".". It's guaranteed that a solution exists.
Output Specification:
Print two space-separated integers — the values of *s* and *t*. If there are several possible solutions you can print any of them.
Demo Input:
['9\n..RRLL...\n', '11\n.RRRLLLLL..\n']
Demo Output:
['3 4\n', '7 5\n']
Note:
The first test sample is the one in the picture. | ```python
n = int(input())
a = list(input())
start = ends = counter = 0
for i in range(n-1):
if a[i] == '.':
continue
elif a[i] == 'R':
start = i + 1
if a[i+1] == 'R':
continue
elif a[i+1] == 'L' or a[i+1] == '.':
ends = i + 2
break
else:
start = i + 1
if a[i+1] == 'L':
continue
elif a[i+1] == 'R' or a[i+1] == '.':
ends = i + 2
break
print(start, ends)
``` | 0 |
|
592 | B | The Monster and the Squirrel | PROGRAMMING | 1,100 | [
"math"
] | null | null | Ari the monster always wakes up very early with the first ray of the sun and the first thing she does is feeding her squirrel.
Ari draws a regular convex polygon on the floor and numbers it's vertices 1,<=2,<=...,<=*n* in clockwise order. Then starting from the vertex 1 she draws a ray in the direction of each other vertex. The ray stops when it reaches a vertex or intersects with another ray drawn before. Ari repeats this process for vertex 2,<=3,<=...,<=*n* (in this particular order). And then she puts a walnut in each region inside the polygon.
Ada the squirrel wants to collect all the walnuts, but she is not allowed to step on the lines drawn by Ari. That means Ada have to perform a small jump if she wants to go from one region to another. Ada can jump from one region P to another region Q if and only if P and Q share a side or a corner.
Assuming that Ada starts from outside of the picture, what is the minimum number of jumps she has to perform in order to collect all the walnuts? | The first and only line of the input contains a single integer *n* (3<=≤<=*n*<=≤<=54321) - the number of vertices of the regular polygon drawn by Ari. | Print the minimum number of jumps Ada should make to collect all the walnuts. Note, that she doesn't need to leave the polygon after. | [
"5\n",
"3\n"
] | [
"9\n",
"1\n"
] | One of the possible solutions for the first sample is shown on the picture above. | 1,000 | [
{
"input": "5",
"output": "9"
},
{
"input": "3",
"output": "1"
},
{
"input": "54321",
"output": "2950553761"
},
{
"input": "4",
"output": "4"
},
{
"input": "6",
"output": "16"
},
{
"input": "7",
"output": "25"
},
{
"input": "8",
"output": "36"
},
{
"input": "9",
"output": "49"
},
{
"input": "10",
"output": "64"
},
{
"input": "54320",
"output": "2950445124"
},
{
"input": "54319",
"output": "2950336489"
},
{
"input": "54318",
"output": "2950227856"
},
{
"input": "54317",
"output": "2950119225"
},
{
"input": "54316",
"output": "2950010596"
},
{
"input": "54315",
"output": "2949901969"
},
{
"input": "54314",
"output": "2949793344"
},
{
"input": "8153",
"output": "66438801"
},
{
"input": "51689",
"output": "2671545969"
},
{
"input": "16659",
"output": "277455649"
},
{
"input": "47389",
"output": "2245527769"
},
{
"input": "314",
"output": "97344"
},
{
"input": "23481",
"output": "551263441"
},
{
"input": "20380",
"output": "415262884"
},
{
"input": "1994",
"output": "3968064"
}
] | 1,446,310,365 | 1,365 | Python 3 | OK | TESTS | 24 | 61 | 0 | #!/usr/bin/env python3
from sys import stdin
n = int(stdin.readline().strip())
out = (n - 2) ** 2
print(out)
| Title: The Monster and the Squirrel
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Ari the monster always wakes up very early with the first ray of the sun and the first thing she does is feeding her squirrel.
Ari draws a regular convex polygon on the floor and numbers it's vertices 1,<=2,<=...,<=*n* in clockwise order. Then starting from the vertex 1 she draws a ray in the direction of each other vertex. The ray stops when it reaches a vertex or intersects with another ray drawn before. Ari repeats this process for vertex 2,<=3,<=...,<=*n* (in this particular order). And then she puts a walnut in each region inside the polygon.
Ada the squirrel wants to collect all the walnuts, but she is not allowed to step on the lines drawn by Ari. That means Ada have to perform a small jump if she wants to go from one region to another. Ada can jump from one region P to another region Q if and only if P and Q share a side or a corner.
Assuming that Ada starts from outside of the picture, what is the minimum number of jumps she has to perform in order to collect all the walnuts?
Input Specification:
The first and only line of the input contains a single integer *n* (3<=≤<=*n*<=≤<=54321) - the number of vertices of the regular polygon drawn by Ari.
Output Specification:
Print the minimum number of jumps Ada should make to collect all the walnuts. Note, that she doesn't need to leave the polygon after.
Demo Input:
['5\n', '3\n']
Demo Output:
['9\n', '1\n']
Note:
One of the possible solutions for the first sample is shown on the picture above. | ```python
#!/usr/bin/env python3
from sys import stdin
n = int(stdin.readline().strip())
out = (n - 2) ** 2
print(out)
``` | 3 |
|
25 | A | IQ test | PROGRAMMING | 1,300 | [
"brute force"
] | A. IQ test | 2 | 256 | Bob is preparing to pass IQ test. The most frequent task in this test is to find out which one of the given *n* numbers differs from the others. Bob observed that one number usually differs from the others in evenness. Help Bob — to check his answers, he needs a program that among the given *n* numbers finds one that is different in evenness. | The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of numbers in the task. The second line contains *n* space-separated natural numbers, not exceeding 100. It is guaranteed, that exactly one of these numbers differs from the others in evenness. | Output index of number that differs from the others in evenness. Numbers are numbered from 1 in the input order. | [
"5\n2 4 7 8 10\n",
"4\n1 2 1 1\n"
] | [
"3\n",
"2\n"
] | none | 0 | [
{
"input": "5\n2 4 7 8 10",
"output": "3"
},
{
"input": "4\n1 2 1 1",
"output": "2"
},
{
"input": "3\n1 2 2",
"output": "1"
},
{
"input": "3\n100 99 100",
"output": "2"
},
{
"input": "3\n5 3 2",
"output": "3"
},
{
"input": "4\n43 28 1 91",
"output": "2"
},
{
"input": "4\n75 13 94 77",
"output": "3"
},
{
"input": "4\n97 8 27 3",
"output": "2"
},
{
"input": "10\n95 51 12 91 85 3 1 31 25 7",
"output": "3"
},
{
"input": "20\n88 96 66 51 14 88 2 92 18 72 18 88 20 30 4 82 90 100 24 46",
"output": "4"
},
{
"input": "30\n20 94 56 50 10 98 52 32 14 22 24 60 4 8 98 46 34 68 82 82 98 90 50 20 78 49 52 94 64 36",
"output": "26"
},
{
"input": "50\n79 27 77 57 37 45 27 49 65 33 57 21 71 19 75 85 65 61 23 97 85 9 23 1 9 3 99 77 77 21 79 69 15 37 15 7 93 81 13 89 91 31 45 93 15 97 55 80 85 83",
"output": "48"
},
{
"input": "60\n46 11 73 65 3 69 3 53 43 53 97 47 55 93 31 75 35 3 9 73 23 31 3 81 91 79 61 21 15 11 11 11 81 7 83 75 39 87 83 59 89 55 93 27 49 67 67 29 1 93 11 17 9 19 35 21 63 31 31 25",
"output": "1"
},
{
"input": "70\n28 42 42 92 64 54 22 38 38 78 62 38 4 38 14 66 4 92 66 58 94 26 4 44 41 88 48 82 44 26 74 44 48 4 16 92 34 38 26 64 94 4 30 78 50 54 12 90 8 16 80 98 28 100 74 50 36 42 92 18 76 98 8 22 2 50 58 50 64 46",
"output": "25"
},
{
"input": "100\n43 35 79 53 13 91 91 45 65 83 57 9 42 39 85 45 71 51 61 59 31 13 63 39 25 21 79 39 91 67 21 61 97 75 93 83 29 79 59 97 11 37 63 51 39 55 91 23 21 17 47 23 35 75 49 5 69 99 5 7 41 17 25 89 15 79 21 63 53 81 43 91 59 91 69 99 85 15 91 51 49 37 65 7 89 81 21 93 61 63 97 93 45 17 13 69 57 25 75 73",
"output": "13"
},
{
"input": "100\n50 24 68 60 70 30 52 22 18 74 68 98 20 82 4 46 26 68 100 78 84 58 74 98 38 88 68 86 64 80 82 100 20 22 98 98 52 6 94 10 48 68 2 18 38 22 22 82 44 20 66 72 36 58 64 6 36 60 4 96 76 64 12 90 10 58 64 60 74 28 90 26 24 60 40 58 2 16 76 48 58 36 82 60 24 44 4 78 28 38 8 12 40 16 38 6 66 24 31 76",
"output": "99"
},
{
"input": "100\n47 48 94 48 14 18 94 36 96 22 12 30 94 20 48 98 40 58 2 94 8 36 98 18 98 68 2 60 76 38 18 100 8 72 100 68 2 86 92 72 58 16 48 14 6 58 72 76 6 88 80 66 20 28 74 62 86 68 90 86 2 56 34 38 56 90 4 8 76 44 32 86 12 98 38 34 54 92 70 94 10 24 82 66 90 58 62 2 32 58 100 22 58 72 2 22 68 72 42 14",
"output": "1"
},
{
"input": "99\n38 20 68 60 84 16 28 88 60 48 80 28 4 92 70 60 46 46 20 34 12 100 76 2 40 10 8 86 6 80 50 66 12 34 14 28 26 70 46 64 34 96 10 90 98 96 56 88 50 74 70 94 2 94 24 66 68 46 22 30 6 10 64 32 88 14 98 100 64 58 50 18 50 50 8 38 8 16 54 2 60 54 62 84 92 98 4 72 66 26 14 88 99 16 10 6 88 56 22",
"output": "93"
},
{
"input": "99\n50 83 43 89 53 47 69 1 5 37 63 87 95 15 55 95 75 89 33 53 89 75 93 75 11 85 49 29 11 97 49 67 87 11 25 37 97 73 67 49 87 43 53 97 43 29 53 33 45 91 37 73 39 49 59 5 21 43 87 35 5 63 89 57 63 47 29 99 19 85 13 13 3 13 43 19 5 9 61 51 51 57 15 89 13 97 41 13 99 79 13 27 97 95 73 33 99 27 23",
"output": "1"
},
{
"input": "98\n61 56 44 30 58 14 20 24 88 28 46 56 96 52 58 42 94 50 46 30 46 80 72 88 68 16 6 60 26 90 10 98 76 20 56 40 30 16 96 20 88 32 62 30 74 58 36 76 60 4 24 36 42 54 24 92 28 14 2 74 86 90 14 52 34 82 40 76 8 64 2 56 10 8 78 16 70 86 70 42 70 74 22 18 76 98 88 28 62 70 36 72 20 68 34 48 80 98",
"output": "1"
},
{
"input": "98\n66 26 46 42 78 32 76 42 26 82 8 12 4 10 24 26 64 44 100 46 94 64 30 18 88 28 8 66 30 82 82 28 74 52 62 80 80 60 94 86 64 32 44 88 92 20 12 74 94 28 34 58 4 22 16 10 94 76 82 58 40 66 22 6 30 32 92 54 16 76 74 98 18 48 48 30 92 2 16 42 84 74 30 60 64 52 50 26 16 86 58 96 79 60 20 62 82 94",
"output": "93"
},
{
"input": "95\n9 31 27 93 17 77 75 9 9 53 89 39 51 99 5 1 11 39 27 49 91 17 27 79 81 71 37 75 35 13 93 4 99 55 85 11 23 57 5 43 5 61 15 35 23 91 3 81 99 85 43 37 39 27 5 67 7 33 75 59 13 71 51 27 15 93 51 63 91 53 43 99 25 47 17 71 81 15 53 31 59 83 41 23 73 25 91 91 13 17 25 13 55 57 29",
"output": "32"
},
{
"input": "100\n91 89 81 45 53 1 41 3 77 93 55 97 55 97 87 27 69 95 73 41 93 21 75 35 53 56 5 51 87 59 91 67 33 3 99 45 83 17 97 47 75 97 7 89 17 99 23 23 81 25 55 97 27 35 69 5 77 35 93 19 55 59 37 21 31 37 49 41 91 53 73 69 7 37 37 39 17 71 7 97 55 17 47 23 15 73 31 39 57 37 9 5 61 41 65 57 77 79 35 47",
"output": "26"
},
{
"input": "99\n38 56 58 98 80 54 26 90 14 16 78 92 52 74 40 30 84 14 44 80 16 90 98 68 26 24 78 72 42 16 84 40 14 44 2 52 50 2 12 96 58 66 8 80 44 52 34 34 72 98 74 4 66 74 56 21 8 38 76 40 10 22 48 32 98 34 12 62 80 68 64 82 22 78 58 74 20 22 48 56 12 38 32 72 6 16 74 24 94 84 26 38 18 24 76 78 98 94 72",
"output": "56"
},
{
"input": "100\n44 40 6 40 56 90 98 8 36 64 76 86 98 76 36 92 6 30 98 70 24 98 96 60 24 82 88 68 86 96 34 42 58 10 40 26 56 10 88 58 70 32 24 28 14 82 52 12 62 36 70 60 52 34 74 30 78 76 10 16 42 94 66 90 70 38 52 12 58 22 98 96 14 68 24 70 4 30 84 98 8 50 14 52 66 34 100 10 28 100 56 48 38 12 38 14 91 80 70 86",
"output": "97"
},
{
"input": "100\n96 62 64 20 90 46 56 90 68 36 30 56 70 28 16 64 94 34 6 32 34 50 94 22 90 32 40 2 72 10 88 38 28 92 20 26 56 80 4 100 100 90 16 74 74 84 8 2 30 20 80 32 16 46 92 56 42 12 96 64 64 42 64 58 50 42 74 28 2 4 36 32 70 50 54 92 70 16 45 76 28 16 18 50 48 2 62 94 4 12 52 52 4 100 70 60 82 62 98 42",
"output": "79"
},
{
"input": "99\n14 26 34 68 90 58 50 36 8 16 18 6 2 74 54 20 36 84 32 50 52 2 26 24 3 64 20 10 54 26 66 44 28 72 4 96 78 90 96 86 68 28 94 4 12 46 100 32 22 36 84 32 44 94 76 94 4 52 12 30 74 4 34 64 58 72 44 16 70 56 54 8 14 74 8 6 58 62 98 54 14 40 80 20 36 72 28 98 20 58 40 52 90 64 22 48 54 70 52",
"output": "25"
},
{
"input": "95\n82 86 30 78 6 46 80 66 74 72 16 24 18 52 52 38 60 36 86 26 62 28 22 46 96 26 94 84 20 46 66 88 76 32 12 86 74 18 34 88 4 48 94 6 58 6 100 82 4 24 88 32 54 98 34 48 6 76 42 88 42 28 100 4 22 2 10 66 82 54 98 20 60 66 38 98 32 47 86 58 6 100 12 46 2 42 8 84 78 28 24 70 34 28 86",
"output": "78"
},
{
"input": "90\n40 50 8 42 76 24 58 42 26 68 20 48 54 12 34 84 14 36 32 88 6 50 96 56 20 92 48 16 40 34 96 46 20 84 30 50 20 98 8 44 96 42 8 76 70 38 84 30 40 88 84 72 2 22 52 58 16 62 100 66 80 40 50 32 14 62 88 72 22 99 76 50 84 82 8 82 98 46 26 40 2 98 18 78 30 72 70 18 34 68",
"output": "70"
},
{
"input": "80\n81 43 87 1 55 43 53 61 27 19 43 13 89 9 33 83 75 55 97 71 91 37 95 5 21 69 81 93 95 69 31 83 55 7 97 7 79 57 8 61 27 85 49 1 15 97 63 79 29 73 41 85 5 41 31 93 67 11 63 59 15 99 91 77 43 69 23 23 81 73 19 1 67 51 1 75 99 67 3 81",
"output": "39"
},
{
"input": "98\n13 83 61 27 35 1 85 95 97 73 95 65 73 45 5 43 27 83 91 19 11 3 85 59 9 39 69 23 45 7 51 85 5 71 5 95 1 51 75 3 43 57 3 11 33 71 21 99 47 41 87 39 71 87 31 85 91 49 83 5 49 85 47 91 55 99 33 23 31 23 23 73 29 77 55 31 25 5 81 49 91 15 15 39 87 5 9 40 69 47 29 33 11 21 49 79 51 83",
"output": "88"
},
{
"input": "3\n100 100 1",
"output": "3"
}
] | 1,539,862,468 | 2,147,483,647 | Python 3 | OK | TESTS | 32 | 218 | 0 | x=int(input())
y=list(map(int,input().split()))
z=[]
q=[]
for i in range(x):
if y[i]%2==0:
z.append(y[i])
else:
q.append(y[i])
if len(z)==1:
print(y.index(z[0])+1)
else:
print(y.index(q[0])+1) | Title: IQ test
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Bob is preparing to pass IQ test. The most frequent task in this test is to find out which one of the given *n* numbers differs from the others. Bob observed that one number usually differs from the others in evenness. Help Bob — to check his answers, he needs a program that among the given *n* numbers finds one that is different in evenness.
Input Specification:
The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of numbers in the task. The second line contains *n* space-separated natural numbers, not exceeding 100. It is guaranteed, that exactly one of these numbers differs from the others in evenness.
Output Specification:
Output index of number that differs from the others in evenness. Numbers are numbered from 1 in the input order.
Demo Input:
['5\n2 4 7 8 10\n', '4\n1 2 1 1\n']
Demo Output:
['3\n', '2\n']
Note:
none | ```python
x=int(input())
y=list(map(int,input().split()))
z=[]
q=[]
for i in range(x):
if y[i]%2==0:
z.append(y[i])
else:
q.append(y[i])
if len(z)==1:
print(y.index(z[0])+1)
else:
print(y.index(q[0])+1)
``` | 3.9455 |
119 | A | Epic Game | PROGRAMMING | 800 | [
"implementation"
] | null | null | Simon and Antisimon play a game. Initially each player receives one fixed positive integer that doesn't change throughout the game. Simon receives number *a* and Antisimon receives number *b*. They also have a heap of *n* stones. The players take turns to make a move and Simon starts. During a move a player should take from the heap the number of stones equal to the greatest common divisor of the fixed number he has received and the number of stones left in the heap. A player loses when he cannot take the required number of stones (i. e. the heap has strictly less stones left than one needs to take).
Your task is to determine by the given *a*, *b* and *n* who wins the game. | The only string contains space-separated integers *a*, *b* and *n* (1<=≤<=*a*,<=*b*,<=*n*<=≤<=100) — the fixed numbers Simon and Antisimon have received correspondingly and the initial number of stones in the pile. | If Simon wins, print "0" (without the quotes), otherwise print "1" (without the quotes). | [
"3 5 9\n",
"1 1 100\n"
] | [
"0",
"1"
] | The greatest common divisor of two non-negative integers *a* and *b* is such maximum positive integer *k*, that *a* is divisible by *k* without remainder and similarly, *b* is divisible by *k* without remainder. Let *gcd*(*a*, *b*) represent the operation of calculating the greatest common divisor of numbers *a* and *b*. Specifically, *gcd*(*x*, 0) = *gcd*(0, *x*) = *x*.
In the first sample the game will go like that:
- Simon should take *gcd*(3, 9) = 3 stones from the heap. After his move the heap has 6 stones left.- Antisimon should take *gcd*(5, 6) = 1 stone from the heap. After his move the heap has 5 stones left.- Simon should take *gcd*(3, 5) = 1 stone from the heap. After his move the heap has 4 stones left.- Antisimon should take *gcd*(5, 4) = 1 stone from the heap. After his move the heap has 3 stones left.- Simon should take *gcd*(3, 3) = 3 stones from the heap. After his move the heap has 0 stones left.- Antisimon should take *gcd*(5, 0) = 5 stones from the heap. As 0 < 5, it is impossible and Antisimon loses.
In the second sample each player during each move takes one stone from the heap. As *n* is even, Antisimon takes the last stone and Simon can't make a move after that. | 500 | [
{
"input": "3 5 9",
"output": "0"
},
{
"input": "1 1 100",
"output": "1"
},
{
"input": "23 12 16",
"output": "1"
},
{
"input": "95 26 29",
"output": "1"
},
{
"input": "73 32 99",
"output": "1"
},
{
"input": "1 1 1",
"output": "0"
},
{
"input": "41 12 65",
"output": "1"
},
{
"input": "13 61 100",
"output": "1"
},
{
"input": "100 100 10",
"output": "0"
},
{
"input": "12 24 26",
"output": "1"
},
{
"input": "73 21 96",
"output": "1"
},
{
"input": "17 22 81",
"output": "1"
},
{
"input": "14 88 97",
"output": "1"
},
{
"input": "42 81 17",
"output": "0"
},
{
"input": "34 36 19",
"output": "1"
},
{
"input": "21 87 12",
"output": "1"
},
{
"input": "97 92 98",
"output": "1"
},
{
"input": "49 25 62",
"output": "1"
},
{
"input": "29 17 44",
"output": "1"
},
{
"input": "10 1 10",
"output": "0"
},
{
"input": "99 98 100",
"output": "1"
},
{
"input": "100 100 1",
"output": "0"
},
{
"input": "11 99 55",
"output": "0"
},
{
"input": "8 13 21",
"output": "0"
},
{
"input": "34 55 89",
"output": "0"
},
{
"input": "21 34 55",
"output": "1"
},
{
"input": "13 21 34",
"output": "1"
},
{
"input": "34 21 55",
"output": "0"
},
{
"input": "10 20 84",
"output": "1"
},
{
"input": "4 2 5",
"output": "0"
},
{
"input": "3 2 3",
"output": "0"
},
{
"input": "31 23 1",
"output": "0"
},
{
"input": "100 100 100",
"output": "0"
},
{
"input": "100 1 100",
"output": "0"
},
{
"input": "1 100 100",
"output": "1"
},
{
"input": "1 1 100",
"output": "1"
},
{
"input": "17 23 100",
"output": "1"
},
{
"input": "99 98 100",
"output": "1"
},
{
"input": "1 1 2",
"output": "1"
},
{
"input": "2 1 1",
"output": "0"
},
{
"input": "1 2 1",
"output": "0"
},
{
"input": "2 1 2",
"output": "0"
},
{
"input": "10 10 3",
"output": "1"
},
{
"input": "2 2 1",
"output": "0"
},
{
"input": "9 5 1",
"output": "0"
}
] | 1,632,822,572 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 30 | 0 | n = int(input())
a = 0
a_1 = 1
b = 0
fipo = [0]
while a <= n :
b = a_1 +a
a = a_1
a_1 = b
fipo . append(a)
l = fipo.index(n)
if n == 3 :
print('1','1','1')
elif n == 2 :
print('0','1','1')
elif n == 1 :
print('0','0','1')
elif n == 0 :
print('0','0','0')
elif n in fipo:
print(fipo[l-4],fipo[l-3],fipo[l-1])
else:
print("I'm too stupid to solve this problem" ) | Title: Epic Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Simon and Antisimon play a game. Initially each player receives one fixed positive integer that doesn't change throughout the game. Simon receives number *a* and Antisimon receives number *b*. They also have a heap of *n* stones. The players take turns to make a move and Simon starts. During a move a player should take from the heap the number of stones equal to the greatest common divisor of the fixed number he has received and the number of stones left in the heap. A player loses when he cannot take the required number of stones (i. e. the heap has strictly less stones left than one needs to take).
Your task is to determine by the given *a*, *b* and *n* who wins the game.
Input Specification:
The only string contains space-separated integers *a*, *b* and *n* (1<=≤<=*a*,<=*b*,<=*n*<=≤<=100) — the fixed numbers Simon and Antisimon have received correspondingly and the initial number of stones in the pile.
Output Specification:
If Simon wins, print "0" (without the quotes), otherwise print "1" (without the quotes).
Demo Input:
['3 5 9\n', '1 1 100\n']
Demo Output:
['0', '1']
Note:
The greatest common divisor of two non-negative integers *a* and *b* is such maximum positive integer *k*, that *a* is divisible by *k* without remainder and similarly, *b* is divisible by *k* without remainder. Let *gcd*(*a*, *b*) represent the operation of calculating the greatest common divisor of numbers *a* and *b*. Specifically, *gcd*(*x*, 0) = *gcd*(0, *x*) = *x*.
In the first sample the game will go like that:
- Simon should take *gcd*(3, 9) = 3 stones from the heap. After his move the heap has 6 stones left.- Antisimon should take *gcd*(5, 6) = 1 stone from the heap. After his move the heap has 5 stones left.- Simon should take *gcd*(3, 5) = 1 stone from the heap. After his move the heap has 4 stones left.- Antisimon should take *gcd*(5, 4) = 1 stone from the heap. After his move the heap has 3 stones left.- Simon should take *gcd*(3, 3) = 3 stones from the heap. After his move the heap has 0 stones left.- Antisimon should take *gcd*(5, 0) = 5 stones from the heap. As 0 < 5, it is impossible and Antisimon loses.
In the second sample each player during each move takes one stone from the heap. As *n* is even, Antisimon takes the last stone and Simon can't make a move after that. | ```python
n = int(input())
a = 0
a_1 = 1
b = 0
fipo = [0]
while a <= n :
b = a_1 +a
a = a_1
a_1 = b
fipo . append(a)
l = fipo.index(n)
if n == 3 :
print('1','1','1')
elif n == 2 :
print('0','1','1')
elif n == 1 :
print('0','0','1')
elif n == 0 :
print('0','0','0')
elif n in fipo:
print(fipo[l-4],fipo[l-3],fipo[l-1])
else:
print("I'm too stupid to solve this problem" )
``` | -1 |
|
770 | A | New Password | PROGRAMMING | 800 | [
"*special",
"implementation"
] | null | null | Innokentiy decides to change the password in the social net "Contact!", but he is too lazy to invent a new password by himself. That is why he needs your help.
Innokentiy decides that new password should satisfy the following conditions:
- the length of the password must be equal to *n*, - the password should consist only of lowercase Latin letters, - the number of distinct symbols in the password must be equal to *k*, - any two consecutive symbols in the password must be distinct.
Your task is to help Innokentiy and to invent a new password which will satisfy all given conditions. | The first line contains two positive integers *n* and *k* (2<=≤<=*n*<=≤<=100, 2<=≤<=*k*<=≤<=*min*(*n*,<=26)) — the length of the password and the number of distinct symbols in it.
Pay attention that a desired new password always exists. | Print any password which satisfies all conditions given by Innokentiy. | [
"4 3\n",
"6 6\n",
"5 2\n"
] | [
"java\n",
"python\n",
"phphp\n"
] | In the first test there is one of the appropriate new passwords — java, because its length is equal to 4 and 3 distinct lowercase letters a, j and v are used in it.
In the second test there is one of the appropriate new passwords — python, because its length is equal to 6 and it consists of 6 distinct lowercase letters.
In the third test there is one of the appropriate new passwords — phphp, because its length is equal to 5 and 2 distinct lowercase letters p and h are used in it.
Pay attention the condition that no two identical symbols are consecutive is correct for all appropriate passwords in tests. | 500 | [
{
"input": "4 3",
"output": "abca"
},
{
"input": "6 6",
"output": "abcdef"
},
{
"input": "5 2",
"output": "ababa"
},
{
"input": "3 2",
"output": "aba"
},
{
"input": "10 2",
"output": "ababababab"
},
{
"input": "26 13",
"output": "abcdefghijklmabcdefghijklm"
},
{
"input": "100 2",
"output": "abababababababababababababababababababababababababababababababababababababababababababababababababab"
},
{
"input": "100 10",
"output": "abcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghij"
},
{
"input": "3 3",
"output": "abc"
},
{
"input": "6 3",
"output": "abcabc"
},
{
"input": "10 3",
"output": "abcabcabca"
},
{
"input": "50 3",
"output": "abcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcab"
},
{
"input": "90 2",
"output": "ababababababababababababababababababababababababababababababababababababababababababababab"
},
{
"input": "6 2",
"output": "ababab"
},
{
"input": "99 3",
"output": "abcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabc"
},
{
"input": "4 2",
"output": "abab"
},
{
"input": "100 3",
"output": "abcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabca"
},
{
"input": "40 22",
"output": "abcdefghijklmnopqrstuvabcdefghijklmnopqr"
},
{
"input": "13 8",
"output": "abcdefghabcde"
},
{
"input": "16 15",
"output": "abcdefghijklmnoa"
},
{
"input": "17 17",
"output": "abcdefghijklmnopq"
},
{
"input": "19 4",
"output": "abcdabcdabcdabcdabc"
},
{
"input": "100 26",
"output": "abcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuv"
},
{
"input": "100 25",
"output": "abcdefghijklmnopqrstuvwxyabcdefghijklmnopqrstuvwxyabcdefghijklmnopqrstuvwxyabcdefghijklmnopqrstuvwxy"
},
{
"input": "26 26",
"output": "abcdefghijklmnopqrstuvwxyz"
},
{
"input": "27 26",
"output": "abcdefghijklmnopqrstuvwxyza"
},
{
"input": "2 2",
"output": "ab"
},
{
"input": "26 25",
"output": "abcdefghijklmnopqrstuvwxya"
},
{
"input": "99 2",
"output": "abababababababababababababababababababababababababababababababababababababababababababababababababa"
},
{
"input": "99 26",
"output": "abcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstu"
},
{
"input": "4 4",
"output": "abcd"
},
{
"input": "5 3",
"output": "abcab"
},
{
"input": "5 4",
"output": "abcda"
},
{
"input": "5 5",
"output": "abcde"
},
{
"input": "24 22",
"output": "abcdefghijklmnopqrstuvab"
},
{
"input": "26 14",
"output": "abcdefghijklmnabcdefghijkl"
},
{
"input": "26 15",
"output": "abcdefghijklmnoabcdefghijk"
},
{
"input": "30 12",
"output": "abcdefghijklabcdefghijklabcdef"
},
{
"input": "35 4",
"output": "abcdabcdabcdabcdabcdabcdabcdabcdabc"
},
{
"input": "79 3",
"output": "abcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabca"
},
{
"input": "79 14",
"output": "abcdefghijklmnabcdefghijklmnabcdefghijklmnabcdefghijklmnabcdefghijklmnabcdefghi"
},
{
"input": "85 13",
"output": "abcdefghijklmabcdefghijklmabcdefghijklmabcdefghijklmabcdefghijklmabcdefghijklmabcdefg"
},
{
"input": "90 25",
"output": "abcdefghijklmnopqrstuvwxyabcdefghijklmnopqrstuvwxyabcdefghijklmnopqrstuvwxyabcdefghijklmno"
},
{
"input": "90 19",
"output": "abcdefghijklmnopqrsabcdefghijklmnopqrsabcdefghijklmnopqrsabcdefghijklmnopqrsabcdefghijklmn"
},
{
"input": "26 24",
"output": "abcdefghijklmnopqrstuvwxab"
},
{
"input": "100 17",
"output": "abcdefghijklmnopqabcdefghijklmnopqabcdefghijklmnopqabcdefghijklmnopqabcdefghijklmnopqabcdefghijklmno"
},
{
"input": "26 2",
"output": "ababababababababababababab"
}
] | 1,652,099,605 | 2,147,483,647 | Python 3 | OK | TESTS | 47 | 62 | 409,600 | from string import ascii_lowercase as alph
n, k = map(int, input().split())
print(''.join(alph[i % k] for i in range(n))) | Title: New Password
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Innokentiy decides to change the password in the social net "Contact!", but he is too lazy to invent a new password by himself. That is why he needs your help.
