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int64 0
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int64 7
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600 | 10 | The crowdedness of the discotheque would never stop our friends from having fun, but a bit more spaciousness won't hurt, will it?
The discotheque can be seen as an infinite xy-plane, in which there are a total of n dancers. Once someone starts moving around, they will move only inside their own movement range, which is a circular area Ci described by a center (xi, yi) and a radius ri. No two ranges' borders have more than one common point, that is for every pair (i, j) (1 ≤ i < j ≤ n) either ranges Ci and Cj are disjoint, or one of them is a subset of the other. Note that it's possible that two ranges' borders share a single common point, but no two dancers have exactly the same ranges.
Tsukihi, being one of them, defines the spaciousness to be the area covered by an odd number of movement ranges of dancers who are moving. An example is shown below, with shaded regions representing the spaciousness if everyone moves at the same time.
<image>
But no one keeps moving for the whole night after all, so the whole night's time is divided into two halves — before midnight and after midnight. Every dancer moves around in one half, while sitting down with friends in the other. The spaciousness of two halves are calculated separately and their sum should, of course, be as large as possible. The following figure shows an optimal solution to the example above.
<image>
By different plans of who dances in the first half and who does in the other, different sums of spaciousness over two halves are achieved. You are to find the largest achievable value of this sum.
Input
The first line of input contains a positive integer n (1 ≤ n ≤ 1 000) — the number of dancers.
The following n lines each describes a dancer: the i-th line among them contains three space-separated integers xi, yi and ri ( - 106 ≤ xi, yi ≤ 106, 1 ≤ ri ≤ 106), describing a circular movement range centered at (xi, yi) with radius ri.
Output
Output one decimal number — the largest achievable sum of spaciousness over two halves of the night.
The output is considered correct if it has a relative or absolute error of at most 10 - 9. Formally, let your answer be a, and the jury's answer be b. Your answer is considered correct if <image>.
Examples
Input
5
2 1 6
0 4 1
2 -1 3
1 -2 1
4 -1 1
Output
138.23007676
Input
8
0 0 1
0 0 2
0 0 3
0 0 4
0 0 5
0 0 6
0 0 7
0 0 8
Output
289.02652413
Note
The first sample corresponds to the illustrations in the legend. | n = int(input())
d = [1] * n
p = [[] for i in range(n)]
def f():
x, y, r = map(int, input().split())
return r * r, x, y
t = sorted(f() for i in range(n))
for i in range(n):
r, x, y = t[i]
for j in range(i + 1, n):
s, a, b = t[j]
if (a - x) ** 2 + (b - y) ** 2 < s:
p[j].append(i)
d[i] = 0
break
def f(i):
s = t[i][0]
q = [(1, j) for j in p[i]]
while q:
d, i = q.pop()
s += d * t[i][0]
q += [(-d, j) for j in p[i]]
return s
print(3.1415926536 * sum(f(i) for i in range(n) if d[i])) | {
"input": [
"5\n2 1 6\n0 4 1\n2 -1 3\n1 -2 1\n4 -1 1\n",
"8\n0 0 1\n0 0 2\n0 0 3\n0 0 4\n0 0 5\n0 0 6\n0 0 7\n0 0 8\n"
],
"output": [
"138.230076757951\n",
"289.026524130261\n"
]
} |
601 | 10 | One quite ordinary day Valera went to school (there's nowhere else he should go on a week day). In a maths lesson his favorite teacher Ms. Evans told students about divisors. Despite the fact that Valera loved math, he didn't find this particular topic interesting. Even more, it seemed so boring that he fell asleep in the middle of a lesson. And only a loud ringing of a school bell could interrupt his sweet dream.
Of course, the valuable material and the teacher's explanations were lost. However, Valera will one way or another have to do the homework. As he does not know the new material absolutely, he cannot do the job himself. That's why he asked you to help. You're his best friend after all, you just cannot refuse to help.
Valera's home task has only one problem, which, though formulated in a very simple way, has not a trivial solution. Its statement looks as follows: if we consider all positive integers in the interval [a;b] then it is required to count the amount of such numbers in this interval that their smallest divisor will be a certain integer k (you do not have to consider divisor equal to one). In other words, you should count the amount of such numbers from the interval [a;b], that are not divisible by any number between 2 and k - 1 and yet are divisible by k.
Input
The first and only line contains three positive integers a, b, k (1 ≤ a ≤ b ≤ 2·109, 2 ≤ k ≤ 2·109).
Output
Print on a single line the answer to the given problem.
Examples
Input
1 10 2
Output
5
Input
12 23 3
Output
2
Input
6 19 5
Output
0
Note
Comments to the samples from the statement:
In the first sample the answer is numbers 2, 4, 6, 8, 10.
In the second one — 15, 21
In the third one there are no such numbers. | def pr(x):
d = 2
while d * d <= x:
if x % d == 0:
return 0
d += 1
return 1
def cnt(n, k):
if not pr(k) or n < k: return 0
n1 = n // k
return n1 - sum(cnt(n1, i) for i in range(2, min(k, n1 + 1)))
a, b, k = map(int, input().split())
ans = cnt(b, k) - cnt(a - 1, k)
print(ans)
| {
"input": [
"12 23 3\n",
"1 10 2\n",
"6 19 5\n"
],
"output": [
"2\n",
"5\n",
"0\n"
]
} |
602 | 11 | You are given a program you want to execute as a set of tasks organized in a dependency graph. The dependency graph is a directed acyclic graph: each task can depend on results of one or several other tasks, and there are no directed circular dependencies between tasks. A task can only be executed if all tasks it depends on have already completed.
Some of the tasks in the graph can only be executed on a coprocessor, and the rest can only be executed on the main processor. In one coprocessor call you can send it a set of tasks which can only be executed on it. For each task of the set, all tasks on which it depends must be either already completed or be included in the set. The main processor starts the program execution and gets the results of tasks executed on the coprocessor automatically.
Find the minimal number of coprocessor calls which are necessary to execute the given program.
Input
The first line contains two space-separated integers N (1 ≤ N ≤ 105) — the total number of tasks given, and M (0 ≤ M ≤ 105) — the total number of dependencies between tasks.
The next line contains N space-separated integers <image>. If Ei = 0, task i can only be executed on the main processor, otherwise it can only be executed on the coprocessor.
The next M lines describe the dependencies between tasks. Each line contains two space-separated integers T1 and T2 and means that task T1 depends on task T2 (T1 ≠ T2). Tasks are indexed from 0 to N - 1. All M pairs (T1, T2) are distinct. It is guaranteed that there are no circular dependencies between tasks.
Output
Output one line containing an integer — the minimal number of coprocessor calls necessary to execute the program.
Examples
Input
4 3
0 1 0 1
0 1
1 2
2 3
Output
2
Input
4 3
1 1 1 0
0 1
0 2
3 0
Output
1
Note
In the first test, tasks 1 and 3 can only be executed on the coprocessor. The dependency graph is linear, so the tasks must be executed in order 3 -> 2 -> 1 -> 0. You have to call coprocessor twice: first you call it for task 3, then you execute task 2 on the main processor, then you call it for for task 1, and finally you execute task 0 on the main processor.
In the second test, tasks 0, 1 and 2 can only be executed on the coprocessor. Tasks 1 and 2 have no dependencies, and task 0 depends on tasks 1 and 2, so all three tasks 0, 1 and 2 can be sent in one coprocessor call. After that task 3 is executed on the main processor. | from collections import deque
import sys
input = sys.stdin.readline
def put():
return map(int, input().split())
def bfs_(main, co, flag):
curr = main if flag else co
while curr:
x = curr.popleft()
for y in graph[x]:
inedge[y]-=1
if inedge[y]==0:
if is_co[y]:co.append(y)
else:main.append(y)
return main,co
def bfs():
main, co, ans = deque(), deque(), 0
for i in range(n):
if inedge[i]==0:
if is_co[i]: co.append(i)
else: main.append(i)
while main or co:
main, co = bfs_(main, co, True)
if co: ans+=1
main, co = bfs_(main, co, False)
return ans
n,m = put()
is_co = list(put())
graph = [[] for i in range(n)]
inedge, vis= [0]*n, [0]*n
for _ in range(m):
x,y = put()
graph[y].append(x)
inedge[x]+=1
print(bfs())
| {
"input": [
"4 3\n1 1 1 0\n0 1\n0 2\n3 0\n",
"4 3\n0 1 0 1\n0 1\n1 2\n2 3\n"
],
"output": [
"1",
"2"
]
} |
603 | 8 | The last stage of Football World Cup is played using the play-off system.
There are n teams left in this stage, they are enumerated from 1 to n. Several rounds are held, in each round the remaining teams are sorted in the order of their ids, then the first in this order plays with the second, the third — with the fourth, the fifth — with the sixth, and so on. It is guaranteed that in each round there is even number of teams. The winner of each game advances to the next round, the loser is eliminated from the tournament, there are no draws. In the last round there is the only game with two remaining teams: the round is called the Final, the winner is called the champion, and the tournament is over.
Arkady wants his two favorite teams to play in the Final. Unfortunately, the team ids are already determined, and it may happen that it is impossible for teams to meet in the Final, because they are to meet in some earlier stage, if they are strong enough. Determine, in which round the teams with ids a and b can meet.
Input
The only line contains three integers n, a and b (2 ≤ n ≤ 256, 1 ≤ a, b ≤ n) — the total number of teams, and the ids of the teams that Arkady is interested in.
It is guaranteed that n is such that in each round an even number of team advance, and that a and b are not equal.
Output
In the only line print "Final!" (without quotes), if teams a and b can meet in the Final.
Otherwise, print a single integer — the number of the round in which teams a and b can meet. The round are enumerated from 1.
Examples
Input
4 1 2
Output
1
Input
8 2 6
Output
Final!
Input
8 7 5
Output
2
Note
In the first example teams 1 and 2 meet in the first round.
In the second example teams 2 and 6 can only meet in the third round, which is the Final, if they win all their opponents in earlier rounds.
In the third example the teams with ids 7 and 5 can meet in the second round, if they win their opponents in the first round. | import math
def rec(a,b,count):
if a//2 == b//2:
return count
else:
return rec(a//2,b//2,count+1)
n,a,b = input().split()
res = rec(int(a)-1,int(b)-1,1)
if res==math.log2(int(n)):
print("Final!")
else:
print(res) | {
"input": [
"4 1 2\n",
"8 7 5\n",
"8 2 6\n"
],
"output": [
"1",
"2",
"Final!"
]
} |
604 | 8 | 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 ai, j = 1 if the i-th switch turns on the j-th lamp and ai, 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
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 ai, 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
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.
Examples
Input
4 5
10101
01000
00111
10000
Output
YES
Input
4 5
10100
01000
00110
00101
Output
NO | def main():
n, m = map(int, input().split())
grid = [input() for _ in range(n)]
counts = [sum(grid[i][j] == '1' for i in range(n)) for j in range(m)]
no_skip = all(any(grid[i][j] == '1' and counts[j] == 1 for j in range(m)) for i in range(n))
print('NO' if no_skip else 'YES')
main()
| {
"input": [
"4 5\n10100\n01000\n00110\n00101\n",
"4 5\n10101\n01000\n00111\n10000\n"
],
"output": [
"NO\n",
"YES\n"
]
} |
605 | 7 | Pavel made a photo of his favourite stars in the sky. His camera takes a photo of all points of the sky that belong to some rectangle with sides parallel to the coordinate axes.
Strictly speaking, it makes a photo of all points with coordinates (x, y), such that x_1 ≤ x ≤ x_2 and y_1 ≤ y ≤ y_2, where (x_1, y_1) and (x_2, y_2) are coordinates of the left bottom and the right top corners of the rectangle being photographed. The area of this rectangle can be zero.
After taking the photo, Pavel wrote down coordinates of n of his favourite stars which appeared in the photo. These points are not necessarily distinct, there can be multiple stars in the same point of the sky.
Pavel has lost his camera recently and wants to buy a similar one. Specifically, he wants to know the dimensions of the photo he took earlier. Unfortunately, the photo is also lost. His notes are also of not much help; numbers are written in random order all over his notepad, so it's impossible to tell which numbers specify coordinates of which points.
Pavel asked you to help him to determine what are the possible dimensions of the photo according to his notes. As there are multiple possible answers, find the dimensions with the minimal possible area of the rectangle.
Input
The first line of the input contains an only integer n (1 ≤ n ≤ 100 000), the number of points in Pavel's records.
The second line contains 2 ⋅ n integers a_1, a_2, ..., a_{2 ⋅ n} (1 ≤ a_i ≤ 10^9), coordinates, written by Pavel in some order.
Output
Print the only integer, the minimal area of the rectangle which could have contained all points from Pavel's records.
Examples
Input
4
4 1 3 2 3 2 1 3
Output
1
Input
3
5 8 5 5 7 5
Output
0
Note
In the first sample stars in Pavel's records can be (1, 3), (1, 3), (2, 3), (2, 4). In this case, the minimal area of the rectangle, which contains all these points is 1 (rectangle with corners at (1, 3) and (2, 4)). | def main():
n = int(input())
numbers = list(map(int, input().split(" ")))
numbers.sort()
r = [numbers[n-1+x]-numbers[x] for x in range(n+1)]
#print(numbers)
#print(r)
print(min([r[0]*r[-1],(numbers[-1]-numbers[0])*min(r)]))
main() | {
"input": [
"4\n4 1 3 2 3 2 1 3\n",
"3\n5 8 5 5 7 5\n"
],
"output": [
"1",
"0"
]
} |
606 | 7 | You have n coins, each of the same value of 1.
Distribute them into packets such that any amount x (1 ≤ x ≤ n) can be formed using some (possibly one or all) number of these packets.
Each packet may only be used entirely or not used at all. No packet may be used more than once in the formation of the single x, however it may be reused for the formation of other x's.
Find the minimum number of packets in such a distribution.
Input
The only line contains a single integer n (1 ≤ n ≤ 10^9) — the number of coins you have.
Output
Output a single integer — the minimum possible number of packets, satisfying the condition above.
Examples
Input
6
Output
3
Input
2
Output
2
Note
In the first example, three packets with 1, 2 and 3 coins can be made to get any amount x (1≤ x≤ 6).
* To get 1 use the packet with 1 coin.
* To get 2 use the packet with 2 coins.
* To get 3 use the packet with 3 coins.
* To get 4 use packets with 1 and 3 coins.
* To get 5 use packets with 2 and 3 coins
* To get 6 use all packets.
In the second example, two packets with 1 and 1 coins can be made to get any amount x (1≤ x≤ 2). | def go():
n = int(input())
return len(bin(n)) - 2
print(go())
| {
"input": [
"6\n",
"2\n"
],
"output": [
"3\n",
"2\n"
]
} |
607 | 8 | Dark Assembly is a governing body in the Netherworld. Here sit the senators who take the most important decisions for the player. For example, to expand the range of the shop or to improve certain characteristics of the character the Dark Assembly's approval is needed.
The Dark Assembly consists of n senators. Each of them is characterized by his level and loyalty to the player. The level is a positive integer which reflects a senator's strength. Loyalty is the probability of a positive decision in the voting, which is measured as a percentage with precision of up to 10%.
Senators make decisions by voting. Each of them makes a positive or negative decision in accordance with their loyalty. If strictly more than half of the senators take a positive decision, the player's proposal is approved.
If the player's proposal is not approved after the voting, then the player may appeal against the decision of the Dark Assembly. To do that, player needs to kill all the senators that voted against (there's nothing wrong in killing senators, they will resurrect later and will treat the player even worse). The probability that a player will be able to kill a certain group of senators is equal to A / (A + B), where A is the sum of levels of all player's characters and B is the sum of levels of all senators in this group. If the player kills all undesired senators, then his proposal is approved.
Senators are very fond of sweets. They can be bribed by giving them candies. For each received candy a senator increases his loyalty to the player by 10%. It's worth to mention that loyalty cannot exceed 100%. The player can take no more than k sweets to the courtroom. Candies should be given to the senators before the start of voting.
Determine the probability that the Dark Assembly approves the player's proposal if the candies are distributed among the senators in the optimal way.
Input
The first line contains three integers n, k and A (1 ≤ n, k ≤ 8, 1 ≤ A ≤ 9999).
Then n lines follow. The i-th of them contains two numbers — bi and li — the i-th senator's level and his loyalty.
The levels of all senators are integers in range from 1 to 9999 (inclusive). The loyalties of all senators are integers in range from 0 to 100 (inclusive) and all of them are divisible by 10.
Output
Print one real number with precision 10 - 6 — the maximal possible probability that the Dark Assembly approves the player's proposal for the best possible distribution of candies among the senators.
Examples
Input
5 6 100
11 80
14 90
23 70
80 30
153 70
Output
1.0000000000
Input
5 3 100
11 80
14 90
23 70
80 30
153 70
Output
0.9628442962
Input
1 3 20
20 20
Output
0.7500000000
Note
In the first sample the best way of candies' distribution is giving them to first three of the senators. It ensures most of votes.
It the second sample player should give all three candies to the fifth senator. | #!/usr/bin/env python3
n, k, A = map(int, input().rstrip().split())
senators = []
mx_bribe = 0
for i in range(n):
lvl, loy = map(int, input().rstrip().split())
senators.append((lvl, loy))
mx_bribe += (100 - loy) // 10
bribe = [0] * n
def calc(votes):
bsum, cnt, p = 0, 0, 1.0
for i, s in enumerate(senators):
if votes & (1 << i):
p *= (s[1] + bribe[i]) / 100
cnt += 1
else:
p *= (100 - s[1] - bribe[i]) / 100
bsum += s[0]
if cnt > (n / 2):
return p
else:
return p * A / (A + bsum)
def dfs(cur, rk):
if cur >= n:
if rk > 0:
return 0.0
sm = 0.0
for i in range(1 << n):
sm += calc(i)
return sm
mx = 0.0
for i in range(rk + 1):
if i * 10 + senators[cur][1] > 100:
break
bribe[cur] = i * 10
tmp = dfs(cur+1, rk-i)
mx = max(tmp, mx)
return mx
print(dfs(0, min(k, mx_bribe)))
| {
"input": [
"5 3 100\n11 80\n14 90\n23 70\n80 30\n153 70\n",
"5 6 100\n11 80\n14 90\n23 70\n80 30\n153 70\n",
"1 3 20\n20 20\n"
],
"output": [
"0.9628442962\n",
"1.0000000000\n",
"0.7500000000\n"
]
} |
608 | 9 | On his free time, Chouti likes doing some housework. He has got one new task, paint some bricks in the yard.
There are n bricks lined in a row on the ground. Chouti has got m paint buckets of different colors at hand, so he painted each brick in one of those m colors.
Having finished painting all bricks, Chouti was satisfied. He stood back and decided to find something fun with these bricks. After some counting, he found there are k bricks with a color different from the color of the brick on its left (the first brick is not counted, for sure).
So as usual, he needs your help in counting how many ways could he paint the bricks. Two ways of painting bricks are different if there is at least one brick painted in different colors in these two ways. Because the answer might be quite big, you only need to output the number of ways modulo 998 244 353.
Input
The first and only line contains three integers n, m and k (1 ≤ n,m ≤ 2000, 0 ≤ k ≤ n-1) — the number of bricks, the number of colors, and the number of bricks, such that its color differs from the color of brick to the left of it.
Output
Print one integer — the number of ways to color bricks modulo 998 244 353.
Examples
Input
3 3 0
Output
3
Input
3 2 1
Output
4
Note
In the first example, since k=0, the color of every brick should be the same, so there will be exactly m=3 ways to color the bricks.
In the second example, suppose the two colors in the buckets are yellow and lime, the following image shows all 4 possible colorings.
<image> | def f(x):
p=1;
for i in range(1,x):
p*=i
return p
n,m,k=map(int,input().split())
print(m*(m-1)**k*f(n)//f(k+1)//f(n-k)%998244353) | {
"input": [
"3 3 0\n",
"3 2 1\n"
],
"output": [
"3\n",
"4\n"
]
} |
609 | 9 | NN is an experienced internet user and that means he spends a lot of time on the social media. Once he found the following image on the Net, which asked him to compare the sizes of inner circles:
<image>
It turned out that the circles are equal. NN was very surprised by this fact, so he decided to create a similar picture himself.
He managed to calculate the number of outer circles n and the radius of the inner circle r. NN thinks that, using this information, you can exactly determine the radius of the outer circles R so that the inner circle touches all of the outer ones externally and each pair of neighboring outer circles also touches each other. While NN tried very hard to guess the required radius, he didn't manage to do that.
Help NN find the required radius for building the required picture.
Input
The first and the only line of the input file contains two numbers n and r (3 ≤ n ≤ 100, 1 ≤ r ≤ 100) — the number of the outer circles and the radius of the inner circle respectively.
Output
Output a single number R — the radius of the outer circle required for building the required picture.
Your answer will be accepted if its relative or absolute error does not exceed 10^{-6}.
Formally, if your answer is a and the jury's answer is b. Your answer is accepted if and only when (|a-b|)/(max(1, |b|)) ≤ 10^{-6}.
Examples
Input
3 1
Output
6.4641016
Input
6 1
Output
1.0000000
Input
100 100
Output
3.2429391 | import math
def f(n, r):
ct = math.sin(math.pi / n)
return r * ct/(1 - ct)
print(f(*map(int,input().split())))
| {
"input": [
"3 1\n",
"100 100\n",
"6 1\n"
],
"output": [
"6.464101615",
"3.242939086",
"1"
]
} |
610 | 7 | This is a simplified version of the task Toy Train. These two versions differ only in the constraints. Hacks for this version are disabled.
Alice received a set of Toy Train™ from Bob. It consists of one train and a connected railway network of n stations, enumerated from 1 through n. The train occupies one station at a time and travels around the network of stations in a circular manner. More precisely, the immediate station that the train will visit after station i is station i+1 if 1 ≤ i < n or station 1 if i = n. It takes the train 1 second to travel to its next station as described.
Bob gave Alice a fun task before he left: to deliver m candies that are initially at some stations to their independent destinations using the train. The candies are enumerated from 1 through m. Candy i (1 ≤ i ≤ m), now at station a_i, should be delivered to station b_i (a_i ≠ b_i).
<image> The blue numbers on the candies correspond to b_i values. The image corresponds to the 1-st example.
The train has infinite capacity, and it is possible to load off any number of candies at a station. However, only at most one candy can be loaded from a station onto the train before it leaves the station. You can choose any candy at this station. The time it takes to move the candies is negligible.
Now, Alice wonders how much time is needed for the train to deliver all candies. Your task is to find, for each station, the minimum time the train would need to deliver all the candies were it to start from there.
Input
The first line contains two space-separated integers n and m (2 ≤ n ≤ 100; 1 ≤ m ≤ 200) — the number of stations and the number of candies, respectively.
The i-th of the following m lines contains two space-separated integers a_i and b_i (1 ≤ a_i, b_i ≤ n; a_i ≠ b_i) — the station that initially contains candy i and the destination station of the candy, respectively.
Output
In the first and only line, print n space-separated integers, the i-th of which is the minimum time, in seconds, the train would need to deliver all the candies were it to start from station i.
Examples
Input
5 7
2 4
5 1
2 3
3 4
4 1
5 3
3 5
Output
10 9 10 10 9
Input
2 3
1 2
1 2
1 2
Output
5 6
Note
Consider the second sample.
If the train started at station 1, the optimal strategy is as follows.
1. Load the first candy onto the train.
2. Proceed to station 2. This step takes 1 second.
3. Deliver the first candy.
4. Proceed to station 1. This step takes 1 second.
5. Load the second candy onto the train.
6. Proceed to station 2. This step takes 1 second.
7. Deliver the second candy.
8. Proceed to station 1. This step takes 1 second.
9. Load the third candy onto the train.
10. Proceed to station 2. This step takes 1 second.
11. Deliver the third candy.
Hence, the train needs 5 seconds to complete the tasks.
If the train were to start at station 2, however, it would need to move to station 1 before it could load the first candy, which would take one additional second. Thus, the answer in this scenario is 5+1 = 6 seconds. | n, m = map(int, input().split())
def dist(a, b):
return (n + b - a) % n
def main():
inp1 = [0] * (n + 1)
inp2 = [n] * (n + 1)
for _ in range(m):
a, b = map(int, input().split())
inp1[a] += 1
inp2[a] = min(inp2[a], dist(a, b))
inp = tuple((((r1 - 1) * n + r2) for r1, r2 in zip(inp1, inp2)))
print(*(max((dist(i, j) + inp[j] for j in range(1, n + 1) if inp[j])) for i in range(1, n + 1)))
main()
| {
"input": [
"5 7\n2 4\n5 1\n2 3\n3 4\n4 1\n5 3\n3 5\n",
"2 3\n1 2\n1 2\n1 2\n"
],
"output": [
"10 9 10 10 9\n",
"5 6\n"
]
} |
611 | 7 | We're giving away nice huge bags containing number tiles! A bag we want to present to you contains n tiles. Each of them has a single number written on it — either 1 or 2.
However, there is one condition you must fulfill in order to receive the prize. You will need to put all the tiles from the bag in a sequence, in any order you wish. We will then compute the sums of all prefixes in the sequence, and then count how many of these sums are prime numbers. If you want to keep the prize, you will need to maximize the number of primes you get.
Can you win the prize? Hurry up, the bags are waiting!
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 200 000) — the number of number tiles in the bag. The following line contains n space-separated integers a_1, a_2, ..., a_n (a_i ∈ \{1, 2\}) — the values written on the tiles.
Output
Output a permutation b_1, b_2, ..., b_n of the input sequence (a_1, a_2, ..., a_n) maximizing the number of the prefix sums being prime numbers. If there are multiple optimal permutations, output any.
Examples
Input
5
1 2 1 2 1
Output
1 1 1 2 2
Input
9
1 1 2 1 1 1 2 1 1
Output
1 1 1 2 1 1 1 2 1
Note
The first solution produces the prefix sums 1, \mathbf{\color{blue}{2}}, \mathbf{\color{blue}{3}}, \mathbf{\color{blue}{5}}, \mathbf{\color{blue}{7}} (four primes constructed), while the prefix sums in the second solution are 1, \mathbf{\color{blue}{2}}, \mathbf{\color{blue}{3}}, \mathbf{\color{blue}{5}}, 6, \mathbf{\color{blue}{7}}, 8, 10, \mathbf{\color{blue}{11}} (five primes). Primes are marked bold and blue. In each of these cases, the number of produced primes is maximum possible. | def max_primes(n,l):
l.sort(reverse=True)
l.insert(1,l[-1])
l.pop()
print(*l)
n=int(input())
l=list(map(int,input().split()))
max_primes(n,l) | {
"input": [
"9\n1 1 2 1 1 1 2 1 1\n",
"5\n1 2 1 2 1\n"
],
"output": [
"2 1 2 1 1 1 1 1 1\n",
"2 1 2 1 1\n"
]
} |
612 | 9 | You are given a string s, consisting of small Latin letters. Let's denote the length of the string as |s|. The characters in the string are numbered starting from 1.
Your task is to find out if it is possible to rearrange characters in string s so that for any prime number p ≤ |s| and for any integer i ranging from 1 to |s| / p (inclusive) the following condition was fulfilled sp = sp × i. If the answer is positive, find one way to rearrange the characters.
Input
The only line contains the initial string s, consisting of small Latin letters (1 ≤ |s| ≤ 1000).
Output
If it is possible to rearrange the characters in the string so that the above-mentioned conditions were fulfilled, then print in the first line "YES" (without the quotes) and print on the second line one of the possible resulting strings. If such permutation is impossible to perform, then print the single string "NO".
Examples
Input
abc
Output
YES
abc
Input
abcd
Output
NO
Input
xxxyxxx
Output
YES
xxxxxxy
Note
In the first sample any of the six possible strings will do: "abc", "acb", "bac", "bca", "cab" or "cba".
In the second sample no letter permutation will satisfy the condition at p = 2 (s2 = s4).
In the third test any string where character "y" doesn't occupy positions 2, 3, 4, 6 will be valid. | import math
ch='abcdefghijklmnopqrstuvwxyz'
def sieve(n):
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * 2, n + 1, p):
prime[i] = False
p += 1
prime[0]= False
prime[1]= False
s = ['#']+list(input())
lis=[0]*26
n = len(s)-1
prime = [True for i in range(1000 + 1)]
sieve(1000)
ans=['']*(n+1)
aa=[]
aa.append(1)
for i in s[1:]:
lis[ord(i)-ord('a')]+=1
for i in range(n//2+1,n+1,1):
if prime[i]:
aa.append(i)
v = n-len(aa)
th=-1
for i in range(26):
if lis[i]>=v:
th=i
if th==-1:
print("NO")
exit()
for i in range(2,n+1):
if i not in aa:
ans[i]=ch[th]
lis[th]-=1
j=0
#print(ans,aa,lis)
for i in aa:
while j<26 and lis[j]<=0:
j+=1
ans[i]=ch[j]
lis[j]-=1
# print(ans,lis)
print("YES")
print(*ans[1:],sep='')
| {
"input": [
"abcd\n",
"abc\n",
"xxxyxxx\n"
],
"output": [
"NO\n",
"YES\nabc",
"YES\nxxxxxxy\n"
]
} |
613 | 7 | Let's call a positive integer composite if it has at least one divisor other than 1 and itself. For example:
* the following numbers are composite: 1024, 4, 6, 9;
* the following numbers are not composite: 13, 1, 2, 3, 37.
You are given a positive integer n. Find two composite integers a,b such that a-b=n.
It can be proven that solution always exists.
Input
The input contains one integer n (1 ≤ n ≤ 10^7): the given integer.
Output
Print two composite integers a,b (2 ≤ a, b ≤ 10^9, a-b=n).
It can be proven, that solution always exists.
If there are several possible solutions, you can print any.
Examples
Input
1
Output
9 8
Input
512
Output
4608 4096 | def answer():
a = int(input())
print(9*a,8*a)
answer() | {
"input": [
"512\n",
"1\n"
],
"output": [
"516 4\n",
"9 8\n"
]
} |
614 | 7 | Let's define a number ebne (even but not even) if and only if its sum of digits is divisible by 2 but the number itself is not divisible by 2. For example, 13, 1227, 185217 are ebne numbers, while 12, 2, 177013, 265918 are not. If you're still unsure what ebne numbers are, you can look at the sample notes for more clarification.
You are given a non-negative integer s, consisting of n digits. You can delete some digits (they are not necessary consecutive/successive) to make the given number ebne. You cannot change the order of the digits, that is, after deleting the digits the remaining digits collapse. The resulting number shouldn't contain leading zeros. You can delete any number of digits between 0 (do not delete any digits at all) and n-1.
For example, if you are given s=222373204424185217171912 then one of possible ways to make it ebne is: 222373204424185217171912 → 2237344218521717191. The sum of digits of 2237344218521717191 is equal to 70 and is divisible by 2, but number itself is not divisible by 2: it means that the resulting number is ebne.
Find any resulting number that is ebne. If it's impossible to create an ebne number from the given number report about it.
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. The description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 3000) — the number of digits in the original number.
The second line of each test case contains a non-negative integer number s, consisting of n digits.
It is guaranteed that s does not contain leading zeros and the sum of n over all test cases does not exceed 3000.