Innokentiy decides that new password should satisfy the following conditions:
- the length of the password must be equal to *n*, - the password should consist only of lowercase Latin letters, - the number of distinct symbols in the password must be equal to *k*, - any two consecutive symbols in the password must be distinct.
Your task is to help Innokentiy and to invent a new password which will satisfy all given conditions.
Input Specification:
The first line contains two positive integers *n* and *k* (2<=≤<=*n*<=≤<=100, 2<=≤<=*k*<=≤<=*min*(*n*,<=26)) — the length of the password and the number of distinct symbols in it.
Pay attention that a desired new password always exists.
Output Specification:
Print any password which satisfies all conditions given by Innokentiy.
Demo Input:
['4 3\n', '6 6\n', '5 2\n']
Demo Output:
['java\n', 'python\n', 'phphp\n']
Note:
In the first test there is one of the appropriate new passwords — java, because its length is equal to 4 and 3 distinct lowercase letters a, j and v are used in it.
In the second test there is one of the appropriate new passwords — python, because its length is equal to 6 and it consists of 6 distinct lowercase letters.
In the third test there is one of the appropriate new passwords — phphp, because its length is equal to 5 and 2 distinct lowercase letters p and h are used in it.
Pay attention the condition that no two identical symbols are consecutive is correct for all appropriate passwords in tests. | ```python
from string import ascii_lowercase as alph
n, k = map(int, input().split())
print(''.join(alph[i % k] for i in range(n)))
``` | 3 |
|
217 | A | Ice Skating | PROGRAMMING | 1,200 | [
"brute force",
"dfs and similar",
"dsu",
"graphs"
] | null | null | Bajtek is learning to skate on ice. He's a beginner, so his only mode of transportation is pushing off from a snow drift to the north, east, south or west and sliding until he lands in another snow drift. He has noticed that in this way it's impossible to get from some snow drifts to some other by any sequence of moves. He now wants to heap up some additional snow drifts, so that he can get from any snow drift to any other one. He asked you to find the minimal number of snow drifts that need to be created.
We assume that Bajtek can only heap up snow drifts at integer coordinates. | The first line of input contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of snow drifts. Each of the following *n* lines contains two integers *x**i* and *y**i* (1<=≤<=*x**i*,<=*y**i*<=≤<=1000) — the coordinates of the *i*-th snow drift.
Note that the north direction coinсides with the direction of *Oy* axis, so the east direction coinсides with the direction of the *Ox* axis. All snow drift's locations are distinct. | Output the minimal number of snow drifts that need to be created in order for Bajtek to be able to reach any snow drift from any other one. | [
"2\n2 1\n1 2\n",
"2\n2 1\n4 1\n"
] | [
"1\n",
"0\n"
] | none | 500 | [
{
"input": "2\n2 1\n1 2",
"output": "1"
},
{
"input": "2\n2 1\n4 1",
"output": "0"
},
{
"input": "24\n171 35\n261 20\n4 206\n501 446\n961 912\n581 748\n946 978\n463 514\n841 889\n341 466\n842 967\n54 102\n235 261\n925 889\n682 672\n623 636\n268 94\n635 710\n474 510\n697 794\n586 663\n182 184\n806 663\n468 459",
"output": "21"
},
{
"input": "17\n660 646\n440 442\n689 618\n441 415\n922 865\n950 972\n312 366\n203 229\n873 860\n219 199\n344 308\n169 176\n961 992\n153 84\n201 230\n987 938\n834 815",
"output": "16"
},
{
"input": "11\n798 845\n722 911\n374 270\n629 537\n748 856\n831 885\n486 641\n751 829\n609 492\n98 27\n654 663",
"output": "10"
},
{
"input": "1\n321 88",
"output": "0"
},
{
"input": "9\n811 859\n656 676\n76 141\n945 951\n497 455\n18 55\n335 294\n267 275\n656 689",
"output": "7"
},
{
"input": "7\n948 946\n130 130\n761 758\n941 938\n971 971\n387 385\n509 510",
"output": "6"
},
{
"input": "6\n535 699\n217 337\n508 780\n180 292\n393 112\n732 888",
"output": "5"
},
{
"input": "14\n25 23\n499 406\n193 266\n823 751\n219 227\n101 138\n978 992\n43 74\n997 932\n237 189\n634 538\n774 740\n842 767\n742 802",
"output": "13"
},
{
"input": "12\n548 506\n151 198\n370 380\n655 694\n654 690\n407 370\n518 497\n819 827\n765 751\n802 771\n741 752\n653 662",
"output": "11"
},
{
"input": "40\n685 711\n433 403\n703 710\n491 485\n616 619\n288 282\n884 871\n367 352\n500 511\n977 982\n51 31\n576 564\n508 519\n755 762\n22 20\n368 353\n232 225\n953 955\n452 436\n311 330\n967 988\n369 364\n791 803\n150 149\n651 661\n118 93\n398 387\n748 766\n852 852\n230 228\n555 545\n515 519\n667 678\n867 862\n134 146\n859 863\n96 99\n486 469\n303 296\n780 786",
"output": "38"
},
{
"input": "3\n175 201\n907 909\n388 360",
"output": "2"
},
{
"input": "7\n312 298\n86 78\n73 97\n619 594\n403 451\n538 528\n71 86",
"output": "6"
},
{
"input": "19\n802 820\n368 248\n758 794\n455 378\n876 888\n771 814\n245 177\n586 555\n844 842\n364 360\n820 856\n731 624\n982 975\n825 856\n122 121\n862 896\n42 4\n792 841\n828 820",
"output": "16"
},
{
"input": "32\n643 877\n842 614\n387 176\n99 338\n894 798\n652 728\n611 648\n622 694\n579 781\n243 46\n322 305\n198 438\n708 579\n246 325\n536 459\n874 593\n120 277\n989 907\n223 110\n35 130\n761 692\n690 661\n518 766\n226 93\n678 597\n725 617\n661 574\n775 496\n56 416\n14 189\n358 359\n898 901",
"output": "31"
},
{
"input": "32\n325 327\n20 22\n72 74\n935 933\n664 663\n726 729\n785 784\n170 171\n315 314\n577 580\n984 987\n313 317\n434 435\n962 961\n55 54\n46 44\n743 742\n434 433\n617 612\n332 332\n883 886\n940 936\n793 792\n645 644\n611 607\n418 418\n465 465\n219 218\n167 164\n56 54\n403 405\n210 210",
"output": "29"
},
{
"input": "32\n652 712\n260 241\n27 154\n188 16\n521 351\n518 356\n452 540\n790 827\n339 396\n336 551\n897 930\n828 627\n27 168\n180 113\n134 67\n794 671\n812 711\n100 241\n686 813\n138 289\n384 506\n884 932\n913 959\n470 508\n730 734\n373 478\n788 862\n392 426\n148 68\n113 49\n713 852\n924 894",
"output": "29"
},
{
"input": "14\n685 808\n542 677\n712 747\n832 852\n187 410\n399 338\n626 556\n530 635\n267 145\n215 209\n559 684\n944 949\n753 596\n601 823",
"output": "13"
},
{
"input": "5\n175 158\n16 2\n397 381\n668 686\n957 945",
"output": "4"
},
{
"input": "5\n312 284\n490 509\n730 747\n504 497\n782 793",
"output": "4"
},
{
"input": "2\n802 903\n476 348",
"output": "1"
},
{
"input": "4\n325 343\n425 442\n785 798\n275 270",
"output": "3"
},
{
"input": "28\n462 483\n411 401\n118 94\n111 127\n5 6\n70 52\n893 910\n73 63\n818 818\n182 201\n642 633\n900 886\n893 886\n684 700\n157 173\n953 953\n671 660\n224 225\n832 801\n152 157\n601 585\n115 101\n739 722\n611 606\n659 642\n461 469\n702 689\n649 653",
"output": "25"
},
{
"input": "36\n952 981\n885 900\n803 790\n107 129\n670 654\n143 132\n66 58\n813 819\n849 837\n165 198\n247 228\n15 39\n619 618\n105 138\n868 855\n965 957\n293 298\n613 599\n227 212\n745 754\n723 704\n877 858\n503 487\n678 697\n592 595\n155 135\n962 982\n93 89\n660 673\n225 212\n967 987\n690 680\n804 813\n489 518\n240 221\n111 124",
"output": "34"
},
{
"input": "30\n89 3\n167 156\n784 849\n943 937\n144 95\n24 159\n80 120\n657 683\n585 596\n43 147\n909 964\n131 84\n345 389\n333 321\n91 126\n274 325\n859 723\n866 922\n622 595\n690 752\n902 944\n127 170\n426 383\n905 925\n172 284\n793 810\n414 510\n890 884\n123 24\n267 255",
"output": "29"
},
{
"input": "5\n664 666\n951 941\n739 742\n844 842\n2 2",
"output": "4"
},
{
"input": "3\n939 867\n411 427\n757 708",
"output": "2"
},
{
"input": "36\n429 424\n885 972\n442 386\n512 511\n751 759\n4 115\n461 497\n496 408\n8 23\n542 562\n296 331\n448 492\n412 395\n109 166\n622 640\n379 355\n251 262\n564 586\n66 115\n275 291\n666 611\n629 534\n510 567\n635 666\n738 803\n420 369\n92 17\n101 144\n141 92\n258 258\n184 235\n492 456\n311 210\n394 357\n531 512\n634 636",
"output": "34"
},
{
"input": "29\n462 519\n871 825\n127 335\n156 93\n576 612\n885 830\n634 779\n340 105\n744 795\n716 474\n93 139\n563 805\n137 276\n177 101\n333 14\n391 437\n873 588\n817 518\n460 597\n572 670\n140 303\n392 441\n273 120\n862 578\n670 639\n410 161\n544 577\n193 116\n252 195",
"output": "28"
},
{
"input": "23\n952 907\n345 356\n812 807\n344 328\n242 268\n254 280\n1000 990\n80 78\n424 396\n595 608\n755 813\n383 380\n55 56\n598 633\n203 211\n508 476\n600 593\n206 192\n855 882\n517 462\n967 994\n642 657\n493 488",
"output": "22"
},
{
"input": "10\n579 816\n806 590\n830 787\n120 278\n677 800\n16 67\n188 251\n559 560\n87 67\n104 235",
"output": "8"
},
{
"input": "23\n420 424\n280 303\n515 511\n956 948\n799 803\n441 455\n362 369\n299 289\n823 813\n982 967\n876 878\n185 157\n529 551\n964 989\n655 656\n1 21\n114 112\n45 56\n935 937\n1000 997\n934 942\n360 366\n648 621",
"output": "22"
},
{
"input": "23\n102 84\n562 608\n200 127\n952 999\n465 496\n322 367\n728 690\n143 147\n855 867\n861 866\n26 59\n300 273\n255 351\n192 246\n70 111\n365 277\n32 104\n298 319\n330 354\n241 141\n56 125\n315 298\n412 461",
"output": "22"
},
{
"input": "7\n429 506\n346 307\n99 171\n853 916\n322 263\n115 157\n906 924",
"output": "6"
},
{
"input": "3\n1 1\n2 1\n2 2",
"output": "0"
},
{
"input": "4\n1 1\n1 2\n2 1\n2 2",
"output": "0"
},
{
"input": "5\n1 1\n1 2\n2 2\n3 1\n3 3",
"output": "0"
},
{
"input": "6\n1 1\n1 2\n2 2\n3 1\n3 2\n3 3",
"output": "0"
},
{
"input": "20\n1 1\n2 2\n3 3\n3 9\n4 4\n5 2\n5 5\n5 7\n5 8\n6 2\n6 6\n6 9\n7 7\n8 8\n9 4\n9 7\n9 9\n10 2\n10 9\n10 10",
"output": "1"
},
{
"input": "21\n1 1\n1 9\n2 1\n2 2\n2 5\n2 6\n2 9\n3 3\n3 8\n4 1\n4 4\n5 5\n5 8\n6 6\n7 7\n8 8\n9 9\n10 4\n10 10\n11 5\n11 11",
"output": "1"
},
{
"input": "22\n1 1\n1 3\n1 4\n1 8\n1 9\n1 11\n2 2\n3 3\n4 4\n4 5\n5 5\n6 6\n6 8\n7 7\n8 3\n8 4\n8 8\n9 9\n10 10\n11 4\n11 9\n11 11",
"output": "3"
},
{
"input": "50\n1 1\n2 2\n2 9\n3 3\n4 4\n4 9\n4 16\n4 24\n5 5\n6 6\n7 7\n8 8\n8 9\n8 20\n9 9\n10 10\n11 11\n12 12\n13 13\n14 7\n14 14\n14 16\n14 25\n15 4\n15 6\n15 15\n15 22\n16 6\n16 16\n17 17\n18 18\n19 6\n19 19\n20 20\n21 21\n22 6\n22 22\n23 23\n24 6\n24 7\n24 8\n24 9\n24 24\n25 1\n25 3\n25 5\n25 7\n25 23\n25 24\n25 25",
"output": "7"
},
{
"input": "55\n1 1\n1 14\n2 2\n2 19\n3 1\n3 3\n3 8\n3 14\n3 23\n4 1\n4 4\n5 5\n5 8\n5 15\n6 2\n6 3\n6 4\n6 6\n7 7\n8 8\n8 21\n9 9\n10 1\n10 10\n11 9\n11 11\n12 12\n13 13\n14 14\n15 15\n15 24\n16 5\n16 16\n17 5\n17 10\n17 17\n17 18\n17 22\n17 27\n18 18\n19 19\n20 20\n21 20\n21 21\n22 22\n23 23\n24 14\n24 24\n25 25\n26 8\n26 11\n26 26\n27 3\n27 27\n28 28",
"output": "5"
},
{
"input": "3\n1 2\n2 1\n2 2",
"output": "0"
},
{
"input": "6\n4 4\n3 4\n5 4\n4 5\n4 3\n3 1",
"output": "0"
},
{
"input": "4\n1 1\n1 2\n2 1\n2 2",
"output": "0"
},
{
"input": "3\n1 1\n2 2\n1 2",
"output": "0"
},
{
"input": "8\n1 3\n1 1\n4 1\n2 2\n2 5\n5 9\n5 1\n5 4",
"output": "1"
},
{
"input": "10\n1 1\n1 2\n1 3\n1 4\n5 5\n6 6\n7 7\n8 8\n9 9\n100 100",
"output": "6"
},
{
"input": "7\n1 1\n2 2\n3 3\n4 4\n1 2\n2 3\n3 4",
"output": "0"
},
{
"input": "6\n1 1\n2 1\n2 2\n2 4\n4 3\n2 3",
"output": "0"
},
{
"input": "4\n3 1\n2 1\n2 2\n1 2",
"output": "0"
},
{
"input": "6\n1 1\n2 2\n2 1\n2 4\n4 3\n2 3",
"output": "0"
},
{
"input": "3\n1 2\n1 3\n1 4",
"output": "0"
},
{
"input": "4\n1 1\n2 2\n1 2\n2 1",
"output": "0"
},
{
"input": "4\n1 3\n2 1\n3 2\n3 1",
"output": "1"
},
{
"input": "7\n1 1\n1 2\n2 2\n3 3\n3 4\n4 4\n1 4",
"output": "0"
},
{
"input": "21\n12 12\n13 12\n12 11\n13 13\n10 10\n11 10\n11 11\n501 500\n501 501\n503 502\n500 500\n503 503\n502 501\n502 502\n700 700\n702 702\n703 702\n701 701\n702 701\n703 703\n701 700",
"output": "2"
},
{
"input": "6\n1 11\n6 8\n11 10\n1 10\n11 11\n6 9",
"output": "1"
},
{
"input": "4\n1 1\n2 2\n3 2\n3 1",
"output": "0"
},
{
"input": "3\n1 2\n3 4\n3 2",
"output": "0"
},
{
"input": "3\n1 1\n1 2\n2 2",
"output": "0"
},
{
"input": "4\n5 5\n5 4\n6 3\n6 4",
"output": "0"
},
{
"input": "3\n1 1\n2 2\n2 1",
"output": "0"
}
] | 1,538,729,906 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 35 | 248 | 0 | def findSet(u):
if parents[u] != u:
parents[u] = findSet(parents[u])
return parents[u]
def unionSet(u, v):
up = findSet(u)
vp = findSet(v)
if u[0] == v[0] and u[1] == v[1]:
return
global count
count -= 1
if ranks[up] > ranks[vp]:
parents[vp] = up
elif ranks[up] < ranks[up]:
parents[up] = vp
else:
parents[up] = vp
ranks[vp] += 1
n = int(input())
nodes = []
count = n
for i in range(n):
x, y = map(int, input().split())
nodes.append((x, y))
parents = {(x, y) : (x, y) for x, y in nodes}
ranks = {(x, y): 0 for x, y in nodes}
for i in range(len(nodes) - 1):
for j in range(i + 1, len(nodes)):
if nodes[i][0] == nodes[j][0] or nodes[i][1] == nodes[j][1]:
unionSet(nodes[i], nodes[j])
print(count - 1) | Title: Ice Skating
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Bajtek is learning to skate on ice. He's a beginner, so his only mode of transportation is pushing off from a snow drift to the north, east, south or west and sliding until he lands in another snow drift. He has noticed that in this way it's impossible to get from some snow drifts to some other by any sequence of moves. He now wants to heap up some additional snow drifts, so that he can get from any snow drift to any other one. He asked you to find the minimal number of snow drifts that need to be created.
We assume that Bajtek can only heap up snow drifts at integer coordinates.
Input Specification:
The first line of input contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of snow drifts. Each of the following *n* lines contains two integers *x**i* and *y**i* (1<=≤<=*x**i*,<=*y**i*<=≤<=1000) — the coordinates of the *i*-th snow drift.
Note that the north direction coinсides with the direction of *Oy* axis, so the east direction coinсides with the direction of the *Ox* axis. All snow drift's locations are distinct.
Output Specification:
Output the minimal number of snow drifts that need to be created in order for Bajtek to be able to reach any snow drift from any other one.
Demo Input:
['2\n2 1\n1 2\n', '2\n2 1\n4 1\n']
Demo Output:
['1\n', '0\n']
Note:
none | ```python
def findSet(u):
if parents[u] != u:
parents[u] = findSet(parents[u])
return parents[u]
def unionSet(u, v):
up = findSet(u)
vp = findSet(v)
if u[0] == v[0] and u[1] == v[1]:
return
global count
count -= 1
if ranks[up] > ranks[vp]:
parents[vp] = up
elif ranks[up] < ranks[up]:
parents[up] = vp
else:
parents[up] = vp
ranks[vp] += 1
n = int(input())
nodes = []
count = n
for i in range(n):
x, y = map(int, input().split())
nodes.append((x, y))
parents = {(x, y) : (x, y) for x, y in nodes}
ranks = {(x, y): 0 for x, y in nodes}
for i in range(len(nodes) - 1):
for j in range(i + 1, len(nodes)):
if nodes[i][0] == nodes[j][0] or nodes[i][1] == nodes[j][1]:
unionSet(nodes[i], nodes[j])
print(count - 1)
``` | 0 |
|
767 | A | Snacktower | PROGRAMMING | 1,100 | [
"data structures",
"implementation"
] | null | null | According to an old legeng, a long time ago Ankh-Morpork residents did something wrong to miss Fortune, and she cursed them. She said that at some time *n* snacks of distinct sizes will fall on the city, and the residents should build a Snacktower of them by placing snacks one on another. Of course, big snacks should be at the bottom of the tower, while small snacks should be at the top.
Years passed, and once different snacks started to fall onto the city, and the residents began to build the Snacktower.
However, they faced some troubles. Each day exactly one snack fell onto the city, but their order was strange. So, at some days the residents weren't able to put the new stack on the top of the Snacktower: they had to wait until all the bigger snacks fell. Of course, in order to not to anger miss Fortune again, the residents placed each snack on the top of the tower immediately as they could do it.
Write a program that models the behavior of Ankh-Morpork residents. | The first line contains single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the total number of snacks.
The second line contains *n* integers, the *i*-th of them equals the size of the snack which fell on the *i*-th day. Sizes are distinct integers from 1 to *n*. | Print *n* lines. On the *i*-th of them print the sizes of the snacks which the residents placed on the top of the Snacktower on the *i*-th day in the order they will do that. If no snack is placed on some day, leave the corresponding line empty. | [
"3\n3 1 2\n",
"5\n4 5 1 2 3\n"
] | [
"3\n \n2 1",
"5 4\n \n \n3 2 1\n"
] | In the example a snack of size 3 fell on the first day, and the residents immediately placed it. On the second day a snack of size 1 fell, and the residents weren't able to place it because they were missing the snack of size 2. On the third day a snack of size 2 fell, and the residents immediately placed it. Right after that they placed the snack of size 1 which had fallen before. | 500 | [
{
"input": "3\n3 1 2",
"output": "3 \n\n2 1 "
},
{
"input": "5\n4 5 1 2 3",
"output": "5 4 \n\n\n3 2 1 "
},
{
"input": "1\n1",
"output": "1 "
},
{
"input": "2\n1 2",
"output": "2 1 "
},
{
"input": "10\n5 1 6 2 8 3 4 10 9 7",
"output": "10 \n9 8 \n7 6 5 4 3 2 1 "
},
{
"input": "30\n16 10 4 29 5 28 12 21 11 30 18 6 14 3 17 22 20 15 9 1 27 19 24 26 13 25 2 23 8 7",
"output": "30 29 28 \n\n\n\n\n\n\n\n\n\n\n27 \n\n\n26 \n\n25 24 \n\n23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 \n8 \n7 6 5 4 3 2 1 "
},
{
"input": "100\n98 52 63 2 18 96 31 58 84 40 41 45 66 100 46 71 26 48 81 20 73 91 68 76 13 93 17 29 64 95 79 21 55 75 19 85 54 51 89 78 15 87 43 59 36 1 90 35 65 56 62 28 86 5 82 49 3 99 33 9 92 32 74 69 27 22 77 16 44 94 34 6 57 70 23 12 61 25 8 11 67 47 83 88 10 14 30 7 97 60 42 37 24 38 53 50 4 80 72 39",
"output": "100 \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n99 98 \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n97 96 95 94 93 92 91 90 89 88 87 86 85 84 83 82 81 \n\n\n\n\n\n\n\n\n80 79 78 77 76 75 74 73 \n72 71 70 69 68 67 66 65 64 63 62 61 60 59 58 57 56 55 54 53 52 51 50 49 48 47 46 45 44 43 42 41 40 \n39 38 37 36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 "
},
{
"input": "2\n2 1",
"output": "2 \n1 "
}
] | 1,696,585,822 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 1 | 31 | 0 | n_of_snakes = int(input())
sizes = input().split(" ")
days = [None for int in sizes]
def generate_descending_numbers_string(start, end):
start, end = max(start, end), min(start, end)
# Generate a list of numbers in descending order
numbers = [str(x) for x in range(start, end - 1, -1)]
# Join the numbers into a single string with spaces
result = ' '.join(numbers)
return result
done = n_of_snakes
for idx, item in enumerate(sizes):
if (int(item) >= (int(n_of_snakes) - idx)):
days[idx] = generate_descending_numbers_string(int(done), abs(int(item) - int(int(n_of_snakes) - int(item))))
done = int(item) - 1
else:
days[idx] = None
for i, item in enumerate(days):
if (item == None):
print("")
else:
print(item)
| Title: Snacktower
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
According to an old legeng, a long time ago Ankh-Morpork residents did something wrong to miss Fortune, and she cursed them. She said that at some time *n* snacks of distinct sizes will fall on the city, and the residents should build a Snacktower of them by placing snacks one on another. Of course, big snacks should be at the bottom of the tower, while small snacks should be at the top.
Years passed, and once different snacks started to fall onto the city, and the residents began to build the Snacktower.
However, they faced some troubles. Each day exactly one snack fell onto the city, but their order was strange. So, at some days the residents weren't able to put the new stack on the top of the Snacktower: they had to wait until all the bigger snacks fell. Of course, in order to not to anger miss Fortune again, the residents placed each snack on the top of the tower immediately as they could do it.
Write a program that models the behavior of Ankh-Morpork residents.
Input Specification:
The first line contains single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the total number of snacks.
The second line contains *n* integers, the *i*-th of them equals the size of the snack which fell on the *i*-th day. Sizes are distinct integers from 1 to *n*.
Output Specification:
Print *n* lines. On the *i*-th of them print the sizes of the snacks which the residents placed on the top of the Snacktower on the *i*-th day in the order they will do that. If no snack is placed on some day, leave the corresponding line empty.
Demo Input:
['3\n3 1 2\n', '5\n4 5 1 2 3\n']
Demo Output:
['3\n\xa0\n2 1', '5 4\n\xa0\n\xa0\n3 2 1\n']
Note:
In the example a snack of size 3 fell on the first day, and the residents immediately placed it. On the second day a snack of size 1 fell, and the residents weren't able to place it because they were missing the snack of size 2. On the third day a snack of size 2 fell, and the residents immediately placed it. Right after that they placed the snack of size 1 which had fallen before. | ```python
n_of_snakes = int(input())
sizes = input().split(" ")
days = [None for int in sizes]
def generate_descending_numbers_string(start, end):
start, end = max(start, end), min(start, end)
# Generate a list of numbers in descending order
numbers = [str(x) for x in range(start, end - 1, -1)]
# Join the numbers into a single string with spaces
result = ' '.join(numbers)
return result
done = n_of_snakes
for idx, item in enumerate(sizes):
if (int(item) >= (int(n_of_snakes) - idx)):
days[idx] = generate_descending_numbers_string(int(done), abs(int(item) - int(int(n_of_snakes) - int(item))))
done = int(item) - 1
else:
days[idx] = None
for i, item in enumerate(days):
if (item == None):
print("")
else:
print(item)
``` | 0 |
|
0 | none | none | none | 0 | [
"none"
] | null | null | You are given an array *a* with *n* distinct integers. Construct an array *b* by permuting *a* such that for every non-empty subset of indices *S*<==<={*x*1,<=*x*2,<=...,<=*x**k*} (1<=≤<=*x**i*<=≤<=*n*, 0<=<<=*k*<=<<=*n*) the sums of elements on that positions in *a* and *b* are different, i. e. | The first line contains one integer *n* (1<=≤<=*n*<=≤<=22) — the size of the array.
The second line contains *n* space-separated distinct integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=109) — the elements of the array. | If there is no such array *b*, print -1.
Otherwise in the only line print *n* space-separated integers *b*1,<=*b*2,<=...,<=*b**n*. Note that *b* must be a permutation of *a*.
If there are multiple answers, print any of them. | [
"2\n1 2\n",
"4\n1000 100 10 1\n"
] | [
"2 1 \n",
"100 1 1000 10\n"
] | An array *x* is a permutation of *y*, if we can shuffle elements of *y* such that it will coincide with *x*.