Output
For each test case given in the input print the answer in the following format:
* If it is impossible to create an ebne number, print "-1" (without quotes);
* Otherwise, print the resulting number after deleting some, possibly zero, but not all digits. This number should be ebne. If there are multiple answers, you can print any of them. Note that answers with leading zeros or empty strings are not accepted. It's not necessary to minimize or maximize the number of deleted digits.
Example
Input
4
4
1227
1
0
6
177013
24
222373204424185217171912
Output
1227
-1
17703
2237344218521717191
Note
In the first test case of the example, 1227 is already an ebne number (as 1 + 2 + 2 + 7 = 12, 12 is divisible by 2, while in the same time, 1227 is not divisible by 2) so we don't need to delete any digits. Answers such as 127 and 17 will also be accepted.
In the second test case of the example, it is clearly impossible to create an ebne number from the given number.
In the third test case of the example, there are many ebne numbers we can obtain by deleting, for example, 1 digit such as 17703, 77013 or 17013. Answers such as 1701 or 770 will not be accepted as they are not ebne numbers. Answer 013 will not be accepted as it contains leading zeroes.
Explanation:
* 1 + 7 + 7 + 0 + 3 = 18. As 18 is divisible by 2 while 17703 is not divisible by 2, we can see that 17703 is an ebne number. Same with 77013 and 17013;
* 1 + 7 + 0 + 1 = 9. Because 9 is not divisible by 2, 1701 is not an ebne number;
* 7 + 7 + 0 = 14. This time, 14 is divisible by 2 but 770 is also divisible by 2, therefore, 770 is not an ebne number.
In the last test case of the example, one of many other possible answers is given. Another possible answer is: 222373204424185217171912 → 22237320442418521717191 (delete the last digit). | def solve(s):
res = ''.join(list(filter(lambda i: int(i) % 2 == 1, s)))[:2]
return -1 if len(res) < 2 else res
n = int(input())
for i in range(n):
input()
print(solve(input()))
| {
"input": [
"4\n4\n1227\n1\n0\n6\n177013\n24\n222373204424185217171912\n"
],
"output": [
"17\n-1\n17\n37\n"
]
} |
615 | 8 | You are given an array a of length n.
You are also given a set of distinct positions p_1, p_2, ..., p_m, where 1 ≤ p_i < n. The position p_i means that you can swap elements a[p_i] and a[p_i + 1]. You can apply this operation any number of times for each of the given positions.
Your task is to determine if it is possible to sort the initial array in non-decreasing order (a_1 ≤ a_2 ≤ ... ≤ a_n) using only allowed swaps.
For example, if a = [3, 2, 1] and p = [1, 2], then we can first swap elements a[2] and a[3] (because position 2 is contained in the given set p). We get the array a = [3, 1, 2]. Then we swap a[1] and a[2] (position 1 is also contained in p). We get the array a = [1, 3, 2]. Finally, we swap a[2] and a[3] again and get the array a = [1, 2, 3], sorted in non-decreasing order.
You can see that if a = [4, 1, 2, 3] and p = [3, 2] then you cannot sort the array.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
Then t test cases follow. The first line of each test case contains two integers n and m (1 ≤ m < n ≤ 100) — the number of elements in a and the number of elements in p. The second line of the test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100). The third line of the test case contains m integers p_1, p_2, ..., p_m (1 ≤ p_i < n, all p_i are distinct) — the set of positions described in the problem statement.
Output
For each test case, print the answer — "YES" (without quotes) if you can sort the initial array in non-decreasing order (a_1 ≤ a_2 ≤ ... ≤ a_n) using only allowed swaps. Otherwise, print "NO".
Example
Input
6
3 2
3 2 1
1 2
4 2
4 1 2 3
3 2
5 1
1 2 3 4 5
1
4 2
2 1 4 3
1 3
4 2
4 3 2 1
1 3
5 2
2 1 2 3 3
1 4
Output
YES
NO
YES
YES
NO
YES |
import math
def rints():
return list(map(int,input().split()))
t = int(input())
for _ in range(t):
n,m = rints()
a = rints()
p = rints()
for _ in range(100):
for pi in p:
if a[pi-1] > a[pi]:
a[pi], a[pi-1] = a[pi-1], a[pi]
print("YES" if sorted(a[:]) == a else "NO")
| {
"input": [
"6\n3 2\n3 2 1\n1 2\n4 2\n4 1 2 3\n3 2\n5 1\n1 2 3 4 5\n1\n4 2\n2 1 4 3\n1 3\n4 2\n4 3 2 1\n1 3\n5 2\n2 1 2 3 3\n1 4\n"
],
"output": [
"YES\nNO\nYES\nYES\nNO\nYES\n"
]
} |
616 | 12 | Kate has a set S of n integers \{1, ..., n\} .
She thinks that imperfection of a subset M ⊆ S is equal to the maximum of gcd(a, b) over all pairs (a, b) such that both a and b are in M and a ≠ b.
Kate is a very neat girl and for each k ∈ \{2, ..., n\} she wants to find a subset that has the smallest imperfection among all subsets in S of size k. There can be more than one subset with the smallest imperfection and the same size, but you don't need to worry about it. Kate wants to find all the subsets herself, but she needs your help to find the smallest possible imperfection for each size k, will name it I_k.
Please, help Kate to find I_2, I_3, ..., I_n.
Input
The first and only line in the input consists of only one integer n (2≤ n ≤ 5 ⋅ 10^5) — the size of the given set S.
Output
Output contains only one line that includes n - 1 integers: I_2, I_3, ..., I_n.
Examples
Input
2
Output
1
Input
3
Output
1 1
Note
First sample: answer is 1, because gcd(1, 2) = 1.
Second sample: there are subsets of S with sizes 2, 3 with imperfection equal to 1. For example, \{2,3\} and \{1, 2, 3\}. | def f(n):
d = [1]*(n+1)
for i in range(2, int(n**0.5)+1):
if d[i]>1:
continue
for j in range(2*i, n+1, i):
d[j] = max(d[j], j//i)
return d
n = int(input())
d = f(n)[1:]
d.sort()
print(*d[1:])
| {
"input": [
"3\n",
"2\n"
],
"output": [
"1 1",
"1"
]
} |
617 | 9 | The statement of this problem is the same as the statement of problem C2. The only difference is that, in problem C1, n is always even, and in C2, n is always odd.
You are given a regular polygon with 2 ⋅ n vertices (it's convex and has equal sides and equal angles) and all its sides have length 1. Let's name it as 2n-gon.
Your task is to find the square of the minimum size such that you can embed 2n-gon in the square. Embedding 2n-gon in the square means that you need to place 2n-gon in the square in such way that each point which lies inside or on a border of 2n-gon should also lie inside or on a border of the square.
You can rotate 2n-gon and/or the square.
Input
The first line contains a single integer T (1 ≤ T ≤ 200) — the number of test cases.
Next T lines contain descriptions of test cases — one per line. Each line contains single even integer n (2 ≤ n ≤ 200). Don't forget you need to embed 2n-gon, not an n-gon.
Output
Print T real numbers — one per test case. For each test case, print the minimum length of a side of the square 2n-gon can be embedded in. Your answer will be considered correct if its absolute or relative error doesn't exceed 10^{-6}.
Example
Input
3
2
4
200
Output
1.000000000
2.414213562
127.321336469 | import math as np
def fun():
n = int(input())
print(1/np.tan(np.pi/(2*n)))
T = int(input())
for _ in range(1,T+1):
fun() | {
"input": [
"3\n2\n4\n200\n"
],
"output": [
"1.0000000000000002\n2.414213562373095\n127.32133646887215\n"
]
} |
618 | 7 | You are given three integers x, y and n. Your task is to find the maximum integer k such that 0 ≤ k ≤ n that k mod x = y, where mod is modulo operation. Many programming languages use percent operator % to implement it.
In other words, with given x, y and n you need to find the maximum possible integer from 0 to n that has the remainder y modulo x.
You have to answer t independent test cases. It is guaranteed that such k exists for each test case.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 5 ⋅ 10^4) — the number of test cases. The next t lines contain test cases.
The only line of the test case contains three integers x, y and n (2 ≤ x ≤ 10^9;~ 0 ≤ y < x;~ y ≤ n ≤ 10^9).
It can be shown that such k always exists under the given constraints.
Output
For each test case, print the answer — maximum non-negative integer k such that 0 ≤ k ≤ n and k mod x = y. It is guaranteed that the answer always exists.
Example
Input
7
7 5 12345
5 0 4
10 5 15
17 8 54321
499999993 9 1000000000
10 5 187
2 0 999999999
Output
12339
0
15
54306
999999995
185
999999998
Note
In the first test case of the example, the answer is 12339 = 7 ⋅ 1762 + 5 (thus, 12339 mod 7 = 5). It is obvious that there is no greater integer not exceeding 12345 which has the remainder 5 modulo 7. | def main():
(x,y,n) = map(int, input().split(' '))
print (((n-y)//x)*x+y)
for _ in range(int(input())):
main()
| {
"input": [
"7\n7 5 12345\n5 0 4\n10 5 15\n17 8 54321\n499999993 9 1000000000\n10 5 187\n2 0 999999999\n"
],
"output": [
"12339\n0\n15\n54306\n999999995\n185\n999999998\n"
]
} |
619 | 8 | Let's call a list of positive integers a_0, a_1, ..., a_{n-1} a power sequence if there is a positive integer c, so that for every 0 ≤ i ≤ n-1 then a_i = c^i.
Given a list of n positive integers a_0, a_1, ..., a_{n-1}, you are allowed to:
* Reorder the list (i.e. pick a permutation p of \{0,1,...,n - 1\} and change a_i to a_{p_i}), then
* Do the following operation any number of times: pick an index i and change a_i to a_i - 1 or a_i + 1 (i.e. increment or decrement a_i by 1) with a cost of 1.
Find the minimum cost to transform a_0, a_1, ..., a_{n-1} into a power sequence.
Input
The first line contains an integer n (3 ≤ n ≤ 10^5).
The second line contains n integers a_0, a_1, ..., a_{n-1} (1 ≤ a_i ≤ 10^9).
Output
Print the minimum cost to transform a_0, a_1, ..., a_{n-1} into a power sequence.
Examples
Input
3
1 3 2
Output
1
Input
3
1000000000 1000000000 1000000000
Output
1999982505
Note
In the first example, we first reorder \{1, 3, 2\} into \{1, 2, 3\}, then increment a_2 to 4 with cost 1 to get a power sequence \{1, 2, 4\}. | n,s=open(0)
arr=sorted(map(int,s.split()))
n=int(n)
def mys(x):
s=0
for i in range(n):
s+=abs(arr[i]-x**i)
return s
mmax=mys(1)
mmin=mys(1)
t=2
while t**(n-1)<=(mmax+arr[n-1]):
mmin=min(mmin,mys(t))
t+=1
print(mmin) | {
"input": [
"3\n1 3 2\n",
"3\n1000000000 1000000000 1000000000\n"
],
"output": [
"1\n",
"1999982505\n"
]
} |
620 | 9 | This is the hard version of the problem. The difference between the versions is that the easy version has no swap operations. You can make hacks only if all versions of the problem are solved.
Pikachu is a cute and friendly pokémon living in the wild pikachu herd.
But it has become known recently that infamous team R wanted to steal all these pokémon! Pokémon trainer Andrew decided to help Pikachu to build a pokémon army to resist.
First, Andrew counted all the pokémon — there were exactly n pikachu. The strength of the i-th pokémon is equal to a_i, and all these numbers are distinct.
As an army, Andrew can choose any non-empty subsequence of pokemons. In other words, Andrew chooses some array b from k indices such that 1 ≤ b_1 < b_2 < ... < b_k ≤ n, and his army will consist of pokémons with forces a_{b_1}, a_{b_2}, ..., a_{b_k}.
The strength of the army is equal to the alternating sum of elements of the subsequence; that is, a_{b_1} - a_{b_2} + a_{b_3} - a_{b_4} + ....
Andrew is experimenting with pokémon order. He performs q operations. In i-th operation Andrew swaps l_i-th and r_i-th pokémon.
Andrew wants to know the maximal stregth of the army he can achieve with the initial pokémon placement. He also needs to know the maximal strength after each operation.
Help Andrew and the pokémon, or team R will realize their tricky plan!
Input
Each test contains multiple test cases.
The first line contains one positive integer t (1 ≤ t ≤ 10^3) denoting the number of test cases. Description of the test cases follows.
The first line of each test case contains two integers n and q (1 ≤ n ≤ 3 ⋅ 10^5, 0 ≤ q ≤ 3 ⋅ 10^5) denoting the number of pokémon and number of operations respectively.
The second line contains n distinct positive integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n) denoting the strengths of the pokémon.
i-th of the last q lines contains two positive integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) denoting the indices of pokémon that were swapped in the i-th operation.
It is guaranteed that the sum of n over all test cases does not exceed 3 ⋅ 10^5, and the sum of q over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case, print q+1 integers: the maximal strength of army before the swaps and after each swap.
Example
Input
3
3 1
1 3 2
1 2
2 2
1 2
1 2
1 2
7 5
1 2 5 4 3 6 7
1 2
6 7
3 4
1 2
2 3
Output
3
4
2
2
2
9
10
10
10
9
11
Note
Let's look at the third test case:
Initially we can build an army in such way: [1 2 5 4 3 6 7], its strength will be 5-3+7=9.
After first operation we can build an army in such way: [2 1 5 4 3 6 7], its strength will be 2-1+5-3+7=10.
After second operation we can build an army in such way: [2 1 5 4 3 7 6], its strength will be 2-1+5-3+7=10.
After third operation we can build an army in such way: [2 1 4 5 3 7 6], its strength will be 2-1+5-3+7=10.
After forth operation we can build an army in such way: [1 2 4 5 3 7 6], its strength will be 5-3+7=9.
After all operations we can build an army in such way: [1 4 2 5 3 7 6], its strength will be 4-2+5-3+7=11. | import sys
import array
input=sys.stdin.readline
t=int(input())
for _ in range(t):
n,q=map(int,input().split())
a=array.array("i",[-1]+list(map(int,input().split()))+[-1])
def f(i):
if i<=0 or i>=n+1:
return 0
if a[i-1]<a[i]>a[i+1]:
return a[i]
if a[i-1]>a[i]<a[i+1]:
return -a[i]
return 0
ans=0
for i in range(n):
ans+=f(i+1)
print(ans)
for __ in range(q):
l,r=map(int,input().split())
u,v=a[l],a[r]
ans-=f(l-1)+f(l)+f(l+1)
a[l]=v
ans+=f(l-1)+f(l)+f(l+1)
ans-=f(r-1)+f(r)+f(r+1)
a[r]=u
ans+=f(r-1)+f(r)+f(r+1)
print(ans) | {
"input": [
"3\n3 1\n1 3 2\n1 2\n2 2\n1 2\n1 2\n1 2\n7 5\n1 2 5 4 3 6 7\n1 2\n6 7\n3 4\n1 2\n2 3\n"
],
"output": [
"3\n4\n2\n2\n2\n9\n10\n10\n10\n9\n11\n"
]
} |
621 | 8 | Athenaeus has just finished creating his latest musical composition and will present it tomorrow to the people of Athens. Unfortunately, the melody is rather dull and highly likely won't be met with a warm reception.
His song consists of n notes, which we will treat as positive integers. The diversity of a song is the number of different notes it contains. As a patron of music, Euterpe watches over composers and guides them throughout the process of creating new melodies. She decided to help Athenaeus by changing his song to make it more diverse.
Being a minor goddess, she cannot arbitrarily change the song. Instead, for each of the n notes in the song, she can either leave it as it is or increase it by 1.
Given the song as a sequence of integers describing the notes, find out the maximal, achievable diversity.
Input
The input consists of multiple test cases. The first line contains an integer t (1 ≤ t ≤ 10 000) — the number of test cases. Then t test cases follow, each one is described in two lines.
In the first line of each test case there is a single integer n (1 ≤ n ≤ 10^5) denoting the length of the song. The next line contains a sequence of n integers x_1, x_2, …, x_n (1 ≤ x_1 ≤ x_2 ≤ … ≤ x_n ≤ 2 ⋅ n), describing the song.
The sum of n over all test cases does not exceed 10^5.
Output
For each test case, you should output a single line containing precisely one integer, the maximal diversity of the song, i.e. the maximal possible number of different elements in the final sequence.
Example
Input
5
6
1 2 2 2 5 6
2
4 4
6
1 1 3 4 4 5
1
1
6
1 1 1 2 2 2
Output
5
2
6
1
3
Note
In the first test case, Euterpe can increase the second, fifth and sixth element to obtain the sequence 1, \underline{3}, 2, 2, \underline{6}, \underline{7}, which has 5 different elements (increased elements are underlined).
In the second test case, Euterpe can increase the first element to obtain the sequence \underline{5}, 4, which has 2 different elements.
In the third test case, Euterpe can increase the second, fifth and sixth element to obtain the sequence 1, \underline{2}, 3, 4, \underline{5}, \underline{6}, which has 6 different elements. | def solve():
n=int(input())
A=[int(x) for x in input().split()]
cnt = [0 for x in range(max(A)+3)]
for x in A: cnt[x]=cnt[x]+1
ans=0
for i in range(max(A)+2):
if cnt[i]>1:
cnt[i+1]=cnt[i+1]+1
if cnt[i]:ans=ans+1
print(ans)
import sys
#sys.stdout=open("oup.txt","w")
T=int(input())
for i in range(T):
solve()
| {
"input": [
"5\n6\n1 2 2 2 5 6\n2\n4 4\n6\n1 1 3 4 4 5\n1\n1\n6\n1 1 1 2 2 2\n"
],
"output": [
"\n5\n2\n6\n1\n3\n"
]
} |
622 | 9 | You are given a positive integer x. Check whether the number x is representable as the sum of the cubes of two positive integers.
Formally, you need to check if there are two integers a and b (1 ≤ a, b) such that a^3+b^3=x.
For example, if x = 35, then the numbers a=2 and b=3 are suitable (2^3+3^3=8+27=35). If x=4, then no pair of numbers a and b is suitable.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases. Then t test cases follow.
Each test case contains one integer x (1 ≤ x ≤ 10^{12}).
Please note, that the input for some test cases won't fit into 32-bit integer type, so you should use at least 64-bit integer type in your programming language.
Output
For each test case, output on a separate line:
* "YES" if x is representable as the sum of the cubes of two positive integers.
* "NO" otherwise.
You can output "YES" and "NO" in any case (for example, the strings yEs, yes, Yes and YES will be recognized as positive).
Example
Input
7
1
2
4
34
35
16
703657519796
Output
NO
YES
NO
NO
YES
YES
YES
Note
The number 1 is not representable as the sum of two cubes.
The number 2 is represented as 1^3+1^3.
The number 4 is not representable as the sum of two cubes.
The number 34 is not representable as the sum of two cubes.
The number 35 is represented as 2^3+3^3.
The number 16 is represented as 2^3+2^3.
The number 703657519796 is represented as 5779^3+7993^3. | def cube(n):
i=1
while(i**3<n):
if(round((n-i**3)**(1/3))**3==(n-i**3)):
return 'Yes'
i+=1
return 'No'
t=int(input())
for _ in range(t):
n=int(input())
print(cube(n)) | {
"input": [
"7\n1\n2\n4\n34\n35\n16\n703657519796\n"
],
"output": [
"\nNO\nYES\nNO\nNO\nYES\nYES\nYES\n"
]
} |
623 | 12 | Polycarp is wondering about buying a new computer, which costs c tugriks. To do this, he wants to get a job as a programmer in a big company.
There are n positions in Polycarp's company, numbered starting from one. An employee in position i earns a[i] tugriks every day. The higher the position number, the more tugriks the employee receives. Initially, Polycarp gets a position with the number 1 and has 0 tugriks.
Each day Polycarp can do one of two things:
* If Polycarp is in the position of x, then he can earn a[x] tugriks.
* If Polycarp is in the position of x (x < n) and has at least b[x] tugriks, then he can spend b[x] tugriks on an online course and move to the position x+1.
For example, if n=4, c=15, a=[1, 3, 10, 11], b=[1, 2, 7], then Polycarp can act like this:
* On the first day, Polycarp is in the 1-st position and earns 1 tugrik. Now he has 1 tugrik;
* On the second day, Polycarp is in the 1-st position and move to the 2-nd position. Now he has 0 tugriks;
* On the third day, Polycarp is in the 2-nd position and earns 3 tugriks. Now he has 3 tugriks;
* On the fourth day, Polycarp is in the 2-nd position and is transferred to the 3-rd position. Now he has 1 tugriks;
* On the fifth day, Polycarp is in the 3-rd position and earns 10 tugriks. Now he has 11 tugriks;
* On the sixth day, Polycarp is in the 3-rd position and earns 10 tugriks. Now he has 21 tugriks;
* Six days later, Polycarp can buy himself a new computer.
Find the minimum number of days after which Polycarp will be able to buy himself a new computer.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4). Then t test cases follow.
The first line of each test case contains two integers n and c (2 ≤ n ≤ 2 ⋅ 10^5, 1 ≤ c ≤ 10^9) — the number of positions in the company and the cost of a new computer.
The second line of each test case contains n integers a_1 ≤ a_2 ≤ … ≤ a_n (1 ≤ a_i ≤ 10^9).
The third line of each test case contains n - 1 integer b_1, b_2, …, b_{n-1} (1 ≤ b_i ≤ 10^9).
It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case, output the minimum number of days after which Polycarp will be able to buy a new computer.
Example
Input
3
4 15
1 3 10 11
1 2 7
4 100
1 5 10 50
3 14 12
2 1000000000
1 1
1
Output
6
13
1000000000 | from math import ceil
def f(arr,b,c):
b.append(0)
currd=0
currm=0
ans=float("inf")
j=0
for i in arr:
ans=min(ans,max(ceil((c-currm)/i),0)+currd)
t=max(ceil((b[j]-currm)/i),0)
currd+=t+1
currm+=t*i-b[j]
j += 1
return ans
for i in range(int(input())):
a,b=map(int,input().strip().split())
lst=list(map(int,input().strip().split()))
lst2=list(map(int,input().strip().split()))
print(f(lst,lst2,b)) | {
"input": [
"3\n4 15\n1 3 10 11\n1 2 7\n4 100\n1 5 10 50\n3 14 12\n2 1000000000\n1 1\n1\n"
],
"output": [
"\n6\n13\n1000000000\n"
]
} |
624 | 7 | 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
The first line contains a single integer n (1 ≤ n ≤ 106) — the number of marbles in Polycarpus's sequence.
Output
Print a single number — the answer to the problem modulo 1000000007 (109 + 7).
Examples
Input
3
Output
6
Input
4
Output
11
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. | from queue import PriorityQueue
from queue import Queue
import math
from collections import *
import sys
import operator as op
from functools import reduce
# sys.setrecursionlimit(10 ** 6)
MOD = int(1e9 + 7)
input = sys.stdin.readline
def ii(): return list(map(int, input().strip().split()))
def ist(): return input().strip().split()
n, = ii()
arr = [0] * max(3, n+1)
arr[1], arr[2] = 1, 3
for k in range(3, n+1):
arr[k] = (arr[k-1] * 2 - arr[k-3]) % MOD
print(arr[n])
| {
"input": [
"3\n",
"4\n"
],
"output": [
"6",
"11"
]
} |
625 | 9 | There are n cities numbered from 1 to n in Berland. Some of them are connected by two-way roads. Each road has its own length — an integer number from 1 to 1000. It is known that from each city it is possible to get to any other city by existing roads. Also for each pair of cities it is known the shortest distance between them. Berland Government plans to build k new roads. For each of the planned road it is known its length, and what cities it will connect. To control the correctness of the construction of new roads, after the opening of another road Berland government wants to check the sum of the shortest distances between all pairs of cities. Help them — for a given matrix of shortest distances on the old roads and plans of all new roads, find out how the sum of the shortest distances between all pairs of cities changes after construction of each road.
Input
The first line contains integer n (2 ≤ n ≤ 300) — amount of cities in Berland. Then there follow n lines with n integer numbers each — the matrix of shortest distances. j-th integer in the i-th row — di, j, the shortest distance between cities i and j. It is guaranteed that di, i = 0, di, j = dj, i, and a given matrix is a matrix of shortest distances for some set of two-way roads with integer lengths from 1 to 1000, such that from each city it is possible to get to any other city using these roads.
Next line contains integer k (1 ≤ k ≤ 300) — amount of planned roads. Following k lines contain the description of the planned roads. Each road is described by three space-separated integers ai, bi, ci (1 ≤ ai, bi ≤ n, ai ≠ bi, 1 ≤ ci ≤ 1000) — ai and bi — pair of cities, which the road connects, ci — the length of the road. It can be several roads between a pair of cities, but no road connects the city with itself.
Output
Output k space-separated integers qi (1 ≤ i ≤ k). qi should be equal to the sum of shortest distances between all pairs of cities after the construction of roads with indexes from 1 to i. Roads are numbered from 1 in the input order. Each pair of cities should be taken into account in the sum exactly once, i. e. we count unordered pairs.
Examples
Input
2
0 5
5 0
1
1 2 3
Output
3
Input
3
0 4 5
4 0 9
5 9 0
2
2 3 8
1 2 1
Output
17 12 | from collections import defaultdict
def summ(x):
s=0
for i in range(n):
for j in range(i+1,n):
s+=x[i][j]
return s
INF=10**15
n=int(input())
dis=[]
for _ in range(1,n+1):
dis.append(list(map(int,input().split())))
insum=summ(dis)
k=int(input())
for i in range(k):
a,b,cost=list(map(int,input().split()))
if cost>dis[a-1][b-1]:
print(insum,end=' ')
else:
insum=insum+cost-dis[a-1][b-1]
dis[a-1][b-1]=cost
dis[b-1][a-1]=cost
for o in range(n):
for p in range(o+1,n):
if dis[o][p]>dis[o][a-1]+dis[a-1][b-1]+dis[b-1][p]:
insum=insum+(dis[o][a-1]+dis[a-1][b-1]+dis[b-1][p])-dis[o][p]
dis[o][p]=dis[o][a-1]+dis[a-1][b-1]+dis[b-1][p]
dis[p][o]=dis[o][p]
if dis[o][p]>dis[o][b-1]+dis[a-1][b-1]+dis[a-1][p]:
insum=insum+(dis[o][b-1]+dis[a-1][b-1]+dis[a-1][p])-dis[o][p]
dis[o][p]=dis[o][b-1]+dis[a-1][b-1]+dis[a-1][p]
dis[p][o]=dis[o][p]
print(insum,end=' ')
| {
"input": [
"2\n0 5\n5 0\n1\n1 2 3\n",
"3\n0 4 5\n4 0 9\n5 9 0\n2\n2 3 8\n1 2 1\n"
],
"output": [
"3\n",
"17 12\n"
]
} |
626 | 7 | Vasya has n items lying in a line. The items are consecutively numbered by numbers from 1 to n in such a way that the leftmost item has number 1, the rightmost item has number n. Each item has a weight, the i-th item weights wi kilograms.
Vasya needs to collect all these items, however he won't do it by himself. He uses his brand new robot. The robot has two different arms — the left one and the right one. The robot can consecutively perform the following actions:
1. Take the leftmost item with the left hand and spend wi · l energy units (wi is a weight of the leftmost item, l is some parameter). If the previous action was the same (left-hand), then the robot spends extra Ql energy units;
2. Take the rightmost item with the right hand and spend wj · r energy units (wj is a weight of the rightmost item, r is some parameter). If the previous action was the same (right-hand), then the robot spends extra Qr energy units;
Naturally, Vasya wants to program the robot in a way that the robot spends as little energy as possible. He asked you to solve this problem. Your task is to find the minimum number of energy units robot spends to collect all items.
Input
The first line contains five integers n, l, r, Ql, Qr (1 ≤ n ≤ 105; 1 ≤ l, r ≤ 100; 1 ≤ Ql, Qr ≤ 104).
The second line contains n integers w1, w2, ..., wn (1 ≤ wi ≤ 100).
Output
In the single line print a single number — the answer to the problem.
Examples
Input
3 4 4 19 1
42 3 99
Output
576
Input
4 7 2 3 9
1 2 3 4
Output
34
Note
Consider the first sample. As l = r, we can take an item in turns: first from the left side, then from the right one and last item from the left. In total the robot spends 4·42 + 4·99 + 4·3 = 576 energy units.
The second sample. The optimal solution is to take one item from the right, then one item from the left and two items from the right. In total the robot spends (2·4) + (7·1) + (2·3) + (2·2 + 9) = 34 energy units. | from sys import stdin
from itertools import accumulate
def arr_sum(arr):
return list(accumulate(arr, lambda x, y: x + y))
rints = lambda: [int(x) for x in stdin.readline().split()]
n, l, r, ql, qr = rints()
w, ans = [0] + rints(), float('inf')
mem = arr_sum(w)
for i in range(n + 1):
s1, s2 = mem[i] - mem[0], mem[-1] - mem[i]
tem = l * s1 + r * s2
if (n - i - i - 1) > 0:
tem += (n - i - i - 1) * qr
elif (i - (n - i) - 1) > 0:
tem += (i - (n - i) - 1) * ql
ans = min(ans, tem)
print(ans)
| {
"input": [
"3 4 4 19 1\n42 3 99\n",
"4 7 2 3 9\n1 2 3 4\n"
],
"output": [
"576\n",
"34\n"
]
} |
627 | 8 | Two semifinals have just been in the running tournament. Each semifinal had n participants. There are n participants advancing to the finals, they are chosen as follows: from each semifinal, we choose k people (0 ≤ 2k ≤ n) who showed the best result in their semifinals and all other places in the finals go to the people who haven't ranked in the top k in their semifinal but got to the n - 2k of the best among the others.
The tournament organizers hasn't yet determined the k value, so the participants want to know who else has any chance to get to the finals and who can go home.
Input
The first line contains a single integer n (1 ≤ n ≤ 105) — the number of participants in each semifinal.
Each of the next n lines contains two integers ai and bi (1 ≤ ai, bi ≤ 109) — the results of the i-th participant (the number of milliseconds he needs to cover the semifinals distance) of the first and second semifinals, correspondingly. All results are distinct. Sequences a1, a2, ..., an and b1, b2, ..., bn are sorted in ascending order, i.e. in the order the participants finished in the corresponding semifinal.
Output
Print two strings consisting of n characters, each equals either "0" or "1". The first line should correspond to the participants of the first semifinal, the second line should correspond to the participants of the second semifinal. The i-th character in the j-th line should equal "1" if the i-th participant of the j-th semifinal has any chances to advance to the finals, otherwise it should equal a "0".
Examples
Input
4
9840 9920
9860 9980
9930 10020
10040 10090
Output
1110
1100
Input
4
9900 9850
9940 9930
10000 10020
10060 10110
Output
1100
1100
Note
Consider the first sample. Each semifinal has 4 participants. The results of the first semifinal are 9840, 9860, 9930, 10040. The results of the second semifinal are 9920, 9980, 10020, 10090.
* If k = 0, the finalists are determined by the time only, so players 9840, 9860, 9920 and 9930 advance to the finals.
* If k = 1, the winners from both semifinals move to the finals (with results 9840 and 9920), and the other places are determined by the time (these places go to the sportsmen who run the distance in 9860 and 9930 milliseconds).