Note that the empty subset and the subset containing all indices are not counted. | 0 | [
{
"input": "2\n1 2",
"output": "2 1 "
},
{
"input": "4\n1000 100 10 1",
"output": "100 1 1000 10"
},
{
"input": "5\n1 3 4 5 2",
"output": "5 2 3 4 1 "
},
{
"input": "1\n10000000",
"output": "10000000 "
},
{
"input": "4\n1 5 8 4",
"output": "8 4 5 1 "
},
{
"input": "3\n1 3 2",
"output": "3 2 1 "
},
{
"input": "4\n3 1 2 4",
"output": "2 4 1 3 "
},
{
"input": "12\n7 1 62 12 3 5 8 9 10 22 23 0",
"output": "5 0 23 10 1 3 7 8 9 12 22 62 "
},
{
"input": "17\n1 3 2 5 4 6 7 8 10 9 13 11 12 14 15 16 18",
"output": "18 2 1 4 3 5 6 7 9 8 12 10 11 13 14 15 16 "
},
{
"input": "22\n1 3 5 7 22 2 4 6 8 9 10 11 12 13 15 14 17 18 16 20 19 23",
"output": "23 2 4 6 20 1 3 5 7 8 9 10 11 12 14 13 16 17 15 19 18 22 "
},
{
"input": "22\n17 6 1 22 9 23 38 40 10 20 29 11 12 39 3 32 26 4 13 36 14 35",
"output": "14 4 40 20 6 22 36 39 9 17 26 10 11 38 1 29 23 3 12 35 13 32 "
},
{
"input": "22\n27 21 12 14 8 40 47 45 24 49 36 37 17 32 42 13 35 10 18 2 5 30",
"output": "24 18 10 13 5 37 45 42 21 47 35 36 14 30 40 12 32 8 17 49 2 27 "
},
{
"input": "22\n33 2 19 26 18 13 27 9 25 35 6 24 20 22 11 5 1 30 17 15 7 29",
"output": "30 1 18 25 17 11 26 7 24 33 5 22 19 20 9 2 35 29 15 13 6 27 "
},
{
"input": "22\n18 37 15 33 35 5 14 1 0 27 22 11 40 20 13 2 30 21 8 25 32 16",
"output": "16 35 14 32 33 2 13 0 40 25 21 8 37 18 11 1 27 20 5 22 30 15 "
},
{
"input": "22\n4 24 22 18 28 3 17 8 29 20 11 15 13 2 19 26 5 36 33 14 30 25",
"output": "3 22 20 17 26 2 15 5 28 19 8 14 11 36 18 25 4 33 30 13 29 24 "
},
{
"input": "22\n28 40 5 38 29 12 21 24 2 33 35 17 30 11 16 0 8 27 34 14 19 36",
"output": "27 38 2 36 28 11 19 21 0 30 34 16 29 8 14 40 5 24 33 12 17 35 "
},
{
"input": "22\n25 12 38 5 6 20 30 27 4 19 8 18 10 17 26 32 43 14 40 35 1 22",
"output": "22 10 35 4 5 19 27 26 1 18 6 17 8 14 25 30 40 12 38 32 43 20 "
},
{
"input": "22\n2 22 21 19 3 25 28 11 10 9 14 37 18 38 15 23 20 34 7 30 31 4",
"output": "38 21 20 18 2 23 25 10 9 7 11 34 15 37 14 22 19 31 4 28 30 3 "
},
{
"input": "22\n7 0 23 37 20 18 46 26 2 24 44 13 47 15 32 5 35 30 39 41 27 10",
"output": "5 47 20 35 18 15 44 24 0 23 41 10 46 13 30 2 32 27 37 39 26 7 "
},
{
"input": "22\n36 5 7 22 33 30 14 8 25 24 28 12 19 29 37 2 20 15 10 17 13 21",
"output": "33 2 5 21 30 29 13 7 24 22 25 10 17 28 36 37 19 14 8 15 12 20 "
},
{
"input": "22\n23 32 13 39 29 41 40 6 21 10 38 42 4 8 20 35 31 26 15 2 17 5",
"output": "21 31 10 38 26 40 39 5 20 8 35 41 2 6 17 32 29 23 13 42 15 4 "
},
{
"input": "22\n41 12 14 36 16 21 0 2 18 22 39 29 40 31 37 25 28 9 4 34 6 43",
"output": "40 9 12 34 14 18 43 0 16 21 37 28 39 29 36 22 25 6 2 31 4 41 "
},
{
"input": "22\n32 43 3 37 29 42 40 12 28 1 14 25 34 46 8 35 5 17 2 23 20 9",
"output": "29 42 2 35 28 40 37 9 25 46 12 23 32 43 5 34 3 14 1 20 17 8 "
},
{
"input": "22\n17 10 24 44 41 33 48 6 30 27 38 19 16 46 22 8 35 13 5 9 4 1",
"output": "16 9 22 41 38 30 46 5 27 24 35 17 13 44 19 6 33 10 4 8 1 48 "
},
{
"input": "22\n16 11 29 30 12 5 3 2 13 6 17 15 9 24 25 35 1 27 0 23 20 33",
"output": "15 9 27 29 11 3 2 1 12 5 16 13 6 23 24 33 0 25 35 20 17 30 "
},
{
"input": "22\n12 38 6 37 14 26 2 0 9 17 28 33 3 11 15 8 31 21 29 34 18 24",
"output": "11 37 3 34 12 24 0 38 8 15 26 31 2 9 14 6 29 18 28 33 17 21 "
},
{
"input": "22\n20 38 26 32 36 8 44 0 40 41 35 21 11 17 29 33 1 42 24 14 5 3",
"output": "17 36 24 29 35 5 42 44 38 40 33 20 8 14 26 32 0 41 21 11 3 1 "
},
{
"input": "22\n7 10 1 25 42 8 39 35 6 19 31 24 16 0 21 32 11 28 13 4 37 22",
"output": "6 8 0 24 39 7 37 32 4 16 28 22 13 42 19 31 10 25 11 1 35 21 "
},
{
"input": "22\n9 13 7 20 38 40 27 12 31 25 1 23 46 35 45 29 19 16 33 4 42 39",
"output": "7 12 4 19 35 39 25 9 29 23 46 20 45 33 42 27 16 13 31 1 40 38 "
},
{
"input": "22\n13 2 10 25 5 34 19 18 16 9 7 22 28 20 31 38 36 35 1 26 6 23",
"output": "10 1 9 23 2 31 18 16 13 7 6 20 26 19 28 36 35 34 38 25 5 22 "
},
{
"input": "22\n106855341 41953605 16663229 140358177 145011760 49391214 42672526 1000000000 173686818 18529133 155326121 177597841 65855243 125680752 111261017 47020618 35558283 100881772 149421816 84207033 181739589 185082482",
"output": "100881772 35558283 1000000000 125680752 140358177 47020618 41953605 185082482 155326121 16663229 149421816 173686818 49391214 111261017 106855341 42672526 18529133 84207033 145011760 65855243 177597841 181739589 "
},
{
"input": "22\n177663922 168256855 139197944 78700101 93490895 127229611 46317725 84284513 48674853 66142856 29224095 1000000000 138390832 117500569 98525700 100418194 44827621 151960474 43225995 16918107 53307514 48861499",
"output": "168256855 151960474 138390832 66142856 84284513 117500569 44827621 78700101 46317725 53307514 16918107 177663922 127229611 100418194 93490895 98525700 43225995 139197944 29224095 1000000000 48861499 48674853 "
},
{
"input": "22\n83255567 39959119 124812899 157774437 12694468 89732189 102545715 67019496 110206980 98186415 63181429 141617294 177406424 195504716 158928060 64956133 67949891 31436243 155002729 1000000000 128745406 52504492",
"output": "67949891 31436243 110206980 155002729 1000000000 83255567 98186415 64956133 102545715 89732189 52504492 128745406 158928060 177406424 157774437 63181429 67019496 12694468 141617294 195504716 124812899 39959119 "
},
{
"input": "22\n138499935 195582510 159774498 12295611 37071371 91641202 167958938 119995178 19438466 182405139 207729895 56797798 79876605 152841775 1000000000 149079380 158867321 154637978 72179187 75460169 145092927 103227705",
"output": "119995178 182405139 158867321 1000000000 19438466 79876605 159774498 103227705 12295611 167958938 195582510 37071371 75460169 149079380 207729895 145092927 154637978 152841775 56797798 72179187 138499935 91641202 "
},
{
"input": "22\n133295371 188010892 71730560 209842234 193069109 184556873 87395258 234247052 230809052 211444018 148989732 17810977 158722706 11753932 100093528 1000000000 43672080 61357581 171830832 13873487 34865589 114340079",
"output": "114340079 184556873 61357581 193069109 188010892 171830832 71730560 230809052 211444018 209842234 133295371 13873487 148989732 1000000000 87395258 234247052 34865589 43672080 158722706 11753932 17810977 100093528 "
},
{
"input": "22\n94506085 195061283 78884975 27418524 41348358 185397891 151515774 66605535 170723638 212843258 218566729 7450050 21809921 1000000000 146101141 132453297 228865386 240705035 57636433 114219677 158240908 228428432",
"output": "78884975 185397891 66605535 21809921 27418524 170723638 146101141 57636433 158240908 195061283 212843258 1000000000 7450050 240705035 132453297 114219677 228428432 228865386 41348358 94506085 151515774 218566729 "
},
{
"input": "22\n116213533 171312666 76695399 60099180 30779320 43431323 146620629 15321904 71245898 94843310 56549974 104020167 84091716 134384095 24383373 83975332 1000000000 101710173 188076412 199811222 153566780 115893674",
"output": "115893674 153566780 71245898 56549974 24383373 30779320 134384095 1000000000 60099180 84091716 43431323 101710173 83975332 116213533 15321904 76695399 199811222 94843310 171312666 188076412 146620629 104020167 "
},
{
"input": "22\n79749952 42551386 1000000000 60427603 50702468 16899307 85913428 116634789 151569595 100251788 152378664 96284924 60769416 136345503 59995727 88224321 29257228 64921932 77805288 126026727 103477637 115959196",
"output": "77805288 29257228 152378664 59995727 42551386 1000000000 79749952 115959196 136345503 96284924 151569595 88224321 60427603 126026727 50702468 85913428 16899307 60769416 64921932 116634789 100251788 103477637 "
},
{
"input": "22\n32119698 129510003 107370317 182795872 160438101 17245069 117836566 141016185 196664039 215252245 170450315 18866624 68629021 47385728 77249092 89835593 132769095 95649030 48749357 126701972 40219294 1000000000",
"output": "18866624 126701972 95649030 170450315 141016185 1000000000 107370317 132769095 182795872 196664039 160438101 17245069 48749357 40219294 68629021 77249092 129510003 89835593 47385728 117836566 32119698 215252245 "
},
{
"input": "22\n148671024 180468173 99388811 78666746 187172484 157360521 112604605 2988530 60271244 163263697 27469084 166381131 1000000000 125847469 137766458 198740424 88387613 15152912 200315776 149201551 45997250 36252057",
"output": "137766458 166381131 88387613 60271244 180468173 149201551 99388811 1000000000 45997250 157360521 15152912 163263697 200315776 112604605 125847469 187172484 78666746 2988530 198740424 148671024 36252057 27469084 "
}
] | 1,510,970,244 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 3 | 62 | 0 | n = int(input())
a = [int(i) for i in input().split()]
if n == 1:
print(a)
else:
a_with_index = [[a[i], i] for i in range(n)]
a_with_index.sort()
ans = [0 for i in range(n)]
for i in range(n):
ans[a_with_index[i][1]] = str(a_with_index[(i + 1) % n][0])
print(" ".join(ans)) | Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given an array *a* with *n* distinct integers. Construct an array *b* by permuting *a* such that for every non-empty subset of indices *S*<==<={*x*1,<=*x*2,<=...,<=*x**k*} (1<=≤<=*x**i*<=≤<=*n*, 0<=<<=*k*<=<<=*n*) the sums of elements on that positions in *a* and *b* are different, i. e.
Input Specification:
The first line contains one integer *n* (1<=≤<=*n*<=≤<=22) — the size of the array.
The second line contains *n* space-separated distinct integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=109) — the elements of the array.
Output Specification:
If there is no such array *b*, print -1.
Otherwise in the only line print *n* space-separated integers *b*1,<=*b*2,<=...,<=*b**n*. Note that *b* must be a permutation of *a*.
If there are multiple answers, print any of them.
Demo Input:
['2\n1 2\n', '4\n1000 100 10 1\n']
Demo Output:
['2 1 \n', '100 1 1000 10\n']
Note:
An array *x* is a permutation of *y*, if we can shuffle elements of *y* such that it will coincide with *x*.
Note that the empty subset and the subset containing all indices are not counted. | ```python
n = int(input())
a = [int(i) for i in input().split()]
if n == 1:
print(a)
else:
a_with_index = [[a[i], i] for i in range(n)]
a_with_index.sort()
ans = [0 for i in range(n)]
for i in range(n):
ans[a_with_index[i][1]] = str(a_with_index[(i + 1) % n][0])
print(" ".join(ans))
``` | -1 |
|
12 | B | Correct Solution? | PROGRAMMING | 1,100 | [
"implementation",
"sortings"
] | B. Correct Solution? | 2 | 256 | One cold winter evening Alice and her older brother Bob was sitting at home near the fireplace and giving each other interesting problems to solve. When it was Alice's turn, she told the number *n* to Bob and said:
—Shuffle the digits in this number in order to obtain the smallest possible number without leading zeroes.
—No problem! — said Bob and immediately gave her an answer.
Alice said a random number, so she doesn't know whether Bob's answer is correct. Help her to find this out, because impatient brother is waiting for the verdict. | The first line contains one integer *n* (0<=≤<=*n*<=≤<=109) without leading zeroes. The second lines contains one integer *m* (0<=≤<=*m*<=≤<=109) — Bob's answer, possibly with leading zeroes. | Print OK if Bob's answer is correct and WRONG_ANSWER otherwise. | [
"3310\n1033\n",
"4\n5\n"
] | [
"OK\n",
"WRONG_ANSWER\n"
] | none | 0 | [
{
"input": "3310\n1033",
"output": "OK"
},
{
"input": "4\n5",
"output": "WRONG_ANSWER"
},
{
"input": "40\n04",
"output": "WRONG_ANSWER"
},
{
"input": "12\n12",
"output": "OK"
},
{
"input": "432\n234",
"output": "OK"
},
{
"input": "17109\n01179",
"output": "WRONG_ANSWER"
},
{
"input": "888\n888",
"output": "OK"
},
{
"input": "912\n9123",
"output": "WRONG_ANSWER"
},
{
"input": "0\n00",
"output": "WRONG_ANSWER"
},
{
"input": "11110\n1111",
"output": "WRONG_ANSWER"
},
{
"input": "7391\n1397",
"output": "WRONG_ANSWER"
},
{
"input": "201\n102",
"output": "OK"
},
{
"input": "111111111\n111111111",
"output": "OK"
},
{
"input": "32352320\n22203335",
"output": "WRONG_ANSWER"
},
{
"input": "1000000000\n1",
"output": "WRONG_ANSWER"
},
{
"input": "994321\n123499",
"output": "OK"
},
{
"input": "10101\n10101",
"output": "WRONG_ANSWER"
},
{
"input": "666\n0666",
"output": "WRONG_ANSWER"
},
{
"input": "2\n02",
"output": "WRONG_ANSWER"
},
{
"input": "313\n133",
"output": "OK"
},
{
"input": "987235645\n234556789",
"output": "OK"
},
{
"input": "90812\n010289",
"output": "WRONG_ANSWER"
},
{
"input": "123\n321",
"output": "WRONG_ANSWER"
},
{
"input": "707\n770",
"output": "WRONG_ANSWER"
},
{
"input": "77774444\n47474747",
"output": "WRONG_ANSWER"
},
{
"input": "1270\n1027",
"output": "OK"
},
{
"input": "320\n23",
"output": "WRONG_ANSWER"
},
{
"input": "123456789\n123456789",
"output": "OK"
},
{
"input": "918273645\n546372819",
"output": "WRONG_ANSWER"
},
{
"input": "300000003\n30000003",
"output": "WRONG_ANSWER"
},
{
"input": "0\n0",
"output": "OK"
},
{
"input": "0\n7",
"output": "WRONG_ANSWER"
}
] | 1,642,896,240 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | #include<cstdio>
#include<algorithm>
using namespace std;
const int MAXN=30;
int a[MAXN],b[MAXN],tot1,tot2,pos=1;
char ch;
inline int read(int* a,int& tot)
{
ch=getchar();
while(ch>='0'&&ch<='9') a[++tot]=(ch-'0'),ch=getchar();
}
int main()
{
read(a,tot1),read(b,tot2);
if(tot1==1&&tot2==tot1)
{
if(a[1]==b[1]) puts("OK");
else puts("WRONG_ANSWER");
return 0;
}
if(tot2!=1) if(b[1]==0){puts("WRONG_ANSWER");return 0;}
sort(a+1,a+1+tot1);
while(a[pos]==0) pos++;
swap(a[1],a[pos]);
if(tot1!=tot2) {puts("WRONG_ANSWER");return 0;}
for(register int i=1;i<=tot1;i++)if(a[i]!=b[i]){puts("WRONG_ANSWER");return 0;}
puts("OK");
return 0;
} | Title: Correct Solution?
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
One cold winter evening Alice and her older brother Bob was sitting at home near the fireplace and giving each other interesting problems to solve. When it was Alice's turn, she told the number *n* to Bob and said:
—Shuffle the digits in this number in order to obtain the smallest possible number without leading zeroes.
—No problem! — said Bob and immediately gave her an answer.
Alice said a random number, so she doesn't know whether Bob's answer is correct. Help her to find this out, because impatient brother is waiting for the verdict.
Input Specification:
The first line contains one integer *n* (0<=≤<=*n*<=≤<=109) without leading zeroes. The second lines contains one integer *m* (0<=≤<=*m*<=≤<=109) — Bob's answer, possibly with leading zeroes.
Output Specification:
Print OK if Bob's answer is correct and WRONG_ANSWER otherwise.
Demo Input:
['3310\n1033\n', '4\n5\n']
Demo Output:
['OK\n', 'WRONG_ANSWER\n']
Note:
none | ```python
#include<cstdio>
#include<algorithm>
using namespace std;
const int MAXN=30;
int a[MAXN],b[MAXN],tot1,tot2,pos=1;
char ch;
inline int read(int* a,int& tot)
{
ch=getchar();
while(ch>='0'&&ch<='9') a[++tot]=(ch-'0'),ch=getchar();
}
int main()
{
read(a,tot1),read(b,tot2);
if(tot1==1&&tot2==tot1)
{
if(a[1]==b[1]) puts("OK");
else puts("WRONG_ANSWER");
return 0;
}
if(tot2!=1) if(b[1]==0){puts("WRONG_ANSWER");return 0;}
sort(a+1,a+1+tot1);
while(a[pos]==0) pos++;
swap(a[1],a[pos]);
if(tot1!=tot2) {puts("WRONG_ANSWER");return 0;}
for(register int i=1;i<=tot1;i++)if(a[i]!=b[i]){puts("WRONG_ANSWER");return 0;}
puts("OK");
return 0;
}
``` | -1 |
270 | A | Fancy Fence | PROGRAMMING | 1,100 | [
"geometry",
"implementation",
"math"
] | null | null | Emuskald needs a fence around his farm, but he is too lazy to build it himself. So he purchased a fence-building robot.
He wants the fence to be a regular polygon. The robot builds the fence along a single path, but it can only make fence corners at a single angle *a*.
Will the robot be able to build the fence Emuskald wants? In other words, is there a regular polygon which angles are equal to *a*? | The first line of input contains an integer *t* (0<=<<=*t*<=<<=180) — the number of tests. Each of the following *t* lines contains a single integer *a* (0<=<<=*a*<=<<=180) — the angle the robot can make corners at measured in degrees. | For each test, output on a single line "YES" (without quotes), if the robot can build a fence Emuskald wants, and "NO" (without quotes), if it is impossible. | [
"3\n30\n60\n90\n"
] | [
"NO\nYES\nYES\n"
] | In the first test case, it is impossible to build the fence, since there is no regular polygon with angle <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/df5f4b07dd5316fde165b43657b2696e2919e791.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second test case, the fence is a regular triangle, and in the last test case — a square. | 500 | [
{
"input": "3\n30\n60\n90",
"output": "NO\nYES\nYES"
},
{
"input": "6\n1\n2\n3\n170\n179\n25",
"output": "NO\nNO\nNO\nYES\nYES\nNO"
}
] | 1,665,559,685 | 2,147,483,647 | PyPy 3 | OK | TESTS | 3 | 124 | 0 | # -*- coding: utf-8 -*-
"""
Created on Wed Oct 12 15:20:42 2022
@author: lenovo
"""
t=int(input())
for i in range(t):
a=int(input())
print('YES' if 360%(180-a)==0 else'NO') | Title: Fancy Fence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Emuskald needs a fence around his farm, but he is too lazy to build it himself. So he purchased a fence-building robot.
He wants the fence to be a regular polygon. The robot builds the fence along a single path, but it can only make fence corners at a single angle *a*.
Will the robot be able to build the fence Emuskald wants? In other words, is there a regular polygon which angles are equal to *a*?
Input Specification:
The first line of input contains an integer *t* (0<=<<=*t*<=<<=180) — the number of tests. Each of the following *t* lines contains a single integer *a* (0<=<<=*a*<=<<=180) — the angle the robot can make corners at measured in degrees.
Output Specification:
For each test, output on a single line "YES" (without quotes), if the robot can build a fence Emuskald wants, and "NO" (without quotes), if it is impossible.
Demo Input:
['3\n30\n60\n90\n']
Demo Output:
['NO\nYES\nYES\n']
Note:
In the first test case, it is impossible to build the fence, since there is no regular polygon with angle <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/df5f4b07dd5316fde165b43657b2696e2919e791.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second test case, the fence is a regular triangle, and in the last test case — a square. | ```python
# -*- coding: utf-8 -*-
"""
Created on Wed Oct 12 15:20:42 2022
@author: lenovo
"""
t=int(input())
for i in range(t):
a=int(input())
print('YES' if 360%(180-a)==0 else'NO')
``` | 3 |
|
26 | A | Almost Prime | PROGRAMMING | 900 | [
"number theory"
] | A. Almost Prime | 2 | 256 | A number is called almost prime if it has exactly two distinct prime divisors. For example, numbers 6, 18, 24 are almost prime, while 4, 8, 9, 42 are not. Find the amount of almost prime numbers which are between 1 and *n*, inclusive. | Input contains one integer number *n* (1<=≤<=*n*<=≤<=3000). | Output the amount of almost prime numbers between 1 and *n*, inclusive. | [
"10\n",
"21\n"
] | [
"2\n",
"8\n"
] | none | 500 | [
{
"input": "10",
"output": "2"
},
{
"input": "21",
"output": "8"
},
{
"input": "1",
"output": "0"
},
{
"input": "2",
"output": "0"
},
{
"input": "4",
"output": "0"
},
{
"input": "3",
"output": "0"
},
{
"input": "8",
"output": "1"
},
{
"input": "19",
"output": "6"
},
{
"input": "40",
"output": "19"
},
{
"input": "77",
"output": "41"
},
{
"input": "222",
"output": "125"
},
{
"input": "987",
"output": "501"
},
{
"input": "1000",
"output": "508"
},
{
"input": "2000",
"output": "958"
},
{
"input": "3000",
"output": "1375"
},
{
"input": "2999",
"output": "1375"
},
{
"input": "2998",
"output": "1375"
},
{
"input": "2997",
"output": "1374"
},
{
"input": "1429",
"output": "706"
},
{
"input": "1673",
"output": "808"
},
{
"input": "1500",
"output": "732"
},
{
"input": "500",
"output": "266"
},
{
"input": "856",
"output": "439"
}
] | 1,616,219,180 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 8 | 186 | 0 | def isp(n) :
if (n <= 1) :
return False
if (n <= 3) :
return True
if (n % 2 == 0 or n % 3 == 0) :
return False
i = 5
while(i * i <= n) :
if (n % i == 0 or n % (i + 2) == 0) :
return False
i = i + 6
return True
n=int(input())
s=int(n**(1/2))
lp=[]
ln=[]
for i in range(1,n+1):
if isp(i):
lp.append(i)
s=0
t=0
for y in range(6,n+1):
for j in range(int(y**(1/2))+1):
if y%lp[j]==0:
s+=1
if s==2:
t+=1
s=0
print(t) | Title: Almost Prime
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
A number is called almost prime if it has exactly two distinct prime divisors. For example, numbers 6, 18, 24 are almost prime, while 4, 8, 9, 42 are not. Find the amount of almost prime numbers which are between 1 and *n*, inclusive.
Input Specification:
Input contains one integer number *n* (1<=≤<=*n*<=≤<=3000).
Output Specification:
Output the amount of almost prime numbers between 1 and *n*, inclusive.
Demo Input:
['10\n', '21\n']
Demo Output:
['2\n', '8\n']
Note:
none | ```python
def isp(n) :
if (n <= 1) :
return False
if (n <= 3) :
return True
if (n % 2 == 0 or n % 3 == 0) :
return False
i = 5
while(i * i <= n) :
if (n % i == 0 or n % (i + 2) == 0) :
return False
i = i + 6
return True
n=int(input())
s=int(n**(1/2))
lp=[]
ln=[]
for i in range(1,n+1):
if isp(i):
lp.append(i)
s=0
t=0
for y in range(6,n+1):
for j in range(int(y**(1/2))+1):
if y%lp[j]==0:
s+=1
if s==2:
t+=1
s=0
print(t)
``` | 0 |
802 | G | Fake News (easy) | PROGRAMMING | 800 | [
"implementation",
"strings"
] | null | null | As it's the first of April, Heidi is suspecting that the news she reads today are fake, and she does not want to look silly in front of all the contestants. She knows that a newspiece is fake if it contains heidi as a subsequence. Help Heidi assess whether the given piece is true, but please be discreet about it... | The first and only line of input contains a single nonempty string *s* of length at most 1000 composed of lowercase letters (a-z). | Output YES if the string *s* contains heidi as a subsequence and NO otherwise. | [
"abcheaibcdi\n",
"hiedi\n"
] | [
"YES",
"NO"
] | A string *s* contains another string *p* as a subsequence if it is possible to delete some characters from *s* and obtain *p*. | 0 | [
{
"input": "abcheaibcdi",
"output": "YES"
},
{
"input": "hiedi",
"output": "NO"
},
{
"input": "ihied",
"output": "NO"
},
{
"input": "diehi",
"output": "NO"
},
{
"input": "deiih",
"output": "NO"
},
{
"input": "iheid",
"output": "NO"
},
{
"input": "eihdi",
"output": "NO"
},
{
"input": "ehdii",
"output": "NO"
},
{
"input": "edhii",
"output": "NO"
},
{
"input": "deiih",
"output": "NO"
},
{
"input": "ehdii",
"output": "NO"
},
{
"input": "eufyajkssayhjhqcwxmctecaeepjwmfoscqprpcxsqfwnlgzsmmuwuoruantipholrauvxydfvftwfzhnckxswussvlidcojiciflpvkcxkkcmmvtfvxrkwcpeelwsuzqgamamdtdgzscmikvojfvqehblmjczkvtdeymgertgkwfwfukafqlfdhtedcctixhyetdypswgagrpyto",
"output": "YES"
},
{
"input": "arfbvxgdvqzuloojjrwoyqqbxamxybaqltfimofulusfebodjkwwrgwcppkwiodtpjaraglyplgerrpqjkpoggjmfxhwtqrijpijrcyxnoodvwpyjfpvqaoazllbrpzananbrvvybboedidtuvqquklkpeflfaltukjhzjgiofombhbmqbihgtapswykfvlgdoapjqntvqsaohmbvnphvyyhvhavslamczuqifxnwknkaenqmlvetrqogqxmlptgrmqvxzdxdmwobjesmgxckpmawtioavwdngyiwkzypfnxcovwzdohshwlavwsthdssiadhiwmhpvgkrbezm",
"output": "YES"
},
{
"input": "zcectngbqnejjjtsfrluummmqabzqbyccshjqbrjthzhlbmzjfxugvjouwhumsgrnopiyakfadjnbsesamhynsbfbfunupwbxvohfmpwlcpxhovwpfpciclatgmiufwdvtsqrsdcymvkldpnhfeisrzhyhhlkwdzthgprvkpyldeysvbmcibqkpudyrraqdlxpjecvwcvuiklcrsbgvqasmxmtxqzmawcjtozioqlfflinnxpeexbzloaeqjvglbdeufultpjqexvjjjkzemtzuzmxvawilcqdrcjzpqyhtwfphuonzwkotthsaxrmwtnlmcdylxqcfffyndqeouztluqwlhnkkvzwcfiscikv",
"output": "YES"
},
{
"input": "plqaykgovxkvsiahdbglktdlhcqwelxxmtlyymrsyubxdskvyjkrowvcbpdofpjqspsrgpakdczletxujzlsegepzleipiyycpinzxgwjsgslnxsotouddgfcybozfpjhhocpybfjbaywsehbcfrayvancbrumdfngqytnhihyxnlvilrqyhnxeckprqafofelospffhtwguzjbbjlzbqrtiielbvzutzgpqxosiaqznndgobcluuqlhmffiowkjdlkokehtjdyjvmxsiyxureflmdomerfekxdvtitvwzmdsdzplkpbtafxqfpudnhfqpoiwvjnylanunmagoweobdvfjgepbsymfutrjarlxclhgavpytiiqwvojrptofuvlohzeguxdsrihsbucelhhuedltnnjgzxwyblbqvnoliiydfinzlogbvucwykryzcyibnniggbkdkdcdgcsbvvnavtyhtkanrblpvomvjs",
"output": "YES"
},
{
"input": "fbldqzggeunkpwcfirxanmntbfrudijltoertsdvcvcmbwodbibsrxendzebvxwydpasaqnisrijctsuatihxxygbeovhxjdptdcppkvfytdpjspvrannxavmkmisqtygntxkdlousdypyfkrpzapysfpdbyprufwzhunlsfugojddkmxzinatiwfxdqmgyrnjnxvrclhxyuwxtshoqdjptmeecvgmrlvuwqtmnfnfeeiwcavwnqmyustawbjodzwsqmnjxhpqmgpysierlwbbdzcwprpsexyvreewcmlbvaiytjlxdqdaqftefdlmtmmjcwvfejshymhnouoshdzqcwzxpzupkbcievodzqkqvyjuuxxwepxjalvkzufnveji",
"output": "YES"
},
{
"input": "htsyljgoelbbuipivuzrhmfpkgderqpoprlxdpasxhpmxvaztccldtmujjzjmcpdvsdghzpretlsyyiljhjznseaacruriufswuvizwwuvdioazophhyytvbiogttnnouauxllbdn",
"output": "YES"
},
{
"input": "ikmxzqdzxqlvgeojsnhqzciujslwjyzzexnregabdqztpplosdakimjxmuqccbnwvzbajoiqgdobccwnrwmixohrbdarhoeeelzbpigiybtesybwefpcfx",
"output": "YES"
},
{
"input": "bpvbpjvbdfiodsmahxpcubjxdykesubnypalhypantshkjffmxjmelblqnjdmtaltneuyudyevkgedkqrdmrfeemgpghwrifcwincfixokfgurhqbcfzeajrgkgpwqwsepudxulywowwxzdxkumsicsvnzfxspmjpaixgejeaoyoibegosqoyoydmphfpbutrrewyjecowjckvpcceoamtfbitdneuwqfvnagswlskmsmkhmxyfsrpqwhxzocyffiumcy",
"output": "YES"
},
{
"input": "vllsexwrazvlfvhvrtqeohvzzresjdiuhomfpgqcxpqdevplecuaepixhlijatxzegciizpvyvxuembiplwklahlqibykfideysjygagjbgqkbhdhkatddcwlxboinfuomnpc",
"output": "YES"
},
{
"input": "pnjdwpxmvfoqkjtbhquqcuredrkwqzzfjmdvpnbqtypzdovemhhclkvigjvtprrpzbrbcbatkucaqteuciuozytsptvsskkeplaxdaqmjkmef",
"output": "NO"
},
{
"input": "jpwfhvlxvsdhtuozvlmnfiotrgapgjxtcsgcjnodcztupysvvvmjpzqkpommadppdrykuqkcpzojcwvlogvkddedwbggkrhuvtsvdiokehlkdlnukcufjvqxnikcdawvexxwffxtriqbdmkahxdtygodzohwtdmmuvmatdkvweqvaehaxiefpevkvqpyxsrhtmgjsdfcwzqobibeduooldrmglbinrepmunizheqzvgqvpdskhxfidxfnbisyizhepwyrcykcmjxnkyfjgrqlkixcvysa",
"output": "YES"
},
{
"input": "aftcrvuumeqbfvaqlltscnuhkpcifrrhnutjinxdhhdbzvizlrapzjdatuaynoplgjketupgaejciosofuhcgcjdcucarfvtsofgubtphijciswsvidnvpztlaarydkeqxzwdhfbmullkimerukusbrdnnujviydldrwhdfllsjtziwfeaiqotbiprespmxjulnyunkdtcghrzvhtcychkwatqqmladxpvmvlkzscthylbzkpgwlzfjqwarqvdeyngekqvrhrftpxnkfcibbowvnqdkulcdydspcubwlgoyinpnzgidbgunparnueddzwtzdiavbprbbg",
"output": "YES"
},
{
"input": "oagjghsidigeh",
"output": "NO"
},
{
"input": "chdhzpfzabupskiusjoefrwmjmqkbmdgboicnszkhdrlegeqjsldurmbshijadlwsycselhlnudndpdhcnhruhhvsgbthpruiqfirxkhpqhzhqdfpyozolbionodypfcqfeqbkcgmqkizgeyyelzeoothexcoaahedgrvoemqcwccbvoeqawqeuusyjxmgjkpfwcdttfmwunzuwvsihliexlzygqcgpbdiawfvqukikhbjerjkyhpcknlndaystrgsinghlmekbvhntcpypmchcwoglsmwwdulqneuabuuuvtyrnjxfcgoothalwkzzfxakneusezgnnepkpipzromqubraiggqndliz",
"output": "YES"
},
{
"input": "lgirxqkrkgjcutpqitmffvbujcljkqardlalyigxorscczuzikoylcxenryhskoavymexysvmhbsvhtycjlmzhijpuvcjshyfeycvvcfyzytzoyvxajpqdjtfiatnvxnyeqtfcagfftafllhhjhplbdsrfpctkqpinpdfrtlzyjllfbeffputywcckupyslkbbzpgcnxgbmhtqeqqehpdaokkjtatrhyiuusjhwgiiiikxpzdueasemosmmccoakafgvxduwiuflovhhfhffgnnjhoperhhjtvocpqytjxkmrknnknqeglffhfuplopmktykxuvcmbwpoeisrlyyhdpxfvzseucofyhziuiikihpqheqdyzwigeaqzhxzvporgisxgvhyicqyejovqloibhbunsvsunpvmdckkbuokitdzleilfwutcvuuytpupizinfjrzhxudsmjcjyfcpfgthujjowdwtgbvi",
"output": "YES"
},
{
"input": "uuehrvufgerqbzyzksmqnewacotuimawhlbycdbsmhshrsbqwybbkwjwsrkwptvlbbwjiivqugzrxxwgidrcrhrwsmwgeoleptfamzefgaeyxouxocrpvomjrazmxrnffdwrrmblgdiabdncvfougtmjgvvazasnygdrigbsrieoonirlivfyodvulouslxosswgpdexuldmkdbpdlgutiotvxjyecbrsvbmqxrlcpcipjjncduyqtohlzybvlemmfdeubihwlwqglkgjvnwrbgydcpwklmjeewqklmqdbajqgrpnynaxfvxjzgibqerxyhnxenrmcdqaaeksbzyrcaepozqpetaurlhjuxxhwppuhgoihxdxbmxeiahyaqkbknktlzkheaarjoqqrsyeducvoygwalgarldcdlqogfvsncejssmx",
"output": "YES"
},
{
"input": "iiopulfjxoitgiusqrhgbkiyzinphjtclodbkkydetylvuimkhdkklmyoacmekdvjpuxcrvqnjhqhhbfenlpzpwijtykqziocilvtpqhxuyrphdlamawjuzgjwiebkqyrzyqgtkcrhntjxqmcgkrqcslogjegfrivzidfedeegwbbsopvxvdoididlpypjogxaodtueebbwuwilacunqpozczcgrpaxxrtogpxgfkudtxchoravrrdtimhenwmnafxaruuojduxxglefejycfcyxllfimkszmbrhcwpnwjqgwvzysmlaaifdxfjjpgjmcksiigpanappjdloiolojmcqbnpnjjzaufdpjaknylmyvolhwypygwepmqwpiglpcnpypnudhdzpdvgzosyjthzcwtytxq",
"output": "YES"
}
] | 1,619,339,875 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 1 | 46 | 0 | S = str(input()) #gets an input as a string value
if S.find('heidi') == -1: #if there is no subset 'heidi',
print("YES") #print YES
else: #if there is 'heidi',
print("NO") #print NO
| Title: Fake News (easy)
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
As it's the first of April, Heidi is suspecting that the news she reads today are fake, and she does not want to look silly in front of all the contestants. She knows that a newspiece is fake if it contains heidi as a subsequence. Help Heidi assess whether the given piece is true, but please be discreet about it...