* If k = 2, then first and second places advance from each seminfial, these are participants with results 9840, 9860, 9920 and 9980 milliseconds. | n = int(input())
a, b = zip(*(map(int, input().split()) for _ in range(n)))
i = j = 0
while i < n and j < n and i + j < n:
if a[i] < b[j]:
i += 1
else:
j += 1
def p(x): return ("1" * max(n // 2, x)).ljust(n, "0")
print(p(i))
print(p(j))
| {
"input": [
"4\n9840 9920\n9860 9980\n9930 10020\n10040 10090\n",
"4\n9900 9850\n9940 9930\n10000 10020\n10060 10110\n"
],
"output": [
"1110\n1100\n",
"1100\n1100\n"
]
} |
628 | 10 | You can find anything whatsoever in our Galaxy! A cubical planet goes round an icosahedral star. Let us introduce a system of axes so that the edges of the cubical planet are parallel to the coordinate axes and two opposite vertices lay in the points (0, 0, 0) and (1, 1, 1). Two flies live on the planet. At the moment they are sitting on two different vertices of the cubical planet. Your task is to determine whether they see each other or not. The flies see each other when the vertices they occupy lie on the same face of the cube.
Input
The first line contains three space-separated integers (0 or 1) — the coordinates of the first fly, the second line analogously contains the coordinates of the second fly.
Output
Output "YES" (without quotes) if the flies see each other. Otherwise, output "NO".
Examples
Input
0 0 0
0 1 0
Output
YES
Input
1 1 0
0 1 0
Output
YES
Input
0 0 0
1 1 1
Output
NO | def f(l1,l2):
return sum([l1[i]==l2[i] for i in range(len(l1))])>0
l1 = list(map(int,input().split()))
l2 = list(map(int,input().split()))
print('YES' if f(l1,l2) else 'NO')
| {
"input": [
"0 0 0\n1 1 1\n",
"0 0 0\n0 1 0\n",
"1 1 0\n0 1 0\n"
],
"output": [
"NO\n",
"YES\n",
"YES\n"
]
} |
629 | 9 | Jzzhu has picked n apples from his big apple tree. All the apples are numbered from 1 to n. Now he wants to sell them to an apple store.
Jzzhu will pack his apples into groups and then sell them. Each group must contain two apples, and the greatest common divisor of numbers of the apples in each group must be greater than 1. Of course, each apple can be part of at most one group.
Jzzhu wonders how to get the maximum possible number of groups. Can you help him?
Input
A single integer n (1 ≤ n ≤ 105), the number of the apples.
Output
The first line must contain a single integer m, representing the maximum number of groups he can get. Each of the next m lines must contain two integers — the numbers of apples in the current group.
If there are several optimal answers you can print any of them.
Examples
Input
6
Output
2
6 3
2 4
Input
9
Output
3
9 3
2 4
6 8
Input
2
Output
0 | #!/usr/bin/env python3
# -*- coding: utf-8 -*-
from math import *
def prm(x):
s = int(sqrt(x+0.5))
for i in range(2, s+1):
if not (x%i): return False
return True
def ap(a):
while(len(a) >= 2):
print(a[-1], a[-2])
a.pop()
a.pop()
n, cnt, ans, vis, ansl, ansr = int(input()), 0, 0, [False]*100100, [], []
prime = filter(prm, range(3, n//2+1))
for x in prime:
tcnt, nlst = 0, []
for i in range(1, n//x+1):
if not vis[i*x]:
vis[i*x] = True
nlst.append(i)
tcnt += 1
ans += tcnt >> 1
cnt += tcnt & 1
if tcnt & 1 :
nlst = nlst[0:1] + nlst[2:]
ansr.append(x<<1)
ansl += list(map(lambda k : x*k, nlst))
for i in range(1, n+1):
if not vis[i] and not (i&1):
ansr.append(i)
cnt += 1
print(ans+(cnt>>1))
ap(ansl)
ap(ansr)
| {
"input": [
"6\n",
"9\n",
"2\n"
],
"output": [
"2\n3 6\n2 4\n",
"3\n3 9\n2 4\n6 8\n",
"0\n"
]
} |
630 | 11 | Vasya is studying in the last class of school and soon he will take exams. He decided to study polynomials. Polynomial is a function P(x) = a0 + a1x1 + ... + anxn. Numbers ai are called coefficients of a polynomial, non-negative integer n is called a degree of a polynomial.
Vasya has made a bet with his friends that he can solve any problem with polynomials. They suggested him the problem: "Determine how many polynomials P(x) exist with integer non-negative coefficients so that <image>, and <image>, where <image> and b are given positive integers"?
Vasya does not like losing bets, but he has no idea how to solve this task, so please help him to solve the problem.
Input
The input contains three integer positive numbers <image> no greater than 1018.
Output
If there is an infinite number of such polynomials, then print "inf" without quotes, otherwise print the reminder of an answer modulo 109 + 7.
Examples
Input
2 2 2
Output
2
Input
2 3 3
Output
1 | def gnb(x, b):
if b == 1: return [x]
g = []
while x:
g.append(x % b)
x //= b
return g
ans = 0
t, a, b = map(int, input().split())
if t == 1 and a == 1 and b == 1:
print("inf")
raise SystemExit
cf = gnb(b, a)
a2 = sum([x * y for x, y in zip(cf, [t**n for n in range(len(cf))])])
if a2 == a: ans += 1
if len(cf) >= 2 and cf[-1] == 1 and cf[-2] == 0:
cf[-1] = 0
cf[-2] = a
a2 = sum([x * y for x, y in zip(cf, [t**n for n in range(len(cf))])])
if a2 == a: ans += 1
#if a == b: ans += 1
print(ans)
| {
"input": [
"2 3 3\n",
"2 2 2\n"
],
"output": [
"1\n",
"2\n"
]
} |
631 | 10 | Ilya got tired of sports programming, left university and got a job in the subway. He was given the task to determine the escalator load factor.
Let's assume that n people stand in the queue for the escalator. At each second one of the two following possibilities takes place: either the first person in the queue enters the escalator with probability p, or the first person in the queue doesn't move with probability (1 - p), paralyzed by his fear of escalators and making the whole queue wait behind him.
Formally speaking, the i-th person in the queue cannot enter the escalator until people with indices from 1 to i - 1 inclusive enter it. In one second only one person can enter the escalator. The escalator is infinite, so if a person enters it, he never leaves it, that is he will be standing on the escalator at any following second. Ilya needs to count the expected value of the number of people standing on the escalator after t seconds.
Your task is to help him solve this complicated task.
Input
The first line of the input contains three numbers n, p, t (1 ≤ n, t ≤ 2000, 0 ≤ p ≤ 1). Numbers n and t are integers, number p is real, given with exactly two digits after the decimal point.
Output
Print a single real number — the expected number of people who will be standing on the escalator after t seconds. The absolute or relative error mustn't exceed 10 - 6.
Examples
Input
1 0.50 1
Output
0.5
Input
1 0.50 4
Output
0.9375
Input
4 0.20 2
Output
0.4 | def main():
tmp = input().split()
n, p, t = int(tmp[0]), float(tmp[1]), int(tmp[2])
q = 1. - p
res = [0.] * (n + 1)
res[0] = 1.
for _ in range(t):
tmp = [x * q for x in res]
tmp[-1] = res[-1]
for i in range(n):
tmp[i + 1] += res[i] * p
res = tmp
print(sum(x * i for i, x in enumerate(res)))
if __name__ == '__main__':
main() | {
"input": [
"1 0.50 1\n",
"4 0.20 2\n",
"1 0.50 4\n"
],
"output": [
"0.500000000000\n",
"0.400000000000\n",
"0.937500000000\n"
]
} |
632 | 10 | In some country there are exactly n cities and m bidirectional roads connecting the cities. Cities are numbered with integers from 1 to n. If cities a and b are connected by a road, then in an hour you can go along this road either from city a to city b, or from city b to city a. The road network is such that from any city you can get to any other one by moving along the roads.
You want to destroy the largest possible number of roads in the country so that the remaining roads would allow you to get from city s1 to city t1 in at most l1 hours and get from city s2 to city t2 in at most l2 hours.
Determine what maximum number of roads you need to destroy in order to meet the condition of your plan. If it is impossible to reach the desired result, print -1.
Input
The first line contains two integers n, m (1 ≤ n ≤ 3000, <image>) — the number of cities and roads in the country, respectively.
Next m lines contain the descriptions of the roads as pairs of integers ai, bi (1 ≤ ai, bi ≤ n, ai ≠ bi). It is guaranteed that the roads that are given in the description can transport you from any city to any other one. It is guaranteed that each pair of cities has at most one road between them.
The last two lines contains three integers each, s1, t1, l1 and s2, t2, l2, respectively (1 ≤ si, ti ≤ n, 0 ≤ li ≤ n).
Output
Print a single number — the answer to the problem. If the it is impossible to meet the conditions, print -1.
Examples
Input
5 4
1 2
2 3
3 4
4 5
1 3 2
3 5 2
Output
0
Input
5 4
1 2
2 3
3 4
4 5
1 3 2
2 4 2
Output
1
Input
5 4
1 2
2 3
3 4
4 5
1 3 2
3 5 1
Output
-1 | from itertools import combinations_with_replacement
from collections import deque
#sys.stdin = open("input_py.txt","r")
n, m = map(int, input().split())
G = [ [] for i in range(n)]
for i in range(m):
x, y = map(int, input().split())
x-=1; y-=1
G[x].append(y)
G[y].append(x)
def BFS(s):
dist = [-1 for i in range(n)]
dist[s] = 0
Q = deque()
Q.append(s)
while len(Q) > 0:
v = Q.popleft()
for to in G[v]:
if dist[to] < 0:
dist[to] = dist[v] + 1
Q.append(to)
return dist
Dist = [BFS(i) for i in range(n)]
s1, t1, l1 = map(int, input(). split())
s2, t2, l2 = map(int, input(). split())
s1-=1; t1-=1; s2-=1; t2-=1
if Dist[s1][t1] > l1 or Dist[s2][t2] > l2:
print(-1)
exit(0)
rest = Dist[s1][t1] + Dist[s2][t2]
for i in range(n):
for j in range(n):
if Dist[i][s1] + Dist[i][j] + Dist[j][t1] <= l1 and Dist[i][s2] + Dist[i][j] + Dist[j][t2] <= l2 :
rest = min(rest, Dist[i][j] + Dist[i][s1] + Dist[i][s2] + Dist[j][t1] + Dist[j][t2])
if Dist[i][s1] + Dist[i][j] + Dist[j][t1] <= l1 and Dist[j][s2] + Dist[i][j] + Dist[i][t2] <= l2 :
rest = min(rest, Dist[i][j] + Dist[j][t1] + Dist[j][s2] + Dist[i][s1] + Dist[i][t2])
print(m-rest) | {
"input": [
"5 4\n1 2\n2 3\n3 4\n4 5\n1 3 2\n3 5 1\n",
"5 4\n1 2\n2 3\n3 4\n4 5\n1 3 2\n3 5 2\n",
"5 4\n1 2\n2 3\n3 4\n4 5\n1 3 2\n2 4 2\n"
],
"output": [
"-1\n",
"0\n",
"1\n"
]
} |
633 | 7 | You are given three sticks with positive integer lengths of a, b, and c centimeters. You can increase length of some of them by some positive integer number of centimeters (different sticks can be increased by a different length), but in total by at most l centimeters. In particular, it is allowed not to increase the length of any stick.
Determine the number of ways to increase the lengths of some sticks so that you can form from them a non-degenerate (that is, having a positive area) triangle. Two ways are considered different, if the length of some stick is increased by different number of centimeters in them.
Input
The single line contains 4 integers a, b, c, l (1 ≤ a, b, c ≤ 3·105, 0 ≤ l ≤ 3·105).
Output
Print a single integer — the number of ways to increase the sizes of the sticks by the total of at most l centimeters, so that you can make a non-degenerate triangle from it.
Examples
Input
1 1 1 2
Output
4
Input
1 2 3 1
Output
2
Input
10 2 1 7
Output
0
Note
In the first sample test you can either not increase any stick or increase any two sticks by 1 centimeter.
In the second sample test you can increase either the first or the second stick by one centimeter. Note that the triangle made from the initial sticks is degenerate and thus, doesn't meet the conditions. | def f(a,b,c,l):
if a<b+c:
return 0
else:
c=min(l,a-b-c)
return (c+1)*(c+2)/2
a,b,c,l = map(int,input().split())
z=(l+1)*(l+2)*(l+3)/6
i=0
while i<=l:
z-=f(a+i,b,c,l-i)+f(b+i,c,a,l-i)+f(c+i,a,b,l-i)
i+=1
print(int(z))
| {
"input": [
"1 1 1 2\n",
"1 2 3 1\n",
"10 2 1 7\n"
],
"output": [
"4\n",
"2\n",
"0\n"
]
} |
634 | 10 | Ari the monster is not an ordinary monster. She is the hidden identity of Super M, the Byteforces’ superhero. Byteforces is a country that consists of n cities, connected by n - 1 bidirectional roads. Every road connects exactly two distinct cities, and the whole road system is designed in a way that one is able to go from any city to any other city using only the given roads. There are m cities being attacked by humans. So Ari... we meant Super M have to immediately go to each of the cities being attacked to scare those bad humans. Super M can pass from one city to another only using the given roads. Moreover, passing through one road takes her exactly one kron - the time unit used in Byteforces.
<image>
However, Super M is not on Byteforces now - she is attending a training camp located in a nearby country Codeforces. Fortunately, there is a special device in Codeforces that allows her to instantly teleport from Codeforces to any city of Byteforces. The way back is too long, so for the purpose of this problem teleportation is used exactly once.
You are to help Super M, by calculating the city in which she should teleport at the beginning in order to end her job in the minimum time (measured in krons). Also, provide her with this time so she can plan her way back to Codeforces.
Input
The first line of the input contains two integers n and m (1 ≤ m ≤ n ≤ 123456) - the number of cities in Byteforces, and the number of cities being attacked respectively.
Then follow n - 1 lines, describing the road system. Each line contains two city numbers ui and vi (1 ≤ ui, vi ≤ n) - the ends of the road i.
The last line contains m distinct integers - numbers of cities being attacked. These numbers are given in no particular order.
Output
First print the number of the city Super M should teleport to. If there are many possible optimal answers, print the one with the lowest city number.
Then print the minimum possible time needed to scare all humans in cities being attacked, measured in Krons.
Note that the correct answer is always unique.
Examples
Input
7 2
1 2
1 3
1 4
3 5
3 6
3 7
2 7
Output
2
3
Input
6 4
1 2
2 3
2 4
4 5
4 6
2 4 5 6
Output
2
4
Note
In the first sample, there are two possibilities to finish the Super M's job in 3 krons. They are:
<image> and <image>.
However, you should choose the first one as it starts in the city with the lower number. | def f(k, h):
print(k, h)
exit()
def g(p):
i = p.pop()
return i, t[i]
import sys
d = list(map(int, sys.stdin.read().split()))
n, m = d[0], d[1]
if n == 1: f(1, 0)
k = 2 * n
t = [[0, set(), i, 0] for i in range(n + 1)]
for i in d[k:]: t[i][0] = 1
for a, b in zip(d[2:k:2], d[3:k:2]):
t[a][1].add(b)
t[b][1].add(a)
p = ([], [])
for x in t:
if len(x[1]) == 1: p[x[0]].append(x[2])
k = s = 1
while p[0]:
i, x = g(p[0])
j, y = g(x[1])
y[1].remove(i)
if len(y[1]) == 1: p[y[0]].append(j)
if len(p[k]) == 1: f(p[k][0], 0)
for i in p[k]: t[i][3] = 1
while 1:
s += 2
i, x = g(p[k])
j, y = g(x[1])
if len(y[1]) == 1: f(min(x[2], y[2]), s - x[3] - y[3])
x[3] += 1
if x[3] > y[3]: y[2:] = x[2:]
elif x[3] == y[3]: y[2] = min(x[2], y[2])
y[1].remove(i)
if len(y[1]) == 1: p[1 - k].append(j)
if not p[k]: k = 1 - k | {
"input": [
"7 2\n1 2\n1 3\n1 4\n3 5\n3 6\n3 7\n2 7\n",
"6 4\n1 2\n2 3\n2 4\n4 5\n4 6\n2 4 5 6\n"
],
"output": [
"2\n3",
"2\n4"
]
} |
635 | 8 | It's the year 4527 and the tanks game that we all know and love still exists. There also exists Great Gena's code, written in 2016. The problem this code solves is: given the number of tanks that go into the battle from each country, find their product. If it is turns to be too large, then the servers might have not enough time to assign tanks into teams and the whole game will collapse!
There are exactly n distinct countries in the world and the i-th country added ai tanks to the game. As the developers of the game are perfectionists, the number of tanks from each country is beautiful. A beautiful number, according to the developers, is such number that its decimal representation consists only of digits '1' and '0', moreover it contains at most one digit '1'. However, due to complaints from players, some number of tanks of one country was removed from the game, hence the number of tanks of this country may not remain beautiful.
Your task is to write the program that solves exactly the same problem in order to verify Gena's code correctness. Just in case.
Input
The first line of the input contains the number of countries n (1 ≤ n ≤ 100 000). The second line contains n non-negative integers ai without leading zeroes — the number of tanks of the i-th country.
It is guaranteed that the second line contains at least n - 1 beautiful numbers and the total length of all these number's representations doesn't exceed 100 000.
Output
Print a single number without leading zeroes — the product of the number of tanks presented by each country.
Examples
Input
3
5 10 1
Output
50
Input
4
1 1 10 11
Output
110
Input
5
0 3 1 100 1
Output
0
Note
In sample 1 numbers 10 and 1 are beautiful, number 5 is not not.
In sample 2 number 11 is not beautiful (contains two '1's), all others are beautiful.
In sample 3 number 3 is not beautiful, all others are beautiful. |
n = int(input())
a = input().split()
def get(l, r):
if l == r:
return int(a[l])
m = (l + r) // 2
return get(l, m) * get(m + 1, r)
print(get(0, n - 1))
| {
"input": [
"3\n5 10 1\n",
"5\n0 3 1 100 1\n",
"4\n1 1 10 11\n"
],
"output": [
"50",
"0\n",
"110"
]
} |
636 | 8 | You are given an undirected graph that consists of n vertices and m edges. Initially, each edge is colored either red or blue. Each turn a player picks a single vertex and switches the color of all edges incident to it. That is, all red edges with an endpoint in this vertex change the color to blue, while all blue edges with an endpoint in this vertex change the color to red.
Find the minimum possible number of moves required to make the colors of all edges equal.
Input
The first line of the input contains two integers n and m (1 ≤ n, m ≤ 100 000) — the number of vertices and edges, respectively.
The following m lines provide the description of the edges, as the i-th of them contains two integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi) — the indices of the vertices connected by the i-th edge, and a character ci (<image>) providing the initial color of this edge. If ci equals 'R', then this edge is initially colored red. Otherwise, ci is equal to 'B' and this edge is initially colored blue. It's guaranteed that there are no self-loops and multiple edges.
Output
If there is no way to make the colors of all edges equal output - 1 in the only line of the output. Otherwise first output k — the minimum number of moves required to achieve the goal, then output k integers a1, a2, ..., ak, where ai is equal to the index of the vertex that should be used at the i-th move.
If there are multiple optimal sequences of moves, output any of them.
Examples
Input
3 3
1 2 B
3 1 R
3 2 B
Output
1
2
Input
6 5
1 3 R
2 3 R
3 4 B
4 5 R
4 6 R
Output
2
3 4
Input
4 5
1 2 R
1 3 R
2 3 B
3 4 B
1 4 B
Output
-1 | from collections import deque
n, m = map(int, input().split())
adj = [[] for i in range(n)]
for i in range(m):
u, v, c = input().split()
u, v = int(u)-1, int(v)-1
adj[u].append((v, c))
adj[v].append((u, c))
visited = S = T = None
def bfs(i, k):
q = deque([(i, 0)])
while q:
u, p = q.pop()
if visited[u] >= 0:
if visited[u] == p: continue
else: return False
visited[u] = p
if p: S.append(u)
else: T.append(u)
for v, c in adj[u]:
nxt = p if c == k else p^1
q.appendleft((v, nxt))
return True
def solve(k):
global visited, S, T
visited = [-1]*n
res = []
for i in range(n):
if visited[i] < 0:
S, T = [], []
if not bfs(i, k):
return [0]*(n+1)
else:
res.extend(S if len(S) < len(T) else T)
return res
res1 = solve("R")
res2 = solve("B")
if min(len(res1), len(res2)) > n:
print (-1)
else:
print (min(len(res1), len(res2)))
print (" ".join(map(lambda x: str(x+1), res1 if len(res1) < len(res2) else res2)))
| {
"input": [
"4 5\n1 2 R\n1 3 R\n2 3 B\n3 4 B\n1 4 B\n",
"6 5\n1 3 R\n2 3 R\n3 4 B\n4 5 R\n4 6 R\n",
"3 3\n1 2 B\n3 1 R\n3 2 B\n"
],
"output": [
"-1",
"2\n3 4 ",
"1\n2 "
]
} |
637 | 10 | For each string s consisting of characters '0' and '1' one can define four integers a00, a01, a10 and a11, where axy is the number of subsequences of length 2 of the string s equal to the sequence {x, y}.
In these problem you are given four integers a00, a01, a10, a11 and have to find any non-empty string s that matches them, or determine that there is no such string. One can prove that if at least one answer exists, there exists an answer of length no more than 1 000 000.
Input
The only line of the input contains four non-negative integers a00, a01, a10 and a11. Each of them doesn't exceed 109.
Output
If there exists a non-empty string that matches four integers from the input, print it in the only line of the output. Otherwise, print "Impossible". The length of your answer must not exceed 1 000 000.
Examples
Input
1 2 3 4
Output
Impossible
Input
1 2 2 1
Output
0110 | root = lambda k : int((1 + (1 + 8*k) ** 0.5)/2)
def check(k):
n = root(k)
return n*(n-1)/2 == k
def solve(a00, a01, a10, a11):
s = a00 + a01 + a10 + a11
if not check(a00) or not check(a11) or not check(s):
return None
x, y, z = root(a00), root(a11), root(s)
if a00 == 0 or a11 == 0:
if s == 0:
return '0'
elif a10 == 0 and a01 == 0:
return '0'*z if a00 > 0 else '1'*z
elif a10 + a01 == z -1:
return '1'*a10 + '0' + '1'*a01 if a11 > 0 else '0'*a01 + '1' + '0'*a10
else:
return None
if x + y != z:
return None
if x == 0:
if a00 == 0 and a01 == 0 and a10 == 0:
return '1'*z
else:
return None
t, k = a10//x, a10%x
if z-t-x > 0:
return '1'*t + '0'*(x-k) + '1' + '0'*k + '1'*(z-t-x-1)
else:
return '1'*t + x*'0'
a00, a01, a10, a11 = [int(x) for x in input().split()]
ans = solve(a00, a01, a10, a11)
if ans is None:
print('Impossible')
else:
print(ans)
| {
"input": [
"1 2 3 4\n",
"1 2 2 1\n"
],
"output": [
"Impossible",
"0110"
]
} |
638 | 7 | One spring day on his way to university Lesha found an array A. Lesha likes to split arrays into several parts. This time Lesha decided to split the array A into several, possibly one, new arrays so that the sum of elements in each of the new arrays is not zero. One more condition is that if we place the new arrays one after another they will form the old array A.
Lesha is tired now so he asked you to split the array. Help Lesha!
Input
The first line contains single integer n (1 ≤ n ≤ 100) — the number of elements in the array A.
The next line contains n integers a1, a2, ..., an ( - 103 ≤ ai ≤ 103) — the elements of the array A.
Output
If it is not possible to split the array A and satisfy all the constraints, print single line containing "NO" (without quotes).
Otherwise in the first line print "YES" (without quotes). In the next line print single integer k — the number of new arrays. In each of the next k lines print two integers li and ri which denote the subarray A[li... ri] of the initial array A being the i-th new array. Integers li, ri should satisfy the following conditions:
* l1 = 1
* rk = n
* ri + 1 = li + 1 for each 1 ≤ i < k.
If there are multiple answers, print any of them.
Examples
Input
3
1 2 -3
Output
YES
2
1 2
3 3
Input
8
9 -12 3 4 -4 -10 7 3
Output
YES
2
1 2
3 8
Input
1
0
Output
NO
Input
4
1 2 3 -5
Output
YES
4
1 1
2 2
3 3
4 4 | def Main(n):
arr = list(map(int, input().split()))
if any(arr):
print('YES')
if sum(arr):
print(1)
print(1,n)
else:
for i, v in enumerate(arr):
if v:
print(2)
print(1,i + 1)
print(i + 2, n)
break
else:
print('NO')
if __name__ == '__main__':
n = int(input())
Main(n) | {
"input": [
"4\n1 2 3 -5\n",
"1\n0\n",
"8\n9 -12 3 4 -4 -10 7 3\n",
"3\n1 2 -3\n"
],
"output": [
"YES\n1\n1 4\n",
"NO\n",
"YES\n2\n1 1\n2 8\n",
"YES\n2\n1 1\n2 3\n"
]
} |
639 | 9 | After Fox Ciel got off a bus, she found that the bus she was on was a wrong bus and she lost her way in a strange town. However, she fortunately met her friend Beaver Taro and asked which way to go to her castle. Taro's response to her was a string s, and she tried to remember the string s correctly.
However, Ciel feels n strings b1, b2, ... , bn are really boring, and unfortunately she dislikes to remember a string that contains a boring substring. To make the thing worse, what she can remember is only the contiguous substring of s.
Determine the longest contiguous substring of s that does not contain any boring string, so that she can remember the longest part of Taro's response.
Input
In the first line there is a string s. The length of s will be between 1 and 105, inclusive.
In the second line there is a single integer n (1 ≤ n ≤ 10). Next n lines, there is a string bi (1 ≤ i ≤ n). Each length of bi will be between 1 and 10, inclusive.
Each character of the given strings will be either a English alphabet (both lowercase and uppercase) or a underscore ('_') or a digit. Assume that these strings are case-sensitive.
Output
Output in the first line two space-separated integers len and pos: the length of the longest contiguous substring of s that does not contain any bi, and the first position of the substring (0-indexed). The position pos must be between 0 and |s| - len inclusive, where |s| is the length of string s.
If there are several solutions, output any.
Examples
Input
Go_straight_along_this_street
5
str
long
tree
biginteger
ellipse
Output
12 4
Input
IhaveNoIdea
9
I
h
a
v
e
N
o
I
d
Output
0 0
Input
unagioisii
2
ioi
unagi
Output
5 5
Note
In the first sample, the solution is traight_alon.
In the second sample, the solution is an empty string, so the output can be «0 0», «0 1», «0 2», and so on.
In the third sample, the solution is either nagio or oisii. | s, n = input(), int(input())
t = [input() for i in range(n)]
def f(i):
global t
for j in range(n):
if i < j:
if len(t[j]) < len(t[i]) and t[j] in t[i]: return False
elif j < i and t[j] in t[i]: return False
return True
t = [t[i] for i in range(n) if f(i)]
n = len(s)
r = [0] * n
for p in t:
i, m = -1, len(p) - 1
while True:
i = s.find(p, i + 1)
if i == -1: break
r[i] += 1
r[i + m] += 2
d, j, q = 0, -1, [-1]
for i in range(n):
if r[i] & 1: q.append(i)
if r[i] & 2:
j = q.pop(0)
if i - j > d: d, k = i - j, j
j = q.pop(0)
if n - j > d: d, k = n - j, j
print(d - 1, k + 1) | {
"input": [
"unagioisii\n2\nioi\nunagi\n",
"Go_straight_along_this_street\n5\nstr\nlong\ntree\nbiginteger\nellipse\n",
"IhaveNoIdea\n9\nI\nh\na\nv\ne\nN\no\nI\nd\n"
],
"output": [
"5 1",
"12 4",
"0 0"
]
} |
640 | 7 | Hideo Kojima has just quit his job at Konami. Now he is going to find a new place to work. Despite being such a well-known person, he still needs a CV to apply for a job.
During all his career Hideo has produced n games. Some of them were successful, some were not. Hideo wants to remove several of them (possibly zero) from his CV to make a better impression on employers. As a result there should be no unsuccessful game which comes right after successful one in his CV.
More formally, you are given an array s1, s2, ..., sn of zeros and ones. Zero corresponds to an unsuccessful game, one — to a successful one. Games are given in order they were produced, and Hideo can't swap these values. He should remove some elements from this array in such a way that no zero comes right after one.
Besides that, Hideo still wants to mention as much games in his CV as possible. Help this genius of a man determine the maximum number of games he can leave in his CV.
Input
The first line contains one integer number n (1 ≤ n ≤ 100).
The second line contains n space-separated integer numbers s1, s2, ..., sn (0 ≤ si ≤ 1). 0 corresponds to an unsuccessful game, 1 — to a successful one.
Output
Print one integer — the maximum number of games Hideo can leave in his CV so that no unsuccessful game comes after a successful one.
Examples
Input
4
1 1 0 1
Output
3
Input
6
0 1 0 0 1 0
Output
4
Input
1
0
Output
1 | def main():
n = int(input())
a = list(map(int, input().split()))
ans = max(sum(a), n - sum(a))
for i in range(n):
ans = max(ans, sum(a[i:]) + i - sum(a[:i]))
print(ans)
main()
| {
"input": [
"6\n0 1 0 0 1 0\n",
"1\n0\n",
"4\n1 1 0 1\n"
],
"output": [
"4\n",
"1\n",
"3\n"
]
} |
641 | 9 | It's another Start[c]up finals, and that means there is pizza to order for the onsite contestants. There are only 2 types of pizza (obviously not, but let's just pretend for the sake of the problem), and all pizzas contain exactly S slices.
It is known that the i-th contestant will eat si slices of pizza, and gain ai happiness for each slice of type 1 pizza they eat, and bi happiness for each slice of type 2 pizza they eat. We can order any number of type 1 and type 2 pizzas, but we want to buy the minimum possible number of pizzas for all of the contestants to be able to eat their required number of slices. Given that restriction, what is the maximum possible total happiness that can be achieved?
Input
The first line of input will contain integers N and S (1 ≤ N ≤ 105, 1 ≤ S ≤ 105), the number of contestants and the number of slices per pizza, respectively. N lines follow.
The i-th such line contains integers si, ai, and bi (1 ≤ si ≤ 105, 1 ≤ ai ≤ 105, 1 ≤ bi ≤ 105), the number of slices the i-th contestant will eat, the happiness they will gain from each type 1 slice they eat, and the happiness they will gain from each type 2 slice they eat, respectively.
Output
Print the maximum total happiness that can be achieved.