Input Specification:
The first and only line of input contains a single nonempty string *s* of length at most 1000 composed of lowercase letters (a-z).
Output Specification:
Output YES if the string *s* contains heidi as a subsequence and NO otherwise.
Demo Input:
['abcheaibcdi\n', 'hiedi\n']
Demo Output:
['YES', 'NO']
Note:
A string *s* contains another string *p* as a subsequence if it is possible to delete some characters from *s* and obtain *p*. | ```python
S = str(input()) #gets an input as a string value
if S.find('heidi') == -1: #if there is no subset 'heidi',
print("YES") #print YES
else: #if there is 'heidi',
print("NO") #print NO
``` | 0 |
|
41 | A | Translation | PROGRAMMING | 800 | [
"implementation",
"strings"
] | A. Translation | 2 | 256 | The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly. | The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols. | If the word *t* is a word *s*, written reversely, print YES, otherwise print NO. | [
"code\nedoc\n",
"abb\naba\n",
"code\ncode\n"
] | [
"YES\n",
"NO\n",
"NO\n"
] | none | 500 | [
{
"input": "code\nedoc",
"output": "YES"
},
{
"input": "abb\naba",
"output": "NO"
},
{
"input": "code\ncode",
"output": "NO"
},
{
"input": "abacaba\nabacaba",
"output": "YES"
},
{
"input": "q\nq",
"output": "YES"
},
{
"input": "asrgdfngfnmfgnhweratgjkk\nasrgdfngfnmfgnhweratgjkk",
"output": "NO"
},
{
"input": "z\na",
"output": "NO"
},
{
"input": "asd\ndsa",
"output": "YES"
},
{
"input": "abcdef\nfecdba",
"output": "NO"
},
{
"input": "ywjjbirapvskozubvxoemscfwl\ngnduubaogtfaiowjizlvjcu",
"output": "NO"
},
{
"input": "mfrmqxtzvgaeuleubcmcxcfqyruwzenguhgrmkuhdgnhgtgkdszwqyd\nmfxufheiperjnhyczclkmzyhcxntdfskzkzdwzzujdinf",
"output": "NO"
},
{
"input": "bnbnemvybqizywlnghlykniaxxxlkhftppbdeqpesrtgkcpoeqowjwhrylpsziiwcldodcoonpimudvrxejjo\ntiynnekmlalogyvrgptbinkoqdwzuiyjlrldxhzjmmp",
"output": "NO"
},
{
"input": "pwlpubwyhzqvcitemnhvvwkmwcaawjvdiwtoxyhbhbxerlypelevasmelpfqwjk\nstruuzebbcenziscuoecywugxncdwzyfozhljjyizpqcgkyonyetarcpwkqhuugsqjuixsxptmbnlfupdcfigacdhhrzb",
"output": "NO"
},
{
"input": "gdvqjoyxnkypfvdxssgrihnwxkeojmnpdeobpecytkbdwujqfjtxsqspxvxpqioyfagzjxupqqzpgnpnpxcuipweunqch\nkkqkiwwasbhezqcfeceyngcyuogrkhqecwsyerdniqiocjehrpkljiljophqhyaiefjpavoom",
"output": "NO"
},
{
"input": "umeszdawsvgkjhlqwzents\nhxqhdungbylhnikwviuh",
"output": "NO"
},
{
"input": "juotpscvyfmgntshcealgbsrwwksgrwnrrbyaqqsxdlzhkbugdyx\nibqvffmfktyipgiopznsqtrtxiijntdbgyy",
"output": "NO"
},
{
"input": "zbwueheveouatecaglziqmudxemhrsozmaujrwlqmppzoumxhamwugedikvkblvmxwuofmpafdprbcftew\nulczwrqhctbtbxrhhodwbcxwimncnexosksujlisgclllxokrsbnozthajnnlilyffmsyko",
"output": "NO"
},
{
"input": "nkgwuugukzcv\nqktnpxedwxpxkrxdvgmfgoxkdfpbzvwsduyiybynbkouonhvmzakeiruhfmvrktghadbfkmwxduoqv",
"output": "NO"
},
{
"input": "incenvizhqpcenhjhehvjvgbsnfixbatrrjstxjzhlmdmxijztphxbrldlqwdfimweepkggzcxsrwelodpnryntepioqpvk\ndhjbjjftlvnxibkklxquwmzhjfvnmwpapdrslioxisbyhhfymyiaqhlgecpxamqnocizwxniubrmpyubvpenoukhcobkdojlybxd",
"output": "NO"
},
{
"input": "w\nw",
"output": "YES"
},
{
"input": "vz\nzv",
"output": "YES"
},
{
"input": "ry\nyr",
"output": "YES"
},
{
"input": "xou\nuox",
"output": "YES"
},
{
"input": "axg\ngax",
"output": "NO"
},
{
"input": "zdsl\nlsdz",
"output": "YES"
},
{
"input": "kudl\nldku",
"output": "NO"
},
{
"input": "zzlzwnqlcl\nlclqnwzlzz",
"output": "YES"
},
{
"input": "vzzgicnzqooejpjzads\nsdazjpjeooqzncigzzv",
"output": "YES"
},
{
"input": "raqhmvmzuwaykjpyxsykr\nxkysrypjkyawuzmvmhqar",
"output": "NO"
},
{
"input": "ngedczubzdcqbxksnxuavdjaqtmdwncjnoaicvmodcqvhfezew\nwezefhvqcdomvciaonjcnwdmtqajdvauxnskxbqcdzbuzcdegn",
"output": "YES"
},
{
"input": "muooqttvrrljcxbroizkymuidvfmhhsjtumksdkcbwwpfqdyvxtrlymofendqvznzlmim\nmimlznzvqdnefomylrtxvydqfpwwbckdskmutjshhmfvdiumykziorbxcjlrrvttqooum",
"output": "YES"
},
{
"input": "vxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaivg\ngviayyikkitmuomcpiakhbxszgbnhvwyzkftwoagzixaearxpjacrnvpvbuzenvovehkmmxvblqyxvctroddksdsgebcmlluqpxv",
"output": "YES"
},
{
"input": "mnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfdc\ncdfmkdgrdptkpewbsqvszipgxvgvuiuzbkkwuowbafkikgvnqdkxnayzdjygvezmtsgywnupocdntipiyiorblqkrzjpzatxahnm",
"output": "NO"
},
{
"input": "dgxmzbqofstzcdgthbaewbwocowvhqpinehpjatnnbrijcolvsatbblsrxabzrpszoiecpwhfjmwuhqrapvtcgvikuxtzbftydkw\nwkdytfbztxukivgctvparqhuwmjfhwpceiozsprzbaxrslbbqasvlocjirbnntajphenipthvwocowbweabhtgdcztsfoqbzmxgd",
"output": "NO"
},
{
"input": "gxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwgeh\nhegwxvocotmzstqfbmpjvijgkcyodlxyjawrpkczpmdspsuhoiruavnnnuwvtwohglkdxjetshkboalvzqbgjgthoteceixioxg",
"output": "YES"
},
{
"input": "sihxuwvmaambplxvjfoskinghzicyfqebjtkysotattkahssumfcgrkheotdxwjckpvapbkaepqrxseyfrwtyaycmrzsrsngkh\nhkgnsrszrmcyaytwrfyesxrqpeakbpavpkcjwxdtoehkrgcfmusshakttatosyktjbeqfycizhgniksofjvxlpbmaamvwuxhis",
"output": "YES"
},
{
"input": "ycnahksbughnonldzrhkysujmylcgcfuludjvjiahtkyzqvkopzqcnwhltbzfugzojqkjjlggmvnultascmygelkiktmfieok\nkoeifmtkiklegkmcsatlunvmggkjjlqjozgufzbtlhwncqzpokvqzykthaijvjdulufcgclymjusyyhrzdlnonhgubskhancy",
"output": "NO"
},
{
"input": "wbqasaehtkfojruzyhrlgwmtyiovmzyfifslvlemhqheyaelzwnthrenjsbmntwaoryzwfbxmscmypvxlfmzpnkkjlvwvmtz\nztmvwvljkknpzmflxvpymcsmxbfwzyroawtnmbsjnerhtnwzleayehqhmelvlsfifyzmvoiytmwglrhyzurjofktheasaqbw",
"output": "YES"
},
{
"input": "imippqurprbhfugngtgifelytadegwrgaefnfhbjjnmzikvjaccotqzemufqieqldgnbmviisgkynzeldlhqxuqphjfmyij\njiymfjhpquxqhldleznykgsiivmbngdlqeiqfumezqtoccajvkizmnjjbhfnfeagrwgedatylefigtgngufhbrpruqppimi",
"output": "YES"
},
{
"input": "bikydffiuisckpvzqlteqfhegsagimodb\nbdomigasgehfqetlqzvpkcsiuiffdykib",
"output": "YES"
}
] | 1,630,481,928 | 2,147,483,647 | Python 3 | OK | TESTS | 40 | 124 | 6,758,400 | word_1 = input()
word_2 = input()
word_2 = word_2[-1:0:-1] + word_2[0]
if word_1 == word_2:
print("YES")
else:
print("NO") | Title: Translation
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly.
Input Specification:
The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols.
Output Specification:
If the word *t* is a word *s*, written reversely, print YES, otherwise print NO.
Demo Input:
['code\nedoc\n', 'abb\naba\n', 'code\ncode\n']
Demo Output:
['YES\n', 'NO\n', 'NO\n']
Note:
none | ```python
word_1 = input()
word_2 = input()
word_2 = word_2[-1:0:-1] + word_2[0]
if word_1 == word_2:
print("YES")
else:
print("NO")
``` | 3.956411 |
7 | D | Palindrome Degree | PROGRAMMING | 2,200 | [
"hashing",
"strings"
] | D. Palindrome Degree | 1 | 256 | String *s* of length *n* is called *k*-palindrome, if it is a palindrome itself, and its prefix and suffix of length are (*k*<=-<=1)-palindromes. By definition, any string (even empty) is 0-palindrome.
Let's call the palindrome degree of string *s* such a maximum number *k*, for which *s* is *k*-palindrome. For example, "abaaba" has degree equals to 3.
You are given a string. Your task is to find the sum of the palindrome degrees of all its prefixes. | The first line of the input data contains a non-empty string, consisting of Latin letters and digits. The length of the string does not exceed 5·106. The string is case-sensitive. | Output the only number — the sum of the polindrome degrees of all the string's prefixes. | [
"a2A\n",
"abacaba\n"
] | [
"1",
"6"
] | none | 0 | [
{
"input": "a2A",
"output": "1"
},
{
"input": "abacaba",
"output": "6"
},
{
"input": "CCeCeCCCee",
"output": "4"
},
{
"input": "opooppppopppopoppopoooppopopooopopppooopppoppoppoppppoooppooooooopppoopoopooooppooooppppppppooopooop",
"output": "3"
},
{
"input": "odribmizzsgholprdsth",
"output": "1"
},
{
"input": "z",
"output": "1"
}
] | 1,501,244,971 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 92 | 4,608,000 | exp=input()
def getv(exp):
l=len(exp)
if l==1:
return 1
for i in range(l//2):
if exp[i]!=exp[l-1-i]:
return 0
return l//2
ans=0
l=len(exp)
while l>0:
tmp=getv(exp)
if tmp==0:
print(1)
exit(0)
ans+=tmp
l=l//2
exp=exp[:l]
print(ans+1)
| Title: Palindrome Degree
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
String *s* of length *n* is called *k*-palindrome, if it is a palindrome itself, and its prefix and suffix of length are (*k*<=-<=1)-palindromes. By definition, any string (even empty) is 0-palindrome.
Let's call the palindrome degree of string *s* such a maximum number *k*, for which *s* is *k*-palindrome. For example, "abaaba" has degree equals to 3.
You are given a string. Your task is to find the sum of the palindrome degrees of all its prefixes.
Input Specification:
The first line of the input data contains a non-empty string, consisting of Latin letters and digits. The length of the string does not exceed 5·106. The string is case-sensitive.
Output Specification:
Output the only number — the sum of the polindrome degrees of all the string's prefixes.
Demo Input:
['a2A\n', 'abacaba\n']
Demo Output:
['1', '6']
Note:
none | ```python
exp=input()
def getv(exp):
l=len(exp)
if l==1:
return 1
for i in range(l//2):
if exp[i]!=exp[l-1-i]:
return 0
return l//2
ans=0
l=len(exp)
while l>0:
tmp=getv(exp)
if tmp==0:
print(1)
exit(0)
ans+=tmp
l=l//2
exp=exp[:l]
print(ans+1)
``` | 0 |
934 | A | A Compatible Pair | PROGRAMMING | 1,400 | [
"brute force",
"games"
] | null | null | Nian is a monster which lives deep in the oceans. Once a year, it shows up on the land, devouring livestock and even people. In order to keep the monster away, people fill their villages with red colour, light, and cracking noise, all of which frighten the monster out of coming.
Little Tommy has *n* lanterns and Big Banban has *m* lanterns. Tommy's lanterns have brightness *a*1,<=*a*2,<=...,<=*a**n*, and Banban's have brightness *b*1,<=*b*2,<=...,<=*b**m* respectively.
Tommy intends to hide one of his lanterns, then Banban picks one of Tommy's non-hidden lanterns and one of his own lanterns to form a pair. The pair's brightness will be the product of the brightness of two lanterns.
Tommy wants to make the product as small as possible, while Banban tries to make it as large as possible.
You are asked to find the brightness of the chosen pair if both of them choose optimally. | The first line contains two space-separated integers *n* and *m* (2<=≤<=*n*,<=*m*<=≤<=50).
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n*.
The third line contains *m* space-separated integers *b*1,<=*b*2,<=...,<=*b**m*.
All the integers range from <=-<=109 to 109. | Print a single integer — the brightness of the chosen pair. | [
"2 2\n20 18\n2 14\n",
"5 3\n-1 0 1 2 3\n-1 0 1\n"
] | [
"252\n",
"2\n"
] | In the first example, Tommy will hide 20 and Banban will choose 18 from Tommy and 14 from himself.
In the second example, Tommy will hide 3 and Banban will choose 2 from Tommy and 1 from himself. | 500 | [
{
"input": "2 2\n20 18\n2 14",
"output": "252"
},
{
"input": "5 3\n-1 0 1 2 3\n-1 0 1",
"output": "2"
},
{
"input": "10 2\n1 6 2 10 2 3 2 10 6 4\n5 7",
"output": "70"
},
{
"input": "50 50\n1 6 2 10 2 3 2 10 6 4 5 0 3 1 7 3 2 4 4 2 1 5 0 6 10 1 8 0 10 9 0 4 10 5 5 7 4 9 9 5 5 2 6 7 9 4 3 7 2 0\n0 5 9 4 4 6 1 8 2 1 6 6 8 6 4 4 7 2 1 8 6 7 4 9 8 3 0 2 0 10 7 1 4 9 4 4 2 5 3 5 1 3 2 4 1 6 5 3 8 6",
"output": "100"
},
{
"input": "5 7\n-130464232 -73113866 -542094710 -53118823 -63528720\n-775179088 631683023 -974858199 -157471745 -629658630 71825477 -6235611",
"output": "127184126241438168"
},
{
"input": "16 15\n-94580188 -713689767 -559972014 -632609438 -930348091 -567718487 -611395744 -819913097 -924009672 -427913920 -812510647 -546415480 -982072775 -693369647 -693004777 -714181162\n-772924706 -202246100 -165871667 -991426281 -490838183 209351416 134956137 -36128588 -754413937 -616596290 696201705 -201191199 967464971 -244181984 -729907974",
"output": "922371547895579571"
},
{
"input": "12 22\n-102896616 -311161241 -67541276 -402842686 -830595520 -813834033 -44046671 -584806552 -598620444 -968935604 -303048547 -545969410\n545786451 262898403 442511997 -441241260 -479587986 -752123290 720443264 500646237 737842681 -571966572 -798463881 -477248830 89875164 410339460 -359022689 -251280099 -441455542 -538431186 -406793869 374561004 -108755237 -440143410",
"output": "663200522440413120"
},
{
"input": "33 14\n-576562007 -218618150 -471719380 -583840778 -256368365 -68451917 -405045344 -775538133 -896830082 -439261765 -947070124 -716577019 -456110999 -689862512 -132480131 -10805271 -518903339 -196240188 -222292638 -828546042 -43887962 -161359263 -281422097 -484060534 963147664 -492377073 -154570101 -52145116 187803553 858844161 66540410 418777176 434025748\n-78301978 -319393213 -12393024 542953412 786804661 845642067 754996432 -985617475 -487171947 56142664 203173079 -268261708 -817080591 -511720682",
"output": "883931400924882950"
},
{
"input": "15 8\n-966400308 -992207261 -302395973 -837980754 -516443826 -492405613 -378127629 -762650324 -519519776 -36132939 -286460372 -351445284 -407653342 -604960925 -523442015\n610042288 27129580 -103108347 -942517864 842060508 -588904868 614786155 37455106",
"output": "910849554065102112"
},
{
"input": "6 30\n-524297819 -947277203 -444186475 -182837689 -385379656 -453917269\n834529938 35245081 663687669 585422565 164412867 850052113 796429008 -307345676 -127653313 426960600 211854713 -733687358 251466836 -33491050 -882811238 455544614 774581544 768447941 -241033484 441104324 -493975870 308277556 275268265 935941507 -152292053 -961509996 -740482111 -954176110 -924254634 -518710544",
"output": "504117593849498724"
},
{
"input": "5 32\n-540510995 -841481393 -94342377 -74818927 -93445356\n686714668 -82581175 736472406 502016312 575563638 -899308712 503504178 -644271272 -437408397 385778869 -746757839 306275973 -663503743 -431116516 -418708278 -515261493 -988182324 900230931 218258353 -714420102 -241118202 294802602 -937785552 -857537498 -723195312 -690515139 -214508504 -44086454 -231621215 -418360090 -810003786 -675944617",
"output": "534123411186652380"
},
{
"input": "32 13\n-999451897 -96946179 -524159869 -906101658 -63367320 -629803888 -968586834 -658416130 -874232857 -926556428 -749908220 -517073321 -659752288 -910152878 -786916085 -607633039 -191428642 -867952926 -873793977 -584331784 -733245792 -779809700 -554228536 -464503499 561577340 258991071 -569805979 -372655165 -106685554 -619607960 188856473 -268960803\n886429660 -587284372 911396803 -462990289 -228681210 -876239914 -822830527 -750131315 -401234943 116991909 -582713480 979631847 813552478",
"output": "848714444125692276"
},
{
"input": "12 25\n-464030345 -914672073 -483242132 -856226270 -925135169 -353124606 -294027092 -619650850 -490724485 -240424784 -483066792 -921640365\n279850608 726838739 -431610610 242749870 -244020223 -396865433 129534799 182767854 -939698671 342579400 330027106 893561388 -263513962 643369418 276245179 -99206565 -473767261 -168908664 -853755837 -270920164 -661186118 199341055 765543053 908211534 -93363867",
"output": "866064226130454915"
},
{
"input": "10 13\n-749120991 -186261632 -335412349 -231354880 -195919225 -808736065 -481883825 -263383991 -664780611 -605377134\n718174936 -140362196 -669193674 -598621021 -464130929 450701419 -331183926 107203430 946959233 -565825915 -558199897 246556991 -666216081",
"output": "501307028237810934"
},
{
"input": "17 13\n-483786205 -947257449 -125949195 -294711143 -420288876 -812462057 -250049555 -911026413 -188146919 -129501682 -869006661 -649643966 -26976411 -275761039 -869067490 -272248209 -342067346\n445539900 529728842 -808170728 673157826 -70778491 642872105 299298867 -76674218 -902394063 377664752 723887448 -121522827 906464625",
"output": "822104826327386019"
},
{
"input": "15 29\n-716525085 -464205793 -577203110 -979997115 -491032521 -70793687 -770595947 -817983495 -767886763 -223333719 -971913221 -944656683 -200397825 -295615495 -945544540\n-877638425 -146878165 523758517 -158778747 -49535534 597311016 77325385 494128313 12111658 -4196724 295706874 477139483 375083042 726254399 -439255703 662913604 -481588088 673747948 -345999555 -723334478 -656721905 276267528 628773156 851420802 -585029291 -643535709 -968999740 -384418713 -510285542",
"output": "941783658451562540"
},
{
"input": "5 7\n-130464232 -73113866 -542094710 -53118823 -63528720\n449942926 482853427 861095072 316710734 194604468 20277633 668816604",
"output": "-1288212069119760"
},
{
"input": "24 24\n-700068683 -418791905 -24650102 -167277317 -182309202 -517748507 -663050677 -854097070 -426998982 -197009558 -101944229 -746589957 -849018439 -774208211 -946709040 -594578249 -276703474 -434567489 -743600446 -625029074 -977300284 -895608684 -878936220 -850670748\n704881272 169877679 705460701 94083210 403943695 987978311 786162506 658067668 697640875 186287 295558596 286470276 251313879 353071193 755450449 173370603 805550377 192465301 168935494 110161743 285139426 985238736 723221868 520679017",
"output": "-18990884587723"
},
{
"input": "39 9\n44558618 981372779 318891054 283079237 285093436 907256321 414759796 652683534 79042330 249010687 7020063 309415438 788425492 138577429 714835649 954204512 795507844 389962019 507308352 408180613 194676444 44962879 922688019 101163040 327953325 560462120 183657590 273616448 226876035 233697890 720185285 689340674 372938362 15088928 283418109 796807778 149989495 694808087 276385512\n-681609072 -210918688 -757170622 -205635977 -597872997 -496188744 -97031207 -311654366 -389141528",
"output": "-1464096896176096"
},
{
"input": "5 7\n869535768 926886134 457905290 946881177 936471280\n-550057074 -517146573 -138904928 -683289266 -805395532 -979722367 -331183396",
"output": "-120782803247464704"
},
{
"input": "24 24\n299931317 581208095 975349898 832722683 817690798 482251493 336949323 145902930 573001018 802990442 898055771 253410043 150981561 225791789 53290960 405421751 723296526 565432511 256399554 374970926 22699716 104391316 121063780 149329252\n-295118728 -830122321 -294539299 -905916790 -596056305 -12021689 -213837494 -341932332 -302359125 -999813713 -704441404 -713529724 -748686121 -646928807 -244549551 -826629397 -194449623 -807534699 -831064506 -889838257 -714860574 -14761264 -276778132 -479320983",
"output": "-640647347631440"
},
{
"input": "14 8\n-1000000000 -1000000000 -1000000000 -1000000000 -1000000000 -1000000000 -1000000000 -1000000000 -1000000000 -1000000000 -1000000000 -1000000000 -1000000000 -1000000000\n1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000",
"output": "-1000000000000000000"
}
] | 1,518,660,421 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 2 | 61 | 19,968,000 | from operator import itemgetter
a = input()
b = input().split(" ")
b = [int(i) for i in b]
c = input().split(" ")
c = [int(i) for i in c]
ans = []
for j in range(len(b)):
for k in range(len(c)):
ans.append([j,b[j]*c[k]])
ans = sorted(ans, key=itemgetter(1))
tmp = ans[-1][0]
ans = []
for j in range(len(b)):
if (j != tmp):
for k in range(len(c)):
ans.append(b[j]*c[k])
print(ans[-1])
| Title: A Compatible Pair
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Nian is a monster which lives deep in the oceans. Once a year, it shows up on the land, devouring livestock and even people. In order to keep the monster away, people fill their villages with red colour, light, and cracking noise, all of which frighten the monster out of coming.
Little Tommy has *n* lanterns and Big Banban has *m* lanterns. Tommy's lanterns have brightness *a*1,<=*a*2,<=...,<=*a**n*, and Banban's have brightness *b*1,<=*b*2,<=...,<=*b**m* respectively.
Tommy intends to hide one of his lanterns, then Banban picks one of Tommy's non-hidden lanterns and one of his own lanterns to form a pair. The pair's brightness will be the product of the brightness of two lanterns.
Tommy wants to make the product as small as possible, while Banban tries to make it as large as possible.
You are asked to find the brightness of the chosen pair if both of them choose optimally.
Input Specification:
The first line contains two space-separated integers *n* and *m* (2<=≤<=*n*,<=*m*<=≤<=50).
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n*.
The third line contains *m* space-separated integers *b*1,<=*b*2,<=...,<=*b**m*.
All the integers range from <=-<=109 to 109.
Output Specification:
Print a single integer — the brightness of the chosen pair.
Demo Input:
['2 2\n20 18\n2 14\n', '5 3\n-1 0 1 2 3\n-1 0 1\n']
Demo Output:
['252\n', '2\n']
Note:
In the first example, Tommy will hide 20 and Banban will choose 18 from Tommy and 14 from himself.
In the second example, Tommy will hide 3 and Banban will choose 2 from Tommy and 1 from himself. | ```python
from operator import itemgetter
a = input()
b = input().split(" ")
b = [int(i) for i in b]
c = input().split(" ")
c = [int(i) for i in c]
ans = []
for j in range(len(b)):
for k in range(len(c)):
ans.append([j,b[j]*c[k]])
ans = sorted(ans, key=itemgetter(1))
tmp = ans[-1][0]
ans = []
for j in range(len(b)):
if (j != tmp):
for k in range(len(c)):
ans.append(b[j]*c[k])
print(ans[-1])
``` | 0 |
|
731 | A | Night at the Museum | PROGRAMMING | 800 | [
"implementation",
"strings"
] | null | null | Grigoriy, like the hero of one famous comedy film, found a job as a night security guard at the museum. At first night he received embosser and was to take stock of the whole exposition.
Embosser is a special devise that allows to "print" the text of a plastic tape. Text is printed sequentially, character by character. The device consists of a wheel with a lowercase English letters written in a circle, static pointer to the current letter and a button that print the chosen letter. At one move it's allowed to rotate the alphabetic wheel one step clockwise or counterclockwise. Initially, static pointer points to letter 'a'. Other letters are located as shown on the picture:
After Grigoriy add new item to the base he has to print its name on the plastic tape and attach it to the corresponding exhibit. It's not required to return the wheel to its initial position with pointer on the letter 'a'.