Examples
Input
3 12
3 5 7
4 6 7
5 9 5
Output
84
Input
6 10
7 4 7
5 8 8
12 5 8
6 11 6
3 3 7
5 9 6
Output
314
Note
In the first example, you only need to buy one pizza. If you buy a type 1 pizza, the total happiness will be 3·5 + 4·6 + 5·9 = 84, and if you buy a type 2 pizza, the total happiness will be 3·7 + 4·7 + 5·5 = 74. | f = lambda: map(int, input().split())
n, s = f()
u, v = [], []
for i in range(n):
d, a, b = f()
if a > b:
u.append([a, b, d])
else:
v.append([b, a, d])
def g(t):
t.sort(key=lambda q: q[1] - q[0])
m = sum(d for a, b, d in t)
k = s * (m // s)
n = m - k
x = y = z = 0
for a, b, d in t:
if k >= d:
k -= d
z += d * a
elif k:
z += k * a
x = (d - k) * a
y = (d - k) * b
k = 0
else:
x += d * a
y += d * b
return x, y, z, n
a, b = g(u), g(v)
d = a[0] + b[0] if a[3] + b[3] > s else max(a[0] + b[1], a[1] + b[0])
print(a[2] + b[2] + d) | {
"input": [
"6 10\n7 4 7\n5 8 8\n12 5 8\n6 11 6\n3 3 7\n5 9 6\n",
"3 12\n3 5 7\n4 6 7\n5 9 5\n"
],
"output": [
"314\n",
"84\n"
]
} |
642 | 11 | You are given a tree (a connected acyclic undirected graph) of n vertices. Vertices are numbered from 1 to n and each vertex is assigned a character from a to t.
A path in the tree is said to be palindromic if at least one permutation of the labels in the path is a palindrome.
For each vertex, output the number of palindromic paths passing through it.
Note: The path from vertex u to vertex v is considered to be the same as the path from vertex v to vertex u, and this path will be counted only once for each of the vertices it passes through.
Input
The first line contains an integer n (2 ≤ n ≤ 2·105) — the number of vertices in the tree.
The next n - 1 lines each contain two integers u and v (1 ≤ u, v ≤ n, u ≠ v) denoting an edge connecting vertex u and vertex v. It is guaranteed that the given graph is a tree.
The next line contains a string consisting of n lowercase characters from a to t where the i-th (1 ≤ i ≤ n) character is the label of vertex i in the tree.
Output
Print n integers in a single line, the i-th of which is the number of palindromic paths passing through vertex i in the tree.
Examples
Input
5
1 2
2 3
3 4
3 5
abcbb
Output
1 3 4 3 3
Input
7
6 2
4 3
3 7
5 2
7 2
1 4
afefdfs
Output
1 4 1 1 2 4 2
Note
In the first sample case, the following paths are palindromic:
2 - 3 - 4
2 - 3 - 5
4 - 3 - 5
Additionally, all paths containing only one vertex are palindromic. Listed below are a few paths in the first sample that are not palindromic:
1 - 2 - 3
1 - 2 - 3 - 4
1 - 2 - 3 - 5 | import sys, os
def centroid_decomp(coupl):
n = len(coupl)
bfs = [n - 1]
for node in bfs:
bfs += coupl[node]
for nei in coupl[node]:
coupl[nei].remove(node)
size = [0] * n
for node in reversed(bfs):
size[node] = 1 + sum(size[child] for child in coupl[node])
def centroid_reroot(root):
while coupl[root]:
child = max(coupl[root], key = size.__getitem__)
if size[child] <= size[root] // 2:
break
size[root] -= size[child]
size[child] += size[root]
coupl[root].remove(child)
coupl[child].append(root)
root = child
return root
bfs = [n - 1]
for node in bfs:
centroid = centroid_reroot(node)
bfs += coupl[centroid]
yield centroid
inp = sys.stdin.buffer.read().split(); ii = 0
n = int(inp[ii]); ii += 1
coupl = [[] for _ in range(n)]
for _ in range(n - 1):
u = int(inp[ii]) - 1; ii += 1
v = int(inp[ii]) - 1; ii += 1
coupl[u].append(v)
coupl[v].append(u)
A = [1 << c - b'a'[0] for c in inp[ii]]; ii += 1
palistates = [0] + [1 << i for i in range(20)]
ans = [0.0] * n
dp = [0.0] * n
val = [0] * n
counter = [0] * (1 << 20)
for centroid in centroid_decomp(coupl):
bfss = []
for root in coupl[centroid]:
bfs = [root]
for node in bfs:
bfs += coupl[node]
bfss.append(bfs)
for node in bfs:
val[node] ^= A[node]
for child in coupl[node]:
val[child] = val[node]
entire_bfs = [centroid]
for bfs in bfss:
entire_bfs += bfs
for node in entire_bfs:
val[node] ^= A[centroid]
counter[val[node]] += 1
for bfs in bfss:
for node in bfs:
counter[val[node]] -= 1
for node in bfs:
v = val[node] ^ A[centroid]
for p in palistates:
dp[node] += counter[v ^ p]
for node in bfs:
counter[val[node]] += 1
for node in reversed(entire_bfs):
dp[node] += sum(dp[child] for child in coupl[node])
dp[centroid] += 1
for p in palistates:
dp[centroid] += counter[p]
dp[centroid] //= 2
for node in entire_bfs:
ans[node] += dp[node]
counter[val[node]] = val[node] = 0
dp[node] = 0.0
os.write(1, b' '.join(str(int(x)).encode('ascii') for x in ans))
| {
"input": [
"7\n6 2\n4 3\n3 7\n5 2\n7 2\n1 4\nafefdfs\n",
"5\n1 2\n2 3\n3 4\n3 5\nabcbb\n"
],
"output": [
"1 4 1 1 2 4 2\n",
"1 3 4 3 3\n"
]
} |
643 | 9 | Julia is going to cook a chicken in the kitchen of her dormitory. To save energy, the stove in the kitchen automatically turns off after k minutes after turning on.
During cooking, Julia goes to the kitchen every d minutes and turns on the stove if it is turned off. While the cooker is turned off, it stays warm. The stove switches on and off instantly.
It is known that the chicken needs t minutes to be cooked on the stove, if it is turned on, and 2t minutes, if it is turned off. You need to find out, how much time will Julia have to cook the chicken, if it is considered that the chicken is cooked evenly, with constant speed when the stove is turned on and at a constant speed when it is turned off.
Input
The single line contains three integers k, d and t (1 ≤ k, d, t ≤ 1018).
Output
Print a single number, the total time of cooking in minutes. The relative or absolute error must not exceed 10 - 9.
Namely, let's assume that your answer is x and the answer of the jury is y. The checker program will consider your answer correct if <image>.
Examples
Input
3 2 6
Output
6.5
Input
4 2 20
Output
20.0
Note
In the first example, the chicken will be cooked for 3 minutes on the turned on stove, after this it will be cooked for <image>. Then the chicken will be cooked for one minute on a turned off stove, it will be cooked for <image>. Thus, after four minutes the chicken will be cooked for <image>. Before the fifth minute Julia will turn on the stove and after 2.5 minutes the chicken will be ready <image>.
In the second example, when the stove is turned off, Julia will immediately turn it on, so the stove will always be turned on and the chicken will be cooked in 20 minutes. | def solve(k, d, t):
q, r = divmod(k, d)
up = d - r if r != 0 else 0
perblock2 = k * 2 + up
qt, rt = divmod(2 * t, perblock2)
if rt <= k * 2:
extra = rt / 2
else:
extra = rt - k
return (k + up) * qt + extra
k, d, t = [int(v) for v in input().split()]
print(solve(k, d, t)) | {
"input": [
"4 2 20\n",
"3 2 6\n"
],
"output": [
"20.0\n",
"6.5\n"
]
} |
644 | 11 | The cities of Byteland and Berland are located on the axis Ox. In addition, on this axis there are also disputed cities, which belong to each of the countries in their opinion. Thus, on the line Ox there are three types of cities:
* the cities of Byteland,
* the cities of Berland,
* disputed cities.
Recently, the project BNET has been launched — a computer network of a new generation. Now the task of the both countries is to connect the cities so that the network of this country is connected.
The countries agreed to connect the pairs of cities with BNET cables in such a way that:
* If you look at the only cities of Byteland and the disputed cities, then in the resulting set of cities, any city should be reachable from any other one by one or more cables,
* If you look at the only cities of Berland and the disputed cities, then in the resulting set of cities, any city should be reachable from any other one by one or more cables.
Thus, it is necessary to choose a set of pairs of cities to connect by cables in such a way that both conditions are satisfied simultaneously. Cables allow bi-directional data transfer. Each cable connects exactly two distinct cities.
The cost of laying a cable from one city to another is equal to the distance between them. Find the minimum total cost of laying a set of cables so that two subsets of cities (Byteland and disputed cities, Berland and disputed cities) are connected.
Each city is a point on the line Ox. It is technically possible to connect the cities a and b with a cable so that the city c (a < c < b) is not connected to this cable, where a, b and c are simultaneously coordinates of the cities a, b and c.
Input
The first line contains a single integer n (2 ≤ n ≤ 2 ⋅ 10^{5}) — the number of cities.
The following n lines contains an integer x_i and the letter c_i (-10^{9} ≤ x_i ≤ 10^{9}) — the coordinate of the city and its type. If the city belongs to Byteland, c_i equals to 'B'. If the city belongs to Berland, c_i equals to «R». If the city is disputed, c_i equals to 'P'.
All cities have distinct coordinates. Guaranteed, that the cities are given in the increasing order of their coordinates.
Output
Print the minimal total length of such set of cables, that if we delete all Berland cities (c_i='R'), it will be possible to find a way from any remaining city to any other remaining city, moving only by cables. Similarly, if we delete all Byteland cities (c_i='B'), it will be possible to find a way from any remaining city to any other remaining city, moving only by cables.
Examples
Input
4
-5 R
0 P
3 P
7 B
Output
12
Input
5
10 R
14 B
16 B
21 R
32 R
Output
24
Note
In the first example, you should connect the first city with the second, the second with the third, and the third with the fourth. The total length of the cables will be 5 + 3 + 4 = 12.
In the second example there are no disputed cities, so you need to connect all the neighboring cities of Byteland and all the neighboring cities of Berland. The cities of Berland have coordinates 10, 21, 32, so to connect them you need two cables of length 11 and 11. The cities of Byteland have coordinates 14 and 16, so to connect them you need one cable of length 2. Thus, the total length of all cables is 11 + 11 + 2 = 24. | def inpmap():
return map(int, input().split())
n = int(input())
b, r, p = None, None, None
res = 0
mr = -1
mb = -1
for i in range(n):
ix, t = input().split()
ix = int(ix)
if t != 'R':
if b is not None:
res += ix - b
mb = max(mb, ix - b)
b = ix
if t != 'B':
if r is not None:
res += ix - r
mr = max(mr, ix - r)
r = ix
if t == 'P':
if p is not None:
if ix - p < mr + mb:
res -= (mr + mb) - (ix - p)
p = ix
mr = mb = 0
print(res)
| {
"input": [
"5\n10 R\n14 B\n16 B\n21 R\n32 R\n",
"4\n-5 R\n0 P\n3 P\n7 B\n"
],
"output": [
"24\n",
"12\n"
]
} |
645 | 7 | Berland Football Cup starts really soon! Commentators from all over the world come to the event.
Organizers have already built n commentary boxes. m regional delegations will come to the Cup. Every delegation should get the same number of the commentary boxes. If any box is left unoccupied then the delegations will be upset. So each box should be occupied by exactly one delegation.
If n is not divisible by m, it is impossible to distribute the boxes to the delegations at the moment.
Organizers can build a new commentary box paying a burles and demolish a commentary box paying b burles. They can both build and demolish boxes arbitrary number of times (each time paying a corresponding fee). It is allowed to demolish all the existing boxes.
What is the minimal amount of burles organizers should pay to satisfy all the delegations (i.e. to make the number of the boxes be divisible by m)?
Input
The only line contains four integer numbers n, m, a and b (1 ≤ n, m ≤ 10^{12}, 1 ≤ a, b ≤ 100), where n is the initial number of the commentary boxes, m is the number of delegations to come, a is the fee to build a box and b is the fee to demolish a box.
Output
Output the minimal amount of burles organizers should pay to satisfy all the delegations (i.e. to make the number of the boxes be divisible by m). It is allowed that the final number of the boxes is equal to 0.
Examples
Input
9 7 3 8
Output
15
Input
2 7 3 7
Output
14
Input
30 6 17 19
Output
0
Note
In the first example organizers can build 5 boxes to make the total of 14 paying 3 burles for the each of them.
In the second example organizers can demolish 2 boxes to make the total of 0 paying 7 burles for the each of them.
In the third example organizers are already able to distribute all the boxes equally among the delegations, each one get 5 boxes. | def i_ints():
return list(map(int, input().split()))
n, m ,a, b = i_ints()
rest = n % m
print(min(rest * b, (m - rest) * a)) | {
"input": [
"30 6 17 19\n",
"2 7 3 7\n",
"9 7 3 8\n"
],
"output": [
"0\n",
"14\n",
"15\n"
]
} |
646 | 9 | IA has so many colorful magnets on her fridge! Exactly one letter is written on each magnet, 'a' or 'b'. She loves to play with them, placing all magnets in a row. However, the girl is quickly bored and usually thinks how to make her entertainment more interesting.
Today, when IA looked at the fridge, she noticed that the word formed by magnets is really messy. "It would look much better when I'll swap some of them!" — thought the girl — "but how to do it?". After a while, she got an idea. IA will look at all prefixes with lengths from 1 to the length of the word and for each prefix she will either reverse this prefix or leave it as it is. She will consider the prefixes in the fixed order: from the shortest to the largest. She wants to get the lexicographically smallest possible word after she considers all prefixes. Can you help her, telling which prefixes should be chosen for reversing?
A string a is lexicographically smaller than a string b if and only if one of the following holds:
* a is a prefix of b, but a ≠ b;
* in the first position where a and b differ, the string a has a letter that appears earlier in the alphabet than the corresponding letter in b.
Input
The first and the only line contains a string s (1 ≤ |s| ≤ 1000), describing the initial string formed by magnets. The string s consists only of characters 'a' and 'b'.
Output
Output exactly |s| integers. If IA should reverse the i-th prefix (that is, the substring from 1 to i), the i-th integer should be equal to 1, and it should be equal to 0 otherwise.
If there are multiple possible sequences leading to the optimal answer, print any of them.
Examples
Input
bbab
Output
0 1 1 0
Input
aaaaa
Output
1 0 0 0 1
Note
In the first example, IA can reverse the second and the third prefix and get a string "abbb". She cannot get better result, since it is also lexicographically smallest string obtainable by permuting characters of the initial string.
In the second example, she can reverse any subset of prefixes — all letters are 'a'. | import sys
import math
def main():
s = next(sys.stdin).strip()
s += 'b'
res = [int(i != j) for i,j in zip(s,s[1:])]
print(' '.join(map(str,res)))
if __name__ == '__main__':
main() | {
"input": [
"bbab\n",
"aaaaa\n"
],
"output": [
"0 1 1 0\n",
"0 0 0 0 1\n"
]
} |
647 | 7 | Vova plans to go to the conference by train. Initially, the train is at the point 1 and the destination point of the path is the point L. The speed of the train is 1 length unit per minute (i.e. at the first minute the train is at the point 1, at the second minute — at the point 2 and so on).
There are lanterns on the path. They are placed at the points with coordinates divisible by v (i.e. the first lantern is at the point v, the second is at the point 2v and so on).
There is also exactly one standing train which occupies all the points from l to r inclusive.
Vova can see the lantern at the point p if p is divisible by v and there is no standing train at this position (p not∈ [l; r]). Thus, if the point with the lantern is one of the points covered by the standing train, Vova can't see this lantern.
Your problem is to say the number of lanterns Vova will see during the path. Vova plans to go to t different conferences, so you should answer t independent queries.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of queries.
Then t lines follow. The i-th line contains four integers L_i, v_i, l_i, r_i (1 ≤ L, v ≤ 10^9, 1 ≤ l ≤ r ≤ L) — destination point of the i-th path, the period of the lantern appearance and the segment occupied by the standing train.
Output
Print t lines. The i-th line should contain one integer — the answer for the i-th query.
Example
Input
4
10 2 3 7
100 51 51 51
1234 1 100 199
1000000000 1 1 1000000000
Output
3
0
1134
0
Note
For the first example query, the answer is 3. There are lanterns at positions 2, 4, 6, 8 and 10, but Vova didn't see the lanterns at positions 4 and 6 because of the standing train.
For the second example query, the answer is 0 because the only lantern is at the point 51 and there is also a standing train at this point.
For the third example query, the answer is 1134 because there are 1234 lanterns, but Vova didn't see the lanterns from the position 100 to the position 199 inclusive.
For the fourth example query, the answer is 0 because the standing train covers the whole path. | def cal(lo,v):
return lo//v
q=int(input())
for _ in range(0,q):
L,v,l,r=map(int,input().split())
print(cal(L,v)-(cal(r,v)-cal(l-1,v))) | {
"input": [
"4\n10 2 3 7\n100 51 51 51\n1234 1 100 199\n1000000000 1 1 1000000000\n"
],
"output": [
"3\n0\n1134\n0\n"
]
} |
648 | 12 | Vasya wants to buy himself a nice new car. Unfortunately, he lacks some money. Currently he has exactly 0 burles.
However, the local bank has n credit offers. Each offer can be described with three numbers a_i, b_i and k_i. Offers are numbered from 1 to n. If Vasya takes the i-th offer, then the bank gives him a_i burles at the beginning of the month and then Vasya pays bank b_i burles at the end of each month for the next k_i months (including the month he activated the offer). Vasya can take the offers any order he wants.
Each month Vasya can take no more than one credit offer. Also each credit offer can not be used more than once. Several credits can be active at the same time. It implies that Vasya pays bank the sum of b_i over all the i of active credits at the end of each month.
Vasya wants to buy a car in the middle of some month. He just takes all the money he currently has and buys the car of that exact price.
Vasya don't really care what he'll have to pay the bank back after he buys a car. He just goes out of the country on his car so that the bank can't find him anymore.
What is the maximum price that car can have?
Input
The first line contains one integer n (1 ≤ n ≤ 500) — the number of credit offers.
Each of the next n lines contains three integers a_i, b_i and k_i (1 ≤ a_i, b_i, k_i ≤ 10^9).
Output
Print one integer — the maximum price of the car.
Examples
Input
4
10 9 2
20 33 1
30 115 1
5 3 2
Output
32
Input
3
40 1 2
1000 1100 5
300 2 1
Output
1337
Note
In the first example, the following sequence of offers taken is optimal: 4 → 3.
The amount of burles Vasya has changes the following way: 5 → 32 → -86 → .... He takes the money he has in the middle of the second month (32 burles) and buys the car.
The negative amount of money means that Vasya has to pay the bank that amount of burles.
In the second example, the following sequence of offers taken is optimal: 3 → 1 → 2.
The amount of burles Vasya has changes the following way: 0 → 300 → 338 → 1337 → 236 → -866 → .... | n = int(input())
a = [tuple(map(int, input().split())) for i in range(n)]
a = [(y, x, k) for x, y, k in a]
a.sort(reverse=True)
dp = [[-1] * (n + 1) for i in range(n)]
def f(i, j):
if i < 0 or j < -1: return 0
if dp[i][j] == -1:
y, x, k = a[i]
dp[i][j] = f(i - 1, j) + max(0, x - k * y)
if 0 <= j < k: dp[i][j] = max(dp[i][j], x - j * y + f(i - 1, j - 1))
return dp[i][j]
print(max(f(n - 1, j) for j in range(-1, n)))
| {
"input": [
"4\n10 9 2\n20 33 1\n30 115 1\n5 3 2\n",
"3\n40 1 2\n1000 1100 5\n300 2 1\n"
],
"output": [
"32\n",
"1337\n"
]
} |
649 | 7 | Dora loves adventures quite a lot. During some journey she encountered an amazing city, which is formed by n streets along the Eastern direction and m streets across the Southern direction. Naturally, this city has nm intersections. At any intersection of i-th Eastern street and j-th Southern street there is a monumental skyscraper. Dora instantly became curious and decided to explore the heights of the city buildings.
When Dora passes through the intersection of the i-th Eastern and j-th Southern street she examines those two streets. After Dora learns the heights of all the skyscrapers on those two streets she wonders: how one should reassign heights to the skyscrapers on those two streets, so that the maximum height would be as small as possible and the result of comparing the heights of any two skyscrapers on one street wouldn't change.
Formally, on every of nm intersections Dora solves an independent problem. She sees n + m - 1 skyscrapers and for each of them she knows its real height. Moreover, any two heights can be compared to get a result "greater", "smaller" or "equal". Now Dora wants to select some integer x and assign every skyscraper a height from 1 to x. When assigning heights, Dora wants to preserve the relative order of the skyscrapers in both streets. That is, the result of any comparison of heights of two skyscrapers in the current Eastern street shouldn't change and the result of any comparison of heights of two skyscrapers in current Southern street shouldn't change as well. Note that skyscrapers located on the Southern street are not compared with skyscrapers located on the Eastern street only. However, the skyscraper located at the streets intersection can be compared with both Southern and Eastern skyscrapers. For every intersection Dora wants to independently calculate the minimum possible x.
For example, if the intersection and the two streets corresponding to it look as follows:
<image>
Then it is optimal to replace the heights of the skyscrapers as follows (note that all comparisons "less", "equal", "greater" inside the Eastern street and inside the Southern street are preserved)
<image>
The largest used number is 5, hence the answer for this intersection would be 5.
Help Dora to compute the answers for each intersection.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 1000) — the number of streets going in the Eastern direction and the number of the streets going in Southern direction.
Each of the following n lines contains m integers a_{i,1}, a_{i,2}, ..., a_{i,m} (1 ≤ a_{i,j} ≤ 10^9). The integer a_{i,j}, located on j-th position in the i-th line denotes the height of the skyscraper at the intersection of the i-th Eastern street and j-th Southern direction.
Output
Print n lines containing m integers each. The integer x_{i,j}, located on j-th position inside the i-th line is an answer for the problem at the intersection of i-th Eastern street and j-th Southern street.
Examples
Input
2 3
1 2 1
2 1 2
Output
2 2 2
2 2 2
Input
2 2
1 2
3 4
Output
2 3
3 2
Note
In the first example, it's not possible to decrease the maximum used height for the problem at any intersection, hence we don't have to change any heights.
In the second example, the answers are as follows:
* For the intersection of the first line and the first column <image>
* For the intersection of the first line and the second column <image>
* For the intersection of the second line and the first column <image>
* For the intersection of the second line and the second column <image> | n, m = list(map(int, input().split()))
gor = []
ver = []
o = []
for i in range(n):
gor.append(list(map(int, input().split())))
o.append([])
for i in range(m):
ver.append([])
for i in range(m):
for g in gor:
ver[i].append(g[i])
ver = list(map(lambda x: sorted(list(set(x))), ver))
def f(vg, vr, o):
vr1 = vr.index(o)
vg1 = vg.index(o)
if vr1 > vg1:
return max([len(vr), len(vg) + vr1 - vg1])
return max([len(vg), len(vr) + vg1 - vr1])
for i, vg in enumerate(gor):
vg1 = sorted(list(set(vg)))
for j, vr in enumerate(vg):
o[i].append(f(vg1, ver[j], vr))
for i in o:
print(' '.join(list(map(str, i))))
| {
"input": [
"2 2\n1 2\n3 4\n",
"2 3\n1 2 1\n2 1 2\n"
],
"output": [
"2 3 \n3 2 \n",
"2 2 2 \n2 2 2 \n"
]
} |
650 | 9 | Ivan is going to sleep now and wants to set his alarm clock. There will be many necessary events tomorrow, the i-th of them will start during the x_i-th minute. Ivan doesn't want to skip any of the events, so he has to set his alarm clock in such a way that it rings during minutes x_1, x_2, ..., x_n, so he will be awake during each of these minutes (note that it does not matter if his alarm clock will ring during any other minute).
Ivan can choose two properties for the alarm clock — the first minute it will ring (let's denote it as y) and the interval between two consecutive signals (let's denote it by p). After the clock is set, it will ring during minutes y, y + p, y + 2p, y + 3p and so on.
Ivan can choose any minute as the first one, but he cannot choose any arbitrary value of p. He has to pick it among the given values p_1, p_2, ..., p_m (his phone does not support any other options for this setting).
So Ivan has to choose the first minute y when the alarm clock should start ringing and the interval between two consecutive signals p_j in such a way that it will ring during all given minutes x_1, x_2, ..., x_n (and it does not matter if his alarm clock will ring in any other minutes).
Your task is to tell the first minute y and the index j such that if Ivan sets his alarm clock with properties y and p_j it will ring during all given minutes x_1, x_2, ..., x_n or say that it is impossible to choose such values of the given properties. If there are multiple answers, you can print any.
Input
The first line of the input contains two integers n and m (2 ≤ n ≤ 3 ⋅ 10^5, 1 ≤ m ≤ 3 ⋅ 10^5) — the number of events and the number of possible settings for the interval between signals.
The second line of the input contains n integers x_1, x_2, ..., x_n (1 ≤ x_i ≤ 10^{18}), where x_i is the minute when i-th event starts. It is guaranteed that all x_i are given in increasing order (i. e. the condition x_1 < x_2 < ... < x_n holds).
The third line of the input contains m integers p_1, p_2, ..., p_m (1 ≤ p_j ≤ 10^{18}), where p_j is the j-th option for the interval between two consecutive signals.
Output
If it's impossible to choose such values y and j so all constraints are satisfied, print "NO" in the first line.
Otherwise print "YES" in the first line. Then print two integers y (1 ≤ y ≤ 10^{18}) and j (1 ≤ j ≤ m) in the second line, where y is the first minute Ivan's alarm clock should start ringing and j is the index of the option for the interval between two consecutive signals (options are numbered from 1 to m in the order they are given input). These values should be chosen in such a way that the alarm clock will ring during all given minutes x_1, x_2, ..., x_n. If there are multiple answers, you can print any.
Examples
Input
3 5
3 12 18
2 6 5 3 3
Output
YES
3 4
Input
4 2
1 5 17 19
4 5
Output
NO
Input
4 2
1 5 17 19
2 1
Output
YES
1 1 | import math
read = lambda: map(int, input().split())
n,m = read(); a = list(read()); b = list(read())
gcd = 0
for i in a[1:]: gcd = math.gcd(gcd, i - a[0])
def solve():
for i in range(m):
if gcd % b[i] == 0:
print('YES\n', a[0],i+1)
return
print('NO')
solve()
| {
"input": [
"3 5\n3 12 18\n2 6 5 3 3\n",
"4 2\n1 5 17 19\n2 1\n",
"4 2\n1 5 17 19\n4 5\n"
],
"output": [
"YES\n 3 4\n",
"YES\n 1 1\n",
"NO\n"
]
} |
651 | 8 | Let's write all the positive integer numbers one after another from 1 without any delimiters (i.e. as a single string). It will be the infinite sequence starting with 123456789101112131415161718192021222324252627282930313233343536...
Your task is to print the k-th digit of this sequence.
Input
The first and only line contains integer k (1 ≤ k ≤ 10^{12}) — the position to process (1-based index).
Output
Print the k-th digit of the resulting infinite sequence.
Examples
Input
7
Output
7
Input
21
Output
5 | def mp():
return map(int, input().split())
def f(i):
return (10 ** i - 10 ** (i - 1)) * i
n = int(input())
i = 1
sum = 0
while n - f(i) >= 0:
n -= f(i)
sum += f(i) // i
i += 1
print(str(sum + (n + i - 1) // i)[n % i - 1]) | {
"input": [
"21\n",
"7\n"
],
"output": [
"5\n",
"7\n"
]
} |
652 | 8 | Polycarp analyzes the prices of the new berPhone. At his disposal are the prices for n last days: a_1, a_2, ..., a_n, where a_i is the price of berPhone on the day i.
Polycarp considers the price on the day i to be bad if later (that is, a day with a greater number) berPhone was sold at a lower price. For example, if n=6 and a=[3, 9, 4, 6, 7, 5], then the number of days with a bad price is 3 — these are days 2 (a_2=9), 4 (a_4=6) and 5 (a_5=7).
Print the number of days with a bad price.
You have to answer t independent data sets.
Input
The first line contains an integer t (1 ≤ t ≤ 10000) — the number of sets of input data in the test. Input data sets must be processed independently, one after another.
Each input data set consists of two lines. The first line contains an integer n (1 ≤ n ≤ 150000) — the number of days. The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^6), where a_i is the price on the i-th day.
It is guaranteed that the sum of n over all data sets in the test does not exceed 150000.
Output
Print t integers, the j-th of which should be equal to the number of days with a bad price in the j-th input data set.
Example
Input
5
6
3 9 4 6 7 5
1
1000000
2
2 1
10
31 41 59 26 53 58 97 93 23 84
7
3 2 1 2 3 4 5
Output
3
0
1
8
2 | def main():
n=int(input())
arr = list(map(int,input().split()))
min = arr[n-1]
co = 0
for i in range(n-1,-1,-1):
if min < arr[i]:
co +=1
else:
min = arr[i]
print(co)
for _ in range(int(input())):
main()
| {
"input": [
"5\n6\n3 9 4 6 7 5\n1\n1000000\n2\n2 1\n10\n31 41 59 26 53 58 97 93 23 84\n7\n3 2 1 2 3 4 5\n"
],
"output": [
"3\n0\n1\n8\n2\n"
]
} |
653 | 9 | In order to do some research, n^2 labs are built on different heights of a mountain. Let's enumerate them with integers from 1 to n^2, such that the lab with the number 1 is at the lowest place, the lab with the number 2 is at the second-lowest place, …, the lab with the number n^2 is at the highest place.
To transport water between the labs, pipes are built between every pair of labs. A pipe can transport at most one unit of water at a time from the lab with the number u to the lab with the number v if u > v.
Now the labs need to be divided into n groups, each group should contain exactly n labs. The labs from different groups can transport water to each other. The sum of units of water that can be sent from a group A to a group B is equal to the number of pairs of labs (u, v) such that the lab with the number u is from the group A, the lab with the number v is from the group B and u > v. Let's denote this value as f(A,B) (i.e. f(A,B) is the sum of units of water that can be sent from a group A to a group B).
For example, if n=3 and there are 3 groups X, Y and Z: X = \{1, 5, 6\}, Y = \{2, 4, 9\} and Z = \{3, 7, 8\}. In this case, the values of f are equal to:
* f(X,Y)=4 because of 5 → 2, 5 → 4, 6 → 2, 6 → 4,
* f(X,Z)=2 because of 5 → 3, 6 → 3,
* f(Y,X)=5 because of 2 → 1, 4 → 1, 9 → 1, 9 → 5, 9 → 6,
* f(Y,Z)=4 because of 4 → 3, 9 → 3, 9 → 7, 9 → 8,
* f(Z,X)=7 because of 3 → 1, 7 → 1, 7 → 5, 7 → 6, 8 → 1, 8 → 5, 8 → 6,
* f(Z,Y)=5 because of 3 → 2, 7 → 2, 7 → 4, 8 → 2, 8 → 4.
Please, divide labs into n groups with size n, such that the value min f(A,B) over all possible pairs of groups A and B (A ≠ B) is maximal.
In other words, divide labs into n groups with size n, such that minimum number of the sum of units of water that can be transported from a group A to a group B for every pair of different groups A and B (A ≠ B) as big as possible.
Note, that the example above doesn't demonstrate an optimal division, but it demonstrates how to calculate the values f for some division.
If there are many optimal divisions, you can find any.
Input
The only line contains one number n (2 ≤ n ≤ 300).
Output
Output n lines:
In the i-th line print n numbers, the numbers of labs of the i-th group, in any order you want.
If there are multiple answers, that maximize the minimum number of the sum of units of water that can be transported from one group the another, you can print any.
Example
Input
3
Output
2 8 5
9 3 4
7 6 1
Note
In the first test we can divide 9 labs into groups \{2, 8, 5\}, \{9, 3, 4\}, \{7, 6, 1\}.
From the first group to the second group we can transport 4 units of water (8 → 3, 8 → 4, 5 → 3, 5 → 4).
From the first group to the third group we can transport 5 units of water (2 → 1, 8 → 7, 8 → 6, 8 → 1, 5 → 1).
From the second group to the first group we can transport 5 units of water (9 → 2, 9 → 8, 9 → 5, 3 → 2, 4 → 2).
From the second group to the third group we can transport 5 units of water (9 → 7, 9 → 6, 9 → 1, 3 → 1, 4 → 1).
From the third group to the first group we can transport 4 units of water (7 → 2, 7 → 5, 6 → 2, 6 → 5).
From the third group to the second group we can transport 4 units of water (7 → 3, 7 → 4, 6 → 3, 6 → 4).
The minimal number of the sum of units of water, that can be transported from one group to another is equal to 4. It can be proved, that it is impossible to make a better division. | def create(n):
res = [[0] * n for _ in range(n)]
cur = 1
for j in range(n):
for i in (range(n - 1, -1, -1) if j & 1 else range(n)):
res[i][j] = cur
cur += 1
return res
n = int(input())
res = create(n)
for l in res: print(*l) | {
"input": [
"3\n"
],
"output": [
"1 6 7 \n2 5 8 \n3 4 9 \n"
]
} |
654 | 8 | This is the easier version of the problem. In this version, 1 ≤ n ≤ 10^5 and 0 ≤ a_i ≤ 1. You can hack this problem only if you solve and lock both problems.
Christmas is coming, and our protagonist, Bob, is preparing a spectacular present for his long-time best friend Alice. This year, he decides to prepare n boxes of chocolate, numbered from 1 to n. Initially, the i-th box contains a_i chocolate pieces.
Since Bob is a typical nice guy, he will not send Alice n empty boxes. In other words, at least one of a_1, a_2, …, a_n is positive. Since Alice dislikes coprime sets, she will be happy only if there exists some integer k > 1 such that the number of pieces in each box is divisible by k. Note that Alice won't mind if there exists some empty boxes.
Charlie, Alice's boyfriend, also is Bob's second best friend, so he decides to help Bob by rearranging the chocolate pieces. In one second, Charlie can pick up a piece in box i and put it into either box i-1 or box i+1 (if such boxes exist). Of course, he wants to help his friend as quickly as possible. Therefore, he asks you to calculate the minimum number of seconds he would need to make Alice happy.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of chocolate boxes.
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 1) — the number of chocolate pieces in the i-th box.
It is guaranteed that at least one of a_1, a_2, …, a_n is positive.
Output
If there is no way for Charlie to make Alice happy, print -1.
Otherwise, print a single integer x — the minimum number of seconds for Charlie to help Bob make Alice happy.
Examples
Input
3
1 0 1
Output
2
Input
1
1
Output
-1 | from itertools import accumulate
def divisors(n):
tab = []
for i in range(2, int(n**.5) + 2):
if n % i == 0:
while n % i == 0:
n //= i
tab.append(i)
if n > 1:
tab.append(n)
return tab
n = int(input())
a = list(map(int, input().split()))
s = sum(a)
tab = divisors(s)
if s == 1:
print(-1)
else:
m = 10**18
pref = list(accumulate(a))
for k in tab:
x = 0
for i in range(n - 1):
x += min(pref[i] % k, k - (pref[i] % k))
m = min(m, x)
print(m) | {
"input": [
"1\n1\n",
"3\n1 0 1\n"
],
"output": [
"-1",
"2\n"
]
} |
655 | 7 | Polycarp has built his own web service. Being a modern web service it includes login feature. And that always implies password security problems.
Polycarp decided to store the hash of the password, generated by the following algorithm:
1. take the password p, consisting of lowercase Latin letters, and shuffle the letters randomly in it to obtain p' (p' can still be equal to p);
2. generate two random strings, consisting of lowercase Latin letters, s_1 and s_2 (any of these strings can be empty);
3. the resulting hash h = s_1 + p' + s_2, where addition is string concatenation.
For example, let the password p = "abacaba". Then p' can be equal to "aabcaab". Random strings s1 = "zyx" and s2 = "kjh". Then h = "zyxaabcaabkjh".
Note that no letters could be deleted or added to p to obtain p', only the order could be changed.
Now Polycarp asks you to help him to implement the password check module. Given the password p and the hash h, check that h can be the hash for the password p.
Your program should answer t independent test cases.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The first line of each test case contains a non-empty string p, consisting of lowercase Latin letters. The length of p does not exceed 100.
The second line of each test case contains a non-empty string h, consisting of lowercase Latin letters. The length of h does not exceed 100.
Output
For each test case print the answer to it — "YES" if the given hash h could be obtained from the given password p or "NO" otherwise.
Example
Input
5
abacaba
zyxaabcaabkjh
onetwothree
threetwoone
one
zzonneyy
one
none
twenty
ten
Output
YES
YES
NO
YES
NO
Note
The first test case is explained in the statement.
In the second test case both s_1 and s_2 are empty and p'= "threetwoone" is p shuffled.
In the third test case the hash could not be obtained from the password.
In the fourth test case s_1= "n", s_2 is empty and p'= "one" is p shuffled (even thought it stayed the same).
In the fifth test case the hash could not be obtained from the password. | def ans(p, h):
n = len(p)
for i in range(len(h)-len(p)+1):
if sorted(h[i:i+n]) == sorted(p):
return "YES"
return "NO"
for u in range(int(input())):
p=input()
h = input()
print(ans(p,h)) | {
"input": [
"5\nabacaba\nzyxaabcaabkjh\nonetwothree\nthreetwoone\none\nzzonneyy\none\nnone\ntwenty\nten\n"
],
"output": [
"YES\nYES\nNO\nYES\nNO\n"
]
} |
656 | 8 | Tanya wants to go on a journey across the cities of Berland. There are n cities situated along the main railroad line of Berland, and these cities are numbered from 1 to n.
Tanya plans her journey as follows. First of all, she will choose some city c_1 to start her journey. She will visit it, and after that go to some other city c_2 > c_1, then to some other city c_3 > c_2, and so on, until she chooses to end her journey in some city c_k > c_{k - 1}. So, the sequence of visited cities [c_1, c_2, ..., c_k] should be strictly increasing.
There are some additional constraints on the sequence of cities Tanya visits. Each city i has a beauty value b_i associated with it. If there is only one city in Tanya's journey, these beauty values imply no additional constraints. But if there are multiple cities in the sequence, then for any pair of adjacent cities c_i and c_{i + 1}, the condition c_{i + 1} - c_i = b_{c_{i + 1}} - b_{c_i} must hold.
For example, if n = 8 and b = [3, 4, 4, 6, 6, 7, 8, 9], there are several three possible ways to plan a journey:
* c = [1, 2, 4];
* c = [3, 5, 6, 8];
* c = [7] (a journey consisting of one city is also valid).
There are some additional ways to plan a journey that are not listed above.
Tanya wants her journey to be as beautiful as possible. The beauty value of the whole journey is the sum of beauty values over all visited cities. Can you help her to choose the optimal plan, that is, to maximize the beauty value of the journey?
Input
The first line contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of cities in Berland.
The second line contains n integers b_1, b_2, ..., b_n (1 ≤ b_i ≤ 4 ⋅ 10^5), where b_i is the beauty value of the i-th city.
Output
Print one integer — the maximum beauty of a journey Tanya can choose.
Examples
Input
6
10 7 1 9 10 15
Output
26
Input
1
400000
Output
400000
Input
7
8 9 26 11 12 29 14
Output
55
Note
The optimal journey plan in the first example is c = [2, 4, 5].
The optimal journey plan in the second example is c = [1].
The optimal journey plan in the third example is c = [3, 6]. | def go():
n = int(input())
b = list(map(int, input().split()))
d={}
for i,bb in enumerate(b):
d[bb-i] = d.get(bb-i,0)+bb
return max(d.values())
print (go()) | {
"input": [
"7\n8 9 26 11 12 29 14\n",
"6\n10 7 1 9 10 15\n",
"1\n400000\n"
],
"output": [
"55\n",
"26\n",
"400000\n"
]
} |
657 | 7 | Denis was very sad after Nastya rejected him. So he decided to walk through the gateways to have some fun. And luck smiled at him! When he entered the first courtyard, he met a strange man who was selling something.
Denis bought a mysterious item and it was... Random permutation generator! Denis could not believed his luck.
When he arrived home, he began to study how his generator works and learned the algorithm. The process of generating a permutation consists of n steps. At the i-th step, a place is chosen for the number i (1 ≤ i ≤ n). The position for the number i is defined as follows:
* For all j from 1 to n, we calculate r_j — the minimum index such that j ≤ r_j ≤ n, and the position r_j is not yet occupied in the permutation. If there are no such positions, then we assume that the value of r_j is not defined.
* For all t from 1 to n, we calculate count_t — the number of positions 1 ≤ j ≤ n such that r_j is defined and r_j = t.
* Consider the positions that are still not occupied by permutation and among those we consider the positions for which the value in the count array is maximum.
* The generator selects one of these positions for the number i. The generator can choose any position.
Let's have a look at the operation of the algorithm in the following example:
<image>
Let n = 5 and the algorithm has already arranged the numbers 1, 2, 3 in the permutation. Consider how the generator will choose a position for the number 4:
* The values of r will be r = [3, 3, 3, 4, ×], where × means an indefinite value.
* Then the count values will be count = [0, 0, 3, 1, 0].
* There are only two unoccupied positions in the permutation: 3 and 4. The value in the count array for position 3 is 3, for position 4 it is 1.
* The maximum value is reached only for position 3, so the algorithm will uniquely select this position for number 4.
Satisfied with his purchase, Denis went home. For several days without a break, he generated permutations. He believes that he can come up with random permutations no worse than a generator. After that, he wrote out the first permutation that came to mind p_1, p_2, …, p_n and decided to find out if it could be obtained as a result of the generator.
Unfortunately, this task was too difficult for him, and he asked you for help. It is necessary to define whether the written permutation could be obtained using the described algorithm if the generator always selects the position Denis needs.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases. Then the descriptions of the test cases follow.
The first line of the test case contains a single integer n (1 ≤ n ≤ 10^5) — the size of the permutation.
The second line of the test case contains n different integers p_1, p_2, …, p_n (1 ≤ p_i ≤ n) — the permutation written by Denis.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
Print "Yes" if this permutation could be obtained as a result of the generator. Otherwise, print "No".
All letters can be displayed in any case.
Example
Input
5
5
2 3 4 5 1
1
1
3
1 3 2
4
4 2 3 1
5
1 5 2 4 3
Output
Yes
Yes
No
Yes
No
Note
Let's simulate the operation of the generator in the first test.
At the 1 step, r = [1, 2, 3, 4, 5], count = [1, 1, 1, 1, 1]. The maximum value is reached in any free position, so the generator can choose a random position from 1 to 5. In our example, it chose 5.
At the 2 step, r = [1, 2, 3, 4, ×], count = [1, 1, 1, 1, 0]. The maximum value is reached in positions from 1 to 4, so the generator can choose a random position among them. In our example, it chose 1.
At the 3 step, r = [2, 2, 3, 4, ×], count = [0, 2, 1, 1, 0]. The maximum value is 2 and is reached only at the 2 position, so the generator will choose this position.
At the 4 step, r = [3, 3, 3, 4, ×], count = [0, 0, 3, 1, 0]. The maximum value is 3 and is reached only at the 3 position, so the generator will choose this position.
At the 5 step, r = [4, 4, 4, 4, ×], count = [0, 0, 0, 4, 0]. The maximum value is 4 and is reached only at the 4 position, so the generator will choose this position.
In total, we got a permutation of 2, 3, 4, 5, 1, that is, a generator could generate it. | t = int(input())
def sol():
n = int(input())
a = list(map(int, input().split()))
for i in range(n-1):
if a[i] + 1 != a[i+1] and a[i] < a[i+1]:
return False
return True
for i in range(t):
print('Yes' if sol() else 'No') | {
"input": [
"5\n5\n2 3 4 5 1\n1\n1\n3\n1 3 2\n4\n4 2 3 1\n5\n1 5 2 4 3\n"
],
"output": [
"Yes\nYes\nNo\nYes\nNo\n"
]
} |
658 | 7 | Johnny has recently found an ancient, broken computer. The machine has only one register, which allows one to put in there one variable. Then in one operation, you can shift its bits left or right by at most three positions. The right shift is forbidden if it cuts off some ones. So, in fact, in one operation, you can multiply or divide your number by 2, 4 or 8, and division is only allowed if the number is divisible by the chosen divisor.
Formally, if the register contains a positive integer x, in one operation it can be replaced by one of the following:
* x ⋅ 2
* x ⋅ 4
* x ⋅ 8
* x / 2, if x is divisible by 2
* x / 4, if x is divisible by 4
* x / 8, if x is divisible by 8
For example, if x = 6, in one operation it can be replaced by 12, 24, 48 or 3. Value 6 isn't divisible by 4 or 8, so there're only four variants of replacement.
Now Johnny wonders how many operations he needs to perform if he puts a in the register and wants to get b at the end.
Input
The input consists of multiple test cases. The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases. The following t lines contain a description of test cases.
The first and only line in each test case contains integers a and b (1 ≤ a, b ≤ 10^{18}) — the initial and target value of the variable, respectively.
Output
Output t lines, each line should contain one integer denoting the minimum number of operations Johnny needs to perform. If Johnny cannot get b at the end, then write -1.
Example
Input
10
10 5
11 44
17 21
1 1
96 3
2 128
1001 1100611139403776
1000000000000000000 1000000000000000000
7 1
10 8
Output
1
1
-1
0
2
2
14
0
-1
-1
Note
In the first test case, Johnny can reach 5 from 10 by using the shift to the right by one (i.e. divide by 2).
In the second test case, Johnny can reach 44 from 11 by using the shift to the left by two (i.e. multiply by 4).
In the third test case, it is impossible for Johnny to reach 21 from 17.
In the fourth test case, initial and target values are equal, so Johnny has to do 0 operations.
In the fifth test case, Johnny can reach 3 from 96 by using two shifts to the right: one by 2, and another by 3 (i.e. divide by 4 and by 8). | def main():
for _ in range(int(input())):
a, b = map(bin, map(int, input().split()))
print((abs(len(a) - len(b)) + 2) // 3 if a.rstrip('0') == b.rstrip('0') else -1)
if __name__ == '__main__':
main()
| {
"input": [
"10\n10 5\n11 44\n17 21\n1 1\n96 3\n2 128\n1001 1100611139403776\n1000000000000000000 1000000000000000000\n7 1\n10 8\n"
],
"output": [
"1\n1\n-1\n0\n2\n2\n14\n0\n-1\n-1\n"
]
} |
659 | 9 | This is the easy version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 3n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 1000) — the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 1000.
Output
For each test case, output an integer k (0≤ k≤ 3n), followed by k integers p_1,…,p_k (1≤ p_i≤ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01→ 11→ 00→ 10.
In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged. | def solve(n, a, b):
cmds = []
for i in range(n):
if a[i] != b[i]:
cmds.extend([i + 1, 1, i + 1])
return cmds
t = int(input())
for case in range(t):
x = solve(int(input()), input(), input())
print(len(x), *x) | {
"input": [
"5\n2\n01\n10\n5\n01011\n11100\n2\n01\n01\n10\n0110011011\n1000110100\n1\n0\n1\n"
],
"output": [
"3 1 2 1 \n3 1 5 2 \n0 \n7 1 10 8 7 1 2 1 \n1 1 \n"
]
} |
660 | 7 | You are given an array a of length n, and an integer x. You can perform the following operation as many times as you would like (possibly zero): replace two adjacent elements of the array by their sum. For example, if the initial array was [3, 6, 9], in a single operation one can replace the last two elements by their sum, yielding an array [3, 15], or replace the first two elements to get an array [9, 9]. Note that the size of the array decreases after each operation.
The beauty of an array b=[b_1, …, b_k] is defined as ∑_{i=1}^k \left⌈ (b_i)/(x) \right⌉, which means that we divide each element by x, round it up to the nearest integer, and sum up the resulting values. For example, if x = 3, and the array is [4, 11, 6], the beauty of the array is equal to \left⌈ 4/3 \right⌉ + \left⌈ 11/3 \right⌉ + \left⌈ 6/3 \right⌉ = 2 + 4 + 2 = 8.
Please determine the minimum and the maximum beauty you can get by performing some operations on the original array.
Input
The first input line contains a single integer t — the number of test cases (1 ≤ t ≤ 1000).
The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5, 1 ≤ x ≤ 10^9).
The next line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9), the elements of the array a.
It is guaranteed that the sum of values of n over all test cases does not exceed 10^5.
Output
For each test case output two integers — the minimal and the maximal possible beauty.
Example
Input
2
3 3
3 6 9
3 3
6 4 11
Output
6 6
7 8
Note
In the first test case the beauty of the array does not change if we perform any operations.
In the second example we can leave the array unchanged to attain the maximum beauty, and to get the minimum beauty one can replace two elements 4 and 11 with their sum, yielding an array [6, 15], which has its beauty equal to 7. | #fi = open("task1.txt")
#def input():
# return fi.readline()
T = int(input())
for _ in range(T):
n, x = [int(y) for y in input().split()]
b = [int(y) for y in input().split()]
print(sum(b)//x+(1 if sum(b)%x>0 else 0),sum([(y//x)+(1 if y%x>0 else 0) for y in b])) | {
"input": [
"2\n3 3\n3 6 9\n3 3\n6 4 11\n"
],
"output": [
"\n6 6\n7 8\n"
]
} |
661 | 7 | The \text{gcdSum} of a positive integer is the gcd of that integer with its sum of digits. Formally, \text{gcdSum}(x) = gcd(x, sum of digits of x) for a positive integer x. gcd(a, b) denotes the greatest common divisor of a and b — the largest integer d such that both integers a and b are divisible by d.
For example: \text{gcdSum}(762) = gcd(762, 7 + 6 + 2)=gcd(762,15) = 3.
Given an integer n, find the smallest integer x ≥ n such that \text{gcdSum}(x) > 1.
Input
The first line of input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases.
Then t lines follow, each containing a single integer n (1 ≤ n ≤ 10^{18}).
All test cases in one test are different.
Output
Output t lines, where the i-th line is a single integer containing the answer to the i-th test case.
Example
Input
3
11
31
75
Output
12
33
75
Note
Let us explain the three test cases in the sample.
Test case 1: n = 11:
\text{gcdSum}(11) = gcd(11, 1 + 1) = gcd(11,\ 2) = 1.
\text{gcdSum}(12) = gcd(12, 1 + 2) = gcd(12,\ 3) = 3.
So the smallest number ≥ 11 whose gcdSum > 1 is 12.
Test case 2: n = 31:
\text{gcdSum}(31) = gcd(31, 3 + 1) = gcd(31,\ 4) = 1.
\text{gcdSum}(32) = gcd(32, 3 + 2) = gcd(32,\ 5) = 1.
\text{gcdSum}(33) = gcd(33, 3 + 3) = gcd(33,\ 6) = 3.
So the smallest number ≥ 31 whose gcdSum > 1 is 33.
Test case 3: \ n = 75:
\text{gcdSum}(75) = gcd(75, 7 + 5) = gcd(75,\ 12) = 3.
The \text{gcdSum} of 75 is already > 1. Hence, it is the answer. | import math
def f(n):
return sum(int(i) for i in str(n))
for _ in range(int(input())):
n = int(input())
while 1:
if math.gcd(n, f(n)) > 1:
print(n)
break
n += 1 | {
"input": [
"3\n11\n31\n75\n"
],
"output": [
"\n12\n33\n75\n"
]
} |
662 | 11 | On a strip of land of length n there are k air conditioners: the i-th air conditioner is placed in cell a_i (1 ≤ a_i ≤ n). Two or more air conditioners cannot be placed in the same cell (i.e. all a_i are distinct).
Each air conditioner is characterized by one parameter: temperature. The i-th air conditioner is set to the temperature t_i.
<image> Example of strip of length n=6, where k=2, a=[2,5] and t=[14,16].
For each cell i (1 ≤ i ≤ n) find it's temperature, that can be calculated by the formula $$$min_{1 ≤ j ≤ k}(t_j + |a_j - i|),$$$
where |a_j - i| denotes absolute value of the difference a_j - i.
In other words, the temperature in cell i is equal to the minimum among the temperatures of air conditioners, increased by the distance from it to the cell i.
Let's look at an example. Consider that n=6, k=2, the first air conditioner is placed in cell a_1=2 and is set to the temperature t_1=14 and the second air conditioner is placed in cell a_2=5 and is set to the temperature t_2=16. In that case temperatures in cells are:
1. temperature in cell 1 is: min(14 + |2 - 1|, 16 + |5 - 1|)=min(14 + 1, 16 + 4)=min(15, 20)=15;
2. temperature in cell 2 is: min(14 + |2 - 2|, 16 + |5 - 2|)=min(14 + 0, 16 + 3)=min(14, 19)=14;
3. temperature in cell 3 is: min(14 + |2 - 3|, 16 + |5 - 3|)=min(14 + 1, 16 + 2)=min(15, 18)=15;
4. temperature in cell 4 is: min(14 + |2 - 4|, 16 + |5 - 4|)=min(14 + 2, 16 + 1)=min(16, 17)=16;
5. temperature in cell 5 is: min(14 + |2 - 5|, 16 + |5 - 5|)=min(14 + 3, 16 + 0)=min(17, 16)=16;
6. temperature in cell 6 is: min(14 + |2 - 6|, 16 + |5 - 6|)=min(14 + 4, 16 + 1)=min(18, 17)=17.
For each cell from 1 to n find the temperature in it.
Input
The first line contains one integer q (1 ≤ q ≤ 10^4) — the number of test cases in the input. Then test cases follow. Before each test case, there is an empty line.
Each test case contains three lines. The first line contains two integers n (1 ≤ n ≤ 3 ⋅ 10^5) and k (1 ≤ k ≤ n) — the length of the strip of land and the number of air conditioners respectively.
The second line contains k integers a_1, a_2, …, a_k (1 ≤ a_i ≤ n) — positions of air conditioners on the strip of land.
The third line contains k integers t_1, t_2, …, t_k (1 ≤ t_i ≤ 10^9) — temperatures of air conditioners.
It is guaranteed that the sum of n over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case output n integers separated by space: temperatures of air in cells.
Example
Input
5
6 2
2 5
14 16
10 1
7
30
5 5
3 1 4 2 5
3 1 4 2 5
7 1
1
1000000000
6 3
6 1 3
5 5 5
Output
15 14 15 16 16 17
36 35 34 33 32 31 30 31 32 33
1 2 3 4 5
1000000000 1000000001 1000000002 1000000003 1000000004 1000000005 1000000006
5 6 5 6 6 5 | def solve():
input()
n, k = [int(i) for i in input().split()]
a = [int(i) for i in input().split()]
t = [int(i) for i in input().split()]
dp = [10 ** 10 for i in range(n + 1)]
for i in range(k):
dp[a[i]] = t[i]
for i in range(1, len(dp)):
dp[i] = min(dp[i - 1] + 1, dp[i])
for i in range(len(dp) - 2, 0, -1):
dp[i] = min(dp[i + 1] + 1, dp[i])
print(*dp[1:])
for _ in range(int(input())):
solve() | {
"input": [
"5\n\n6 2\n2 5\n14 16\n\n10 1\n7\n30\n\n5 5\n3 1 4 2 5\n3 1 4 2 5\n\n7 1\n1\n1000000000\n\n6 3\n6 1 3\n5 5 5\n"
],
"output": [
"15 14 15 16 16 17 \n36 35 34 33 32 31 30 31 32 33 \n1 2 3 4 5 \n1000000000 1000000001 1000000002 1000000003 1000000004 1000000005 1000000006 \n5 6 5 6 6 5 \n"
]
} |
663 | 8 | Vasya has been playing Plane of Tanks with his friends the whole year. Now it is time to divide the participants into several categories depending on their results.
A player is given a non-negative integer number of points in each round of the Plane of Tanks. Vasya wrote results for each round of the last year. He has n records in total.
In order to determine a player's category consider the best result obtained by the player and the best results of other players. The player belongs to category:
* "noob" — if more than 50% of players have better results;
* "random" — if his result is not worse than the result that 50% of players have, but more than 20% of players have better results;
* "average" — if his result is not worse than the result that 80% of players have, but more than 10% of players have better results;
* "hardcore" — if his result is not worse than the result that 90% of players have, but more than 1% of players have better results;
* "pro" — if his result is not worse than the result that 99% of players have.
When the percentage is calculated the player himself is taken into account. That means that if two players played the game and the first one gained 100 points and the second one 1000 points, then the first player's result is not worse than the result that 50% of players have, and the second one is not worse than the result that 100% of players have.
Vasya gave you the last year Plane of Tanks results. Help Vasya determine each player's category.
Input
The first line contains the only integer number n (1 ≤ n ≤ 1000) — a number of records with the players' results.
Each of the next n lines contains a player's name and the amount of points, obtained by the player for the round, separated with a space. The name contains not less than 1 and no more than 10 characters. The name consists of lowercase Latin letters only. It is guaranteed that any two different players have different names. The amount of points, obtained by the player for the round, is a non-negative integer number and does not exceed 1000.
Output
Print on the first line the number m — the number of players, who participated in one round at least.
Each one of the next m lines should contain a player name and a category he belongs to, separated with space. Category can be one of the following: "noob", "random", "average", "hardcore" or "pro" (without quotes). The name of each player should be printed only once. Player names with respective categories can be printed in an arbitrary order.
Examples
Input
5
vasya 100
vasya 200
artem 100
kolya 200
igor 250
Output
4
artem noob
igor pro
kolya random
vasya random
Input
3
vasya 200
kolya 1000
vasya 1000
Output
2
kolya pro
vasya pro
Note
In the first example the best result, obtained by artem is not worse than the result that 25% of players have (his own result), so he belongs to category "noob". vasya and kolya have best results not worse than the results that 75% players have (both of them and artem), so they belong to category "random". igor has best result not worse than the result that 100% of players have (all other players and himself), so he belongs to category "pro".
In the second example both players have the same amount of points, so they have results not worse than 100% players have, so they belong to category "pro". | def t(g, n):
for x, y in ((50, 'noob'), (20, 'random'), (10, 'average'), (1, 'hardcore')):
if g > x * n // 100:
return y
return 'pro'
p, st = {}, {}
for i in range(int(input())):
n, s = input().split()
if n not in p or int(s) > p[n]:
p[n] = int(s)
for i, si in enumerate(sorted(p.values(), reverse=True)):
if si not in st:
st[si] = t(i, len(p))
print(len(p))
print(*sorted(k + ' ' + st[p[k]] for k in p), sep='\n')
| {
"input": [
"5\nvasya 100\nvasya 200\nartem 100\nkolya 200\nigor 250\n",
"3\nvasya 200\nkolya 1000\nvasya 1000\n"
],
"output": [
"4\nartem noob\nigor pro\nkolya random\nvasya random\n",
"2\nkolya pro\nvasya pro\n"
]
} |
664 | 9 | Vasya is developing his own programming language VPL (Vasya Programming Language). Right now he is busy making the system of exceptions. He thinks that the system of exceptions must function like that.
The exceptions are processed by try-catch-blocks. There are two operators that work with the blocks:
1. The try operator. It opens a new try-catch-block.
2. The catch(<exception_type>, <message>) operator. It closes the try-catch-block that was started last and haven't yet been closed. This block can be activated only via exception of type <exception_type>. When we activate this block, the screen displays the <message>. If at the given moment there is no open try-catch-block, then we can't use the catch operator.
The exceptions can occur in the program in only one case: when we use the throw operator. The throw(<exception_type>) operator creates the exception of the given type.
Let's suggest that as a result of using some throw operator the program created an exception of type a. In this case a try-catch-block is activated, such that this block's try operator was described in the program earlier than the used throw operator. Also, this block's catch operator was given an exception type a as a parameter and this block's catch operator is described later that the used throw operator. If there are several such try-catch-blocks, then the system activates the block whose catch operator occurs earlier than others. If no try-catch-block was activated, then the screen displays message "Unhandled Exception".
To test the system, Vasya wrote a program that contains only try, catch and throw operators, one line contains no more than one operator, the whole program contains exactly one throw operator.
Your task is: given a program in VPL, determine, what message will be displayed on the screen.
Input
The first line contains a single integer: n (1 ≤ n ≤ 105) the number of lines in the program. Next n lines contain the program in language VPL. Each line contains no more than one operator. It means that input file can contain empty lines and lines, consisting only of spaces.
The program contains only operators try, catch and throw. It is guaranteed that the program is correct. It means that each started try-catch-block was closed, the catch operators aren't used unless there is an open try-catch-block. The program has exactly one throw operator. The program may have spaces at the beginning of a line, at the end of a line, before and after a bracket, a comma or a quote mark.
The exception type is a nonempty string, that consists only of upper and lower case english letters. The length of the string does not exceed 20 symbols. Message is a nonempty string, that consists only of upper and lower case english letters, digits and spaces. Message is surrounded with quote marks. Quote marks shouldn't be printed. The length of the string does not exceed 20 symbols.
Length of any line in the input file does not exceed 50 symbols.
Output
Print the message the screen will show after the given program is executed.
Examples
Input
8
try
try
throw ( AE )
catch ( BE, "BE in line 3")
try
catch(AE, "AE in line 5")
catch(AE,"AE somewhere")
Output
AE somewhere
Input
8
try
try
throw ( AE )
catch ( AE, "AE in line 3")
try
catch(BE, "BE in line 5")
catch(AE,"AE somewhere")
Output
AE in line 3
Input
8
try
try
throw ( CE )
catch ( BE, "BE in line 3")
try
catch(AE, "AE in line 5")
catch(AE,"AE somewhere")
Output
Unhandled Exception
Note
In the first sample there are 2 try-catch-blocks such that try operator is described earlier than throw operator and catch operator is described later than throw operator: try-catch(BE,"BE in line 3") and try-catch(AE,"AE somewhere"). Exception type is AE, so the second block will be activated, because operator catch(AE,"AE somewhere") has exception type AE as parameter and operator catch(BE,"BE in line 3") has exception type BE.