Our hero is afraid that some exhibits may become alive and start to attack him, so he wants to print the names as fast as possible. Help him, for the given string find the minimum number of rotations of the wheel required to print it. | The only line of input contains the name of some exhibit — the non-empty string consisting of no more than 100 characters. It's guaranteed that the string consists of only lowercase English letters. | Print one integer — the minimum number of rotations of the wheel, required to print the name given in the input. | [
"zeus\n",
"map\n",
"ares\n"
] | [
"18\n",
"35\n",
"34\n"
] | To print the string from the first sample it would be optimal to perform the following sequence of rotations:
1. from 'a' to 'z' (1 rotation counterclockwise), 1. from 'z' to 'e' (5 clockwise rotations), 1. from 'e' to 'u' (10 rotations counterclockwise), 1. from 'u' to 's' (2 counterclockwise rotations). | 500 | [
{
"input": "zeus",
"output": "18"
},
{
"input": "map",
"output": "35"
},
{
"input": "ares",
"output": "34"
},
{
"input": "l",
"output": "11"
},
{
"input": "abcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuv",
"output": "99"
},
{
"input": "gngvi",
"output": "44"
},
{
"input": "aaaaa",
"output": "0"
},
{
"input": "a",
"output": "0"
},
{
"input": "z",
"output": "1"
},
{
"input": "vyadeehhikklnoqrs",
"output": "28"
},
{
"input": "jjiihhhhgggfedcccbazyxx",
"output": "21"
},
{
"input": "fyyptqqxuciqvwdewyppjdzur",
"output": "117"
},
{
"input": "fqcnzmzmbobmancqcoalzmanaobpdse",
"output": "368"
},
{
"input": "zzzzzaaaaaaazzzzzzaaaaaaazzzzzzaaaazzzza",
"output": "8"
},
{
"input": "aucnwhfixuruefkypvrvnvznwtjgwlghoqtisbkhuwxmgzuljvqhmnwzisnsgjhivnjmbknptxatdkelhzkhsuxzrmlcpeoyukiy",
"output": "644"
},
{
"input": "sssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssss",
"output": "8"
},
{
"input": "nypjygrdtpzpigzyrisqeqfriwgwlengnezppgttgtndbrryjdl",
"output": "421"
},
{
"input": "pnllnnmmmmoqqqqqrrtssssuuvtsrpopqoonllmonnnpppopnonoopooqpnopppqppqstuuuwwwwvxzxzzaa",
"output": "84"
},
{
"input": "btaoahqgxnfsdmzsjxgvdwjukcvereqeskrdufqfqgzqfsftdqcthtkcnaipftcnco",
"output": "666"
},
{
"input": "eeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeerrrrrrrrrrrrrrrrwwwwwwwwww",
"output": "22"
},
{
"input": "uyknzcrwjyzmscqucclvacmorepdgmnyhmakmmnygqwglrxkxhkpansbmruwxdeoprxzmpsvwackopujxbbkpwyeggsvjykpxh",
"output": "643"
},
{
"input": "gzwpooohffcxwtpjgfzwtooiccxsrrokezutoojdzwsrmmhecaxwrojcbyrqlfdwwrliiib",
"output": "245"
},
{
"input": "dbvnkktasjdwqsrzfwwtmjgbcxggdxsoeilecihduypktkkbwfbruxzzhlttrssicgdwqruddwrlbtxgmhdbatzvdxbbro",
"output": "468"
},
{
"input": "mdtvowlktxzzbuaeiuebfeorgbdczauxsovbucactkvyvemsknsjfhifqgycqredzchipmkvzbxdjkcbyukomjlzvxzoswumned",
"output": "523"
},
{
"input": "kkkkkkkaaaaxxaaaaaaaxxxxxxxxaaaaaaxaaaaaaaaaakkkkkkkkkaaaaaaannnnnxxxxkkkkkkkkaannnnnnna",
"output": "130"
},
{
"input": "dffiknqqrsvwzcdgjkmpqtuwxadfhkkkmpqrtwxyadfggjmpppsuuwyyzcdgghhknnpsvvvwwwyabccffiloqruwwyyzabeeehh",
"output": "163"
},
{
"input": "qpppmmkjihgecbyvvsppnnnkjiffeebaaywutrrqpmkjhgddbzzzywtssssqnmmljheddbbaxvusrqonmlifedbbzyywwtqnkheb",
"output": "155"
},
{
"input": "wvvwwwvvwxxxyyyxxwwvwwvuttttttuvvwxxwxxyxxwwwwwvvuttssrssstsssssrqpqqppqrssrsrrssrssssrrsrqqrrqpppqp",
"output": "57"
},
{
"input": "dqcpcobpcobnznamznamzlykxkxlxlylzmaobnaobpbnanbpcoaobnboaoboanzlymzmykylymylzlylymanboanaocqdqesfrfs",
"output": "1236"
},
{
"input": "nnnnnnnnnnnnnnnnnnnnaaaaaaaaaaaaaaaaaaaakkkkkkkkkkkkkkkkkkkkkkaaaaaaaaaaaaaaaaaaaaxxxxxxxxxxxxxxxxxx",
"output": "49"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "0"
},
{
"input": "cgilqsuwzaffilptwwbgmnttyyejkorxzflqvzbddhmnrvxchijpuwaeiimosxyycejlpquuwbfkpvbgijkqvxybdjjjptxcfkqt",
"output": "331"
},
{
"input": "ufsepwgtzgtgjssxaitgpailuvgqweoppszjwhoxdhhhpwwdorwfrdjwcdekxiktwziqwbkvbknrtvajpyeqbjvhiikxxaejjpte",
"output": "692"
},
{
"input": "uhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuhuh",
"output": "1293"
},
{
"input": "vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvgggggggggggggggggggggggggggggggggggggggggggggggggg",
"output": "16"
},
{
"input": "lyidmjyzbszgiwkxhhpnnthfwcvvstueionspfrvqgkvngmwyhezlosrpdnbvtcjjxxsykixwnepbumaacdzadlqhnjlcejovple",
"output": "616"
},
{
"input": "etzqqbaveffalkdguunfmyyrzkccnxmlluxeasqmopxzfvlkbhipqdwjgrttoemruohgwukfisdhznqyvhswbbypoxgtxyappcrl",
"output": "605"
},
{
"input": "lizussgedcbdjhrbeskhgatyozvwwekanlggcstijrniivupmcoofbaxfqrxddyzzptwxcftlhajsmmkkriarrqtkoauhcqefyud",
"output": "549"
},
{
"input": "dvjuvgfdogpknmbowlsfjzcimnygbtjiucyeeroqwhmzwpjqxlbjkqawrdtmvxbiqufllfuqibxvmtdrwaqkjblxqjpwzmhwqore",
"output": "688"
},
{
"input": "eeycuijtbgynmiczjfslwobmnkpgodfgvujvduyfeqchuaoktqrrairkkmmsjahltfcxwtpzzyddxrqfxabfoocmpuviinrjitsc",
"output": "604"
},
{
"input": "cgglnakewwvzoytaghksebrhjdbcdegssuzilrcppayxtgxopybbwshvyqnzhdsifkuwghourmeottrgjwdqpihbklvfzxpomqsa",
"output": "572"
},
{
"input": "aexullmxncckzryymfnuugdklaffevabqqztelpvojecljnhqldazdcaamubpenwxikysxxjjctvbndprsolzehywmgnvkgqvrfp",
"output": "609"
},
{
"input": "psnoieutsvvcwfhtnnphhxkwigzsbzyjmdiyl",
"output": "223"
},
{
"input": "aa",
"output": "0"
}
] | 1,655,084,839 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 4 | 61 | 409,600 | import string
name = str(input())
alphabet = string.ascii_lowercase
initialRotate = 14
currentChar = 'a'
count = 0
for n in name:
rotate = alphabet[initialRotate:] + alphabet[:initialRotate]
idxCurrentChar = rotate.index(currentChar)
idxTargetChart = rotate.index(n)
if idxTargetChart < idxCurrentChar:
operator = -1
else:
operator = 1
tempCount = 0
for i in range(idxCurrentChar, idxTargetChart, operator):
tempCount += 1
if operator == -1:
initialRotate -= tempCount
else:
initialRotate += tempCount
currentChar = n
count += tempCount
print(count) | Title: Night at the Museum
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Grigoriy, like the hero of one famous comedy film, found a job as a night security guard at the museum. At first night he received embosser and was to take stock of the whole exposition.
Embosser is a special devise that allows to "print" the text of a plastic tape. Text is printed sequentially, character by character. The device consists of a wheel with a lowercase English letters written in a circle, static pointer to the current letter and a button that print the chosen letter. At one move it's allowed to rotate the alphabetic wheel one step clockwise or counterclockwise. Initially, static pointer points to letter 'a'. Other letters are located as shown on the picture:
After Grigoriy add new item to the base he has to print its name on the plastic tape and attach it to the corresponding exhibit. It's not required to return the wheel to its initial position with pointer on the letter 'a'.
Our hero is afraid that some exhibits may become alive and start to attack him, so he wants to print the names as fast as possible. Help him, for the given string find the minimum number of rotations of the wheel required to print it.
Input Specification:
The only line of input contains the name of some exhibit — the non-empty string consisting of no more than 100 characters. It's guaranteed that the string consists of only lowercase English letters.
Output Specification:
Print one integer — the minimum number of rotations of the wheel, required to print the name given in the input.
Demo Input:
['zeus\n', 'map\n', 'ares\n']
Demo Output:
['18\n', '35\n', '34\n']
Note:
To print the string from the first sample it would be optimal to perform the following sequence of rotations:
1. from 'a' to 'z' (1 rotation counterclockwise), 1. from 'z' to 'e' (5 clockwise rotations), 1. from 'e' to 'u' (10 rotations counterclockwise), 1. from 'u' to 's' (2 counterclockwise rotations). | ```python
import string
name = str(input())
alphabet = string.ascii_lowercase
initialRotate = 14
currentChar = 'a'
count = 0
for n in name:
rotate = alphabet[initialRotate:] + alphabet[:initialRotate]
idxCurrentChar = rotate.index(currentChar)
idxTargetChart = rotate.index(n)
if idxTargetChart < idxCurrentChar:
operator = -1
else:
operator = 1
tempCount = 0
for i in range(idxCurrentChar, idxTargetChart, operator):
tempCount += 1
if operator == -1:
initialRotate -= tempCount
else:
initialRotate += tempCount
currentChar = n
count += tempCount
print(count)
``` | 0 |
|
776 | A | A Serial Killer | PROGRAMMING | 900 | [
"brute force",
"implementation",
"strings"
] | null | null | Our beloved detective, Sherlock is currently trying to catch a serial killer who kills a person each day. Using his powers of deduction, he came to know that the killer has a strategy for selecting his next victim.
The killer starts with two potential victims on his first day, selects one of these two, kills selected victim and replaces him with a new person. He repeats this procedure each day. This way, each day he has two potential victims to choose from. Sherlock knows the initial two potential victims. Also, he knows the murder that happened on a particular day and the new person who replaced this victim.
You need to help him get all the pairs of potential victims at each day so that Sherlock can observe some pattern. | First line of input contains two names (length of each of them doesn't exceed 10), the two initials potential victims. Next line contains integer *n* (1<=≤<=*n*<=≤<=1000), the number of days.
Next *n* lines contains two names (length of each of them doesn't exceed 10), first being the person murdered on this day and the second being the one who replaced that person.
The input format is consistent, that is, a person murdered is guaranteed to be from the two potential victims at that time. Also, all the names are guaranteed to be distinct and consists of lowercase English letters. | Output *n*<=+<=1 lines, the *i*-th line should contain the two persons from which the killer selects for the *i*-th murder. The (*n*<=+<=1)-th line should contain the two persons from which the next victim is selected. In each line, the two names can be printed in any order. | [
"ross rachel\n4\nross joey\nrachel phoebe\nphoebe monica\nmonica chandler\n",
"icm codeforces\n1\ncodeforces technex\n"
] | [
"ross rachel\njoey rachel\njoey phoebe\njoey monica\njoey chandler\n",
"icm codeforces\nicm technex\n"
] | In first example, the killer starts with ross and rachel.
- After day 1, ross is killed and joey appears. - After day 2, rachel is killed and phoebe appears. - After day 3, phoebe is killed and monica appears. - After day 4, monica is killed and chandler appears. | 500 | [
{
"input": "ross rachel\n4\nross joey\nrachel phoebe\nphoebe monica\nmonica chandler",
"output": "ross rachel\njoey rachel\njoey phoebe\njoey monica\njoey chandler"
},
{
"input": "icm codeforces\n1\ncodeforces technex",
"output": "icm codeforces\nicm technex"
},
{
"input": "a b\n3\na c\nb d\nd e",
"output": "a b\nc b\nc d\nc e"
},
{
"input": "ze udggmyop\n4\nze szhrbmft\nudggmyop mjorab\nszhrbmft ojdtfnzxj\nojdtfnzxj yjlkg",
"output": "ze udggmyop\nszhrbmft udggmyop\nszhrbmft mjorab\nojdtfnzxj mjorab\nyjlkg mjorab"
},
{
"input": "q s\n10\nq b\nb j\ns g\nj f\nf m\ng c\nc a\nm d\nd z\nz o",
"output": "q s\nb s\nj s\nj g\nf g\nm g\nm c\nm a\nd a\nz a\no a"
},
{
"input": "iii iiiiii\n7\niii iiiiiiiiii\niiiiiiiiii iiii\niiii i\niiiiii iiiiiiii\niiiiiiii iiiiiiiii\ni iiiii\niiiii ii",
"output": "iii iiiiii\niiiiiiiiii iiiiii\niiii iiiiii\ni iiiiii\ni iiiiiiii\ni iiiiiiiii\niiiii iiiiiiiii\nii iiiiiiiii"
},
{
"input": "bwyplnjn zkms\n26\nzkms nzmcsytxh\nnzmcsytxh yujsb\nbwyplnjn gtbzhudpb\ngtbzhudpb hpk\nyujsb xvy\nhpk wrwnfokml\nwrwnfokml ndouuikw\nndouuikw ucgrja\nucgrja tgfmpldz\nxvy nycrfphn\nnycrfphn quvs\nquvs htdy\nhtdy k\ntgfmpldz xtdpkxm\nxtdpkxm suwqxs\nk fv\nsuwqxs qckllwy\nqckllwy diun\nfv lefa\nlefa gdoqjysx\ndiun dhpz\ngdoqjysx bdmqdyt\ndhpz dgz\ndgz v\nbdmqdyt aswy\naswy ydkayhlrnm",
"output": "bwyplnjn zkms\nbwyplnjn nzmcsytxh\nbwyplnjn yujsb\ngtbzhudpb yujsb\nhpk yujsb\nhpk xvy\nwrwnfokml xvy\nndouuikw xvy\nucgrja xvy\ntgfmpldz xvy\ntgfmpldz nycrfphn\ntgfmpldz quvs\ntgfmpldz htdy\ntgfmpldz k\nxtdpkxm k\nsuwqxs k\nsuwqxs fv\nqckllwy fv\ndiun fv\ndiun lefa\ndiun gdoqjysx\ndhpz gdoqjysx\ndhpz bdmqdyt\ndgz bdmqdyt\nv bdmqdyt\nv aswy\nv ydkayhlrnm"
},
{
"input": "wxz hbeqwqp\n7\nhbeqwqp cpieghnszh\ncpieghnszh tlqrpd\ntlqrpd ttwrtio\nttwrtio xapvds\nxapvds zk\nwxz yryk\nzk b",
"output": "wxz hbeqwqp\nwxz cpieghnszh\nwxz tlqrpd\nwxz ttwrtio\nwxz xapvds\nwxz zk\nyryk zk\nyryk b"
},
{
"input": "wced gnsgv\n23\ngnsgv japawpaf\njapawpaf nnvpeu\nnnvpeu a\na ddupputljq\nddupputljq qyhnvbh\nqyhnvbh pqwijl\nwced khuvs\nkhuvs bjkh\npqwijl ysacmboc\nbjkh srf\nsrf jknoz\njknoz hodf\nysacmboc xqtkoyh\nhodf rfp\nxqtkoyh bivgnwqvoe\nbivgnwqvoe nknf\nnknf wuig\nrfp e\ne bqqknq\nwuig sznhhhu\nbqqknq dhrtdld\ndhrtdld n\nsznhhhu bguylf",
"output": "wced gnsgv\nwced japawpaf\nwced nnvpeu\nwced a\nwced ddupputljq\nwced qyhnvbh\nwced pqwijl\nkhuvs pqwijl\nbjkh pqwijl\nbjkh ysacmboc\nsrf ysacmboc\njknoz ysacmboc\nhodf ysacmboc\nhodf xqtkoyh\nrfp xqtkoyh\nrfp bivgnwqvoe\nrfp nknf\nrfp wuig\ne wuig\nbqqknq wuig\nbqqknq sznhhhu\ndhrtdld sznhhhu\nn sznhhhu\nn bguylf"
},
{
"input": "qqqqqqqqqq qqqqqqqq\n3\nqqqqqqqq qqqqqqqqq\nqqqqqqqqq qqqqq\nqqqqq q",
"output": "qqqqqqqqqq qqqqqqqq\nqqqqqqqqqq qqqqqqqqq\nqqqqqqqqqq qqqqq\nqqqqqqqqqq q"
},
{
"input": "wwwww w\n8\nwwwww wwwwwwww\nwwwwwwww wwwwwwwww\nwwwwwwwww wwwwwwwwww\nw www\nwwwwwwwwww wwww\nwwww ww\nwww wwwwww\nwwwwww wwwwwww",
"output": "wwwww w\nwwwwwwww w\nwwwwwwwww w\nwwwwwwwwww w\nwwwwwwwwww www\nwwww www\nww www\nww wwwwww\nww wwwwwww"
},
{
"input": "k d\n17\nk l\nd v\nv z\nl r\nz i\nr s\ns p\np w\nw j\nj h\ni c\nh m\nm q\nc o\no g\nq x\nx n",
"output": "k d\nl d\nl v\nl z\nr z\nr i\ns i\np i\nw i\nj i\nh i\nh c\nm c\nq c\nq o\nq g\nx g\nn g"
}
] | 1,628,630,589 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 46 | 6,758,400 | a, b=input().split()
n=input()
n=int(n)
print(a+" "+b)
for i in range(n):
c, d=input().split()
if(c==a):
print(b+" "+d)
if(c==b):
print(a+" "+d)
a=c
b=d
| Title: A Serial Killer
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Our beloved detective, Sherlock is currently trying to catch a serial killer who kills a person each day. Using his powers of deduction, he came to know that the killer has a strategy for selecting his next victim.
The killer starts with two potential victims on his first day, selects one of these two, kills selected victim and replaces him with a new person. He repeats this procedure each day. This way, each day he has two potential victims to choose from. Sherlock knows the initial two potential victims. Also, he knows the murder that happened on a particular day and the new person who replaced this victim.
You need to help him get all the pairs of potential victims at each day so that Sherlock can observe some pattern.
Input Specification:
First line of input contains two names (length of each of them doesn't exceed 10), the two initials potential victims. Next line contains integer *n* (1<=≤<=*n*<=≤<=1000), the number of days.
Next *n* lines contains two names (length of each of them doesn't exceed 10), first being the person murdered on this day and the second being the one who replaced that person.
The input format is consistent, that is, a person murdered is guaranteed to be from the two potential victims at that time. Also, all the names are guaranteed to be distinct and consists of lowercase English letters.
Output Specification:
Output *n*<=+<=1 lines, the *i*-th line should contain the two persons from which the killer selects for the *i*-th murder. The (*n*<=+<=1)-th line should contain the two persons from which the next victim is selected. In each line, the two names can be printed in any order.
Demo Input:
['ross rachel\n4\nross joey\nrachel phoebe\nphoebe monica\nmonica chandler\n', 'icm codeforces\n1\ncodeforces technex\n']
Demo Output:
['ross rachel\njoey rachel\njoey phoebe\njoey monica\njoey chandler\n', 'icm codeforces\nicm technex\n']
Note:
In first example, the killer starts with ross and rachel.
- After day 1, ross is killed and joey appears. - After day 2, rachel is killed and phoebe appears. - After day 3, phoebe is killed and monica appears. - After day 4, monica is killed and chandler appears. | ```python
a, b=input().split()
n=input()
n=int(n)
print(a+" "+b)
for i in range(n):
c, d=input().split()
if(c==a):
print(b+" "+d)
if(c==b):
print(a+" "+d)
a=c
b=d
``` | 0 |
|
835 | A | Key races | PROGRAMMING | 800 | [
"math"
] | null | null | Two boys decided to compete in text typing on the site "Key races". During the competition, they have to type a text consisting of *s* characters. The first participant types one character in *v*1 milliseconds and has ping *t*1 milliseconds. The second participant types one character in *v*2 milliseconds and has ping *t*2 milliseconds.
If connection ping (delay) is *t* milliseconds, the competition passes for a participant as follows:
1. Exactly after *t* milliseconds after the start of the competition the participant receives the text to be entered. 1. Right after that he starts to type it. 1. Exactly *t* milliseconds after he ends typing all the text, the site receives information about it.
The winner is the participant whose information on the success comes earlier. If the information comes from both participants at the same time, it is considered that there is a draw.
Given the length of the text and the information about participants, determine the result of the game. | The first line contains five integers *s*, *v*1, *v*2, *t*1, *t*2 (1<=≤<=*s*,<=*v*1,<=*v*2,<=*t*1,<=*t*2<=≤<=1000) — the number of characters in the text, the time of typing one character for the first participant, the time of typing one character for the the second participant, the ping of the first participant and the ping of the second participant. | If the first participant wins, print "First". If the second participant wins, print "Second". In case of a draw print "Friendship". | [
"5 1 2 1 2\n",
"3 3 1 1 1\n",
"4 5 3 1 5\n"
] | [
"First\n",
"Second\n",
"Friendship\n"
] | In the first example, information on the success of the first participant comes in 7 milliseconds, of the second participant — in 14 milliseconds. So, the first wins.
In the second example, information on the success of the first participant comes in 11 milliseconds, of the second participant — in 5 milliseconds. So, the second wins.
In the third example, information on the success of the first participant comes in 22 milliseconds, of the second participant — in 22 milliseconds. So, it is be a draw. | 500 | [
{
"input": "5 1 2 1 2",
"output": "First"
},
{
"input": "3 3 1 1 1",
"output": "Second"
},
{
"input": "4 5 3 1 5",
"output": "Friendship"
},
{
"input": "1000 1000 1000 1000 1000",
"output": "Friendship"
},
{
"input": "1 1 1 1 1",
"output": "Friendship"
},
{
"input": "8 8 1 1 1",
"output": "Second"
},
{
"input": "15 14 32 65 28",
"output": "First"
},
{
"input": "894 197 325 232 902",
"output": "First"
},
{
"input": "1 2 8 8 5",
"output": "Friendship"
},
{
"input": "37 261 207 1 1000",
"output": "Friendship"
},
{
"input": "29 344 406 900 1",
"output": "Friendship"
},
{
"input": "1 2 8 9 8",
"output": "First"
},
{
"input": "2 9 8 8 9",
"output": "Friendship"
},
{
"input": "213 480 811 134 745",
"output": "First"
},
{
"input": "2 313 856 964 421",
"output": "Friendship"
},
{
"input": "1 10 2 6 10",
"output": "Friendship"
},
{
"input": "2 7 6 2 3",
"output": "Friendship"
},
{
"input": "637 324 69 612 998",
"output": "Second"
},
{
"input": "13 849 819 723 918",
"output": "Friendship"
},
{
"input": "9 5 7 8 7",
"output": "First"
},
{
"input": "6 5 7 10 4",
"output": "Friendship"
},
{
"input": "61 464 623 89 548",
"output": "First"
},
{
"input": "641 31 29 161 802",
"output": "Friendship"
},
{
"input": "3 3 1 6 9",
"output": "Friendship"
},
{
"input": "2 3 9 8 2",
"output": "Friendship"
},
{
"input": "485 117 368 567 609",
"output": "First"
},
{
"input": "4 202 512 995 375",
"output": "Friendship"
},
{
"input": "424 41 41 909 909",
"output": "Friendship"
},
{
"input": "884 913 263 641 265",
"output": "Second"
},
{
"input": "12 462 8 311 327",
"output": "Second"
},
{
"input": "436 306 266 493 580",
"output": "Second"
},
{
"input": "69 1 2 1 2",
"output": "First"
}
] | 1,562,564,175 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 202 | 1,638,400 | import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,queue,copy,numpy
sys.setrecursionlimit(10**7)
inf=10**20
mod=10**9+7
dd=[(-1,0),(0,1),(1,0),(0,-1)]
ddn=[(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def I(): return int(sys.stdin.readline())
def LS(): return sys.stdin.readline().split()
def S(): return input()
def main():
a,b,c,d,e=LI()
x=a*b+2*d
y=a*c+2*e
if x==y:
return 'Friendship'
elif x<y:
return 'First'
return 'Second'
print(main())
| Title: Key races
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Two boys decided to compete in text typing on the site "Key races". During the competition, they have to type a text consisting of *s* characters. The first participant types one character in *v*1 milliseconds and has ping *t*1 milliseconds. The second participant types one character in *v*2 milliseconds and has ping *t*2 milliseconds.
If connection ping (delay) is *t* milliseconds, the competition passes for a participant as follows:
1. Exactly after *t* milliseconds after the start of the competition the participant receives the text to be entered. 1. Right after that he starts to type it. 1. Exactly *t* milliseconds after he ends typing all the text, the site receives information about it.
The winner is the participant whose information on the success comes earlier. If the information comes from both participants at the same time, it is considered that there is a draw.
Given the length of the text and the information about participants, determine the result of the game.
Input Specification:
The first line contains five integers *s*, *v*1, *v*2, *t*1, *t*2 (1<=≤<=*s*,<=*v*1,<=*v*2,<=*t*1,<=*t*2<=≤<=1000) — the number of characters in the text, the time of typing one character for the first participant, the time of typing one character for the the second participant, the ping of the first participant and the ping of the second participant.
Output Specification:
If the first participant wins, print "First". If the second participant wins, print "Second". In case of a draw print "Friendship".
Demo Input:
['5 1 2 1 2\n', '3 3 1 1 1\n', '4 5 3 1 5\n']
Demo Output:
['First\n', 'Second\n', 'Friendship\n']
Note:
In the first example, information on the success of the first participant comes in 7 milliseconds, of the second participant — in 14 milliseconds. So, the first wins.
In the second example, information on the success of the first participant comes in 11 milliseconds, of the second participant — in 5 milliseconds. So, the second wins.
In the third example, information on the success of the first participant comes in 22 milliseconds, of the second participant — in 22 milliseconds. So, it is be a draw. | ```python
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,queue,copy,numpy
sys.setrecursionlimit(10**7)
inf=10**20
mod=10**9+7
dd=[(-1,0),(0,1),(1,0),(0,-1)]
ddn=[(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def I(): return int(sys.stdin.readline())
def LS(): return sys.stdin.readline().split()
def S(): return input()
def main():
a,b,c,d,e=LI()
x=a*b+2*d
y=a*c+2*e
if x==y:
return 'Friendship'
elif x<y:
return 'First'
return 'Second'
print(main())
``` | -1 |
|
69 | A | Young Physicist | PROGRAMMING | 1,000 | [
"implementation",
"math"
] | A. Young Physicist | 2 | 256 | A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces. | The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100). | Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not. | [
"3\n4 1 7\n-2 4 -1\n1 -5 -3\n",
"3\n3 -1 7\n-5 2 -4\n2 -1 -3\n"
] | [
"NO",
"YES"
] | none | 500 | [
{
"input": "3\n4 1 7\n-2 4 -1\n1 -5 -3",
"output": "NO"
},
{
"input": "3\n3 -1 7\n-5 2 -4\n2 -1 -3",
"output": "YES"
},
{
"input": "10\n21 32 -46\n43 -35 21\n42 2 -50\n22 40 20\n-27 -9 38\n-4 1 1\n-40 6 -31\n-13 -2 34\n-21 34 -12\n-32 -29 41",
"output": "NO"
},
{
"input": "10\n25 -33 43\n-27 -42 28\n-35 -20 19\n41 -42 -1\n49 -39 -4\n-49 -22 7\n-19 29 41\n8 -27 -43\n8 34 9\n-11 -3 33",
"output": "NO"
},
{
"input": "10\n-6 21 18\n20 -11 -8\n37 -11 41\n-5 8 33\n29 23 32\n30 -33 -11\n39 -49 -36\n28 34 -49\n22 29 -34\n-18 -6 7",
"output": "NO"
},
{
"input": "10\n47 -2 -27\n0 26 -14\n5 -12 33\n2 18 3\n45 -30 -49\n4 -18 8\n-46 -44 -41\n-22 -10 -40\n-35 -21 26\n33 20 38",
"output": "NO"
},
{
"input": "13\n-3 -36 -46\n-11 -50 37\n42 -11 -15\n9 42 44\n-29 -12 24\n3 9 -40\n-35 13 50\n14 43 18\n-13 8 24\n-48 -15 10\n50 9 -50\n21 0 -50\n0 0 -6",
"output": "YES"
},
{
"input": "14\n43 23 17\n4 17 44\n5 -5 -16\n-43 -7 -6\n47 -48 12\n50 47 -45\n2 14 43\n37 -30 15\n4 -17 -11\n17 9 -45\n-50 -3 -8\n-50 0 0\n-50 0 0\n-16 0 0",
"output": "YES"
},
{
"input": "13\n29 49 -11\n38 -11 -20\n25 1 -40\n-11 28 11\n23 -19 1\n45 -41 -17\n-3 0 -19\n-13 -33 49\n-30 0 28\n34 17 45\n-50 9 -27\n-50 0 0\n-37 0 0",
"output": "YES"
},
{
"input": "12\n3 28 -35\n-32 -44 -17\n9 -25 -6\n-42 -22 20\n-19 15 38\n-21 38 48\n-1 -37 -28\n-10 -13 -50\n-5 21 29\n34 28 50\n50 11 -49\n34 0 0",
"output": "YES"
},
{
"input": "37\n-64 -79 26\n-22 59 93\n-5 39 -12\n77 -9 76\n55 -86 57\n83 100 -97\n-70 94 84\n-14 46 -94\n26 72 35\n14 78 -62\n17 82 92\n-57 11 91\n23 15 92\n-80 -1 1\n12 39 18\n-23 -99 -75\n-34 50 19\n-39 84 -7\n45 -30 -39\n-60 49 37\n45 -16 -72\n33 -51 -56\n-48 28 5\n97 91 88\n45 -82 -11\n-21 -15 -90\n-53 73 -26\n-74 85 -90\n-40 23 38\n100 -13 49\n32 -100 -100\n0 -100 -70\n0 -100 0\n0 -100 0\n0 -100 0\n0 -100 0\n0 -37 0",
"output": "YES"
},
{
"input": "4\n68 3 100\n68 21 -100\n-100 -24 0\n-36 0 0",
"output": "YES"
},
{
"input": "33\n-1 -46 -12\n45 -16 -21\n-11 45 -21\n-60 -42 -93\n-22 -45 93\n37 96 85\n-76 26 83\n-4 9 55\n7 -52 -9\n66 8 -85\n-100 -54 11\n-29 59 74\n-24 12 2\n-56 81 85\n-92 69 -52\n-26 -97 91\n54 59 -51\n58 21 -57\n7 68 56\n-47 -20 -51\n-59 77 -13\n-85 27 91\n79 60 -56\n66 -80 5\n21 -99 42\n-31 -29 98\n66 93 76\n-49 45 61\n100 -100 -100\n100 -100 -100\n66 -75 -100\n0 0 -100\n0 0 -87",
"output": "YES"
},
{
"input": "3\n1 2 3\n3 2 1\n0 0 0",
"output": "NO"
},
{
"input": "2\n5 -23 12\n0 0 0",
"output": "NO"
},
{
"input": "1\n0 0 0",
"output": "YES"
},
{
"input": "1\n1 -2 0",
"output": "NO"
},
{
"input": "2\n-23 77 -86\n23 -77 86",
"output": "YES"
},
{
"input": "26\n86 7 20\n-57 -64 39\n-45 6 -93\n-44 -21 100\n-11 -49 21\n73 -71 -80\n-2 -89 56\n-65 -2 7\n5 14 84\n57 41 13\n-12 69 54\n40 -25 27\n-17 -59 0\n64 -91 -30\n-53 9 42\n-54 -8 14\n-35 82 27\n-48 -59 -80\n88 70 79\n94 57 97\n44 63 25\n84 -90 -40\n-100 100 -100\n-92 100 -100\n0 10 -100\n0 0 -82",
"output": "YES"
},
{
"input": "42\n11 27 92\n-18 -56 -57\n1 71 81\n33 -92 30\n82 83 49\n-87 -61 -1\n-49 45 49\n73 26 15\n-22 22 -77\n29 -93 87\n-68 44 -90\n-4 -84 20\n85 67 -6\n-39 26 77\n-28 -64 20\n65 -97 24\n-72 -39 51\n35 -75 -91\n39 -44 -8\n-25 -27 -57\n91 8 -46\n-98 -94 56\n94 -60 59\n-9 -95 18\n-53 -37 98\n-8 -94 -84\n-52 55 60\n15 -14 37\n65 -43 -25\n94 12 66\n-8 -19 -83\n29 81 -78\n-58 57 33\n24 86 -84\n-53 32 -88\n-14 7 3\n89 97 -53\n-5 -28 -91\n-100 100 -6\n-84 100 0\n0 100 0\n0 70 0",
"output": "YES"
},
{
"input": "3\n96 49 -12\n2 -66 28\n-98 17 -16",
"output": "YES"
},
{
"input": "5\n70 -46 86\n-100 94 24\n-27 63 -63\n57 -100 -47\n0 -11 0",
"output": "YES"
},
{
"input": "18\n-86 -28 70\n-31 -89 42\n31 -48 -55\n95 -17 -43\n24 -95 -85\n-21 -14 31\n68 -18 81\n13 31 60\n-15 28 99\n-42 15 9\n28 -61 -62\n-16 71 29\n-28 75 -48\n-77 -67 36\n-100 83 89\n100 100 -100\n57 34 -100\n0 0 -53",
"output": "YES"
},
{
"input": "44\n52 -54 -29\n-82 -5 -94\n-54 43 43\n91 16 71\n7 80 -91\n3 15 29\n-99 -6 -77\n-3 -77 -64\n73 67 34\n25 -10 -18\n-29 91 63\n-72 86 -16\n-68 85 -81\n-3 36 44\n-74 -14 -80\n34 -96 -97\n-76 -78 -33\n-24 44 -58\n98 12 77\n95 -63 -6\n-51 3 -90\n-92 -10 72\n7 3 -68\n57 -53 71\n29 57 -48\n35 -60 10\n79 -70 -61\n-20 77 55\n-86 -15 -35\n84 -88 -18\n100 -42 77\n-20 46 8\n-41 -43 -65\n38 -98 -23\n-100 65 45\n-7 -91 -63\n46 88 -85\n48 59 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 1",
"output": "YES"
},
{
"input": "18\n-14 -64 -91\n-8 -66 -86\n-23 92 -40\n6 -3 -53\n57 41 78\n-79 42 -22\n-88 -17 45\n4 -45 44\n83 -18 -25\n34 86 -92\n75 -30 12\n44 99 11\n-67 -13 72\n22 83 -56\n-37 71 72\n-9 -100 100\n0 -100 31\n0 -58 0",
"output": "YES"
},
{
"input": "23\n-70 37 78\n42 84 6\n28 -94 -24\n-49 76 95\n-67 18 84\n-53 78 -5\n65 -63 -64\n-66 -64 -69\n81 -93 95\n10 52 -79\n-89 -61 -64\n-64 -47 43\n-81 -35 55\n80 82 73\n-60 -81 -18\n49 -10 -19\n-58 70 12\n-24 -15 -93\n98 -93 -54\n-28 -75 11\n100 100 -63\n100 100 0\n56 34 0",
"output": "YES"
},
{
"input": "3\n0 2 -2\n1 -1 3\n-3 0 0",
"output": "NO"
}
] | 1,591,045,045 | 2,147,483,647 | Python 3 | OK | TESTS | 81 | 218 | 0 | no = int(input())
final_sum = 0
vectors = []
for i in range(no):
vector = list(map(int, input().split()))
vectors.append(vector)
x_count = y_count = z_count = 0
for vector in vectors:
x_count += vector[0]
y_count += vector[1]
z_count += vector[2]
if(x_count == y_count == z_count == 0):
print("YES")
else:
print("NO")
| Title: Young Physicist
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
Input Specification:
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100).