In the second sample there are 2 try-catch-blocks such that try operator is described earlier than throw operator and catch operator is described later than throw operator: try-catch(AE,"AE in line 3") and try-catch(AE,"AE somewhere"). Exception type is AE, so both blocks can be activated, but only the first one will be activated, because operator catch(AE,"AE in line 3") is described earlier than catch(AE,"AE somewhere")
In the third sample there is no blocks that can be activated by an exception of type CE. | from sys import stdin, stdout
n = int(stdin.readline())
label = ''
cnt = 0
def first(s):
f = ''
for a in s:
if 'z' >= a >= 'a':
f += a
else:
break
return f
for i in range(n):
s = stdin.readline().strip()
if first(s) == 'throw':
label = s[s.index('(') + 1:s.index(')')].strip()
elif first(s) == 'try' and label:
cnt += 1
elif first(s) == 'catch' and cnt:
cnt -= 1
elif first(s) == 'catch' and s[s.index('(') + 1:s.index(',')].strip() == label:
f = s[s.index(',') + 1:s.index(')')].strip()
f = f.strip('"').strip()
stdout.write(f)
break
else:
stdout.write('Unhandled Exception') | {
"input": [
"8\ntry\n try\n throw ( AE ) \n catch ( BE, \"BE in line 3\")\n\n try\n catch(AE, \"AE in line 5\") \ncatch(AE,\"AE somewhere\")\n",
"8\ntry\n try\n throw ( AE ) \n catch ( AE, \"AE in line 3\")\n\n try\n catch(BE, \"BE in line 5\") \ncatch(AE,\"AE somewhere\")\n",
"8\ntry\n try\n throw ( CE ) \n catch ( BE, \"BE in line 3\")\n\n try\n catch(AE, \"AE in line 5\") \ncatch(AE,\"AE somewhere\")\n"
],
"output": [
"AE somewhere\n",
"AE in line 3\n",
"Unhandled Exception\n"
]
} |
665 | 9 | A colored stripe is represented by a horizontal row of n square cells, each cell is pained one of k colors. Your task is to repaint the minimum number of cells so that no two neighbouring cells are of the same color. You can use any color from 1 to k to repaint the cells.
Input
The first input line contains two integers n and k (1 ≤ n ≤ 5·105; 2 ≤ k ≤ 26). The second line contains n uppercase English letters. Letter "A" stands for the first color, letter "B" stands for the second color and so on. The first k English letters may be used. Each letter represents the color of the corresponding cell of the stripe.
Output
Print a single integer — the required minimum number of repaintings. In the second line print any possible variant of the repainted stripe.
Examples
Input
6 3
ABBACC
Output
2
ABCACA
Input
3 2
BBB
Output
1
BAB | n, k = map(int, input().split())
s = list(input())
if k > 2:
c = set(ord("A") + i for i in range(26))
r = 0
for i in range(1,len(s)):
if s[i-1] == s[i]:
r += 1
if i + 1 < n:
s[i] = "A" if "A" not in {s[i-1], s[i+1]} else ('B' if "B" not in {s[i], s[i+1]} else "C")
else:
s[i] = "A" if s[i-1] != "A" else "B"
print(r)
print("".join(s))
else:
x, y = "AB" * n, "BA" * n
def check(x, y):
return sum(x[i] != y[i] for i in range(n))
check_1 = check(x, s)
check_2 = check(y, s)
z = [x,y][check_1 > check_2]
print(min(check_1, check_2))
print(z[:n]) | {
"input": [
"3 2\nBBB\n",
"6 3\nABBACC\n"
],
"output": [
"1\nBAB",
"2\nABCACA\n"
]
} |
666 | 9 | Manao has invented a new mathematical term — a beautiful set of points. He calls a set of points on a plane beautiful if it meets the following conditions:
1. The coordinates of each point in the set are integers.
2. For any two points from the set, the distance between them is a non-integer.
Consider all points (x, y) which satisfy the inequations: 0 ≤ x ≤ n; 0 ≤ y ≤ m; x + y > 0. Choose their subset of maximum size such that it is also a beautiful set of points.
Input
The single line contains two space-separated integers n and m (1 ≤ n, m ≤ 100).
Output
In the first line print a single integer — the size k of the found beautiful set. In each of the next k lines print a pair of space-separated integers — the x- and y- coordinates, respectively, of a point from the set.
If there are several optimal solutions, you may print any of them.
Examples
Input
2 2
Output
3
0 1
1 2
2 0
Input
4 3
Output
4
0 3
2 1
3 0
4 2
Note
Consider the first sample. The distance between points (0, 1) and (1, 2) equals <image>, between (0, 1) and (2, 0) — <image>, between (1, 2) and (2, 0) — <image>. Thus, these points form a beautiful set. You cannot form a beautiful set with more than three points out of the given points. Note that this is not the only solution. | def solve(n,m):
a=min(n+1,m+1)
print(a)
for i in range(a):
print(i,a-1-i)
A=list(map(int,input().split()))
solve(A[0],A[1])
| {
"input": [
"2 2\n",
"4 3\n"
],
"output": [
"3\n0 2\n1 1\n2 0\n",
"4\n0 3\n1 2\n2 1\n3 0\n"
]
} |
667 | 9 | Xenia has a set of weights and pan scales. Each weight has an integer weight from 1 to 10 kilos. Xenia is going to play with scales and weights a little. For this, she puts weights on the scalepans, one by one. The first weight goes on the left scalepan, the second weight goes on the right scalepan, the third one goes on the left scalepan, the fourth one goes on the right scalepan and so on. Xenia wants to put the total of m weights on the scalepans.
Simply putting weights on the scales is not interesting, so Xenia has set some rules. First, she does not put on the scales two consecutive weights of the same weight. That is, the weight that goes i-th should be different from the (i + 1)-th weight for any i (1 ≤ i < m). Second, every time Xenia puts a weight on some scalepan, she wants this scalepan to outweigh the other one. That is, the sum of the weights on the corresponding scalepan must be strictly greater than the sum on the other pan.
You are given all types of weights available for Xenia. You can assume that the girl has an infinite number of weights of each specified type. Your task is to help Xenia lay m weights on the scales or to say that it can't be done.
Input
The first line contains a string consisting of exactly ten zeroes and ones: the i-th (i ≥ 1) character in the line equals "1" if Xenia has i kilo weights, otherwise the character equals "0". The second line contains integer m (1 ≤ m ≤ 1000).
Output
In the first line print "YES", if there is a way to put m weights on the scales by all rules. Otherwise, print in the first line "NO". If you can put m weights on the scales, then print in the next line m integers — the weights' weights in the order you put them on the scales.
If there are multiple solutions, you can print any of them.
Examples
Input
0000000101
3
Output
YES
8 10 8
Input
1000000000
2
Output
NO | import math
from decimal import *
getcontext().prec=30
x=input()
m=int(input())
ans=[0]*1010
def dfs(curr,bal):
if curr>m:
return 1
for i in range(bal+1,11):
if ans[curr-1]!=i and x[i-1]=='1':
ans[curr]=i
if dfs(curr+1,i-bal)==1:
return 1
return 0
if dfs(1,0)==1:
print("YES")
print(*ans[1:m+1])
else :
print("NO")
| {
"input": [
"0000000101\n3\n",
"1000000000\n2\n"
],
"output": [
"YES\n8 10 8\n",
"NO\n"
]
} |
668 | 9 | Recently, the bear started studying data structures and faced the following problem.
You are given a sequence of integers x1, x2, ..., xn of length n and m queries, each of them is characterized by two integers li, ri. Let's introduce f(p) to represent the number of such indexes k, that xk is divisible by p. The answer to the query li, ri is the sum: <image>, where S(li, ri) is a set of prime numbers from segment [li, ri] (both borders are included in the segment).
Help the bear cope with the problem.
Input
The first line contains integer n (1 ≤ n ≤ 106). The second line contains n integers x1, x2, ..., xn (2 ≤ xi ≤ 107). The numbers are not necessarily distinct.
The third line contains integer m (1 ≤ m ≤ 50000). Each of the following m lines contains a pair of space-separated integers, li and ri (2 ≤ li ≤ ri ≤ 2·109) — the numbers that characterize the current query.
Output
Print m integers — the answers to the queries on the order the queries appear in the input.
Examples
Input
6
5 5 7 10 14 15
3
2 11
3 12
4 4
Output
9
7
0
Input
7
2 3 5 7 11 4 8
2
8 10
2 123
Output
0
7
Note
Consider the first sample. Overall, the first sample has 3 queries.
1. The first query l = 2, r = 11 comes. You need to count f(2) + f(3) + f(5) + f(7) + f(11) = 2 + 1 + 4 + 2 + 0 = 9.
2. The second query comes l = 3, r = 12. You need to count f(3) + f(5) + f(7) + f(11) = 1 + 4 + 2 + 0 = 7.
3. The third query comes l = 4, r = 4. As this interval has no prime numbers, then the sum equals 0. | import math, sys
input = sys.stdin.buffer.readline
def ints():
return map(int, input().split())
n = int(input())
x = list(ints())
MAX = max(x) + 1
freq = [0] * MAX
for i in x:
freq[i] += 1
sieve = [False] * MAX
f = [0] * MAX
for i in range(2, MAX):
if sieve[i]:
continue
for j in range(i, MAX, i):
sieve[j] = True
f[i] += freq[j]
for i in range(2, MAX):
f[i] += f[i-1]
m = int(input())
for i in range(m):
l, r = ints()
if l >= MAX:
print(0)
elif r >= MAX:
print(f[-1] - f[l-1])
else:
print(f[r] - f[l-1]) | {
"input": [
"7\n2 3 5 7 11 4 8\n2\n8 10\n2 123\n",
"6\n5 5 7 10 14 15\n3\n2 11\n3 12\n4 4\n"
],
"output": [
"0\n7\n",
"9\n7\n0\n"
]
} |
669 | 8 | Little Chris is very keen on his toy blocks. His teacher, however, wants Chris to solve more problems, so he decided to play a trick on Chris.
There are exactly s blocks in Chris's set, each block has a unique number from 1 to s. Chris's teacher picks a subset of blocks X and keeps it to himself. He will give them back only if Chris can pick such a non-empty subset Y from the remaining blocks, that the equality holds:
<image> "Are you kidding me?", asks Chris.
For example, consider a case where s = 8 and Chris's teacher took the blocks with numbers 1, 4 and 5. One way for Chris to choose a set is to pick the blocks with numbers 3 and 6, see figure. Then the required sums would be equal: (1 - 1) + (4 - 1) + (5 - 1) = (8 - 3) + (8 - 6) = 7.
<image>
However, now Chris has exactly s = 106 blocks. Given the set X of blocks his teacher chooses, help Chris to find the required set Y!
Input
The first line of input contains a single integer n (1 ≤ n ≤ 5·105), the number of blocks in the set X. The next line contains n distinct space-separated integers x1, x2, ..., xn (1 ≤ xi ≤ 106), the numbers of the blocks in X.
Note: since the size of the input and output could be very large, don't use slow output techniques in your language. For example, do not use input and output streams (cin, cout) in C++.
Output
In the first line of output print a single integer m (1 ≤ m ≤ 106 - n), the number of blocks in the set Y. In the next line output m distinct space-separated integers y1, y2, ..., ym (1 ≤ yi ≤ 106), such that the required equality holds. The sets X and Y should not intersect, i.e. xi ≠ yj for all i, j (1 ≤ i ≤ n; 1 ≤ j ≤ m). It is guaranteed that at least one solution always exists. If there are multiple solutions, output any of them.
Examples
Input
3
1 4 5
Output
2
999993 1000000
Input
1
1
Output
1
1000000 | import sys
def solve():
n = int(input())
s = 1000000
xset = set(map(int, input().split()))
res = set()
wantother = 0
for i in range(1, s + 1):
opposite = s - i + 1
if i in xset:
if opposite not in xset:
res.add(opposite)
else:
wantother+=1
wantother /= 2
for i in range(1, s + 1):
if wantother == 0: break
opposite = s - i + 1
if i not in res and i not in xset and opposite not in xset:
res.add(i)
res.add(opposite)
wantother-=1
print(len(res))
return " ".join(map(str, res))
if sys.hexversion == 50594544 : sys.stdin = open("test.txt")
print(solve()) | {
"input": [
"1\n1\n",
"3\n1 4 5\n"
],
"output": [
"1\n1000000 ",
"3\n1000000 999997 999996 "
]
} |
670 | 9 | Ryouko is an extremely forgetful girl, she could even forget something that has just happened. So in order to remember, she takes a notebook with her, called Ryouko's Memory Note. She writes what she sees and what she hears on the notebook, and the notebook became her memory.
Though Ryouko is forgetful, she is also born with superb analyzing abilities. However, analyzing depends greatly on gathered information, in other words, memory. So she has to shuffle through her notebook whenever she needs to analyze, which is tough work.
Ryouko's notebook consists of n pages, numbered from 1 to n. To make life (and this problem) easier, we consider that to turn from page x to page y, |x - y| pages should be turned. During analyzing, Ryouko needs m pieces of information, the i-th piece of information is on page ai. Information must be read from the notebook in order, so the total number of pages that Ryouko needs to turn is <image>.
Ryouko wants to decrease the number of pages that need to be turned. In order to achieve this, she can merge two pages of her notebook. If Ryouko merges page x to page y, she would copy all the information on page x to y (1 ≤ x, y ≤ n), and consequently, all elements in sequence a that was x would become y. Note that x can be equal to y, in which case no changes take place.
Please tell Ryouko the minimum number of pages that she needs to turn. Note she can apply the described operation at most once before the reading. Note that the answer can exceed 32-bit integers.
Input
The first line of input contains two integers n and m (1 ≤ n, m ≤ 105).
The next line contains m integers separated by spaces: a1, a2, ..., am (1 ≤ ai ≤ n).
Output
Print a single integer — the minimum number of pages Ryouko needs to turn.
Examples
Input
4 6
1 2 3 4 3 2
Output
3
Input
10 5
9 4 3 8 8
Output
6
Note
In the first sample, the optimal solution is to merge page 4 to 3, after merging sequence a becomes {1, 2, 3, 3, 3, 2}, so the number of pages Ryouko needs to turn is |1 - 2| + |2 - 3| + |3 - 3| + |3 - 3| + |3 - 2| = 3.
In the second sample, optimal solution is achieved by merging page 9 to 4. | n, m = map(int, input().split())
arr = list(map(int, input().split()))
price = sum(abs(a - b) for a, b in zip(arr, arr[1:]))
from collections import defaultdict
adj = defaultdict(list)
for a,b in zip(arr, arr[1:]):
if a != b:
adj[a].append(b)
adj[b].append(a)
def median(xs):
return sorted(xs)[len(xs)//2]
res = 0
for i, xs in adj.items():
m = median(xs)
a = sum(abs(i - x) for x in xs)
b = sum(abs(m - x) for x in xs)
res = max(res, a - b)
print(price - res)
| {
"input": [
"10 5\n9 4 3 8 8\n",
"4 6\n1 2 3 4 3 2\n"
],
"output": [
" 6\n",
" 3\n"
]
} |
671 | 9 | You have r red, g green and b blue balloons. To decorate a single table for the banquet you need exactly three balloons. Three balloons attached to some table shouldn't have the same color. What maximum number t of tables can be decorated if we know number of balloons of each color?
Your task is to write a program that for given values r, g and b will find the maximum number t of tables, that can be decorated in the required manner.
Input
The single line contains three integers r, g and b (0 ≤ r, g, b ≤ 2·109) — the number of red, green and blue baloons respectively. The numbers are separated by exactly one space.
Output
Print a single integer t — the maximum number of tables that can be decorated in the required manner.
Examples
Input
5 4 3
Output
4
Input
1 1 1
Output
1
Input
2 3 3
Output
2
Note
In the first sample you can decorate the tables with the following balloon sets: "rgg", "gbb", "brr", "rrg", where "r", "g" and "b" represent the red, green and blue balls, respectively. | def get_answer(a, b, c):
return min((a+b+c)//3, a+b+c-max(a,b,c))
r, g, b = map(int, input().split())
print(get_answer(r, g, b)) | {
"input": [
"1 1 1\n",
"5 4 3\n",
"2 3 3\n"
],
"output": [
"1\n",
"4\n",
"2\n"
]
} |
672 | 11 | Celebrating the new year, many people post videos of falling dominoes; Here's a list of them: https://www.youtube.com/results?search_query=New+Years+Dominos
User ainta, who lives in a 2D world, is going to post a video as well.
There are n dominoes on a 2D Cartesian plane. i-th domino (1 ≤ i ≤ n) can be represented as a line segment which is parallel to the y-axis and whose length is li. The lower point of the domino is on the x-axis. Let's denote the x-coordinate of the i-th domino as pi. Dominoes are placed one after another, so p1 < p2 < ... < pn - 1 < pn holds.
User ainta wants to take a video of falling dominoes. To make dominoes fall, he can push a single domino to the right. Then, the domino will fall down drawing a circle-shaped orbit until the line segment totally overlaps with the x-axis.
<image>
Also, if the s-th domino touches the t-th domino while falling down, the t-th domino will also fall down towards the right, following the same procedure above. Domino s touches domino t if and only if the segment representing s and t intersects.
<image>
See the picture above. If he pushes the leftmost domino to the right, it falls down, touching dominoes (A), (B) and (C). As a result, dominoes (A), (B), (C) will also fall towards the right. However, domino (D) won't be affected by pushing the leftmost domino, but eventually it will fall because it is touched by domino (C) for the first time.
<image>
The picture above is an example of falling dominoes. Each red circle denotes a touch of two dominoes.
User ainta has q plans of posting the video. j-th of them starts with pushing the xj-th domino, and lasts until the yj-th domino falls. But sometimes, it could be impossible to achieve such plan, so he has to lengthen some dominoes. It costs one dollar to increase the length of a single domino by 1. User ainta wants to know, for each plan, the minimum cost needed to achieve it. Plans are processed independently, i. e. if domino's length is increased in some plan, it doesn't affect its length in other plans. Set of dominos that will fall except xj-th domino and yj-th domino doesn't matter, but the initial push should be on domino xj.
Input
The first line contains an integer n (2 ≤ n ≤ 2 × 105)— the number of dominoes.
Next n lines describe the dominoes. The i-th line (1 ≤ i ≤ n) contains two space-separated integers pi, li (1 ≤ pi, li ≤ 109)— the x-coordinate and the length of the i-th domino. It is guaranteed that p1 < p2 < ... < pn - 1 < pn.
The next line contains an integer q (1 ≤ q ≤ 2 × 105) — the number of plans.
Next q lines describe the plans. The j-th line (1 ≤ j ≤ q) contains two space-separated integers xj, yj (1 ≤ xj < yj ≤ n). It means the j-th plan is, to push the xj-th domino, and shoot a video until the yj-th domino falls.
Output
For each plan, print a line containing the minimum cost needed to achieve it. If no cost is needed, print 0.
Examples
Input
6
1 5
3 3
4 4
9 2
10 1
12 1
4
1 2
2 4
2 5
2 6
Output
0
1
1
2
Note
Consider the example. The dominoes are set like the picture below.
<image>
Let's take a look at the 4th plan. To make the 6th domino fall by pushing the 2nd domino, the length of the 3rd domino (whose x-coordinate is 4) should be increased by 1, and the 5th domino (whose x-coordinate is 9) should be increased by 1 (other option is to increase 4th domino instead of 5th also by 1). Then, the dominoes will fall like in the picture below. Each cross denotes a touch between two dominoes.
<image> <image> <image> <image> <image> | '''
from bisect import bisect,bisect_left
from collections import *
from math import gcd,ceil,sqrt,floor,inf
from heapq import *
from itertools import *
from operator import add,mul,sub,xor,truediv,floordiv
from functools import *'''
#------------------------------------------------------------------------
import os
import sys
from io import BytesIO, IOBase
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#------------------------------------------------------------------------
def RL(): return map(int, sys.stdin.readline().split())
def RLL(): return list(map(int, sys.stdin.readline().split()))
def N(): return int(input())
def A(n):return [0]*n
def AI(n,x): return [x]*n
def A2(n,m): return [[0]*m for i in range(n)]
def G(n): return [[] for i in range(n)]
def GP(it): return [[ch,len(list(g))] for ch,g in groupby(it)]
#------------------------------------------------------------------------
from types import GeneratorType
def bootstrap(f, stack=[]):
def wrappedfunc(*args, **kwargs):
if stack:
return f(*args, **kwargs)
else:
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
else:
stack.pop()
if not stack:
break
to = stack[-1].send(to)
return to
return wrappedfunc
mod=10**9+7
farr=[1]
ifa=[]
def fact(x,mod=0):
if mod:
while x>=len(farr):
farr.append(farr[-1]*len(farr)%mod)
else:
while x>=len(farr):
farr.append(farr[-1]*len(farr))
return farr[x]
def ifact(x,mod):
global ifa
fact(x,mod)
ifa.append(pow(farr[-1],mod-2,mod))
for i in range(x,0,-1):
ifa.append(ifa[-1]*i%mod)
ifa.reverse()
def per(i,j,mod=0):
if i<j: return 0
if not mod:
return fact(i)//fact(i-j)
return farr[i]*ifa[i-j]%mod
def com(i,j,mod=0):
if i<j: return 0
if not mod:
return per(i,j)//fact(j)
return per(i,j,mod)*ifa[j]%mod
def catalan(n):
return com(2*n,n)//(n+1)
def isprime(n):
for i in range(2,int(n**0.5)+1):
if n%i==0:
return False
return True
def floorsum(a,b,c,n):#sum((a*i+b)//c for i in range(n+1))
if a==0:return b//c*(n+1)
if a>=c or b>=c: return floorsum(a%c,b%c,c,n)+b//c*(n+1)+a//c*n*(n+1)//2
m=(a*n+b)//c
return n*m-floorsum(c,c-b-1,a,m-1)
def inverse(a,m):
a%=m
if a<=1: return a
return ((1-inverse(m,a)*m)//a)%m
def lowbit(n):
return n&-n
class BIT:
def __init__(self,arr):
self.arr=arr
self.n=len(arr)-1
def update(self,x,v):
while x<=self.n:
self.arr[x]+=v
x+=x&-x
def query(self,x):
ans=0
while x:
ans+=self.arr[x]
x&=x-1
return ans
class ST:
def __init__(self,arr):#n!=0
n=len(arr)
mx=n.bit_length()#取不到
self.st=[[0]*mx for i in range(n)]
for i in range(n):
self.st[i][0]=arr[i]
for j in range(1,mx):
for i in range(n-(1<<j)+1):
self.st[i][j]=max(self.st[i][j-1],self.st[i+(1<<j-1)][j-1])
def query(self,l,r):
if l>r:return -inf
s=(r+1-l).bit_length()-1
return max(self.st[l][s],self.st[r-(1<<s)+1][s])
class DSU:#容量+路径压缩
def __init__(self,n):
self.c=[-1]*n
def same(self,x,y):
return self.find(x)==self.find(y)
def find(self,x):
if self.c[x]<0:
return x
self.c[x]=self.find(self.c[x])
return self.c[x]
def union(self,u,v):
u,v=self.find(u),self.find(v)
if u==v:
return False
if self.c[u]>self.c[v]:
u,v=v,u
self.c[u]+=self.c[v]
self.c[v]=u
return True
def size(self,x): return -self.c[self.find(x)]
class UFS:#秩+路径
def __init__(self,n):
self.parent=[i for i in range(n)]
self.ranks=[0]*n
def find(self,x):
if x!=self.parent[x]:
self.parent[x]=self.find(self.parent[x])
return self.parent[x]
def union(self,u,v):
pu,pv=self.find(u),self.find(v)
if pu==pv:
return False
if self.ranks[pu]>=self.ranks[pv]:
self.parent[pv]=pu
if self.ranks[pv]==self.ranks[pu]:
self.ranks[pu]+=1
else:
self.parent[pu]=pv
def Prime(n):
c=0
prime=[]
flag=[0]*(n+1)
for i in range(2,n+1):
if not flag[i]:
prime.append(i)
c+=1
for j in range(c):
if i*prime[j]>n: break
flag[i*prime[j]]=prime[j]
if i%prime[j]==0: break
return flag
def dij(s,graph):
d={}
d[s]=0
heap=[(0,s)]
seen=set()
while heap:
dis,u=heappop(heap)
if u in seen:
continue
seen.add(u)
for v,w in graph[u]:
if v not in d or d[v]>d[u]+w:
d[v]=d[u]+w
heappush(heap,(d[v],v))
return d
def bell(s,g):#bellman-Ford
dis=AI(n,inf)
dis[s]=0
for i in range(n-1):
for u,v,w in edge:
if dis[v]>dis[u]+w:
dis[v]=dis[u]+w
change=A(n)
for i in range(n):
for u,v,w in edge:
if dis[v]>dis[u]+w:
dis[v]=dis[u]+w
change[v]=1
return dis
def lcm(a,b): return a*b//gcd(a,b)
def lis(nums):
res=[]
for k in nums:
i=bisect.bisect_left(res,k)
if i==len(res):
res.append(k)
else:
res[i]=k
return len(res)
def RP(nums):#逆序对
n = len(nums)
s=set(nums)
d={}
for i,k in enumerate(sorted(s),1):
d[k]=i
bi=BIT([0]*(len(s)+1))
ans=0
for i in range(n-1,-1,-1):
ans+=bi.query(d[nums[i]]-1)
bi.update(d[nums[i]],1)
return ans
class DLN:
def __init__(self,val):
self.val=val
self.pre=None
self.next=None
def nb(i,j,n,m):
for ni,nj in [[i+1,j],[i-1,j],[i,j-1],[i,j+1]]:
if 0<=ni<n and 0<=nj<m:
yield ni,nj
def topo(n):
q=deque()
res=[]
for i in range(1,n+1):
if ind[i]==0:
q.append(i)
res.append(i)
while q:
u=q.popleft()
for v in g[u]:
ind[v]-=1
if ind[v]==0:
q.append(v)
res.append(v)
return res
@bootstrap
def gdfs(r,p):
for ch in g[r]:
if ch!=p:
yield gdfs(ch,r)
yield None
class SMT:
def __init__(self,n,query_fn,I):
self.h = 1
while 1 << self.h < n:
self.h += 1
self.n=1<<self.h
self.qf=query_fn
self.tree = [I] * (self.n<<1)
self.lazy = [0] * self.n
def build(self, v):
for k in range(len(v)):
self.tree[k + self.n] = v[k]
for k in range(self.n - 1, 0, -1):
self.tree[k] = self.qf(self.tree[2*k],self.tree[2 * k + 1])
def apply(self, x, val):
self.tree[x] = val
if x < self.n:
self.lazy[x] = 1
def pushup(self, x):
self.tree[x]=self.qf(self.tree[2*x],self.tree[2*x+1])
def push(self, x):#pushdown
if self.lazy[x]:
self.apply(x * 2,0)
self.apply(x * 2+ 1,0)
self.lazy[x] = 0
def set(self,p,x):
p+=self.n
for i in range(self.h,0,-1):self.push(p>>i)
self.tree[p]=x
for i in range(1,self.h+1):self.pushup(p>>i)
def get(self,p):
p+=self.n
for i in range(self.h,0,-1):self.push(p>>i)
return self.tree[p]
def update(self, l, r, h):#左闭右闭
l += self.n
r += self.n
l0=l
r0=r
for i in range(self.h,0,-1):
if (((l>>i)<<i)!=l):self.push(l>>i)
if ((((r+1)>>i)<<i)!=r+1):self.push(r>>i)
while l <= r:
if l & 1:
self.apply(l, h)
l += 1
if not r & 1:
self.apply(r, h)
r -= 1
l>>=1
r>>=1
for i in range(1,self.h+1):
if (((l0>>i)<<i)!=l0):self.pushup(l0>>i)
if ((((r0+1)>>i)<<i)!=r0+1):
self.pushup(r0>>i)
def query(self, l, r):
l += self.n
r += self.n
for i in range(self.h,0,-1):
if (((l>>i)<<i)!=l):self.push(l>>i)
if ((((r+1)>>i)<<i)!=r+1):self.push(r>>i)
ans = 0
while l <= r:
if l & 1:
ans = self.qf(ans, self.tree[l])
l += 1
if not r & 1:
ans = self.qf(ans, self.tree[r])
r -= 1
l >>= 1; r >>= 1
return ans
def add(x,y):
return x+y
t=1
for i in range(t):
n=N()
s=set()
x=A(n)
y=A(n)
for i in range(n):
p,l=RL()
x[i]=p
y[i]=p+l
s.add(p)
s.add(y[i])
s=sorted(s)
d={x:i for i,x in enumerate(s)}
k=len(s)-1
arr=[s[i]-s[i-1] for i in range(1,k+1)]
smt=SMT(k,add,0)
smt.build(arr)
g=G(n)
q=N()
ans=A(q)
for i in range(q):
a,b=RL()
a-=1
b-=1
g[a].append((i,b))
for i in range(n-1,-1,-1):
l=d[x[i]]
r=d[y[i]]
smt.update(l,r-1,0)
for j,cur in g[i]:
cur=d[x[cur]]
ans[j]=smt.query(l,cur-1)
for x in ans:
print(x)
'''
sys.setrecursionlimit(200000)
import threading
threading.stack_size(10**8)
t=threading.Thr
ead(target=main)
t.start()
t.join()
'''
| {
"input": [
"6\n1 5\n3 3\n4 4\n9 2\n10 1\n12 1\n4\n1 2\n2 4\n2 5\n2 6\n"
],
"output": [
"0\n1\n1\n2\n"
]
} |
673 | 9 | A sweet little monster Om Nom loves candies very much. One day he found himself in a rather tricky situation that required him to think a bit in order to enjoy candies the most. Would you succeed with the same task if you were on his place?
<image>
One day, when he came to his friend Evan, Om Nom didn't find him at home but he found two bags with candies. The first was full of blue candies and the second bag was full of red candies. Om Nom knows that each red candy weighs Wr grams and each blue candy weighs Wb grams. Eating a single red candy gives Om Nom Hr joy units and eating a single blue candy gives Om Nom Hb joy units.
Candies are the most important thing in the world, but on the other hand overeating is not good. Om Nom knows if he eats more than C grams of candies, he will get sick. Om Nom thinks that it isn't proper to leave candy leftovers, so he can only eat a whole candy. Om Nom is a great mathematician and he quickly determined how many candies of what type he should eat in order to get the maximum number of joy units. Can you repeat his achievement? You can assume that each bag contains more candies that Om Nom can eat.
Input
The single line contains five integers C, Hr, Hb, Wr, Wb (1 ≤ C, Hr, Hb, Wr, Wb ≤ 109).
Output
Print a single integer — the maximum number of joy units that Om Nom can get.
Examples
Input
10 3 5 2 3
Output
16
Note
In the sample test Om Nom can eat two candies of each type and thus get 16 joy units. | def main():
c, hr, hb, wr, wb = map(int, input().split())
ans = 0
for i in range(10**6 + 1):
if i * wr <= c:
ans = max(ans, i * hr + ((c - i * wr) // wb) * hb)
if i * wb <= c:
ans = max(ans, i * hb + ((c - i * wb) // wr) * hr)
print(ans)
main() | {
"input": [
"10 3 5 2 3\n"
],
"output": [
"16"
]
} |
674 | 10 | We all know that GukiZ often plays with arrays.
Now he is thinking about this problem: how many arrays a, of length n, with non-negative elements strictly less then 2l meet the following condition: <image>? Here operation <image> means bitwise AND (in Pascal it is equivalent to and, in C/C++/Java/Python it is equivalent to &), operation <image> means bitwise OR (in Pascal it is equivalent to <image>, in C/C++/Java/Python it is equivalent to |).
Because the answer can be quite large, calculate it modulo m. This time GukiZ hasn't come up with solution, and needs you to help him!
Input
First and the only line of input contains four integers n, k, l, m (2 ≤ n ≤ 1018, 0 ≤ k ≤ 1018, 0 ≤ l ≤ 64, 1 ≤ m ≤ 109 + 7).
Output
In the single line print the number of arrays satisfying the condition above modulo m.