Output Specification:
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
Demo Input:
['3\n4 1 7\n-2 4 -1\n1 -5 -3\n', '3\n3 -1 7\n-5 2 -4\n2 -1 -3\n']
Demo Output:
['NO', 'YES']
Note:
none | ```python
no = int(input())
final_sum = 0
vectors = []
for i in range(no):
vector = list(map(int, input().split()))
vectors.append(vector)
x_count = y_count = z_count = 0
for vector in vectors:
x_count += vector[0]
y_count += vector[1]
z_count += vector[2]
if(x_count == y_count == z_count == 0):
print("YES")
else:
print("NO")
``` | 3.9455 |
610 | B | Vika and Squares | PROGRAMMING | 1,300 | [
"constructive algorithms",
"implementation"
] | null | null | Vika has *n* jars with paints of distinct colors. All the jars are numbered from 1 to *n* and the *i*-th jar contains *a**i* liters of paint of color *i*.
Vika also has an infinitely long rectangular piece of paper of width 1, consisting of squares of size 1<=×<=1. Squares are numbered 1, 2, 3 and so on. Vika decided that she will start painting squares one by one from left to right, starting from the square number 1 and some arbitrary color. If the square was painted in color *x*, then the next square will be painted in color *x*<=+<=1. In case of *x*<==<=*n*, next square is painted in color 1. If there is no more paint of the color Vika wants to use now, then she stops.
Square is always painted in only one color, and it takes exactly 1 liter of paint. Your task is to calculate the maximum number of squares that might be painted, if Vika chooses right color to paint the first square. | The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of jars with colors Vika has.
The second line of the input contains a sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109), where *a**i* is equal to the number of liters of paint in the *i*-th jar, i.e. the number of liters of color *i* that Vika has. | The only line of the output should contain a single integer — the maximum number of squares that Vika can paint if she follows the rules described above. | [
"5\n2 4 2 3 3\n",
"3\n5 5 5\n",
"6\n10 10 10 1 10 10\n"
] | [
"12\n",
"15\n",
"11\n"
] | In the first sample the best strategy is to start painting using color 4. Then the squares will be painted in the following colors (from left to right): 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5.
In the second sample Vika can start to paint using any color.
In the third sample Vika should start painting using color number 5. | 1,000 | [
{
"input": "5\n2 4 2 3 3",
"output": "12"
},
{
"input": "3\n5 5 5",
"output": "15"
},
{
"input": "6\n10 10 10 1 10 10",
"output": "11"
},
{
"input": "1\n167959139",
"output": "167959139"
},
{
"input": "10\n896619242 805194919 844752453 848347723 816995848 856813612 805194919 833406689 816255448 805194919",
"output": "8051949194"
},
{
"input": "2\n2 3",
"output": "5"
},
{
"input": "2\n10 10",
"output": "20"
},
{
"input": "2\n1111 1110",
"output": "2221"
},
{
"input": "3\n100 101 100",
"output": "301"
},
{
"input": "3\n100 100 101",
"output": "301"
},
{
"input": "3\n101 100 100",
"output": "301"
},
{
"input": "4\n2 3 2 2",
"output": "9"
},
{
"input": "4\n1 1 1 1",
"output": "4"
},
{
"input": "4\n100 100 100 101",
"output": "401"
},
{
"input": "4\n101 100 100 101",
"output": "402"
},
{
"input": "4\n100 101 101 100",
"output": "402"
},
{
"input": "4\n200 201 200 201",
"output": "801"
},
{
"input": "4\n201 200 201 200",
"output": "801"
},
{
"input": "4\n300 302 302 300",
"output": "1202"
},
{
"input": "8\n5 4 3 2 1 1 1 1",
"output": "12"
},
{
"input": "8\n2 2 1 2 2 1 2 2",
"output": "12"
},
{
"input": "10\n2 1 2 1 2 2 2 2 2 1",
"output": "15"
},
{
"input": "13\n2 2 2 1 1 1 1 1 1 1 2 2 2",
"output": "19"
}
] | 1,689,268,958 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 15 | 0 | n = int(input())
arr = []
for i in range(n):
arr.append(int(input()))
min_val = min(arr)
first = -1
last = -1
for i in range(n):
if first == -1 and arr[i] == min_val:
first = i
last = i
elif last != -1 and arr[i] == min_val:
last = i
ans = 0
for i in range(n):
if first <= i <= last:
ans += min_val
else:
ans += min_val + 1
print(ans)
| Title: Vika and Squares
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vika has *n* jars with paints of distinct colors. All the jars are numbered from 1 to *n* and the *i*-th jar contains *a**i* liters of paint of color *i*.
Vika also has an infinitely long rectangular piece of paper of width 1, consisting of squares of size 1<=×<=1. Squares are numbered 1, 2, 3 and so on. Vika decided that she will start painting squares one by one from left to right, starting from the square number 1 and some arbitrary color. If the square was painted in color *x*, then the next square will be painted in color *x*<=+<=1. In case of *x*<==<=*n*, next square is painted in color 1. If there is no more paint of the color Vika wants to use now, then she stops.
Square is always painted in only one color, and it takes exactly 1 liter of paint. Your task is to calculate the maximum number of squares that might be painted, if Vika chooses right color to paint the first square.
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of jars with colors Vika has.
The second line of the input contains a sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109), where *a**i* is equal to the number of liters of paint in the *i*-th jar, i.e. the number of liters of color *i* that Vika has.
Output Specification:
The only line of the output should contain a single integer — the maximum number of squares that Vika can paint if she follows the rules described above.
Demo Input:
['5\n2 4 2 3 3\n', '3\n5 5 5\n', '6\n10 10 10 1 10 10\n']
Demo Output:
['12\n', '15\n', '11\n']
Note:
In the first sample the best strategy is to start painting using color 4. Then the squares will be painted in the following colors (from left to right): 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5.
In the second sample Vika can start to paint using any color.
In the third sample Vika should start painting using color number 5. | ```python
n = int(input())
arr = []
for i in range(n):
arr.append(int(input()))
min_val = min(arr)
first = -1
last = -1
for i in range(n):
if first == -1 and arr[i] == min_val:
first = i
last = i
elif last != -1 and arr[i] == min_val:
last = i
ans = 0
for i in range(n):
if first <= i <= last:
ans += min_val
else:
ans += min_val + 1
print(ans)
``` | -1 |
|
80 | A | Panoramix's Prediction | PROGRAMMING | 800 | [
"brute force"
] | A. Panoramix's Prediction | 2 | 256 | A prime number is a number which has exactly two distinct divisors: one and itself. For example, numbers 2, 7, 3 are prime, and 1, 6, 4 are not.
The next prime number after *x* is the smallest prime number greater than *x*. For example, the next prime number after 2 is 3, and the next prime number after 3 is 5. Note that there is exactly one next prime number after each number. So 5 is not the next prime number for 2.
One cold April morning Panoramix predicted that soon Kakofonix will break free from his straitjacket, and this will be a black day for the residents of the Gallic countryside.
Panoramix's prophecy tells that if some day Asterix and Obelix beat exactly *x* Roman soldiers, where *x* is a prime number, and next day they beat exactly *y* Roman soldiers, where *y* is the next prime number after *x*, then it's time to wait for Armageddon, for nothing can shut Kakofonix up while he sings his infernal song.
Yesterday the Gauls beat *n* Roman soldiers and it turned out that the number *n* was prime! Today their victims were a troop of *m* Romans (*m*<=><=*n*). Determine whether the Gauls should wait for the black day after today's victory of Asterix and Obelix? | The first and only input line contains two positive integers — *n* and *m* (2<=≤<=*n*<=<<=*m*<=≤<=50). It is guaranteed that *n* is prime.
Pretests contain all the cases with restrictions 2<=≤<=*n*<=<<=*m*<=≤<=4. | Print YES, if *m* is the next prime number after *n*, or NO otherwise. | [
"3 5\n",
"7 11\n",
"7 9\n"
] | [
"YES",
"YES",
"NO"
] | none | 500 | [
{
"input": "3 5",
"output": "YES"
},
{
"input": "7 11",
"output": "YES"
},
{
"input": "7 9",
"output": "NO"
},
{
"input": "2 3",
"output": "YES"
},
{
"input": "2 4",
"output": "NO"
},
{
"input": "3 4",
"output": "NO"
},
{
"input": "3 5",
"output": "YES"
},
{
"input": "5 7",
"output": "YES"
},
{
"input": "7 11",
"output": "YES"
},
{
"input": "11 13",
"output": "YES"
},
{
"input": "13 17",
"output": "YES"
},
{
"input": "17 19",
"output": "YES"
},
{
"input": "19 23",
"output": "YES"
},
{
"input": "23 29",
"output": "YES"
},
{
"input": "29 31",
"output": "YES"
},
{
"input": "31 37",
"output": "YES"
},
{
"input": "37 41",
"output": "YES"
},
{
"input": "41 43",
"output": "YES"
},
{
"input": "43 47",
"output": "YES"
},
{
"input": "2 5",
"output": "NO"
},
{
"input": "2 7",
"output": "NO"
},
{
"input": "2 6",
"output": "NO"
},
{
"input": "2 11",
"output": "NO"
},
{
"input": "3 6",
"output": "NO"
},
{
"input": "3 7",
"output": "NO"
},
{
"input": "3 9",
"output": "NO"
},
{
"input": "5 6",
"output": "NO"
},
{
"input": "5 9",
"output": "NO"
},
{
"input": "5 11",
"output": "NO"
},
{
"input": "5 13",
"output": "NO"
},
{
"input": "5 15",
"output": "NO"
},
{
"input": "7 8",
"output": "NO"
},
{
"input": "7 13",
"output": "NO"
},
{
"input": "13 15",
"output": "NO"
},
{
"input": "13 17",
"output": "YES"
},
{
"input": "19 21",
"output": "NO"
},
{
"input": "13 20",
"output": "NO"
},
{
"input": "41 49",
"output": "NO"
},
{
"input": "43 49",
"output": "NO"
},
{
"input": "47 50",
"output": "NO"
},
{
"input": "47 49",
"output": "NO"
},
{
"input": "47 48",
"output": "NO"
},
{
"input": "23 25",
"output": "NO"
},
{
"input": "2 50",
"output": "NO"
},
{
"input": "31 33",
"output": "NO"
}
] | 1,690,273,542 | 2,147,483,647 | Python 3 | OK | TESTS | 45 | 92 | 0 | s, k = list(map(int, input().split()))
count, prime = 0, 0
for i in range(2, (k // 2) + 1):
if k % i == 0:
count += 1
if count > 0:
print("NO")
exit()
count, prime = 0, 0
for j in range(s, k + 1):
count = 0
for i in range(2, (j // 2) + 1):
if j % i == 0:
count += 1
if count == 0:
prime += 1
if prime == 2:
print("YES")
else:
print("NO")
| Title: Panoramix's Prediction
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
A prime number is a number which has exactly two distinct divisors: one and itself. For example, numbers 2, 7, 3 are prime, and 1, 6, 4 are not.
The next prime number after *x* is the smallest prime number greater than *x*. For example, the next prime number after 2 is 3, and the next prime number after 3 is 5. Note that there is exactly one next prime number after each number. So 5 is not the next prime number for 2.
One cold April morning Panoramix predicted that soon Kakofonix will break free from his straitjacket, and this will be a black day for the residents of the Gallic countryside.
Panoramix's prophecy tells that if some day Asterix and Obelix beat exactly *x* Roman soldiers, where *x* is a prime number, and next day they beat exactly *y* Roman soldiers, where *y* is the next prime number after *x*, then it's time to wait for Armageddon, for nothing can shut Kakofonix up while he sings his infernal song.
Yesterday the Gauls beat *n* Roman soldiers and it turned out that the number *n* was prime! Today their victims were a troop of *m* Romans (*m*<=><=*n*). Determine whether the Gauls should wait for the black day after today's victory of Asterix and Obelix?
Input Specification:
The first and only input line contains two positive integers — *n* and *m* (2<=≤<=*n*<=<<=*m*<=≤<=50). It is guaranteed that *n* is prime.
Pretests contain all the cases with restrictions 2<=≤<=*n*<=<<=*m*<=≤<=4.
Output Specification:
Print YES, if *m* is the next prime number after *n*, or NO otherwise.
Demo Input:
['3 5\n', '7 11\n', '7 9\n']
Demo Output:
['YES', 'YES', 'NO']
Note:
none | ```python
s, k = list(map(int, input().split()))
count, prime = 0, 0
for i in range(2, (k // 2) + 1):
if k % i == 0:
count += 1
if count > 0:
print("NO")
exit()
count, prime = 0, 0
for j in range(s, k + 1):
count = 0
for i in range(2, (j // 2) + 1):
if j % i == 0:
count += 1
if count == 0:
prime += 1
if prime == 2:
print("YES")
else:
print("NO")
``` | 3.977 |
855 | B | Marvolo Gaunt's Ring | PROGRAMMING | 1,500 | [
"brute force",
"data structures",
"dp"
] | null | null | Professor Dumbledore is helping Harry destroy the Horcruxes. He went to Gaunt Shack as he suspected a Horcrux to be present there. He saw Marvolo Gaunt's Ring and identified it as a Horcrux. Although he destroyed it, he is still affected by its curse. Professor Snape is helping Dumbledore remove the curse. For this, he wants to give Dumbledore exactly *x* drops of the potion he made.
Value of *x* is calculated as maximum of *p*·*a**i*<=+<=*q*·*a**j*<=+<=*r*·*a**k* for given *p*,<=*q*,<=*r* and array *a*1,<=*a*2,<=... *a**n* such that 1<=≤<=*i*<=≤<=*j*<=≤<=*k*<=≤<=*n*. Help Snape find the value of *x*. Do note that the value of *x* may be negative. | First line of input contains 4 integers *n*,<=*p*,<=*q*,<=*r* (<=-<=109<=≤<=*p*,<=*q*,<=*r*<=≤<=109,<=1<=≤<=*n*<=≤<=105).
Next line of input contains *n* space separated integers *a*1,<=*a*2,<=... *a**n* (<=-<=109<=≤<=*a**i*<=≤<=109). | Output a single integer the maximum value of *p*·*a**i*<=+<=*q*·*a**j*<=+<=*r*·*a**k* that can be obtained provided 1<=≤<=*i*<=≤<=*j*<=≤<=*k*<=≤<=*n*. | [
"5 1 2 3\n1 2 3 4 5\n",
"5 1 2 -3\n-1 -2 -3 -4 -5\n"
] | [
"30\n",
"12\n"
] | In the first sample case, we can take *i* = *j* = *k* = 5, thus making the answer as 1·5 + 2·5 + 3·5 = 30.
In second sample case, selecting *i* = *j* = 1 and *k* = 5 gives the answer 12. | 1,000 | [
{
"input": "5 1 2 3\n1 2 3 4 5",
"output": "30"
},
{
"input": "5 1 2 -3\n-1 -2 -3 -4 -5",
"output": "12"
},
{
"input": "5 886327859 82309257 -68295239\n-731225382 354766539 -48222231 -474691998 360965777",
"output": "376059240645059046"
},
{
"input": "4 -96405765 -495906217 625385006\n-509961652 392159235 -577128498 -744548876",
"output": "547306902373544674"
},
{
"input": "43 959134961 -868367850 142426380\n921743429 63959718 -797293233 122041422 -407576197 700139744 299598010 168207043 362252658 591926075 941946099 812263640 -76679927 -824267725 89529990 -73303355 83596189 -982699817 -235197848 654773327 125211479 -497091570 -2301804 203486596 -126652024 309810546 -581289415 -740125230 64425927 -501018049 304730559 34930193 -762964086 723645139 -826821494 495947907 816331024 9932423 -876541603 -782692568 322360800 841436938 40787162",
"output": "1876641179289775029"
},
{
"input": "1 0 0 0\n0",
"output": "0"
},
{
"input": "1 1000000000 1000000000 1000000000\n1000000000",
"output": "3000000000000000000"
},
{
"input": "1 -1000000000 -1000000000 1000000000\n1000000000",
"output": "-1000000000000000000"
},
{
"input": "1 -1000000000 -1000000000 -1000000000\n1000000000",
"output": "-3000000000000000000"
},
{
"input": "3 1000000000 1000000000 1000000000\n-1000000000 -1000000000 -1000000000",
"output": "-3000000000000000000"
},
{
"input": "1 1 1 1\n-1",
"output": "-3"
},
{
"input": "1 -1 -1 -1\n1",
"output": "-3"
},
{
"input": "1 1000000000 1000000000 1000000000\n-1000000000",
"output": "-3000000000000000000"
},
{
"input": "1 1 2 3\n-1",
"output": "-6"
},
{
"input": "3 -1000000000 -1000000000 -1000000000\n1000000000 1000000000 1000000000",
"output": "-3000000000000000000"
},
{
"input": "2 -1000000000 -1000000000 -1000000000\n1000000000 1000000000",
"output": "-3000000000000000000"
},
{
"input": "3 1 1 1\n-1 -1 -1",
"output": "-3"
},
{
"input": "1 -1000000000 0 0\n1000000000",
"output": "-1000000000000000000"
},
{
"input": "1 -100 -100 -100\n100",
"output": "-30000"
},
{
"input": "5 -1000000000 -1000000000 -1000000000\n1000000000 1000000000 1000000000 1000000000 1000000000",
"output": "-3000000000000000000"
},
{
"input": "1 999999999 999999999 999999999\n-999999999",
"output": "-2999999994000000003"
},
{
"input": "3 -1000000000 -1000000000 1\n1000000000 1000000000 1000000000",
"output": "-1999999999000000000"
},
{
"input": "3 -2 3 -2\n1 2 1",
"output": "2"
},
{
"input": "2 1 -1 1\n1 -1",
"output": "1"
},
{
"input": "1 -1000000000 1 -1000000000\n1000000000",
"output": "-1999999999000000000"
},
{
"input": "1 1000000000 1000000000 -1000000000\n-1000000000",
"output": "-1000000000000000000"
},
{
"input": "1 -1000000000 -1000000000 0\n1000000000",
"output": "-2000000000000000000"
}
] | 1,653,416,375 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 30 | 0 | max_integer, map1 = max, map
n, p, q, r = map1(int, input().strip().split())
list1 = [int(x) for x in input().strip().split()]
prefix_max, suffix_max = [0]*n, [0]*n, float('-inf')
prefix_max[0] = p*list1[0]
for i in range(1, n):
prefix_max[i] = max_integer(prefix_max[i-1], p*list1[i])
suffix_max[n-1] = r*list1[n-1]
for i in range(n-2, -1, -1):
suffix_max[i] = max_integer(suffix_max[i+1], r*list1[i])
for i in range(n):
result = max_integer(result, (prefix_max[i] + (q*list1[i]) + suffix_max[i]))
print(f"{result}") | Title: Marvolo Gaunt's Ring
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Professor Dumbledore is helping Harry destroy the Horcruxes. He went to Gaunt Shack as he suspected a Horcrux to be present there. He saw Marvolo Gaunt's Ring and identified it as a Horcrux. Although he destroyed it, he is still affected by its curse. Professor Snape is helping Dumbledore remove the curse. For this, he wants to give Dumbledore exactly *x* drops of the potion he made.
Value of *x* is calculated as maximum of *p*·*a**i*<=+<=*q*·*a**j*<=+<=*r*·*a**k* for given *p*,<=*q*,<=*r* and array *a*1,<=*a*2,<=... *a**n* such that 1<=≤<=*i*<=≤<=*j*<=≤<=*k*<=≤<=*n*. Help Snape find the value of *x*. Do note that the value of *x* may be negative.
Input Specification:
First line of input contains 4 integers *n*,<=*p*,<=*q*,<=*r* (<=-<=109<=≤<=*p*,<=*q*,<=*r*<=≤<=109,<=1<=≤<=*n*<=≤<=105).
Next line of input contains *n* space separated integers *a*1,<=*a*2,<=... *a**n* (<=-<=109<=≤<=*a**i*<=≤<=109).
Output Specification:
Output a single integer the maximum value of *p*·*a**i*<=+<=*q*·*a**j*<=+<=*r*·*a**k* that can be obtained provided 1<=≤<=*i*<=≤<=*j*<=≤<=*k*<=≤<=*n*.
Demo Input:
['5 1 2 3\n1 2 3 4 5\n', '5 1 2 -3\n-1 -2 -3 -4 -5\n']
Demo Output:
['30\n', '12\n']
Note:
In the first sample case, we can take *i* = *j* = *k* = 5, thus making the answer as 1·5 + 2·5 + 3·5 = 30.
In second sample case, selecting *i* = *j* = 1 and *k* = 5 gives the answer 12. | ```python
max_integer, map1 = max, map
n, p, q, r = map1(int, input().strip().split())
list1 = [int(x) for x in input().strip().split()]
prefix_max, suffix_max = [0]*n, [0]*n, float('-inf')
prefix_max[0] = p*list1[0]
for i in range(1, n):
prefix_max[i] = max_integer(prefix_max[i-1], p*list1[i])
suffix_max[n-1] = r*list1[n-1]
for i in range(n-2, -1, -1):
suffix_max[i] = max_integer(suffix_max[i+1], r*list1[i])
for i in range(n):
result = max_integer(result, (prefix_max[i] + (q*list1[i]) + suffix_max[i]))
print(f"{result}")
``` | -1 |
|
976 | E | Well played! | PROGRAMMING | 2,100 | [
"greedy",
"sortings"
] | null | null | Recently Max has got himself into popular CCG "BrainStone". As "BrainStone" is a pretty intellectual game, Max has to solve numerous hard problems during the gameplay. Here is one of them:
Max owns *n* creatures, *i*-th of them can be described with two numbers — its health *hp**i* and its damage *dmg**i*. Max also has two types of spells in stock:
1. Doubles health of the creature (*hp**i* := *hp**i*·2); 1. Assigns value of health of the creature to its damage (*dmg**i* := *hp**i*).
Spell of first type can be used no more than *a* times in total, of the second type — no more than *b* times in total. Spell can be used on a certain creature multiple times. Spells can be used in arbitrary order. It isn't necessary to use all the spells.
Max is really busy preparing for his final exams, so he asks you to determine what is the maximal total damage of all creatures he can achieve if he uses spells in most optimal way. | The first line contains three integers *n*, *a*, *b* (1<=≤<=*n*<=≤<=2·105, 0<=≤<=*a*<=≤<=20, 0<=≤<=*b*<=≤<=2·105) — the number of creatures, spells of the first type and spells of the second type, respectively.
The *i*-th of the next *n* lines contain two number *hp**i* and *dmg**i* (1<=≤<=*hp**i*,<=*dmg**i*<=≤<=109) — description of the *i*-th creature. | Print single integer — maximum total damage creatures can deal. | [
"2 1 1\n10 15\n6 1\n",
"3 0 3\n10 8\n7 11\n5 2\n"
] | [
"27\n",
"26\n"
] | In the first example Max should use the spell of the first type on the second creature, then the spell of the second type on the same creature. Then total damage will be equal to 15 + 6·2 = 27.
In the second example Max should use the spell of the second type on the first creature, then the spell of the second type on the third creature. Total damage will be equal to 10 + 11 + 5 = 26. | 0 | [
{
"input": "2 1 1\n10 15\n6 1",
"output": "27"
},
{
"input": "3 0 3\n10 8\n7 11\n5 2",
"output": "26"
},
{
"input": "1 0 0\n2 1",
"output": "1"
},
{
"input": "1 0 200000\n1 2",
"output": "2"
},
{
"input": "7 5 7\n29 25\n84 28\n34 34\n14 76\n85 9\n40 57\n99 88",
"output": "3533"
},
{
"input": "7 6 7\n11 75\n61 90\n22 14\n100 36\n29 48\n69 52\n16 3",
"output": "6720"
},
{
"input": "7 8 7\n88 29\n30 44\n14 1\n83 95\n73 88\n10 42\n29 26",
"output": "22840"
},
{
"input": "12 7 7\n78 189\n614 271\n981 510\n37 762\n803 106\n78 369\n787 54\n768 159\n238 111\n107 54\n207 72\n485 593",
"output": "130952"
},
{
"input": "12 20 4\n852 935\n583 820\n969 197\n219 918\n547 842\n615 163\n704 377\n326 482\n183 833\n884 994\n886 581\n909 450",
"output": "1016078777"
},
{
"input": "2 13 2\n208637 682633\n393097 724045",
"output": "3220933257"
},
{
"input": "1 0 200000\n42 1",
"output": "42"
},
{
"input": "1 6 200000\n42 1",
"output": "2688"
},
{
"input": "1 0 200000\n1 42",
"output": "42"
},
{
"input": "1 6 200000\n1 42",
"output": "64"
},
{
"input": "3 1 1\n10 9\n8 6\n7 5",
"output": "31"
},
{
"input": "1 1 0\n10 1",
"output": "1"
},
{
"input": "1 1 0\n3 4",
"output": "4"
},
{
"input": "3 20 0\n1 5\n5 1\n5 1",
"output": "7"
},
{
"input": "2 5 1\n10 1\n20 20",
"output": "641"
},
{
"input": "3 20 0\n3 2\n4 3\n5 4",
"output": "9"
},
{
"input": "2 1 0\n10 15\n6 1",
"output": "16"
},
{
"input": "5 10 0\n20 1\n22 1\n30 1\n30 5\n40 6",
"output": "14"
},
{
"input": "1 20 0\n1 5",
"output": "5"
},
{
"input": "2 3 14\n28 5\n32 47",
"output": "284"
},
{
"input": "3 1 2\n20 10\n5 1\n25 25",
"output": "71"
},
{
"input": "2 3 3\n28 5\n32 47",
"output": "284"
},
{
"input": "2 2 1\n10 15\n6 1",
"output": "41"
},
{
"input": "2 1 2\n20 1\n22 23",
"output": "64"
},
{
"input": "10 7 2\n8 6\n5 5\n3 7\n7 7\n3 8\n6 1\n10 9\n4 6\n9 5\n7 9",
"output": "1339"
},
{
"input": "3 8 1\n6 6\n7 9\n2 5",
"output": "1803"
},
{
"input": "10 4 4\n5 5\n8 1\n10 10\n3 1\n7 10\n1 7\n8 7\n5 9\n3 3\n1 1",
"output": "214"
},
{
"input": "4 8 3\n1 6\n10 10\n4 8\n9 4",
"output": "2583"
},
{
"input": "8 18 1\n8 6\n6 8\n1 7\n7 2\n5 1\n10 5\n8 3\n9 3",
"output": "2621470"
},
{
"input": "2 11 1\n1 4\n1 5",
"output": "2053"
},
{
"input": "2 19 2\n9 3\n7 2",
"output": "4718599"
},
{
"input": "5 13 0\n4 4\n8 10\n1 8\n3 9\n4 6",
"output": "37"
},
{
"input": "5 8 0\n10 7\n6 6\n6 5\n7 9\n10 7",
"output": "34"
},
{
"input": "5 20 2\n1 10\n7 8\n10 1\n6 5\n2 1",
"output": "10485785"
},
{
"input": "2 1 0\n5 6\n8 8",
"output": "14"
},
{
"input": "7 3 5\n2 6\n5 9\n5 5\n4 10\n5 7\n7 8\n3 10",
"output": "103"
},
{
"input": "10 9 0\n620118469 704168608\n528098892 341451371\n15150469 449838744\n960504540 722185004\n271837337 344050133\n940943201 419522619\n85569623 788965215\n161962866 563795701\n943389281 445744350\n610994199 473866838",
"output": "5253588583"
},
{
"input": "10 11 1\n7 3\n9 4\n1 5\n10 3\n6 1\n10 7\n8 5\n7 6\n1 4\n9 9",
"output": "20524"
},
{
"input": "2 1 200000\n44 42\n1000 1001",
"output": "2044"
},
{
"input": "5 12 2\n5 9\n8 9\n4 1\n2 5\n1 8",
"output": "32794"
},
{
"input": "4 8 1\n9 9\n7 6\n2 4\n6 10",
"output": "2324"
},
{
"input": "2 1 1\n292725479 742903381\n239450691 307356865",
"output": "1221804763"
}
] | 1,525,152,943 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 2 | 93 | 7,065,600 | import sys
if locals()['__file__'][-2:] == 'py':
sys.stdin = open('in.txt', 'r')
# rd = lambda: map(int, input().split())
# n, a, b = rd()
# p = []
# b = min(b, n)
# s = 0
# ans = 0
# for i in range(n):
# hp, dmg = rd()
# p.append([max(hp - dmg, 0), hp, dmg])
# s += dmg
# if b:
# p.sort(reverse=True)
# t = 0
# for i in range(b):
# t += p[i][0]
# for i in range(b):
# ans = max(ans, t - p[i][0] + (p[i][1] << a) - p[i][2])
# for i in range(b, n):
# ans = max(ans, t - p[b - 1][0] + (p[i][1] << a) - p[i][2])
# print(ans + s)
rd = lambda: list(map(int, input().split()))
n, a, b = rd()
b = min(b, n)
f = lambda x: max(0, x[0] - x[1])
g = lambda x: (x[0] << a) - x[1]
p = [rd() for i in range(n)]
s = sum(x[1] for x in p)
if b:
p.sort(key=f, reverse=True)
e = 0
t = sum(f(x) for x in p[:b])
for i in range(b):
e = max(e, t - f(p[i]) + g(p[i]))
for i in range(b, n):
e = max(e, t - f(p[b - 1]) + g(p[i]))
print(s + e)
# python3
# from sys import stdin
# def main():
# def parseints(line): return tuple(map(int, line.split()))
# def bdiff(creature): return max(0, creature[0] - creature[1])
# n, a, b = parseints(input())
# hand = list(map(parseints, stdin.readlines()))
# ans = sum(creature[1] for creature in hand) # default damage
# if b:
# hand.sort(key=bdiff)
# best = 0
# if n > b:
# lost = bdiff(hand[n - b])
# for creature in hand[:n-b]:
# best = max(best, (creature[0] << a) - creature[1] - lost)
# for creature in hand[max(0,n-b):]:
# best = max(best, (creature[0] << a) - max(creature))
# ans += bdiff(creature)
# ans += best
# print(ans)
# main() | Title: Well played!
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Recently Max has got himself into popular CCG "BrainStone". As "BrainStone" is a pretty intellectual game, Max has to solve numerous hard problems during the gameplay. Here is one of them:
Max owns *n* creatures, *i*-th of them can be described with two numbers — its health *hp**i* and its damage *dmg**i*. Max also has two types of spells in stock:
1. Doubles health of the creature (*hp**i* := *hp**i*·2); 1. Assigns value of health of the creature to its damage (*dmg**i* := *hp**i*).
Spell of first type can be used no more than *a* times in total, of the second type — no more than *b* times in total. Spell can be used on a certain creature multiple times. Spells can be used in arbitrary order. It isn't necessary to use all the spells.
Max is really busy preparing for his final exams, so he asks you to determine what is the maximal total damage of all creatures he can achieve if he uses spells in most optimal way.
Input Specification:
The first line contains three integers *n*, *a*, *b* (1<=≤<=*n*<=≤<=2·105, 0<=≤<=*a*<=≤<=20, 0<=≤<=*b*<=≤<=2·105) — the number of creatures, spells of the first type and spells of the second type, respectively.
The *i*-th of the next *n* lines contain two number *hp**i* and *dmg**i* (1<=≤<=*hp**i*,<=*dmg**i*<=≤<=109) — description of the *i*-th creature.
Output Specification:
Print single integer — maximum total damage creatures can deal.
Demo Input:
['2 1 1\n10 15\n6 1\n', '3 0 3\n10 8\n7 11\n5 2\n']
Demo Output:
['27\n', '26\n']
Note:
In the first example Max should use the spell of the first type on the second creature, then the spell of the second type on the same creature. Then total damage will be equal to 15 + 6·2 = 27.
In the second example Max should use the spell of the second type on the first creature, then the spell of the second type on the third creature. Total damage will be equal to 10 + 11 + 5 = 26. | ```python
import sys
if locals()['__file__'][-2:] == 'py':
sys.stdin = open('in.txt', 'r')
# rd = lambda: map(int, input().split())
# n, a, b = rd()
# p = []
# b = min(b, n)
# s = 0
# ans = 0
# for i in range(n):
# hp, dmg = rd()
# p.append([max(hp - dmg, 0), hp, dmg])
# s += dmg
# if b:
# p.sort(reverse=True)
# t = 0
# for i in range(b):
# t += p[i][0]
# for i in range(b):
# ans = max(ans, t - p[i][0] + (p[i][1] << a) - p[i][2])
# for i in range(b, n):
# ans = max(ans, t - p[b - 1][0] + (p[i][1] << a) - p[i][2])
# print(ans + s)
rd = lambda: list(map(int, input().split()))
n, a, b = rd()
b = min(b, n)
f = lambda x: max(0, x[0] - x[1])
g = lambda x: (x[0] << a) - x[1]
p = [rd() for i in range(n)]
s = sum(x[1] for x in p)
if b:
p.sort(key=f, reverse=True)
e = 0
t = sum(f(x) for x in p[:b])
for i in range(b):
e = max(e, t - f(p[i]) + g(p[i]))
for i in range(b, n):
e = max(e, t - f(p[b - 1]) + g(p[i]))
print(s + e)
# python3
# from sys import stdin
# def main():
# def parseints(line): return tuple(map(int, line.split()))
# def bdiff(creature): return max(0, creature[0] - creature[1])
# n, a, b = parseints(input())
# hand = list(map(parseints, stdin.readlines()))
# ans = sum(creature[1] for creature in hand) # default damage
# if b:
# hand.sort(key=bdiff)
# best = 0
# if n > b:
# lost = bdiff(hand[n - b])
# for creature in hand[:n-b]:
# best = max(best, (creature[0] << a) - creature[1] - lost)
# for creature in hand[max(0,n-b):]:
# best = max(best, (creature[0] << a) - max(creature))
# ans += bdiff(creature)
# ans += best
# print(ans)
# main()
``` | -1 |
|
614 | A | Link/Cut Tree | PROGRAMMING | 1,500 | [
"brute force",
"implementation"
] | null | null | Programmer Rostislav got seriously interested in the Link/Cut Tree data structure, which is based on Splay trees. Specifically, he is now studying the *expose* procedure.
Unfortunately, Rostislav is unable to understand the definition of this procedure, so he decided to ask programmer Serezha to help him. Serezha agreed to help if Rostislav solves a simple task (and if he doesn't, then why would he need Splay trees anyway?)
Given integers *l*, *r* and *k*, you need to print all powers of number *k* within range from *l* to *r* inclusive. However, Rostislav doesn't want to spent time doing this, as he got interested in playing a network game called Agar with Gleb. Help him! | The first line of the input contains three space-separated integers *l*, *r* and *k* (1<=≤<=*l*<=≤<=*r*<=≤<=1018, 2<=≤<=*k*<=≤<=109). | Print all powers of number *k*, that lie within range from *l* to *r* in the increasing order. If there are no such numbers, print "-1" (without the quotes). | [
"1 10 2\n",
"2 4 5\n"
] | [
"1 2 4 8 ",
"-1"
] | Note to the first sample: numbers 2<sup class="upper-index">0</sup> = 1, 2<sup class="upper-index">1</sup> = 2, 2<sup class="upper-index">2</sup> = 4, 2<sup class="upper-index">3</sup> = 8 lie within the specified range. The number 2<sup class="upper-index">4</sup> = 16 is greater then 10, thus it shouldn't be printed. | 500 | [
{
"input": "1 10 2",
"output": "1 2 4 8 "
},
{
"input": "2 4 5",
"output": "-1"
},
{
"input": "18102 43332383920 28554",
"output": "28554 815330916 "
},
{
"input": "19562 31702689720 17701",
"output": "313325401 "
},
{
"input": "11729 55221128400 313",
"output": "97969 30664297 9597924961 "
},
{
"input": "5482 100347128000 342",
"output": "116964 40001688 13680577296 "
},
{
"input": "3680 37745933600 10",
"output": "10000 100000 1000000 10000000 100000000 1000000000 10000000000 "
},
{
"input": "17098 191120104800 43",
"output": "79507 3418801 147008443 6321363049 "
},
{
"input": "10462 418807699200 2",
"output": "16384 32768 65536 131072 262144 524288 1048576 2097152 4194304 8388608 16777216 33554432 67108864 134217728 268435456 536870912 1073741824 2147483648 4294967296 8589934592 17179869184 34359738368 68719476736 137438953472 274877906944 "
},
{
"input": "30061 641846400000 3",
"output": "59049 177147 531441 1594323 4782969 14348907 43046721 129140163 387420489 1162261467 3486784401 10460353203 31381059609 94143178827 282429536481 "
},
{
"input": "1 1000000000000000000 2",
"output": "1 2 4 8 16 32 64 128 256 512 1024 2048 4096 8192 16384 32768 65536 131072 262144 524288 1048576 2097152 4194304 8388608 16777216 33554432 67108864 134217728 268435456 536870912 1073741824 2147483648 4294967296 8589934592 17179869184 34359738368 68719476736 137438953472 274877906944 549755813888 1099511627776 2199023255552 4398046511104 8796093022208 17592186044416 35184372088832 70368744177664 140737488355328 281474976710656 562949953421312 1125899906842624 2251799813685248 4503599627370496 900719925474099..."
},
{
"input": "32 2498039712000 4",
"output": "64 256 1024 4096 16384 65536 262144 1048576 4194304 16777216 67108864 268435456 1073741824 4294967296 17179869184 68719476736 274877906944 1099511627776 "
},
{
"input": "1 2576683920000 2",
"output": "1 2 4 8 16 32 64 128 256 512 1024 2048 4096 8192 16384 32768 65536 131072 262144 524288 1048576 2097152 4194304 8388608 16777216 33554432 67108864 134217728 268435456 536870912 1073741824 2147483648 4294967296 8589934592 17179869184 34359738368 68719476736 137438953472 274877906944 549755813888 1099511627776 2199023255552 "
},
{
"input": "5 25 5",
"output": "5 25 "
},
{
"input": "1 90 90",
"output": "1 90 "
},
{
"input": "95 2200128528000 68",
"output": "4624 314432 21381376 1453933568 98867482624 "
},
{
"input": "64 426314644000 53",
"output": "2809 148877 7890481 418195493 22164361129 "
},
{
"input": "198765 198765 198765",
"output": "198765 "
},
{
"input": "42 2845016496000 12",
"output": "144 1728 20736 248832 2985984 35831808 429981696 5159780352 61917364224 743008370688 "
},
{
"input": "6 6 3",
"output": "-1"
},
{
"input": "1 10 11",
"output": "1 "
},
{
"input": "2 10 11",
"output": "-1"
},
{
"input": "87 160 41",
"output": "-1"
},
{
"input": "237171123124584251 923523399718980912 7150",
"output": "-1"
},
{
"input": "101021572000739548 453766043506276015 8898",
"output": "-1"
},
{
"input": "366070689449360724 928290634811046396 8230",
"output": "-1"
},
{
"input": "438133886369772308 942612870269666780 7193",
"output": "-1"
},
{
"input": "10 10 10",
"output": "10 "
},
{
"input": "16 16 256",
"output": "-1"
},
{
"input": "1 1000000000000000000 1000000000",
"output": "1 1000000000 1000000000000000000 "
},
{
"input": "1000000000000000000 1000000000000000000 1000000000",
"output": "1000000000000000000 "
},
{
"input": "1000000000 1000000000000000000 1000000000",
"output": "1000000000 1000000000000000000 "
},
{
"input": "1 1 4",
"output": "1 "
},
{
"input": "1 999999999999999999 1000000000",
"output": "1 1000000000 "
},
{
"input": "1 1000000000000000000 999999990",
"output": "1 999999990 999999980000000100 "
},
{
"input": "1 1000000000000000000 999999984",
"output": "1 999999984 999999968000000256 "
},
{
"input": "1 1000000000000000000 324325",
"output": "1 324325 105186705625 34114678301828125 "
},
{
"input": "1 1000000000000000000 999999523",
"output": "1 999999523 999999046000227529 "
},
{
"input": "1 243 3",
"output": "1 3 9 27 81 243 "
},
{
"input": "62769392426654367 567152589733560993 688813",
"output": "326816522793383797 "
},
{
"input": "1 1000000000000000000 690852001",
"output": "1 690852001 477276487285704001 "
},
{
"input": "1 1000000000000000000 918745157",
"output": "1 918745157 844092663510954649 "
},
{
"input": "1 1000000000000000000 131299843",
"output": "1 131299843 17239648771824649 "
},
{
"input": "2861381721051425 2861381721051425 1234",
"output": "-1"
}
] | 1,489,423,409 | 2,147,483,647 | Python 3 | OK | TESTS | 44 | 62 | 4,608,000 | l,r,k=map(int,input().split())
f=0
t=1
while (t<=r):
if(t>=l):
f=1
print(t)
t*=k;
if (not(f)):
print(-1) | Title: Link/Cut Tree
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Programmer Rostislav got seriously interested in the Link/Cut Tree data structure, which is based on Splay trees. Specifically, he is now studying the *expose* procedure.
Unfortunately, Rostislav is unable to understand the definition of this procedure, so he decided to ask programmer Serezha to help him. Serezha agreed to help if Rostislav solves a simple task (and if he doesn't, then why would he need Splay trees anyway?)
Given integers *l*, *r* and *k*, you need to print all powers of number *k* within range from *l* to *r* inclusive. However, Rostislav doesn't want to spent time doing this, as he got interested in playing a network game called Agar with Gleb. Help him!
Input Specification:
The first line of the input contains three space-separated integers *l*, *r* and *k* (1<=≤<=*l*<=≤<=*r*<=≤<=1018, 2<=≤<=*k*<=≤<=109).
Output Specification:
Print all powers of number *k*, that lie within range from *l* to *r* in the increasing order. If there are no such numbers, print "-1" (without the quotes).
Demo Input:
['1 10 2\n', '2 4 5\n']
Demo Output:
['1 2 4 8 ', '-1']
Note:
Note to the first sample: numbers 2<sup class="upper-index">0</sup> = 1, 2<sup class="upper-index">1</sup> = 2, 2<sup class="upper-index">2</sup> = 4, 2<sup class="upper-index">3</sup> = 8 lie within the specified range. The number 2<sup class="upper-index">4</sup> = 16 is greater then 10, thus it shouldn't be printed. | ```python
l,r,k=map(int,input().split())
f=0
t=1
while (t<=r):
if(t>=l):
f=1
print(t)
t*=k;
if (not(f)):
print(-1)
``` | 3 |
|
116 | A | Tram | PROGRAMMING | 800 | [
"implementation"
] | null | null | Linear Kingdom has exactly one tram line. It has *n* stops, numbered from 1 to *n* in the order of tram's movement. At the *i*-th stop *a**i* passengers exit the tram, while *b**i* passengers enter it. The tram is empty before it arrives at the first stop. Also, when the tram arrives at the last stop, all passengers exit so that it becomes empty.
Your task is to calculate the tram's minimum capacity such that the number of people inside the tram at any time never exceeds this capacity. Note that at each stop all exiting passengers exit before any entering passenger enters the tram. | The first line contains a single number *n* (2<=≤<=*n*<=≤<=1000) — the number of the tram's stops.
Then *n* lines follow, each contains two integers *a**i* and *b**i* (0<=≤<=*a**i*,<=*b**i*<=≤<=1000) — the number of passengers that exits the tram at the *i*-th stop, and the number of passengers that enter the tram at the *i*-th stop. The stops are given from the first to the last stop in the order of tram's movement.
- The number of people who exit at a given stop does not exceed the total number of people in the tram immediately before it arrives at the stop. More formally, . This particularly means that *a*1<==<=0. - At the last stop, all the passengers exit the tram and it becomes empty. More formally, . - No passenger will enter the train at the last stop. That is, *b**n*<==<=0. | Print a single integer denoting the minimum possible capacity of the tram (0 is allowed). | [
"4\n0 3\n2 5\n4 2\n4 0\n"
] | [
"6\n"
] | For the first example, a capacity of 6 is sufficient:
- At the first stop, the number of passengers inside the tram before arriving is 0. Then, 3 passengers enter the tram, and the number of passengers inside the tram becomes 3. - At the second stop, 2 passengers exit the tram (1 passenger remains inside). Then, 5 passengers enter the tram. There are 6 passengers inside the tram now. - At the third stop, 4 passengers exit the tram (2 passengers remain inside). Then, 2 passengers enter the tram. There are 4 passengers inside the tram now. - Finally, all the remaining passengers inside the tram exit the tram at the last stop. There are no passenger inside the tram now, which is in line with the constraints.
Since the number of passengers inside the tram never exceeds 6, a capacity of 6 is sufficient. Furthermore it is not possible for the tram to have a capacity less than 6. Hence, 6 is the correct answer. | 500 | [
{
"input": "4\n0 3\n2 5\n4 2\n4 0",
"output": "6"
},
{
"input": "5\n0 4\n4 6\n6 5\n5 4\n4 0",
"output": "6"
},
{
"input": "10\n0 5\n1 7\n10 8\n5 3\n0 5\n3 3\n8 8\n0 6\n10 1\n9 0",
"output": "18"
},
{
"input": "3\n0 1\n1 1\n1 0",
"output": "1"
},
{
"input": "4\n0 1\n0 1\n1 0\n1 0",
"output": "2"
},
{
"input": "3\n0 0\n0 0\n0 0",
"output": "0"
},
{
"input": "3\n0 1000\n1000 1000\n1000 0",
"output": "1000"
},
{
"input": "5\n0 73\n73 189\n189 766\n766 0\n0 0",
"output": "766"
},
{
"input": "5\n0 0\n0 0\n0 0\n0 1\n1 0",
"output": "1"
},
{
"input": "5\n0 917\n917 923\n904 992\n1000 0\n11 0",
"output": "1011"
},
{
"input": "5\n0 1\n1 2\n2 1\n1 2\n2 0",
"output": "2"
},
{
"input": "5\n0 0\n0 0\n0 0\n0 0\n0 0",
"output": "0"
},
{
"input": "20\n0 7\n2 1\n2 2\n5 7\n2 6\n6 10\n2 4\n0 4\n7 4\n8 0\n10 6\n2 1\n6 1\n1 7\n0 3\n8 7\n6 3\n6 3\n1 1\n3 0",
"output": "22"
},
{
"input": "5\n0 1000\n1000 1000\n1000 1000\n1000 1000\n1000 0",
"output": "1000"
},
{
"input": "10\n0 592\n258 598\n389 203\n249 836\n196 635\n478 482\n994 987\n1000 0\n769 0\n0 0",
"output": "1776"
},
{
"input": "10\n0 1\n1 0\n0 0\n0 0\n0 0\n0 1\n1 1\n0 1\n1 0\n1 0",
"output": "2"
},
{
"input": "10\n0 926\n926 938\n938 931\n931 964\n937 989\n983 936\n908 949\n997 932\n945 988\n988 0",
"output": "1016"
},
{
"input": "10\n0 1\n1 2\n1 2\n2 2\n2 2\n2 2\n1 1\n1 1\n2 1\n2 0",
"output": "3"
},
{
"input": "10\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0",
"output": "0"
},
{
"input": "10\n0 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 0",
"output": "1000"
},
{
"input": "50\n0 332\n332 268\n268 56\n56 711\n420 180\n160 834\n149 341\n373 777\n763 93\n994 407\n86 803\n700 132\n471 608\n429 467\n75 5\n638 305\n405 853\n316 478\n643 163\n18 131\n648 241\n241 766\n316 847\n640 380\n923 759\n789 41\n125 421\n421 9\n9 388\n388 829\n408 108\n462 856\n816 411\n518 688\n290 7\n405 912\n397 772\n396 652\n394 146\n27 648\n462 617\n514 433\n780 35\n710 705\n460 390\n194 508\n643 56\n172 469\n1000 0\n194 0",
"output": "2071"
},
{
"input": "50\n0 0\n0 1\n1 1\n0 1\n0 0\n1 0\n0 0\n1 0\n0 0\n0 0\n0 0\n0 0\n0 1\n0 0\n0 0\n0 1\n1 0\n0 1\n0 0\n1 1\n1 0\n0 1\n0 0\n1 1\n0 1\n1 0\n1 1\n1 0\n0 0\n1 1\n1 0\n0 1\n0 0\n0 1\n1 1\n1 1\n1 1\n1 0\n1 1\n1 0\n0 1\n1 0\n0 0\n0 1\n1 1\n1 1\n0 1\n0 0\n1 0\n1 0",
"output": "3"
},
{
"input": "50\n0 926\n926 971\n915 980\n920 965\n954 944\n928 952\n955 980\n916 980\n906 935\n944 913\n905 923\n912 922\n965 934\n912 900\n946 930\n931 983\n979 905\n925 969\n924 926\n910 914\n921 977\n934 979\n962 986\n942 909\n976 903\n982 982\n991 941\n954 929\n902 980\n947 983\n919 924\n917 943\n916 905\n907 913\n964 977\n984 904\n905 999\n950 970\n986 906\n993 970\n960 994\n963 983\n918 986\n980 900\n931 986\n993 997\n941 909\n907 909\n1000 0\n278 0",
"output": "1329"
},
{
"input": "2\n0 863\n863 0",
"output": "863"
},
{
"input": "50\n0 1\n1 2\n2 2\n1 1\n1 1\n1 2\n1 2\n1 1\n1 2\n1 1\n1 1\n1 2\n1 2\n1 1\n2 1\n2 2\n1 2\n2 2\n1 2\n2 1\n2 1\n2 2\n2 1\n1 2\n1 2\n2 1\n1 1\n2 2\n1 1\n2 1\n2 2\n2 1\n1 2\n2 2\n1 2\n1 1\n1 1\n2 1\n2 1\n2 2\n2 1\n2 1\n1 2\n1 2\n1 2\n1 2\n2 0\n2 0\n2 0\n0 0",
"output": "8"
},
{
"input": "50\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0",
"output": "0"
},
{
"input": "100\n0 1\n0 0\n0 0\n1 0\n0 0\n0 1\n0 1\n1 1\n0 0\n0 0\n1 1\n0 0\n1 1\n0 1\n1 1\n0 1\n1 1\n1 0\n1 0\n0 0\n1 0\n0 1\n1 0\n0 0\n0 0\n1 1\n1 1\n0 1\n0 0\n1 0\n1 1\n0 1\n1 0\n1 1\n0 1\n1 1\n1 0\n0 0\n0 0\n0 1\n0 0\n0 1\n1 1\n0 0\n1 1\n1 1\n0 0\n0 1\n1 0\n0 1\n0 0\n0 1\n0 1\n1 1\n1 1\n1 1\n0 0\n0 0\n1 1\n0 1\n0 1\n1 0\n0 0\n0 0\n1 1\n0 1\n0 1\n1 1\n1 1\n0 1\n1 1\n1 1\n0 0\n1 0\n0 1\n0 0\n0 0\n1 1\n1 1\n1 1\n1 1\n0 1\n1 0\n1 0\n1 0\n1 0\n1 0\n0 0\n1 0\n1 0\n0 0\n1 0\n0 0\n0 1\n1 0\n0 1\n1 0\n1 0\n1 0\n1 0",
"output": "11"
},
{
"input": "100\n0 2\n1 2\n2 1\n1 2\n1 2\n2 1\n2 2\n1 1\n1 1\n2 1\n1 2\n2 1\n1 2\n2 2\n2 2\n2 2\n1 2\n2 2\n2 1\n1 1\n1 1\n1 1\n2 2\n1 2\n2 2\n1 1\n1 1\n1 1\n1 1\n2 2\n1 2\n2 1\n1 1\n2 2\n1 1\n2 1\n1 1\n2 2\n2 1\n1 2\n1 1\n1 2\n2 1\n2 2\n1 1\n2 1\n1 1\n2 1\n1 1\n1 2\n2 2\n2 2\n1 1\n2 2\n1 2\n2 1\n2 1\n1 1\n1 1\n1 2\n1 2\n1 1\n1 1\n2 1\n1 2\n1 2\n2 1\n2 2\n2 2\n2 2\n2 1\n2 2\n1 1\n1 2\n1 2\n1 1\n2 2\n2 2\n1 1\n2 1\n1 1\n1 2\n1 2\n1 2\n1 1\n1 1\n2 2\n1 2\n2 1\n2 1\n2 1\n1 2\n1 2\n1 1\n2 2\n1 2\n2 0\n2 0\n2 0\n1 0",
"output": "7"
},
{
"input": "100\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0",
"output": "0"
},
{
"input": "2\n0 1\n1 0",
"output": "1"
},
{
"input": "2\n0 900\n900 0",
"output": "900"
},
{
"input": "2\n0 1\n1 0",
"output": "1"
},
{
"input": "2\n0 0\n0 0",
"output": "0"
},
{
"input": "2\n0 1000\n1000 0",
"output": "1000"
},
{
"input": "3\n0 802\n175 188\n815 0",
"output": "815"
},
{
"input": "3\n0 910\n910 976\n976 0",
"output": "976"
},
{
"input": "3\n0 2\n2 1\n1 0",
"output": "2"
}
] | 1,699,518,117 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 4 | 92 | 0 | n=int(input())
a,b=map(int,input().split())
m=0;cur=b
for i in range(n-1):
x,y=map(int,input().split())
if cur+y-x>m:
m=cur+y-x
cur=cur+y-x
print(m) | Title: Tram
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Linear Kingdom has exactly one tram line. It has *n* stops, numbered from 1 to *n* in the order of tram's movement. At the *i*-th stop *a**i* passengers exit the tram, while *b**i* passengers enter it. The tram is empty before it arrives at the first stop. Also, when the tram arrives at the last stop, all passengers exit so that it becomes empty.
Your task is to calculate the tram's minimum capacity such that the number of people inside the tram at any time never exceeds this capacity. Note that at each stop all exiting passengers exit before any entering passenger enters the tram.
Input Specification:
The first line contains a single number *n* (2<=≤<=*n*<=≤<=1000) — the number of the tram's stops.
Then *n* lines follow, each contains two integers *a**i* and *b**i* (0<=≤<=*a**i*,<=*b**i*<=≤<=1000) — the number of passengers that exits the tram at the *i*-th stop, and the number of passengers that enter the tram at the *i*-th stop. The stops are given from the first to the last stop in the order of tram's movement.
- The number of people who exit at a given stop does not exceed the total number of people in the tram immediately before it arrives at the stop. More formally, . This particularly means that *a*1<==<=0. - At the last stop, all the passengers exit the tram and it becomes empty. More formally, . - No passenger will enter the train at the last stop. That is, *b**n*<==<=0.
Output Specification:
Print a single integer denoting the minimum possible capacity of the tram (0 is allowed).
Demo Input:
['4\n0 3\n2 5\n4 2\n4 0\n']
Demo Output:
['6\n']
Note:
For the first example, a capacity of 6 is sufficient:
- At the first stop, the number of passengers inside the tram before arriving is 0. Then, 3 passengers enter the tram, and the number of passengers inside the tram becomes 3. - At the second stop, 2 passengers exit the tram (1 passenger remains inside). Then, 5 passengers enter the tram. There are 6 passengers inside the tram now. - At the third stop, 4 passengers exit the tram (2 passengers remain inside). Then, 2 passengers enter the tram. There are 4 passengers inside the tram now. - Finally, all the remaining passengers inside the tram exit the tram at the last stop. There are no passenger inside the tram now, which is in line with the constraints.