Examples
Input
2 1 2 10
Output
3
Input
2 1 1 3
Output
1
Input
3 3 2 10
Output
9
Note
In the first sample, satisfying arrays are {1, 1}, {3, 1}, {1, 3}.
In the second sample, only satisfying array is {1, 1}.
In the third sample, satisfying arrays are {0, 3, 3}, {1, 3, 2}, {1, 3, 3}, {2, 3, 1}, {2, 3, 3}, {3, 3, 0}, {3, 3, 1}, {3, 3, 2}, {3, 3, 3}. | n,k,l,m=map(int,input().split())
fc={1:1,2:1}
def f(n):
if n not in fc: k=n//2;fc[n]=(f(k+1)**2+f(k)**2 if n%2 else f(k)*(2*f(k+1)-f(k)))%m
return fc[n]
s=k<2**l
for i in range(l): s*=pow(2,n,m)-f(n+2) if (k>>i)%2 else f(n+2)
print(s%m)
| {
"input": [
"2 1 2 10\n",
"2 1 1 3\n",
"3 3 2 10\n"
],
"output": [
"3\n",
"1\n",
"9\n"
]
} |
675 | 9 | This is yet another problem dealing with regular bracket sequences.
We should remind you that a bracket sequence is called regular, if by inserting «+» and «1» into it we can get a correct mathematical expression. For example, sequences «(())()», «()» and «(()(()))» are regular, while «)(», «(()» and «(()))(» are not.
You are given a string of «(» and «)» characters. You are to find its longest substring that is a regular bracket sequence. You are to find the number of such substrings as well.
Input
The first line of the input file contains a non-empty string, consisting of «(» and «)» characters. Its length does not exceed 106.
Output
Print the length of the longest substring that is a regular bracket sequence, and the number of such substrings. If there are no such substrings, write the only line containing "0 1".
Examples
Input
)((())))(()())
Output
6 2
Input
))(
Output
0 1 | def solve(s):
l=[]
length=ref=0
count=1
pos=-1
for i in range(len(s)):
if s[i]=="(":
l.append(i)
elif l:
l.pop()
ref=i-l[-1] if l else i-pos
if ref>length:
length=ref
count=1
elif ref==length:
count+=1
else:
pos=i
print(length,count)
s=input()
solve(s) | {
"input": [
")((())))(()())\n",
"))(\n"
],
"output": [
"6 2\n",
"0 1\n"
]
} |
676 | 7 | Consider the infinite sequence of integers: 1, 1, 2, 1, 2, 3, 1, 2, 3, 4, 1, 2, 3, 4, 5.... The sequence is built in the following way: at first the number 1 is written out, then the numbers from 1 to 2, then the numbers from 1 to 3, then the numbers from 1 to 4 and so on. Note that the sequence contains numbers, not digits. For example number 10 first appears in the sequence in position 55 (the elements are numerated from one).
Find the number on the n-th position of the sequence.
Input
The only line contains integer n (1 ≤ n ≤ 1014) — the position of the number to find.
Note that the given number is too large, so you should use 64-bit integer type to store it. In C++ you can use the long long integer type and in Java you can use long integer type.
Output
Print the element in the n-th position of the sequence (the elements are numerated from one).
Examples
Input
3
Output
2
Input
5
Output
2
Input
10
Output
4
Input
55
Output
10
Input
56
Output
1 | def main():
n = int(input())
for i in range(1,10**8):
if n-i <= 0:
break
n -= i
print(n)
if __name__ == "__main__":
main()
| {
"input": [
"56\n",
"5\n",
"55\n",
"3\n",
"10\n"
],
"output": [
"1\n",
"2\n",
"10\n",
"2\n",
"4\n"
]
} |
677 | 10 | This problem is given in two versions that differ only by constraints. If you can solve this problem in large constraints, then you can just write a single solution to the both versions. If you find the problem too difficult in large constraints, you can write solution to the simplified version only.
Waking up in the morning, Apollinaria decided to bake cookies. To bake one cookie, she needs n ingredients, and for each ingredient she knows the value ai — how many grams of this ingredient one needs to bake a cookie. To prepare one cookie Apollinaria needs to use all n ingredients.
Apollinaria has bi gram of the i-th ingredient. Also she has k grams of a magic powder. Each gram of magic powder can be turned to exactly 1 gram of any of the n ingredients and can be used for baking cookies.
Your task is to determine the maximum number of cookies, which Apollinaria is able to bake using the ingredients that she has and the magic powder.
Input
The first line of the input contains two positive integers n and k (1 ≤ n, k ≤ 1000) — the number of ingredients and the number of grams of the magic powder.
The second line contains the sequence a1, a2, ..., an (1 ≤ ai ≤ 1000), where the i-th number is equal to the number of grams of the i-th ingredient, needed to bake one cookie.
The third line contains the sequence b1, b2, ..., bn (1 ≤ bi ≤ 1000), where the i-th number is equal to the number of grams of the i-th ingredient, which Apollinaria has.
Output
Print the maximum number of cookies, which Apollinaria will be able to bake using the ingredients that she has and the magic powder.
Examples
Input
3 1
2 1 4
11 3 16
Output
4
Input
4 3
4 3 5 6
11 12 14 20
Output
3
Note
In the first sample it is profitably for Apollinaria to make the existing 1 gram of her magic powder to ingredient with the index 2, then Apollinaria will be able to bake 4 cookies.
In the second sample Apollinaria should turn 1 gram of magic powder to ingredient with the index 1 and 1 gram of magic powder to ingredient with the index 3. Then Apollinaria will be able to bake 3 cookies. The remaining 1 gram of the magic powder can be left, because it can't be used to increase the answer. | def I(): return(list(map(int,input().split())))
n,k=I()
a=I()
b=I()
l=0
# l=104
r=2*(10**9)+1
# r=106
# sumi=sum(a)
# summ=sum(b)
# print(sumi,summ)
while(l<r-1):
# print(l,r)
m=(l+r)//2
s=0
for i in range(n):
s+=max(m*a[i]-b[i],0)
# print(s)
if s>k:
r=m
else:
l=m
print(l)
| {
"input": [
"4 3\n4 3 5 6\n11 12 14 20\n",
"3 1\n2 1 4\n11 3 16\n"
],
"output": [
"3\n",
"4\n"
]
} |
678 | 9 | There are n workers in a company, each of them has a unique id from 1 to n. Exaclty one of them is a chief, his id is s. Each worker except the chief has exactly one immediate superior.
There was a request to each of the workers to tell how how many superiors (not only immediate). Worker's superiors are his immediate superior, the immediate superior of the his immediate superior, and so on. For example, if there are three workers in the company, from which the first is the chief, the second worker's immediate superior is the first, the third worker's immediate superior is the second, then the third worker has two superiors, one of them is immediate and one not immediate. The chief is a superior to all the workers except himself.
Some of the workers were in a hurry and made a mistake. You are to find the minimum number of workers that could make a mistake.
Input
The first line contains two positive integers n and s (1 ≤ n ≤ 2·105, 1 ≤ s ≤ n) — the number of workers and the id of the chief.
The second line contains n integers a1, a2, ..., an (0 ≤ ai ≤ n - 1), where ai is the number of superiors (not only immediate) the worker with id i reported about.
Output
Print the minimum number of workers that could make a mistake.
Examples
Input
3 2
2 0 2
Output
1
Input
5 3
1 0 0 4 1
Output
2
Note
In the first example it is possible that only the first worker made a mistake. Then:
* the immediate superior of the first worker is the second worker,
* the immediate superior of the third worker is the first worker,
* the second worker is the chief. | n, root = map(int, input().split())
a = list(map(int, input().split()))
def push(d, x, val):
if x not in d:
d[x] = 0
d[x]+=val
if d[x]==0:
del d[x]
d = {}
for x in a:
push(d, x, 1)
min_ = 0
root -= 1
inf = 9999999
if a[root] != 0:
min_+=1
push(d, a[root], -1)
push(d, 0, 1)
if 0 in d and d[0] > 1:
add = d[0] - 1
min_+=add
push(d, inf, add)
d[0] = 1
S = [[val, num] for val, num in sorted(d.items(), key = lambda x:x[0])]
#print(min_, S)
cur = -1
i = 0
while i < len(S):
remain = S[i][0] - (cur+1)
while remain > 0:
val, num = S[-1]
if val == S[i][0]:
if val != inf:
min_ += min(remain, num)
break
else:
add = min(num, remain)
remain -= add
if val != inf:
min_ += add
if num == add:
S.pop()
else:
S[-1][1] -= add
cur=S[i][0]
i+=1
print(min_) | {
"input": [
"5 3\n1 0 0 4 1\n",
"3 2\n2 0 2\n"
],
"output": [
"2\n",
"1\n"
]
} |
679 | 8 | Running with barriers on the circle track is very popular in the country where Dasha lives, so no wonder that on her way to classes she saw the following situation:
The track is the circle with length L, in distinct points of which there are n barriers. Athlete always run the track in counterclockwise direction if you look on him from above. All barriers are located at integer distance from each other along the track.
Her friends the parrot Kefa and the leopard Sasha participated in competitions and each of them ran one lap. Each of the friends started from some integral point on the track. Both friends wrote the distance from their start along the track to each of the n barriers. Thus, each of them wrote n integers in the ascending order, each of them was between 0 and L - 1, inclusively.
<image> Consider an example. Let L = 8, blue points are barriers, and green points are Kefa's start (A) and Sasha's start (B). Then Kefa writes down the sequence [2, 4, 6], and Sasha writes down [1, 5, 7].
There are several tracks in the country, all of them have same length and same number of barriers, but the positions of the barriers can differ among different tracks. Now Dasha is interested if it is possible that Kefa and Sasha ran the same track or they participated on different tracks.
Write the program which will check that Kefa's and Sasha's tracks coincide (it means that one can be obtained from the other by changing the start position). Note that they always run the track in one direction — counterclockwise, if you look on a track from above.
Input
The first line contains two integers n and L (1 ≤ n ≤ 50, n ≤ L ≤ 100) — the number of barriers on a track and its length.
The second line contains n distinct integers in the ascending order — the distance from Kefa's start to each barrier in the order of its appearance. All integers are in the range from 0 to L - 1 inclusively.
The second line contains n distinct integers in the ascending order — the distance from Sasha's start to each barrier in the order of its overcoming. All integers are in the range from 0 to L - 1 inclusively.
Output
Print "YES" (without quotes), if Kefa and Sasha ran the coinciding tracks (it means that the position of all barriers coincides, if they start running from the same points on the track). Otherwise print "NO" (without quotes).
Examples
Input
3 8
2 4 6
1 5 7
Output
YES
Input
4 9
2 3 5 8
0 1 3 6
Output
YES
Input
2 4
1 3
1 2
Output
NO
Note
The first test is analyzed in the statement. | k, l = map(int, input().split())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
def solve():
for i in range(l):
if a == sorted((x + i) % l for x in b):
return "YES"
return "NO"
print(solve())
| {
"input": [
"4 9\n2 3 5 8\n0 1 3 6\n",
"2 4\n1 3\n1 2\n",
"3 8\n2 4 6\n1 5 7\n"
],
"output": [
"YES\n",
"NO\n",
"YES\n"
]
} |
680 | 10 |
Input
The only line of the input contains a string of digits. The length of the string is between 1 and 10, inclusive.
Output
Output "Yes" or "No".
Examples
Input
373
Output
Yes
Input
121
Output
No
Input
436
Output
Yes | def k(s):
n = (8, -1, -1, 3, 6, 9, 4, 7, 0, 5)
for i in range(len(s)):
if n[int(s[i])] != int(s[len(s) - i - 1]):
return False
return True
s = input()
if k(s):
print('Yes')
else:
print("No") | {
"input": [
"436\n",
"121\n",
"373\n"
],
"output": [
"YES",
"NO",
"YES"
]
} |
681 | 9 | A few years ago Sajjad left his school and register to another one due to security reasons. Now he wishes to find Amir, one of his schoolmates and good friends.
There are n schools numerated from 1 to n. One can travel between each pair of them, to do so, he needs to buy a ticket. The ticker between schools i and j costs <image> and can be used multiple times. Help Sajjad to find the minimum cost he needs to pay for tickets to visit all schools. He can start and finish in any school.
Input
The first line contains a single integer n (1 ≤ n ≤ 105) — the number of schools.
Output
Print single integer: the minimum cost of tickets needed to visit all schools.
Examples
Input
2
Output
0
Input
10
Output
4
Note
In the first example we can buy a ticket between the schools that costs <image>. | def find(n):
return (n + 1) // 2 - 1
print(find(int(input())))
| {
"input": [
"10\n",
"2\n"
],
"output": [
"4",
"0"
]
} |
682 | 9 | Vladimir wants to modernize partitions in his office. To make the office more comfortable he decided to remove a partition and plant several bamboos in a row. He thinks it would be nice if there are n bamboos in a row, and the i-th from the left is ai meters high.
Vladimir has just planted n bamboos in a row, each of which has height 0 meters right now, but they grow 1 meter each day. In order to make the partition nice Vladimir can cut each bamboo once at any height (no greater that the height of the bamboo), and then the bamboo will stop growing.
Vladimir wants to check the bamboos each d days (i.e. d days after he planted, then after 2d days and so on), and cut the bamboos that reached the required height. Vladimir wants the total length of bamboo parts he will cut off to be no greater than k meters.
What is the maximum value d he can choose so that he can achieve what he wants without cutting off more than k meters of bamboo?
Input
The first line contains two integers n and k (1 ≤ n ≤ 100, 1 ≤ k ≤ 1011) — the number of bamboos and the maximum total length of cut parts, in meters.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109) — the required heights of bamboos, in meters.
Output
Print a single integer — the maximum value of d such that Vladimir can reach his goal.
Examples
Input
3 4
1 3 5
Output
3
Input
3 40
10 30 50
Output
32
Note
In the first example Vladimir can check bamboos each 3 days. Then he will cut the first and the second bamboos after 3 days, and the third bamboo after 6 days. The total length of cut parts is 2 + 0 + 1 = 3 meters. | import itertools
unfold = itertools.chain.from_iterable
def jumps(a):
d = speedup
while d < a - 1:
c = (a + d - 1) // d
d = (a + c - 2) // (c - 1)
yield d
def calc(d):
return sum(d - 1 - (i - 1) % d for i in a)
def ans():
for d, pd in zip(D, D[1:]):
d -= 1
cd = calc(d)
if cd <= k:
return d
if d == pd:
continue
cpd = calc(pd)
if d - pd >= (cd - k) * (d - pd) / (cd - cpd):
return d - (cd - k) * (d - pd) / (cd - cpd)
return 1
n, k = map(int, input().split())
a = list(map(int, input().split()))
speedup = int(max(a) ** 0.5)
D = sorted(set(range(1, speedup + 1)).union([max(a) + k + 1]).union(set(
unfold(map(jumps, a)))), reverse=True)
print('%d' % ans()) | {
"input": [
"3 40\n10 30 50\n",
"3 4\n1 3 5\n"
],
"output": [
"32\n",
"3\n"
]
} |
683 | 10 | Polycarp likes to play with numbers. He takes some integer number x, writes it down on the board, and then performs with it n - 1 operations of the two kinds:
* divide the number x by 3 (x must be divisible by 3);
* multiply the number x by 2.
After each operation, Polycarp writes down the result on the board and replaces x by the result. So there will be n numbers on the board after all.
You are given a sequence of length n — the numbers that Polycarp wrote down. This sequence is given in arbitrary order, i.e. the order of the sequence can mismatch the order of the numbers written on the board.
Your problem is to rearrange (reorder) elements of this sequence in such a way that it can match possible Polycarp's game in the order of the numbers written on the board. I.e. each next number will be exactly two times of the previous number or exactly one third of previous number.
It is guaranteed that the answer exists.
Input
The first line of the input contatins an integer number n (2 ≤ n ≤ 100) — the number of the elements in the sequence. The second line of the input contains n integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 3 ⋅ 10^{18}) — rearranged (reordered) sequence that Polycarp can wrote down on the board.
Output
Print n integer numbers — rearranged (reordered) input sequence that can be the sequence that Polycarp could write down on the board.
It is guaranteed that the answer exists.
Examples
Input
6
4 8 6 3 12 9
Output
9 3 6 12 4 8
Input
4
42 28 84 126
Output
126 42 84 28
Input
2
1000000000000000000 3000000000000000000
Output
3000000000000000000 1000000000000000000
Note
In the first example the given sequence can be rearranged in the following way: [9, 3, 6, 12, 4, 8]. It can match possible Polycarp's game which started with x = 9. | def f(x):
i=0;
while x%3==0:
x//=3
i-=1
while x%2==0:
x//=2
i+=1
return i
input()
a=list(map(int,input().split()))
a= sorted(a,key=f)
print(*a) | {
"input": [
"4\n42 28 84 126\n",
"2\n1000000000000000000 3000000000000000000\n",
"6\n4 8 6 3 12 9\n"
],
"output": [
"126 42 84 28 \n",
"3000000000000000000 1000000000000000000 \n",
"9 3 6 12 4 8 \n"
]
} |
684 | 8 | Let's introduce a number system which is based on a roman digits. There are digits I, V, X, L which correspond to the numbers 1, 5, 10 and 50 respectively. The use of other roman digits is not allowed.
Numbers in this system are written as a sequence of one or more digits. We define the value of the sequence simply as the sum of digits in it.
For example, the number XXXV evaluates to 35 and the number IXI — to 12.
Pay attention to the difference to the traditional roman system — in our system any sequence of digits is valid, moreover the order of digits doesn't matter, for example IX means 11, not 9.
One can notice that this system is ambiguous, and some numbers can be written in many different ways. Your goal is to determine how many distinct integers can be represented by exactly n roman digits I, V, X, L.
Input
The only line of the input file contains a single integer n (1 ≤ n ≤ 10^9) — the number of roman digits to use.
Output
Output a single integer — the number of distinct integers which can be represented using n roman digits exactly.
Examples
Input
1
Output
4
Input
2
Output
10
Input
10
Output
244
Note
In the first sample there are exactly 4 integers which can be represented — I, V, X and L.
In the second sample it is possible to represent integers 2 (II), 6 (VI), 10 (VV), 11 (XI), 15 (XV), 20 (XX), 51 (IL), 55 (VL), 60 (XL) and 100 (LL). | def f(n):
return (n+3)*(n+2)*(n+1)//6;
n=int(input());
ans=f(n);
if n>=6:ans-=f(n-6);
if n>=9:ans-=f(n-9)*2;
if n>=10:ans+=f(n-10);
if n>=14:ans+=f(n-14);
print(ans); | {
"input": [
"2\n",
"10\n",
"1\n"
],
"output": [
"10",
"244",
"4"
]
} |
685 | 10 | There is one apple tree in Arkady's garden. It can be represented as a set of junctions connected with branches so that there is only one way to reach any junctions from any other one using branches. The junctions are enumerated from 1 to n, the junction 1 is called the root.
A subtree of a junction v is a set of junctions u such that the path from u to the root must pass through v. Note that v itself is included in a subtree of v.
A leaf is such a junction that its subtree contains exactly one junction.
The New Year is coming, so Arkady wants to decorate the tree. He will put a light bulb of some color on each leaf junction and then count the number happy junctions. A happy junction is such a junction t that all light bulbs in the subtree of t have different colors.
Arkady is interested in the following question: for each k from 1 to n, what is the minimum number of different colors needed to make the number of happy junctions be greater than or equal to k?
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of junctions in the tree.
The second line contains n - 1 integers p_2, p_3, ..., p_n (1 ≤ p_i < i), where p_i means there is a branch between junctions i and p_i. It is guaranteed that this set of branches forms a tree.
Output
Output n integers. The i-th of them should be the minimum number of colors needed to make the number of happy junctions be at least i.
Examples
Input
3
1 1
Output
1 1 2
Input
5
1 1 3 3
Output
1 1 1 2 3
Note
In the first example for k = 1 and k = 2 we can use only one color: the junctions 2 and 3 will be happy. For k = 3 you have to put the bulbs of different colors to make all the junctions happy.
In the second example for k = 4 you can, for example, put the bulbs of color 1 in junctions 2 and 4, and a bulb of color 2 into junction 5. The happy junctions are the ones with indices 2, 3, 4 and 5 then. | n = int(input())
tr = {}
p = [int(s) for s in input().split()]
for i in range(n-1):
if not tr.get(p[i]-1):
tr[p[i]-1] = []
tr[p[i]-1].append(i+1)
# print(tr)
lc = [-1 for i in range(n)]
def get_lc(i):
if lc[i] == -1:
if tr.get(i):
lc[i] = 0
for j in tr[i]:
lc[i] += get_lc(j)
else:
lc[i] = 1
return lc[i]
for i in range(n-1, -1, -1):
get_lc(i)
print(*sorted(lc)) | {
"input": [
"5\n1 1 3 3\n",
"3\n1 1\n"
],
"output": [
"1 1 1 2 3 \n",
"1 1 2 \n"
]
} |
686 | 11 | Everyone knows that computers become faster and faster. Recently Berland scientists have built a machine that can move itself back in time!
More specifically, it works as follows. It has an infinite grid and a robot which stands on one of the cells. Each cell of the grid can either be empty or contain 0 or 1. The machine also has a program which consists of instructions, which are being handled one by one. Each instruction is represented by exactly one symbol (letter or digit) and takes exactly one unit of time (say, second) to be performed, except the last type of operation (it's described below). Here they are:
* 0 or 1: the robot places this number into the cell he is currently at. If this cell wasn't empty before the operation, its previous number is replaced anyway.
* e: the robot erases the number into the cell he is at.
* l, r, u or d: the robot goes one cell to the left/right/up/down.
* s: the robot stays where he is for a unit of time.
* t: let x be 0, if the cell with the robot is empty, otherwise let x be one more than the digit in this cell (that is, x = 1 if the digit in this cell is 0, and x = 2 if the digit is 1). Then the machine travels x seconds back in time. Note that this doesn't change the instructions order, but it changes the position of the robot and the numbers in the grid as they were x units of time ago. You can consider this instruction to be equivalent to a Ctrl-Z pressed x times.
For example, let the board be completely empty, and the program be sr1t0. Let the robot initially be at (0, 0).
* [now is the moment 0, the command is s]: we do nothing.
* [now is the moment 1, the command is r]: we are now at (1, 0).
* [now is the moment 2, the command is 1]: we are at (1, 0), and this cell contains 1.
* [now is the moment 3, the command is t]: we travel 1 + 1 = 2 moments back, that is, to the moment 1.
* [now is the moment 1, the command is 0]: we are again at (0, 0), and the board is clear again, but after we follow this instruction, this cell has 0 in it. We've just rewritten the history. The consequences of the third instruction have never happened.
Now Berland scientists want to use their machine in practice. For example, they want to be able to add two integers.
Assume that the initial state of the machine is as follows:
* One positive integer is written in binary on the grid in such a way that its right bit is at the cell (0, 1), from left to right from the highest bit to the lowest bit.
* The other positive integer is written in binary on the grid in such a way that its right bit is at the cell (0, 0), from left to right from the highest bit to the lowest bit.
* All the other cells are empty.
* The robot is at (0, 0).
* We consider this state to be always in the past; that is, if you manage to travel to any negative moment, the board was always as described above, and the robot was at (0, 0) for eternity.
You are asked to write a program after which
* The robot stands on a non-empty cell,
* If we read the number starting from the cell with the robot and moving to the right until the first empty cell, this will be a + b in binary, from the highest bit to the lowest bit.
Note that there are no restrictions on other cells. In particular, there may be a digit just to the left to the robot after all instructions.
In each test you are given up to 1000 pairs (a, b), and your program must work for all these pairs. Also since the machine's memory is not very big, your program must consist of no more than 10^5 instructions.
Input
The first line contains the only integer t (1≤ t≤ 1000) standing for the number of testcases. Each of the next t lines consists of two positive integers a and b (1≤ a, b < 2^{30}) in decimal.
Output
Output the only line consisting of no more than 10^5 symbols from 01eslrudt standing for your program.
Note that formally you may output different programs for different tests.
Example
Input
2
123456789 987654321
555555555 555555555
Output
0l1l1l0l0l0l1l1l1l0l1l0l1l1l0l0l0l1l0l1l1l1l0l0l0l1l0l0l0l0l1l0lr | def C(x, y):
return x + '10' + y + 't' + y
def CBF(x, y):
return x + '01' + y + 't' + y
Cr = C('r', 'l')
Cl = C('l', 'r')
Cu = C('u', 'd')
Cd = C('d', 'u')
CBFr = CBF('r', 'l')
CBFl = CBF('l', 'r')
CBFu = CBF('u', 'd')
CBFd = CBF('d', 'u')
def CE(x, y):
return x+x+'0'+x+'1'+y+y+'10'+y+'t'+x+x+'t'+y+x+x+'e'+x+'e'+y+y+y
def MNE(x, y):
return CE(x, y) + x + C(y, x) + 'e' + y
def MNEall(n):
return (MNE('d', 'u') + 'l') * n + 'r' * n + 'u' + (MNE('u', 'd') + 'l') * n + 'r' * n + 'd'
def ADD():
return 'u' + Cu + 'u' + Cu + 'dd' + Cu + 'u' + Cu + 'u' + Cl + '010ltut' + Cd + 'dd'
def ADDc():
return ADD() + '0ulrtd' + ADD() + 'lltr'
def ADDbit():
return 'u' + Cu + 'u' + Cl + 'll0l1rrr' + ADDc() + 'dd' + Cu + 'u' + Cu + 'u' + Cl + ADDc() + 'dd'
def full(n):
return MNEall(n) + 'uuu' + '0l' * (n+1) + 'r' * (n+1) + 'ddd' + (ADDbit() + 'l') * n + 'uuu'
print(full(30)) | {
"input": [
"2\n123456789 987654321\n555555555 555555555\n"
],
"output": [
"ds10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslsrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrds10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslsrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrruus10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslss10utsusdtedslsrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrdl0r1010utdtltrs10ltslsrtersl1100utdtltls10rtsrsltelsls10rtsrsltelsrrl1010utdtltrs10ltslsrtersl1100utdtltls10rtsrsltelsls10rtsrsltelsrrl1010utdtltrs10ltslsrtersl1100utdtltls10rtsrsltelsls10rtsrsltelsrrl1010utdtltrs10ltslsrtersl1100utdtltls10rtsrsltelsls10rtsrsltelsrrl1010utdtltrs10ltslsrtersl1100utdtltls10rtsrsltelsls10rtsrsltelsrrl1010utdtltrs10ltslsrtersl1100utdtltls10rtsrsltelsls10rtsrsltelsrrl1010utdtltrs10ltslsrtersl1100utdtltls10rtsrsltelsls10rtsrsltelsrrl1010utdtltrs10ltslsrtersl1100utdtltls10rtsrsltelsls10rtsrsltelsrrl1010utdtltrs10ltslsrtersl1100utdtltls10rtsrsltelsls10rtsrsltelsrrl1010utdtltrs10ltslsrtersl1100utdtltls10rtsrsltelsls10rtsrsltelsrrl1010utdtltrs10ltslsrtersl1100utdtltls10rtsrsltelsls10rtsrsltelsrrl1010utdtltrs10ltslsrtersl1100utdtltls10rtsrsltelsls10rtsrsltelsrrl1010utdtltrs10ltslsrtersl1100utdtltls10rtsrsltelsls10rtsrsltelsrrl1010utdtltrs10ltslsrtersl1100utdtltls10rtsrsltelsls10rtsrsltelsrrl1010utdtltrs10ltslsrtersl1100utdtltls10rtsrsltelsls10rtsrsltelsrrl1010utdtltrs10ltslsrtersl1100utdtltls10rtsrsltelsls10rtsrsltelsrrl1010utdtltrs10ltslsrtersl1100utdtltls10rtsrsltelsls10rtsrsltelsrrl1010utdtltrs10ltslsrtersl1100utdtltls10rtsrsltelsls10rtsrsltelsrrl1010utdtltrs10ltslsrtersl1100utdtltls10rtsrsltelsls10rtsrsltelsrrl1010utdtltrs10ltslsrtersl1100utdtltls10rtsrsltelsls10rtsrsltelsrrl1010utdtltrs10ltslsrtersl1100utdtltls10rtsrsltelsls10rtsrsltelsrrl1010utdtltrs10ltslsrtersl1100utdtltls10rtsrsltelsls10rtsrsltelsrrl1010utdtltrs10ltslsrtersl1100utdtltls10rtsrsltelsls10rtsrsltelsrrl1010utdtltrs10ltslsrtersl1100utdtltls10rtsrsltelsls10rtsrsltelsrrl1010utdtltrs10ltslsrtersl1100utdtltls10rtsrsltelsls10rtsrsltelsrrl1010utdtltrs10ltslsrtersl1100utdtltls10rtsrsltelsls10rtsrsltelsrrl1010utdtltrs10ltslsrtersl1100utdtltls10rtsrsltelsls10rtsrsltelsrrl1010utdtltrs10ltslsrtersl1100utdtltls10rtsrsltelsls10rtsrsltelsrrl1010utdtltrs10ltslsrtersl1100utdtltls10rtsrsltelsls10rtsrsltelsrrl1010utdtltrs10ltslsrtersl1100utdtltls10rtsrsltelsls10rtsrsltelsrrl1010utdtltrs10ltslsrtersl1100utdtltls10rtsrsltelsls10rtsrsltelsrrl1010utdtltrs10ltslsrtersl1100utdtltls10rtsrsltelsls10rtsrsltelsrrlrr\n"
]
} |
687 | 7 | Alice and Bob are playing a game on a line with n cells. There are n cells labeled from 1 through n. For each i from 1 to n-1, cells i and i+1 are adjacent.
Alice initially has a token on some cell on the line, and Bob tries to guess where it is.
Bob guesses a sequence of line cell numbers x_1, x_2, …, x_k in order. In the i-th question, Bob asks Alice if her token is currently on cell x_i. That is, Alice can answer either "YES" or "NO" to each Bob's question.
At most one time in this process, before or after answering a question, Alice is allowed to move her token from her current cell to some adjacent cell. Alice acted in such a way that she was able to answer "NO" to all of Bob's questions.
Note that Alice can even move her token before answering the first question or after answering the last question. Alice can also choose to not move at all.
You are given n and Bob's questions x_1, …, x_k. You would like to count the number of scenarios that let Alice answer "NO" to all of Bob's questions.
Let (a,b) denote a scenario where Alice starts at cell a and ends at cell b. Two scenarios (a_i, b_i) and (a_j, b_j) are different if a_i ≠ a_j or b_i ≠ b_j.
Input
The first line contains two integers n and k (1 ≤ n,k ≤ 10^5) — the number of cells and the number of questions Bob asked.
The second line contains k integers x_1, x_2, …, x_k (1 ≤ x_i ≤ n) — Bob's questions.
Output
Print a single integer, the number of scenarios that let Alice answer "NO" to all of Bob's questions.
Examples
Input
5 3
5 1 4
Output
9
Input
4 8
1 2 3 4 4 3 2 1
Output
0
Input
100000 1
42
Output
299997
Note
The notation (i,j) denotes a scenario where Alice starts at cell i and ends at cell j.
In the first example, the valid scenarios are (1, 2), (2, 1), (2, 2), (2, 3), (3, 2), (3, 3), (3, 4), (4, 3), (4, 5). For example, (3,4) is valid since Alice can start at cell 3, stay there for the first three questions, then move to cell 4 after the last question.