Since the number of passengers inside the tram never exceeds 6, a capacity of 6 is sufficient. Furthermore it is not possible for the tram to have a capacity less than 6. Hence, 6 is the correct answer. | ```python
n=int(input())
a,b=map(int,input().split())
m=0;cur=b
for i in range(n-1):
x,y=map(int,input().split())
if cur+y-x>m:
m=cur+y-x
cur=cur+y-x
print(m)
``` | 0 |
|
0 | none | none | none | 0 | [
"none"
] | null | null | Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
- the following words have typos: "hellno", "hackcerrs" and "backtothefutttture"; - the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem. | The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters. | Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them. | [
"hellno\n",
"abacaba\n",
"asdfasdf\n"
] | [
"hell no \n",
"abacaba \n",
"asd fasd f \n"
] | none | 0 | [
{
"input": "hellno",
"output": "hell no "
},
{
"input": "abacaba",
"output": "abacaba "
},
{
"input": "asdfasdf",
"output": "asd fasd f "
},
{
"input": "ooo",
"output": "ooo "
},
{
"input": "moyaoborona",
"output": "moyaoborona "
},
{
"input": "jxegxxx",
"output": "jxegx xx "
},
{
"input": "orfyaenanabckumulsboloyhljhacdgcmnooxvxrtuhcslxgslfpnfnyejbxqisxjyoyvcvuddboxkqgbogkfz",
"output": "orf yaenanabc kumuls boloyh lj hacd gc mnooxv xr tuhc sl xg sl fp nf nyejb xqisx jyoyv cvudd boxk qg bogk fz "
},
{
"input": "zxdgmhsjotvajkwshjpvzcuwehpeyfhakhtlvuoftkgdmvpafmxcliqvrztloocziqdkexhzcbdgxaoyvte",
"output": "zx dg mh sjotvajk ws hj pv zcuwehpeyf hakh tl vuoft kg dm vpafm xc liqv rz tloocziqd kexh zc bd gxaoyv te "
},
{
"input": "niblehmwtycadhbfuginpyafszjbucaszihijndzjtuyuaxkrovotshtsajmdcflnfdmahzbvpymiczqqleedpofcnvhieknlz",
"output": "niblehm wt ycadh bfuginp yafs zj bucaszihijn dz jtuyuaxk rovots ht sajm dc fl nf dmahz bv py micz qq leedpofc nv hiekn lz "
},
{
"input": "pqvtgtctpkgjgxnposjqedofficoyznxlerxyqypyzpoehejtjvyafjxjppywwgeakf",
"output": "pq vt gt ct pk gj gx nposj qedofficoyz nx lerx yq yp yz poehejt jv yafj xj pp yw wgeakf "
},
{
"input": "mvjajoyeg",
"output": "mv jajoyeg "
},
{
"input": "dipxocwjosvdaillxolmthjhzhsxskzqslebpixpuhpgeesrkedhohisdsjsrkiktbjzlhectrfcathvewzficirqbdvzq",
"output": "dipxocw josv daill xolm th jh zh sx sk zq slebpixpuhp geesr kedhohisd sj sr kikt bj zl hect rf cath vewz ficirq bd vz q "
},
{
"input": "ibbtvelwjirxqermucqrgmoauonisgmarjxxybllktccdykvef",
"output": "ibb tvelw jirx qermucq rg moauonisg marj xx yb ll kt cc dy kvef "
},
{
"input": "jxevkmrwlomaaahaubvjzqtyfqhqbhpqhomxqpiuersltohinvfyeykmlooujymldjqhgqjkvqknlyj",
"output": "jxevk mr wlomaaahaubv jz qt yf qh qb hp qhomx qpiuers ltohinv fyeyk mlooujy ml dj qh gq jk vq kn ly j "
},
{
"input": "hzxkuwqxonsulnndlhygvmallghjerwp",
"output": "hz xkuwq xonsuln nd lh yg vmall gh jerw p "
},
{
"input": "jbvcsjdyzlzmxwcvmixunfzxidzvwzaqqdhguvelwbdosbd",
"output": "jb vc sj dy zl zm xw cv mixunf zxidz vw zaqq dh guvelw bdosb d "
},
{
"input": "uyrsxaqmtibbxpfabprvnvbinjoxubupvfyjlqnfrfdeptipketwghr",
"output": "uyr sxaqm tibb xp fabp rv nv binjoxubupv fy jl qn fr fdeptipketw gh r "
},
{
"input": "xfcftysljytybkkzkpqdzralahgvbkxdtheqrhfxpecdjqofnyiahggnkiuusalu",
"output": "xf cf ty sl jy ty bk kz kp qd zralahg vb kx dt heqr hf xpecd jqofn yiahg gn kiuusalu "
},
{
"input": "a",
"output": "a "
},
{
"input": "b",
"output": "b "
},
{
"input": "aa",
"output": "aa "
},
{
"input": "ab",
"output": "ab "
},
{
"input": "ba",
"output": "ba "
},
{
"input": "bb",
"output": "bb "
},
{
"input": "aaa",
"output": "aaa "
},
{
"input": "aab",
"output": "aab "
},
{
"input": "aba",
"output": "aba "
},
{
"input": "abb",
"output": "abb "
},
{
"input": "baa",
"output": "baa "
},
{
"input": "bab",
"output": "bab "
},
{
"input": "bba",
"output": "bba "
},
{
"input": "bbb",
"output": "bbb "
},
{
"input": "bbc",
"output": "bb c "
},
{
"input": "bcb",
"output": "bc b "
},
{
"input": "cbb",
"output": "cb b "
},
{
"input": "bababcdfabbcabcdfacbbabcdfacacabcdfacbcabcdfaccbabcdfacaaabcdfabacabcdfabcbabcdfacbaabcdfabaaabcdfabbaabcdfacababcdfabbbabcdfabcaabcdfaaababcdfabccabcdfacccabcdfaacbabcdfaabaabcdfaabcabcdfaaacabcdfaccaabcdfaabbabcdfaaaaabcdfaacaabcdfaacc",
"output": "bababc dfabb cabc dfacb babc dfacacabc dfacb cabc dfacc babc dfacaaabc dfabacabc dfabc babc dfacbaabc dfabaaabc dfabbaabc dfacababc dfabbbabc dfabcaabc dfaaababc dfabc cabc dfacccabc dfaacbabc dfaabaabc dfaabcabc dfaaacabc dfaccaabc dfaabbabc dfaaaaabc dfaacaabc dfaacc "
},
{
"input": "bddabcdfaccdabcdfadddabcdfabbdabcdfacddabcdfacdbabcdfacbbabcdfacbcabcdfacbdabcdfadbbabcdfabdbabcdfabdcabcdfabbcabcdfabccabcdfabbbabcdfaddcabcdfaccbabcdfadbdabcdfacccabcdfadcdabcdfadcbabcdfabcbabcdfadbcabcdfacdcabcdfabcdabcdfadccabcdfaddb",
"output": "bd dabc dfacc dabc dfadddabc dfabb dabc dfacd dabc dfacd babc dfacb babc dfacb cabc dfacb dabc dfadb babc dfabd babc dfabd cabc dfabb cabc dfabc cabc dfabbbabc dfadd cabc dfacc babc dfadb dabc dfacccabc dfadc dabc dfadc babc dfabc babc dfadb cabc dfacd cabc dfabc dabc dfadc cabc dfadd b "
},
{
"input": "helllllooooo",
"output": "helllllooooo "
},
{
"input": "bbbzxxx",
"output": "bbb zx xx "
},
{
"input": "ffff",
"output": "ffff "
},
{
"input": "cdddddddddddddddddd",
"output": "cd ddddddddddddddddd "
},
{
"input": "bbbc",
"output": "bbb c "
},
{
"input": "lll",
"output": "lll "
},
{
"input": "bbbbb",
"output": "bbbbb "
},
{
"input": "llll",
"output": "llll "
},
{
"input": "bbbbbbccc",
"output": "bbbbbb ccc "
},
{
"input": "lllllb",
"output": "lllll b "
},
{
"input": "zzzzzzzzzzzzzzzzzzzzzzzzzzzzzz",
"output": "zzzzzzzzzzzzzzzzzzzzzzzzzzzzzz "
},
{
"input": "lllll",
"output": "lllll "
},
{
"input": "bbbbbbbbbc",
"output": "bbbbbbbbb c "
},
{
"input": "helllllno",
"output": "helllll no "
},
{
"input": "nnnnnnnnnnnn",
"output": "nnnnnnnnnnnn "
},
{
"input": "bbbbbccc",
"output": "bbbbb ccc "
},
{
"input": "zzzzzzzzzzzzzzzzzzzzzzzzzzzzz",
"output": "zzzzzzzzzzzzzzzzzzzzzzzzzzzzz "
},
{
"input": "nnnnnnnnnnnnnnnnnn",
"output": "nnnnnnnnnnnnnnnnnn "
},
{
"input": "zzzzzzzzzzzzzzzzzzzzzzz",
"output": "zzzzzzzzzzzzzzzzzzzzzzz "
},
{
"input": "hhhh",
"output": "hhhh "
},
{
"input": "nnnnnnnnnnnnnnnnnnnnnnnnn",
"output": "nnnnnnnnnnnnnnnnnnnnnnnnn "
},
{
"input": "zzzzzzzzzz",
"output": "zzzzzzzzzz "
},
{
"input": "dddd",
"output": "dddd "
},
{
"input": "heffffffgggggghhhhhh",
"output": "heffffff gggggg hhhhhh "
},
{
"input": "bcddd",
"output": "bc ddd "
},
{
"input": "x",
"output": "x "
},
{
"input": "nnn",
"output": "nnn "
},
{
"input": "xxxxxxxx",
"output": "xxxxxxxx "
},
{
"input": "cclcc",
"output": "cc lc c "
},
{
"input": "tttttttttttttt",
"output": "tttttttttttttt "
},
{
"input": "xxxxxxx",
"output": "xxxxxxx "
},
{
"input": "ccccb",
"output": "cccc b "
},
{
"input": "bcecccc",
"output": "bcecccc "
},
{
"input": "jjja",
"output": "jjja "
},
{
"input": "zzz",
"output": "zzz "
},
{
"input": "xxxxxxxxxzzzzzzzzzzzz",
"output": "xxxxxxxxx zzzzzzzzzzzz "
},
{
"input": "alllewww",
"output": "alllewww "
},
{
"input": "bbbbbbbbb",
"output": "bbbbbbbbb "
},
{
"input": "jjj",
"output": "jjj "
},
{
"input": "bbbbbbbbbbbbbbbbbbbbbbbbb",
"output": "bbbbbbbbbbbbbbbbbbbbbbbbb "
},
{
"input": "kkkkkkkkkklllllllllllllll",
"output": "kkkkkkkkkk lllllllllllllll "
},
{
"input": "helllllllllllo",
"output": "helllllllllllo "
},
{
"input": "ttttsttttt",
"output": "tttt st tttt "
},
{
"input": "tttttttsssssss",
"output": "ttttttt sssssss "
},
{
"input": "assstttttatsfatsfdjfdhtsjdsaatttssssststsss",
"output": "asss tttttats fats fd jf dh ts jd saattt sssss ts ts ss "
},
{
"input": "xxxxx",
"output": "xxxxx "
},
{
"input": "bbbhbbb",
"output": "bbb hb bb "
},
{
"input": "bbbbbbbb",
"output": "bbbbbbbb "
}
] | 1,689,956,231 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 108 | 77 | 2,150,400 | word = input()
n = len(word)
breaks = [False] * n
vowels = set({'a', 'e', 'i', 'o', 'u'})
streak = 1 if word[0] not in vowels else 0
numdiff = streak
for i in range(1, n):
if word[i] not in vowels:
streak += 1
if word[i] != word[i - 1]:
numdiff += 1
else:
streak = 0
numdiff = 0
if streak >= 3 and numdiff >= 2:
breaks[i] = True
streak = 1
numdiff = 1
result = ""
for i in range(n):
if breaks[i]:
result += " "
result += word[i]
print(result)# 1689956231.6888063 | Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
- the following words have typos: "hellno", "hackcerrs" and "backtothefutttture"; - the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input Specification:
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output Specification:
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Demo Input:
['hellno\n', 'abacaba\n', 'asdfasdf\n']
Demo Output:
['hell no \n', 'abacaba \n', 'asd fasd f \n']
Note:
none | ```python
word = input()
n = len(word)
breaks = [False] * n
vowels = set({'a', 'e', 'i', 'o', 'u'})
streak = 1 if word[0] not in vowels else 0
numdiff = streak
for i in range(1, n):
if word[i] not in vowels:
streak += 1
if word[i] != word[i - 1]:
numdiff += 1
else:
streak = 0
numdiff = 0
if streak >= 3 and numdiff >= 2:
breaks[i] = True
streak = 1
numdiff = 1
result = ""
for i in range(n):
if breaks[i]:
result += " "
result += word[i]
print(result)# 1689956231.6888063
``` | 3 |
|
690 | C1 | Brain Network (easy) | PROGRAMMING | 1,300 | [] | null | null | One particularly well-known fact about zombies is that they move and think terribly slowly. While we still don't know why their movements are so sluggish, the problem of laggy thinking has been recently resolved. It turns out that the reason is not (as previously suspected) any kind of brain defect – it's the opposite! Independent researchers confirmed that the nervous system of a zombie is highly complicated – it consists of *n* brains (much like a cow has several stomachs). They are interconnected by brain connectors, which are veins capable of transmitting thoughts between brains. There are two important properties such a brain network should have to function properly:
1. It should be possible to exchange thoughts between any two pairs of brains (perhaps indirectly, through other brains). 1. There should be no redundant brain connectors, that is, removing any brain connector would make property 1 false.
If both properties are satisfied, we say that the nervous system is valid. Unfortunately (?), if the system is not valid, the zombie stops thinking and becomes (even more) dead. Your task is to analyze a given nervous system of a zombie and find out whether it is valid. | The first line of the input contains two space-separated integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=1000) denoting the number of brains (which are conveniently numbered from 1 to *n*) and the number of brain connectors in the nervous system, respectively. In the next *m* lines, descriptions of brain connectors follow. Every connector is given as a pair of brains *a* *b* it connects (1<=≤<=*a*,<=*b*<=≤<=*n*, *a*<=≠<=*b*). | The output consists of one line, containing either yes or no depending on whether the nervous system is valid. | [
"4 4\n1 2\n2 3\n3 1\n4 1\n",
"6 5\n1 2\n2 3\n3 4\n4 5\n3 6\n"
] | [
"no\n",
"yes\n"
] | none | 0 | [
{
"input": "4 4\n1 2\n2 3\n3 1\n4 1",
"output": "no"
},
{
"input": "6 5\n1 2\n2 3\n3 4\n4 5\n3 6",
"output": "yes"
},
{
"input": "2 1\n1 2",
"output": "yes"
},
{
"input": "3 3\n2 1\n1 3\n3 2",
"output": "no"
},
{
"input": "3 2\n1 2\n2 3",
"output": "yes"
},
{
"input": "9 8\n1 2\n2 3\n3 4\n4 1\n5 6\n6 7\n7 8\n8 9",
"output": "no"
},
{
"input": "8 7\n6 2\n1 5\n4 8\n4 7\n6 7\n8 3\n8 1",
"output": "yes"
},
{
"input": "200 5\n93 101\n199 164\n14 94\n115 61\n106 156",
"output": "no"
},
{
"input": "10 9\n6 5\n9 2\n4 7\n2 3\n7 3\n3 4\n10 6\n1 2\n5 8",
"output": "no"
},
{
"input": "10 9\n2 3\n6 8\n10 1\n1 8\n6 7\n8 7\n10 5\n7 10\n2 5",
"output": "no"
},
{
"input": "10 9\n3 2\n4 1\n6 1\n7 1\n9 2\n6 9\n5 2\n7 9\n3 7",
"output": "no"
}
] | 1,670,242,652 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 6 | 46 | 0 | def dfs(cell):
global vis,d
if cell in vis:
return
vis.add(cell)
for eachCell in d[cell]:
dfs(eachCell)
m,n = map(int,input().split())
if (n + 1) != m:
print('no')
else:
d = {}
for __ in range(m-1):
x, y = map(int,input().split())
if x not in d:
d[x] = []
if y not in d:
d[y] = []
d[x].append(y)
d[y].append(x)
vis = set()
dfs(1)
vis = list(vis)
total = list(range(1,m+1))
if vis == total:
print('yes')
else:
print('no')
# print(vis)
| Title: Brain Network (easy)
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One particularly well-known fact about zombies is that they move and think terribly slowly. While we still don't know why their movements are so sluggish, the problem of laggy thinking has been recently resolved. It turns out that the reason is not (as previously suspected) any kind of brain defect – it's the opposite! Independent researchers confirmed that the nervous system of a zombie is highly complicated – it consists of *n* brains (much like a cow has several stomachs). They are interconnected by brain connectors, which are veins capable of transmitting thoughts between brains. There are two important properties such a brain network should have to function properly:
1. It should be possible to exchange thoughts between any two pairs of brains (perhaps indirectly, through other brains). 1. There should be no redundant brain connectors, that is, removing any brain connector would make property 1 false.
If both properties are satisfied, we say that the nervous system is valid. Unfortunately (?), if the system is not valid, the zombie stops thinking and becomes (even more) dead. Your task is to analyze a given nervous system of a zombie and find out whether it is valid.
Input Specification:
The first line of the input contains two space-separated integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=1000) denoting the number of brains (which are conveniently numbered from 1 to *n*) and the number of brain connectors in the nervous system, respectively. In the next *m* lines, descriptions of brain connectors follow. Every connector is given as a pair of brains *a* *b* it connects (1<=≤<=*a*,<=*b*<=≤<=*n*, *a*<=≠<=*b*).
Output Specification:
The output consists of one line, containing either yes or no depending on whether the nervous system is valid.
Demo Input:
['4 4\n1 2\n2 3\n3 1\n4 1\n', '6 5\n1 2\n2 3\n3 4\n4 5\n3 6\n']
Demo Output:
['no\n', 'yes\n']
Note:
none | ```python
def dfs(cell):
global vis,d
if cell in vis:
return
vis.add(cell)
for eachCell in d[cell]:
dfs(eachCell)
m,n = map(int,input().split())
if (n + 1) != m:
print('no')
else:
d = {}
for __ in range(m-1):
x, y = map(int,input().split())
if x not in d:
d[x] = []
if y not in d:
d[y] = []
d[x].append(y)
d[y].append(x)
vis = set()
dfs(1)
vis = list(vis)
total = list(range(1,m+1))
if vis == total:
print('yes')
else:
print('no')
# print(vis)
``` | 0 |
|
680 | B | Bear and Finding Criminals | PROGRAMMING | 1,000 | [
"constructive algorithms",
"implementation"
] | null | null | There are *n* cities in Bearland, numbered 1 through *n*. Cities are arranged in one long row. The distance between cities *i* and *j* is equal to |*i*<=-<=*j*|.
Limak is a police officer. He lives in a city *a*. His job is to catch criminals. It's hard because he doesn't know in which cities criminals are. Though, he knows that there is at most one criminal in each city.
Limak is going to use a BCD (Bear Criminal Detector). The BCD will tell Limak how many criminals there are for every distance from a city *a*. After that, Limak can catch a criminal in each city for which he is sure that there must be a criminal.
You know in which cities criminals are. Count the number of criminals Limak will catch, after he uses the BCD. | The first line of the input contains two integers *n* and *a* (1<=≤<=*a*<=≤<=*n*<=≤<=100) — the number of cities and the index of city where Limak lives.
The second line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (0<=≤<=*t**i*<=≤<=1). There are *t**i* criminals in the *i*-th city. | Print the number of criminals Limak will catch. | [
"6 3\n1 1 1 0 1 0\n",
"5 2\n0 0 0 1 0\n"
] | [
"3\n",
"1\n"
] | In the first sample, there are six cities and Limak lives in the third one (blue arrow below). Criminals are in cities marked red.
Using the BCD gives Limak the following information:
- There is one criminal at distance 0 from the third city — Limak is sure that this criminal is exactly in the third city. - There is one criminal at distance 1 from the third city — Limak doesn't know if a criminal is in the second or fourth city. - There are two criminals at distance 2 from the third city — Limak is sure that there is one criminal in the first city and one in the fifth city. - There are zero criminals for every greater distance.
So, Limak will catch criminals in cities 1, 3 and 5, that is 3 criminals in total.
In the second sample (drawing below), the BCD gives Limak the information that there is one criminal at distance 2 from Limak's city. There is only one city at distance 2 so Limak is sure where a criminal is. | 1,000 | [
{
"input": "6 3\n1 1 1 0 1 0",
"output": "3"
},
{
"input": "5 2\n0 0 0 1 0",
"output": "1"
},
{
"input": "1 1\n1",
"output": "1"
},
{
"input": "1 1\n0",
"output": "0"
},
{
"input": "9 3\n1 1 1 1 1 1 1 1 0",
"output": "8"
},
{
"input": "9 5\n1 0 1 0 1 0 1 0 1",
"output": "5"
},
{
"input": "20 17\n1 1 0 1 1 1 1 0 1 0 1 1 1 0 1 1 0 0 0 0",
"output": "10"
},
{
"input": "100 60\n1 1 1 1 1 1 0 1 0 0 1 1 0 1 1 1 1 1 0 0 1 1 0 0 0 0 0 1 0 1 1 0 1 0 1 0 1 0 1 1 0 0 0 0 0 1 1 1 0 1 1 0 0 0 1 0 0 0 1 1 1 0 1 0 0 1 1 1 0 1 1 1 0 0 1 1 0 1 0 0 0 1 0 0 0 0 0 0 1 1 1 0 0 1 1 1 0 1 0 0",
"output": "27"
},
{
"input": "8 1\n1 0 1 1 0 0 1 0",
"output": "4"
},
{
"input": "11 11\n0 1 0 0 1 1 1 0 0 0 0",
"output": "4"
},
{
"input": "19 10\n0 1 1 0 1 0 0 1 1 0 0 1 0 1 0 0 1 0 1",
"output": "4"
},
{
"input": "100 38\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0",
"output": "0"
},
{
"input": "100 38\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0",
"output": "1"
},
{
"input": "100 38\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0",
"output": "3"
},
{
"input": "99 38\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "25"
},
{
"input": "99 38\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "24"
},
{
"input": "99 38\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "24"
},
{
"input": "98 70\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0",
"output": "41"
},
{
"input": "99 70\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "9"
},
{
"input": "99 60\n0 0 0 1 1 0 1 1 0 0 1 0 0 0 0 0 1 0 1 0 0 0 1 0 1 1 1 0 1 1 1 1 0 0 1 1 1 0 1 1 1 1 1 0 1 1 0 0 0 0 0 1 0 0 1 0 1 1 1 1 1 0 1 0 1 1 0 0 1 0 1 0 0 1 0 0 1 1 1 0 0 0 0 1 1 0 1 1 1 1 0 0 0 0 0 0 1 1 1",
"output": "34"
},
{
"input": "98 24\n0 0 0 1 1 0 1 1 0 0 0 0 1 0 0 1 0 0 1 1 1 0 0 1 0 0 1 0 1 0 0 1 1 0 1 1 1 0 1 0 0 1 0 0 0 1 1 1 1 0 1 1 1 0 1 1 1 1 0 0 1 0 0 0 1 0 1 1 0 1 0 1 1 1 0 0 1 0 0 0 1 1 0 0 1 1 1 1 1 1 0 1 0 1 0 0 1 1",
"output": "39"
},
{
"input": "100 100\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "100"
},
{
"input": "100 1\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0",
"output": "0"
},
{
"input": "2 1\n0 1",
"output": "1"
}
] | 1,619,381,594 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 5 | 61 | 0 | n, a = map(int, input().split())
arr = list(map(int, input().split()))
count = 0
a -= 1
if arr[a] == 1:
count += 1
arr1 = arr[:a]
arr2 = arr[a+1:]
if len(arr1) == len(arr2):
for i in range(len(arr1)):
if arr1[i] & arr2[i] == 1:
count += 2
elif len(arr1) < len(arr2):
for i in range(len(arr1)):
if arr1[i] & arr2[i] == 1:
count += 2
for j in range(i + 1, len(arr2)):
if arr2[j] == 1:
count += 1
else:
for i in range(len(arr2)):
if arr1[i] & arr2[i] == 1:
count += 2
for j in range(i + 1, len(arr1)):
if arr1[j] == 1:
count += 1
print(count) | Title: Bear and Finding Criminals
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* cities in Bearland, numbered 1 through *n*. Cities are arranged in one long row. The distance between cities *i* and *j* is equal to |*i*<=-<=*j*|.
Limak is a police officer. He lives in a city *a*. His job is to catch criminals. It's hard because he doesn't know in which cities criminals are. Though, he knows that there is at most one criminal in each city.
Limak is going to use a BCD (Bear Criminal Detector). The BCD will tell Limak how many criminals there are for every distance from a city *a*. After that, Limak can catch a criminal in each city for which he is sure that there must be a criminal.
You know in which cities criminals are. Count the number of criminals Limak will catch, after he uses the BCD.
Input Specification:
The first line of the input contains two integers *n* and *a* (1<=≤<=*a*<=≤<=*n*<=≤<=100) — the number of cities and the index of city where Limak lives.
The second line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (0<=≤<=*t**i*<=≤<=1). There are *t**i* criminals in the *i*-th city.
Output Specification:
Print the number of criminals Limak will catch.
Demo Input:
['6 3\n1 1 1 0 1 0\n', '5 2\n0 0 0 1 0\n']
Demo Output:
['3\n', '1\n']
Note:
In the first sample, there are six cities and Limak lives in the third one (blue arrow below). Criminals are in cities marked red.
Using the BCD gives Limak the following information:
- There is one criminal at distance 0 from the third city — Limak is sure that this criminal is exactly in the third city. - There is one criminal at distance 1 from the third city — Limak doesn't know if a criminal is in the second or fourth city. - There are two criminals at distance 2 from the third city — Limak is sure that there is one criminal in the first city and one in the fifth city. - There are zero criminals for every greater distance.
So, Limak will catch criminals in cities 1, 3 and 5, that is 3 criminals in total.
In the second sample (drawing below), the BCD gives Limak the information that there is one criminal at distance 2 from Limak's city. There is only one city at distance 2 so Limak is sure where a criminal is. | ```python
n, a = map(int, input().split())
arr = list(map(int, input().split()))
count = 0
a -= 1
if arr[a] == 1:
count += 1
arr1 = arr[:a]
arr2 = arr[a+1:]
if len(arr1) == len(arr2):
for i in range(len(arr1)):
if arr1[i] & arr2[i] == 1:
count += 2
elif len(arr1) < len(arr2):
for i in range(len(arr1)):
if arr1[i] & arr2[i] == 1:
count += 2
for j in range(i + 1, len(arr2)):
if arr2[j] == 1:
count += 1
else:
for i in range(len(arr2)):
if arr1[i] & arr2[i] == 1:
count += 2
for j in range(i + 1, len(arr1)):
if arr1[j] == 1:
count += 1
print(count)
``` | 0 |
|
709 | A | Juicer | PROGRAMMING | 900 | [
"implementation"
] | null | null | Kolya is going to make fresh orange juice. He has *n* oranges of sizes *a*1,<=*a*2,<=...,<=*a**n*. Kolya will put them in the juicer in the fixed order, starting with orange of size *a*1, then orange of size *a*2 and so on. To be put in the juicer the orange must have size not exceeding *b*, so if Kolya sees an orange that is strictly greater he throws it away and continues with the next one.
The juicer has a special section to collect waste. It overflows if Kolya squeezes oranges of the total size strictly greater than *d*. When it happens Kolya empties the waste section (even if there are no more oranges) and continues to squeeze the juice. How many times will he have to empty the waste section? | The first line of the input contains three integers *n*, *b* and *d* (1<=≤<=*n*<=≤<=100<=000, 1<=≤<=*b*<=≤<=*d*<=≤<=1<=000<=000) — the number of oranges, the maximum size of the orange that fits in the juicer and the value *d*, which determines the condition when the waste section should be emptied.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1<=000<=000) — sizes of the oranges listed in the order Kolya is going to try to put them in the juicer. | Print one integer — the number of times Kolya will have to empty the waste section. | [
"2 7 10\n5 6\n",
"1 5 10\n7\n",
"3 10 10\n5 7 7\n",
"1 1 1\n1\n"
] | [
"1\n",
"0\n",
"1\n",
"0\n"
] | In the first sample, Kolya will squeeze the juice from two oranges and empty the waste section afterwards.
In the second sample, the orange won't fit in the juicer so Kolya will have no juice at all. | 500 | [
{
"input": "2 7 10\n5 6",
"output": "1"
},
{
"input": "1 5 10\n7",
"output": "0"
},
{
"input": "3 10 10\n5 7 7",
"output": "1"
},
{
"input": "1 1 1\n1",
"output": "0"
},
{
"input": "2 951637 951638\n44069 951637",
"output": "1"
},
{
"input": "50 100 129\n55 130 91 19 116 3 63 52 104 76 75 27 151 99 149 147 39 148 84 9 132 49 40 112 124 141 144 93 36 32 146 74 48 38 150 55 94 32 107 69 77 81 33 57 62 98 78 127 154 126",
"output": "12"
},
{
"input": "100 1000 1083\n992 616 818 359 609 783 263 989 501 929 362 394 919 1081 870 830 1097 975 62 346 531 367 323 457 707 360 949 334 867 116 478 417 961 963 1029 114 867 1008 988 916 983 1077 959 942 572 961 579 318 721 337 488 717 111 70 416 685 987 130 353 107 61 191 827 849 106 815 211 953 111 398 889 860 801 71 375 320 395 1059 116 222 931 444 582 74 677 655 88 173 686 491 661 186 114 832 615 814 791 464 517 850",
"output": "36"
},
{
"input": "2 6 8\n2 1",
"output": "0"
},
{
"input": "5 15 16\n7 11 5 12 8",
"output": "2"
},
{
"input": "15 759966 759967\n890397 182209 878577 548548 759966 812923 759966 860479 200595 381358 299175 339368 759966 907668 69574",
"output": "4"
},
{
"input": "5 234613 716125\n642626 494941 234613 234613 234613",
"output": "0"
},
{
"input": "50 48547 567054\n529808 597004 242355 559114 78865 537318 631455 733020 655072 645093 309010 855034 306058 625046 524574 834944 27330 664392 443637 821584 338013 490702 289520 675471 885846 258814 134220 571301 84875 94132 200425 928833 375166 521232 317961 175315 947093 89971 322071 174033 48547 998535 954205 704114 943163 438900 48547 538422 48547 48547",
"output": "0"
},
{
"input": "5 10 20\n10 10 10 10 1",
"output": "1"
},
{
"input": "5 10 11\n10 10 10 10 1",
"output": "2"
},
{
"input": "3 10 10\n4 3 3",
"output": "0"
},
{
"input": "3 5 5\n5 5 5",
"output": "1"
},
{
"input": "3 4 14\n5 5 5",
"output": "0"
},
{
"input": "2 7 10\n1234 1234",
"output": "0"
},
{
"input": "1 5 6\n10",
"output": "0"
},
{
"input": "3 4 6\n1 2 3",
"output": "0"
},
{
"input": "5 10 12\n13 13 13 13 13",
"output": "0"
},
{
"input": "3 4 5\n5 7 9",
"output": "0"
},
{
"input": "3 10 100\n5 5 5",
"output": "0"
},
{
"input": "5 1 2\n2 2 2 2 2",
"output": "0"
},
{
"input": "5 5 5\n5 5 5 5 5",
"output": "2"
},
{
"input": "2 2 5\n5 5",
"output": "0"
},
{
"input": "3 1 4\n2 2 2",
"output": "0"
}
] | 1,647,940,353 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 4 | 77 | 7,372,800 | n, max_orange_size, max_total_size = tuple(map(int, input().split()))
orange_sizes = list(map(int, input().split()))
total_size = 0
for orange in orange_sizes:
if orange > max_orange_size:
continue
else:
total_size += orange
# we checked if total_size < max_total_size because if we don't it won't
# give us correct answers for example if we entered 1 5 10 , 7
# as 7 > 5 total_size will remain 0 so in the 2nd if it will enter else block
# and rint 0 - 1 which is -1
if total_size % max_total_size != 0 or total_size < max_total_size:
print(total_size // max_total_size)
else:
print((total_size // max_total_size) - 1)
| Title: Juicer
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Kolya is going to make fresh orange juice. He has *n* oranges of sizes *a*1,<=*a*2,<=...,<=*a**n*. Kolya will put them in the juicer in the fixed order, starting with orange of size *a*1, then orange of size *a*2 and so on. To be put in the juicer the orange must have size not exceeding *b*, so if Kolya sees an orange that is strictly greater he throws it away and continues with the next one.
The juicer has a special section to collect waste. It overflows if Kolya squeezes oranges of the total size strictly greater than *d*. When it happens Kolya empties the waste section (even if there are no more oranges) and continues to squeeze the juice. How many times will he have to empty the waste section?
Input Specification:
The first line of the input contains three integers *n*, *b* and *d* (1<=≤<=*n*<=≤<=100<=000, 1<=≤<=*b*<=≤<=*d*<=≤<=1<=000<=000) — the number of oranges, the maximum size of the orange that fits in the juicer and the value *d*, which determines the condition when the waste section should be emptied.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1<=000<=000) — sizes of the oranges listed in the order Kolya is going to try to put them in the juicer.
Output Specification:
Print one integer — the number of times Kolya will have to empty the waste section.
Demo Input:
['2 7 10\n5 6\n', '1 5 10\n7\n', '3 10 10\n5 7 7\n', '1 1 1\n1\n']
Demo Output:
['1\n', '0\n', '1\n', '0\n']
Note:
In the first sample, Kolya will squeeze the juice from two oranges and empty the waste section afterwards.
In the second sample, the orange won't fit in the juicer so Kolya will have no juice at all. | ```python
n, max_orange_size, max_total_size = tuple(map(int, input().split()))
orange_sizes = list(map(int, input().split()))
total_size = 0
for orange in orange_sizes:
if orange > max_orange_size:
continue
else:
total_size += orange
# we checked if total_size < max_total_size because if we don't it won't
# give us correct answers for example if we entered 1 5 10 , 7
# as 7 > 5 total_size will remain 0 so in the 2nd if it will enter else block
# and rint 0 - 1 which is -1
if total_size % max_total_size != 0 or total_size < max_total_size:
print(total_size // max_total_size)
else:
print((total_size // max_total_size) - 1)
``` | 0 |
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