(4,5) is valid since Alice can start at cell 4, stay there for the first question, the move to cell 5 for the next two questions. Note that (4,5) is only counted once, even though there are different questions that Alice can choose to do the move, but remember, we only count each pair of starting and ending positions once.
In the second example, Alice has no valid scenarios.
In the last example, all (i,j) where |i-j| ≤ 1 except for (42, 42) are valid scenarios. | def ck(p,q):
return 1 if bpos[p]>fpos[q] else 0
n,k=map(int,input().split())
x=list(map(int,input().split()))
fpos=[-1 for i in range(100001)]
bpos=[k+5 for i in range(100001)]
for i in range(k):
fpos[x[i]]=i
bpos[x[k-1-i]]=k-1-i
ans=ck(n,n)
for i in range(1,n):
ans+=ck(i,i)+ck(i,i+1)+ck(i+1,i)
print(ans) | {
"input": [
"5 3\n5 1 4\n",
"100000 1\n42\n",
"4 8\n1 2 3 4 4 3 2 1\n"
],
"output": [
"9\n",
"299997\n",
"0\n"
]
} |
688 | 8 | Toad Rash has a binary string s. A binary string consists only of zeros and ones.
Let n be the length of s.
Rash needs to find the number of such pairs of integers l, r that 1 ≤ l ≤ r ≤ n and there is at least one pair of integers x, k such that 1 ≤ x, k ≤ n, l ≤ x < x + 2k ≤ r, and s_x = s_{x+k} = s_{x+2k}.
Find this number of pairs for Rash.
Input
The first line contains the string s (1 ≤ |s| ≤ 300 000), consisting of zeros and ones.
Output
Output one integer: the number of such pairs of integers l, r that 1 ≤ l ≤ r ≤ n and there is at least one pair of integers x, k such that 1 ≤ x, k ≤ n, l ≤ x < x + 2k ≤ r, and s_x = s_{x+k} = s_{x+2k}.
Examples
Input
010101
Output
3
Input
11001100
Output
0
Note
In the first example, there are three l, r pairs we need to count: 1, 6; 2, 6; and 1, 5.
In the second example, there are no values x, k for the initial string, so the answer is 0. | s=input()
def pri(l,r):
k=1
while r-2*k>=l:
if s[r]!=s[r-k] or s[r-k]!=s[r-2*k]:
k+=1
continue
return False
return True
ans=0
for i in range(len(s)):
j=i
while j<len(s) and pri(i,j):
j+=1
ans+=len(s)-j
print(ans)
| {
"input": [
"010101\n",
"11001100\n"
],
"output": [
"3\n",
"0\n"
]
} |
689 | 11 | Vus the Cossack has a field with dimensions n × m, which consists of "0" and "1". He is building an infinite field from this field. He is doing this in this way:
1. He takes the current field and finds a new inverted field. In other words, the new field will contain "1" only there, where "0" was in the current field, and "0" there, where "1" was.
2. To the current field, he adds the inverted field to the right.
3. To the current field, he adds the inverted field to the bottom.
4. To the current field, he adds the current field to the bottom right.
5. He repeats it.
For example, if the initial field was:
\begin{matrix} 1 & 0 & \\\ 1 & 1 & \\\ \end{matrix}
After the first iteration, the field will be like this:
\begin{matrix} 1 & 0 & 0 & 1 \\\ 1 & 1 & 0 & 0 \\\ 0 & 1 & 1 & 0 \\\ 0 & 0 & 1 & 1 \\\ \end{matrix}
After the second iteration, the field will be like this:
\begin{matrix} 1 & 0 & 0 & 1 & 0 & 1 & 1 & 0 \\\ 1 & 1 & 0 & 0 & 0 & 0 & 1 & 1 \\\ 0 & 1 & 1 & 0 & 1 & 0 & 0 & 1 \\\ 0 & 0 & 1 & 1 & 1 & 1 & 0 & 0 \\\ 0 & 1 & 1 & 0 & 1 & 0 & 0 & 1 \\\ 0 & 0 & 1 & 1 & 1 & 1 & 0 & 0 \\\ 1 & 0 & 0 & 1 & 0 & 1 & 1 & 0 \\\ 1 & 1 & 0& 0 & 0 & 0 & 1 & 1 \\\ \end{matrix}
And so on...
Let's numerate lines from top to bottom from 1 to infinity, and columns from left to right from 1 to infinity. We call the submatrix (x_1, y_1, x_2, y_2) all numbers that have coordinates (x, y) such that x_1 ≤ x ≤ x_2 and y_1 ≤ y ≤ y_2.
The Cossack needs sometimes to find the sum of all the numbers in submatrices. Since he is pretty busy right now, he is asking you to find the answers!
Input
The first line contains three integers n, m, q (1 ≤ n, m ≤ 1 000, 1 ≤ q ≤ 10^5) — the dimensions of the initial matrix and the number of queries.
Each of the next n lines contains m characters c_{ij} (0 ≤ c_{ij} ≤ 1) — the characters in the matrix.
Each of the next q lines contains four integers x_1, y_1, x_2, y_2 (1 ≤ x_1 ≤ x_2 ≤ 10^9, 1 ≤ y_1 ≤ y_2 ≤ 10^9) — the coordinates of the upper left cell and bottom right cell, between which you need to find the sum of all numbers.
Output
For each query, print the answer.
Examples
Input
2 2 5
10
11
1 1 8 8
2 4 5 6
1 2 7 8
3 3 6 8
5 6 7 8
Output
32
5
25
14
4
Input
2 3 7
100
101
4 12 5 17
5 4 9 4
1 4 13 18
12 1 14 9
3 10 7 18
3 15 12 17
8 6 8 12
Output
6
3
98
13
22
15
3
Note
The first example is explained in the legend. | from sys import stdin,stdout
n,m,q = map(int, stdin.readline().split())
mat = [[0]*m for i in range(n)]
for i in range(n):
row = stdin.readline().strip()
for j,c in enumerate(row):
mat[i][j] = 1 if c == '1' else -1
#print(mat)
def get(a,b):
if a < 0 or b < 0:
return 0
x = a^b
ans = 1
while x > 0:
if x % 2 == 1:
ans *= -1
x //= 2
return ans
row_sums = [[0]*(m+1) for i in range(n+1)]
for i in range(n):
for j in range(m):
row_sums[i+1][j+1] = row_sums[i][j+1] + mat[i][j]
#print(row_sums)
mat_sums = [[0]*(m+1) for i in range(n+1)]
for i in range(n):
for j in range(m):
mat_sums[i+1][j+1] = mat_sums[i+1][j] + row_sums[i+1][j+1]
#print(mat_sums)
total = mat_sums[n][m]
def rect_sum(a, b):
if a == 0 or b == 0:
return 0
top_edge = 0
right_edge = 0
small = 0
x = a//n
x_rem = a%n
y = b // m
y_rem = b%m
# print("x", x, "y", y, "x_rem", x_rem, "y_rem", y_rem)
big = 0 if x % 2 == 0 or y % 2 == 0 else total
big *= get(x-1,y-1)
if x % 2 == 1:
right_edge= mat_sums[n][y_rem]
right_edge *= get(x-1,y)
if y % 2 == 1:
top_edge = mat_sums[x_rem][m]
top_edge *= get(x,y-1)
small = mat_sums[x_rem][y_rem]
small *= get(x,y)
# print("big", big, "top", top_edge, "right", right_edge, "small", small)
return top_edge + right_edge+small+big
for it in range(q):
x1,y1,x2,y2 = map(int, stdin.readline().split())
ans = rect_sum(x2,y2) - rect_sum(x1-1, y2) - rect_sum(x2, y1-1) + rect_sum(x1-1,y1-1)
ans = ((x2-x1+1)*(y2-y1+1) + ans)//2
stdout.write(str(ans) + '\n')
| {
"input": [
"2 3 7\n100\n101\n4 12 5 17\n5 4 9 4\n1 4 13 18\n12 1 14 9\n3 10 7 18\n3 15 12 17\n8 6 8 12\n",
"2 2 5\n10\n11\n1 1 8 8\n2 4 5 6\n1 2 7 8\n3 3 6 8\n5 6 7 8\n"
],
"output": [
"6\n3\n98\n13\n22\n15\n3\n",
"32\n5\n25\n14\n4\n"
]
} |
690 | 10 | You are given a tree with n nodes. You have to write non-negative integers on its edges so that the following condition would be satisfied:
For every two nodes i, j, look at the path between them and count the sum of numbers on the edges of this path. Write all obtained sums on the blackboard. Then every integer from 1 to ⌊ (2n^2)/(9) ⌋ has to be written on the blackboard at least once.
It is guaranteed that such an arrangement exists.
Input
The first line contains a single integer n (1 ≤ n ≤ 1000) — the number of nodes.
Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.
Output
Output n-1 lines, each of form u v x (0 ≤ x ≤ 10^6), which will mean that you wrote number x on the edge between u, v.
Set of edges (u, v) has to coincide with the set of edges of the input graph, but you can output edges in any order. You can also output ends of edges in an order different from the order in input.
Examples
Input
3
2 3
2 1
Output
3 2 1
1 2 2
Input
4
2 4
2 3
2 1
Output
4 2 1
3 2 2
1 2 3
Input
5
1 2
1 3
1 4
2 5
Output
2 1 1
5 2 1
3 1 3
4 1 6
Note
In the first example, distance between nodes 1 and 2 is equal to 2, between nodes 2 and 3 to 1, between 1 and 3 to 3.
In the third example, numbers from 1 to 9 (inclusive) will be written on the blackboard, while we need just from 1 to 5 to pass the test. | import math
n = int(input())
if n == 1:
print()
else:
edge = [list(map(int, input().split())) for i in range(1, n) ]
g = {}
for x, y in edge:
if x not in g:
g[x] = []
if y not in g:
g[y] = []
g[x].append(y)
g[y].append(x)
def find_centroid(g):
p = {}
size = {}
p[1] = -1
Q = [1]
i = 0
while i < len(Q):
u = Q[i]
for v in g[u]:
if p[u] == v: continue
p[v] = u
Q.append(v)
i+=1
for u in Q[::-1]:
size[u] = 1
for v in g[u]:
if p[u] == v:
continue
size[u] += size[v]
cur = 1
n = size[cur]
while True:
max_ = n - size[cur]
ind_ = p[cur]
for v in g[cur]:
if v == p[cur]: continue
if size[v] > max_:
max_ = size[v]
ind_ = v
if max_ <= n // 2:
return cur
cur = ind_
def find_center(g):
d = {}
d[1] = 0
Q = [(1, 0)]
while len(Q) > 0:
u, dis = Q.pop(0)
for v in g[u]:
if v not in d:
d[v] = dis +1
Q.append((v, d[v]))
max_length = -1
s = None
for u, dis in d.items():
if dis > max_length:
max_length = dis
s = u
d = {}
pre = {}
d[s] = 0
Q = [(s, 0)]
while len(Q) > 0:
u, dis = Q.pop(0)
for v in g[u]:
if v not in d:
pre[v] = u
d[v] = dis +1
Q.append((v, d[v]))
max_length = -1
e = None
for u, dis in d.items():
if dis > max_length:
max_length = dis
e = u
route = [e]
while pre[route[-1]] != s:
route.append(pre[route[-1]])
print(route)
return route[len(route) // 2]
root = find_centroid(g)
p = {}
size = {}
Q = [root]
p[root] = -1
i = 0
while i < len(Q):
u = Q[i]
for v in g[u]:
if p[u] == v: continue
p[v] = u
Q.append(v)
i+=1
for u in Q[::-1]:
size[u] = 1
for v in g[u]:
if p[u] == v:
continue
size[u] += size[v]
gr = [(u, size[u]) for u in g[root]]
gr = sorted(gr, key=lambda x:x[1])
thres = math.ceil((n-1) / 3)
sum_ = 0
gr1 = []
gr2 = []
i = 0
while sum_ < thres:
gr1.append(gr[i][0])
sum_ += gr[i][1]
i+=1
while i < len(gr):
gr2.append(gr[i][0])
i+=1
def asign(u, W, ew):
if size[u] == 1:
return
cur = 0
for v in g[u]:
if v == p[u]: continue
first = W[cur]
ew.append((u, v, first))
W_ = [x - first for x in W[cur+1: cur+size[v]]]
asign(v, W_, ew)
cur+=size[v]
a, b = 0, 0
for x in gr1:
a += size[x]
for x in gr2:
b += size[x]
arr_1 = [x for x in range(1, a+1)]
arr_2 = [i*(a+1) for i in range(1, b+1)]
ew = []
cur = 0
for u in gr1:
first = arr_1[cur]
ew.append((root, u, first))
W_ = [x - first for x in arr_1[cur+1:cur+size[u]]]
cur += size[u]
#print(u, W_)
asign(u, W_, ew)
cur = 0
for u in gr2:
first = arr_2[cur]
ew.append((root, u, first))
W_ = [x - first for x in arr_2[cur+1:cur+size[u]]]
cur += size[u]
#print(u, W_)
asign(u, W_, ew)
for u, v, w in ew:
print('{} {} {}'.format(u, v, w)) | {
"input": [
"3\n2 3\n2 1\n",
"4\n2 4\n2 3\n2 1\n",
"5\n1 2\n1 3\n1 4\n2 5\n"
],
"output": [
"2 3 1\n2 1 2\n",
"2 4 1\n2 3 2\n2 1 4\n",
"1 2 1\n2 5 1\n1 3 3\n1 4 6\n"
]
} |
691 | 7 | Let's denote correct match equation (we will denote it as CME) an equation a + b = c there all integers a, b and c are greater than zero.
For example, equations 2 + 2 = 4 (||+||=||||) and 1 + 2 = 3 (|+||=|||) are CME but equations 1 + 2 = 4 (|+||=||||), 2 + 2 = 3 (||+||=|||), and 0 + 1 = 1 (+|=|) are not.
Now, you have n matches. You want to assemble a CME using all your matches. Unfortunately, it is possible that you can't assemble the CME using all matches. But you can buy some extra matches and then assemble CME!
For example, if n = 2, you can buy two matches and assemble |+|=||, and if n = 5 you can buy one match and assemble ||+|=|||.
<image>
Calculate the minimum number of matches which you have to buy for assembling CME.
Note, that you have to answer q independent queries.
Input
The first line contains one integer q (1 ≤ q ≤ 100) — the number of queries.
The only line of each query contains one integer n (2 ≤ n ≤ 10^9) — the number of matches.
Output
For each test case print one integer in single line — the minimum number of matches which you have to buy for assembling CME.
Example
Input
4
2
5
8
11
Output
2
1
0
1
Note
The first and second queries are explained in the statement.
In the third query, you can assemble 1 + 3 = 4 (|+|||=||||) without buying matches.
In the fourth query, buy one match and assemble 2 + 4 = 6 (||+||||=||||||). | def f(n):
if n == 2:
return 2
return n%2
for i in range(int(input())):
print(f(int(input())))
| {
"input": [
"4\n2\n5\n8\n11\n"
],
"output": [
"2\n1\n0\n1\n"
]
} |
692 | 8 | Balph is learning to play a game called Buma. In this game, he is given a row of colored balls. He has to choose the color of one new ball and the place to insert it (between two balls, or to the left of all the balls, or to the right of all the balls).
When the ball is inserted the following happens repeatedly: if some segment of balls of the same color became longer as a result of a previous action and its length became at least 3, then all the balls of this segment are eliminated.
Consider, for example, a row of balls 'AAABBBWWBB'. Suppose Balph chooses a ball of color 'W' and the place to insert it after the sixth ball, i. e. to the left of the two 'W's. After Balph inserts this ball, the balls of color 'W' are eliminated, since this segment was made longer and has length 3 now, so the row becomes 'AAABBBBB'. The balls of color 'B' are eliminated now, because the segment of balls of color 'B' became longer and has length 5 now. Thus, the row becomes 'AAA'. However, none of the balls are eliminated now, because there is no elongated segment.
Help Balph count the number of possible ways to choose a color of a new ball and a place to insert it that leads to the elimination of all the balls.
Input
The only line contains a non-empty string of uppercase English letters of length at most 3 ⋅ 10^5. Each letter represents a ball with the corresponding color.
Output
Output the number of ways to choose a color and a position of a new ball in order to eliminate all the balls.
Examples
Input
BBWWBB
Output
3
Input
BWWB
Output
0
Input
BBWBB
Output
0
Input
OOOWWW
Output
0
Input
WWWOOOOOOWWW
Output
7 | def run():
s = input()
i = 0
j = len(s)-1
count = 0
while i<=j:
if s[i] != s[j]:
return 0
c = s[i]
count = 0
while i < len(s) and s[i] == c and i<=j:
i += 1
count += 1
while j >= 0 and s[j] == c and i<=j:
j -= 1
count += 1
if i>j:
count += 1
if count >= 3:
return count
else:
return 0
elif count<3:
return 0
print(run()) | {
"input": [
"OOOWWW\n",
"BBWBB\n",
"BBWWBB\n",
"BWWB\n",
"WWWOOOOOOWWW\n"
],
"output": [
"0\n",
"0\n",
"3\n",
"0\n",
"7\n"
]
} |
693 | 10 | You are given n arrays a_1, a_2, ..., a_n; each array consists of exactly m integers. We denote the y-th element of the x-th array as a_{x, y}.
You have to choose two arrays a_i and a_j (1 ≤ i, j ≤ n, it is possible that i = j). After that, you will obtain a new array b consisting of m integers, such that for every k ∈ [1, m] b_k = max(a_{i, k}, a_{j, k}).
Your goal is to choose i and j so that the value of min _{k = 1}^{m} b_k is maximum possible.
Input
The first line contains two integers n and m (1 ≤ n ≤ 3 ⋅ 10^5, 1 ≤ m ≤ 8) — the number of arrays and the number of elements in each array, respectively.
Then n lines follow, the x-th line contains the array a_x represented by m integers a_{x, 1}, a_{x, 2}, ..., a_{x, m} (0 ≤ a_{x, y} ≤ 10^9).
Output
Print two integers i and j (1 ≤ i, j ≤ n, it is possible that i = j) — the indices of the two arrays you have to choose so that the value of min _{k = 1}^{m} b_k is maximum possible. If there are multiple answers, print any of them.
Example
Input
6 5
5 0 3 1 2
1 8 9 1 3
1 2 3 4 5
9 1 0 3 7
2 3 0 6 3
6 4 1 7 0
Output
1 5 | import math as mt
import sys
input=sys.stdin.readline
I=lambda:list(map(int,input().split()))
n,m=I()
a=[I() for i in range(n)]
ans=[]
lo=0
hi=10**9
def vanguda(mid: int) -> bool:
global ans
f={}
for i in range(n):
bi=0
for j in range(m):
if a[i][j]>=mid:
bi+=1
bi<<=1
f[bi>>1]=i
for aa,bb in f.items():
for cc,dd in f.items():
if aa|cc==(2**m-1):
ans =bb+1,dd+1
return True
return False
while lo<=hi:
mid=(lo+hi)//2
if vanguda(mid):
lo=mid+1
else:
hi=mid-1
print(*ans)
| {
"input": [
"6 5\n5 0 3 1 2\n1 8 9 1 3\n1 2 3 4 5\n9 1 0 3 7\n2 3 0 6 3\n6 4 1 7 0\n"
],
"output": [
"1 5\n"
]
} |
694 | 8 | You are given two positive integers n (1 ≤ n ≤ 10^9) and k (1 ≤ k ≤ 100). Represent the number n as the sum of k positive integers of the same parity (have the same remainder when divided by 2).
In other words, find a_1, a_2, …, a_k such that all a_i>0, n = a_1 + a_2 + … + a_k and either all a_i are even or all a_i are odd at the same time.
If such a representation does not exist, then report it.
Input
The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases in the input. Next, t test cases are given, one per line.
Each test case is two positive integers n (1 ≤ n ≤ 10^9) and k (1 ≤ k ≤ 100).
Output
For each test case print:
* YES and the required values a_i, if the answer exists (if there are several answers, print any of them);
* NO if the answer does not exist.
The letters in the words YES and NO can be printed in any case.
Example
Input
8
10 3
100 4
8 7
97 2
8 8
3 10
5 3
1000000000 9
Output
YES
4 2 4
YES
55 5 5 35
NO
NO
YES
1 1 1 1 1 1 1 1
NO
YES
3 1 1
YES
111111110 111111110 111111110 111111110 111111110 111111110 111111110 111111110 111111120 | def main(n,k):
if (n - 2*(k-1)) %2==0 and n-2*(k-1)>0:
print("YES")
print( (str(2)+" ")*(k-1) + str(n- 2*(k-1)) )
return
if (n - k + 1)%2==1 and n- k+1>0:
print("YES")
print( "1 "*(k-1) + str(n-k+1) )
return
print("NO")
t = int(input())
for i in range(t):
n,k = [int(i) for i in input().split()]
main(n,k) | {
"input": [
"8\n10 3\n100 4\n8 7\n97 2\n8 8\n3 10\n5 3\n1000000000 9\n"
],
"output": [
"yes\n2 2 6\nyes\n1 1 1 97\nno\nno\nyes\n1 1 1 1 1 1 1 1\nno\nyes\n1 1 3\nyes\n2 2 2 2 2 2 2 2 999999984\n"
]
} |
695 | 8 | Jett is tired after destroying the town and she wants to have a rest. She likes high places, that's why for having a rest she wants to get high and she decided to craft staircases.
A staircase is a squared figure that consists of square cells. Each staircase consists of an arbitrary number of stairs. If a staircase has n stairs, then it is made of n columns, the first column is 1 cell high, the second column is 2 cells high, …, the n-th column if n cells high. The lowest cells of all stairs must be in the same row.
A staircase with n stairs is called nice, if it may be covered by n disjoint squares made of cells. All squares should fully consist of cells of a staircase.
This is how a nice covered staircase with 7 stairs looks like: <image>
Find out the maximal number of different nice staircases, that can be built, using no more than x cells, in total. No cell can be used more than once.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The description of each test case contains a single integer x (1 ≤ x ≤ 10^{18}) — the number of cells for building staircases.
Output
For each test case output a single integer — the number of different nice staircases, that can be built, using not more than x cells, in total.
Example
Input
4
1
8
6
1000000000000000000
Output
1
2
1
30
Note
In the first test case, it is possible to build only one staircase, that consists of 1 stair. It's nice. That's why the answer is 1.
In the second test case, it is possible to build two different nice staircases: one consists of 1 stair, and another consists of 3 stairs. This will cost 7 cells. In this case, there is one cell left, but it is not possible to use it for building any nice staircases, that have not been built yet. That's why the answer is 2.
In the third test case, it is possible to build only one of two nice staircases: with 1 stair or with 3 stairs. In the first case, there will be 5 cells left, that may be used only to build a staircase with 2 stairs. This staircase is not nice, and Jett only builds nice staircases. That's why in this case the answer is 1. If Jett builds a staircase with 3 stairs, then there are no more cells left, so the answer is 1 again. | def func(n):
return (n*(n+1))//2
for _ in range(int(input())):
x=int(input())
s=0
i=1
while x-func(i)>=0:
x-=func(i)
i=2*i+1
s+=1
print(s) | {
"input": [
"4\n1\n8\n6\n1000000000000000000\n"
],
"output": [
"1\n2\n1\n30\n"
]
} |
696 | 8 | You are given a string s of even length n. String s is binary, in other words, consists only of 0's and 1's.
String s has exactly n/2 zeroes and n/2 ones (n is even).
In one operation you can reverse any substring of s. A substring of a string is a contiguous subsequence of that string.
What is the minimum number of operations you need to make string s alternating? A string is alternating if s_i ≠ s_{i + 1} for all i. There are two types of alternating strings in general: 01010101... or 10101010...
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains a single integer n (2 ≤ n ≤ 10^5; n is even) — the length of string s.
The second line of each test case contains a binary string s of length n (s_i ∈ {0, 1}). String s has exactly n/2 zeroes and n/2 ones.
It's guaranteed that the total sum of n over test cases doesn't exceed 10^5.
Output
For each test case, print the minimum number of operations to make s alternating.
Example
Input
3
2
10
4
0110
8
11101000
Output
0
1
2
Note
In the first test case, string 10 is already alternating.
In the second test case, we can, for example, reverse the last two elements of s and get: 0110 → 0101.
In the third test case, we can, for example, make the following two operations:
1. 11101000 → 10101100;
2. 10101100 → 10101010. |
def solve():
n=int(input())
s=input()
answer=(n-s.count('10')-s.count('01'))//2
print(answer)
t=int(input())
while t:
solve()
t-=1
| {
"input": [
"3\n2\n10\n4\n0110\n8\n11101000\n"
],
"output": [
"0\n1\n2\n"
]
} |
697 | 12 | Barbara was late for her math class so as a punishment the teacher made her solve the task on a sheet of paper. Barbara looked at the sheet of paper and only saw n numbers a_1, a_2, …, a_n without any mathematical symbols. The teacher explained to Barbara that she has to place the available symbols between the numbers in a way that would make the resulting expression's value as large as possible. To find out which symbols were available the teacher has given Barbara a string s which contained that information.
<image>
It's easy to notice that Barbara has to place n - 1 symbols between numbers in total. The expression must start with a number and all symbols must be allowed (i.e. included in s). Note that multiplication takes precedence over addition or subtraction, addition and subtraction have the same priority and performed from left to right. Help Barbara and create the required expression!
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 10^5) — the amount of numbers on the paper.
The second line of the input contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 9), where a_i is the i-th element of a.
The third line of the input contains the string s (1 ≤ |s| ≤ 3) — symbols allowed in the expression. It is guaranteed that the string may only consist of symbols "-", "+" and "*". It is also guaranteed that all symbols in the string are distinct.
Output
Print n numbers separated by n - 1 symbols — a mathematical expression with the greatest result. If there are multiple equally valid results — output any one of them.
Examples
Input
3
2 2 0
+-*
Output
2*2-0
Input
4
2 1 1 2
+*
Output
2+1+1+2
Note
The following answers also fit the first example: "2+2+0", "2+2-0", "2*2+0". | #!/usr/bin/env python3
from itertools import accumulate, groupby
from functools import reduce
def prod(a):
return reduce(lambda x,y: min(x*y,10**6),a,1)
def solve_positive(a):
if a == '': return '+'
b = [''.join(v) for _,v in groupby(a, key=lambda x: x == '1')]
if b[0][0] == '1':
return '+' * len(b[0]) + solve_positive(a[len(b[0]):])
if b[-1][0] == '1':
return solve_positive(a[:-len(b[-1])]) + '+' * len(b[-1])
p = [prod(map(int,x)) for x in b[::2]]
q = [len(x) for x in b[1::2]]
k = len(p)
if prod(p) >= 10**6:
return '+' + '*' * (len(a)-1) + '+'
dp = [0] * k
go = [k] * k
for i in range(k)[::-1]:
dp[i] = prod(p[i:])
for j in range(i+1,k):
ndp = prod(p[i:j]) + q[j-1] + dp[j]
if ndp > dp[i]:
dp[i], go[i] = ndp, j
offset = [0] + list(accumulate(map(len,b)))
res = ['*'] * (len(a)-1)
i = go[0]
while i < k:
a = offset[2*i-1]-1
b = offset[2*i]
res[a:b] = '+' * (b-a)
i = go[i]
return '+' + ''.join(res) + '+'
def solve(a,ops):
n = len(a)
if len(ops) == 1: return ops * (n-1)
if sorted(ops) == list('+-'): return '+' * (n-1)
if sorted(ops) == list('*-'):
k = a.index('0') if '0' in a else n
if k == 0 or k == n: return '*' * (n-1)
return '*' * (k-1) + '-' + '*' * (n-k-1)
return ''.join(map(solve_positive,a.split('0')))[1:-1]
n = int(input())
a = ''.join(input().split())
ops = input()
b = solve(a,ops) + '\n'
for i in range(n):
print(a[i], end='')
print(b[i], end='')
| {
"input": [
"4\n2 1 1 2\n+*\n",
"3\n2 2 0\n+-*\n"
],
"output": [
"\n2+1+1+2\n",
"\n2*2-0\n"
]
} |
698 | 7 | n heroes fight against each other in the Arena. Initially, the i-th hero has level a_i.
Each minute, a fight between two different heroes occurs. These heroes can be chosen arbitrarily (it's even possible that it is the same two heroes that were fighting during the last minute).
When two heroes of equal levels fight, nobody wins the fight. When two heroes of different levels fight, the one with the higher level wins, and his level increases by 1.
The winner of the tournament is the first hero that wins in at least 100^{500} fights (note that it's possible that the tournament lasts forever if no hero wins this number of fights, then there is no winner). A possible winner is a hero such that there exists a sequence of fights that this hero becomes the winner of the tournament.
Calculate the number of possible winners among n heroes.
Input
The first line contains one integer t (1 ≤ t ≤ 500) — the number of test cases.
Each test case consists of two lines. The first line contains one integer n (2 ≤ n ≤ 100) — the number of heroes. The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100), where a_i is the initial level of the i-th hero.
Output
For each test case, print one integer — the number of possible winners among the given n heroes.
Example
Input
3
3
3 2 2
2
5 5
4
1 3 3 7
Output
1
0
3
Note
In the first test case of the example, the only possible winner is the first hero.
In the second test case of the example, each fight between the heroes results in nobody winning it, so the tournament lasts forever and there is no winner. | def read_ints():
return map(int, input().split())
t_n, = read_ints()
for i_t in range(t_n):
a_n, = read_ints()
a_seq = tuple(read_ints())
result = len(a_seq) - a_seq.count(min(a_seq))
print(result)
| {
"input": [
"3\n3\n3 2 2\n2\n5 5\n4\n1 3 3 7\n"
],
"output": [
"\n1\n0\n3\n"
]
} |
699 | 9 | You are given an array a of n integers. Find the number of pairs (i, j) (1 ≤ i < j ≤ n) where the sum of a_i + a_j is greater than or equal to l and less than or equal to r (that is, l ≤ a_i + a_j ≤ r).
For example, if n = 3, a = [5, 1, 2], l = 4 and r = 7, then two pairs are suitable:
* i=1 and j=2 (4 ≤ 5 + 1 ≤ 7);
* i=1 and j=3 (4 ≤ 5 + 2 ≤ 7).
Input
The first line contains an integer t (1 ≤ t ≤ 10^4). Then t test cases follow.
The first line of each test case contains three integers n, l, r (1 ≤ n ≤ 2 ⋅ 10^5, 1 ≤ l ≤ r ≤ 10^9) — the length of the array and the limits on the sum in the pair.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9).
It is guaranteed that the sum of n overall test cases does not exceed 2 ⋅ 10^5.
Output
For each test case, output a single integer — the number of index pairs (i, j) (i < j), such that l ≤ a_i + a_j ≤ r.
Example
Input
4
3 4 7
5 1 2
5 5 8
5 1 2 4 3
4 100 1000
1 1 1 1
5 9 13
2 5 5 1 1
Output
2
7
0
1 | from bisect import bisect_right
def solve(v,x):
pcnt=0
for i,e in enumerate(v):
j=bisect_right(v,x-e)
if i<j:
pcnt+=j-i-1
return pcnt
for _ in range(int(input())):
n,l,r=map(int,input().split())
v=list(map(int,input().split()))
v=sorted(v)
print(solve(v,r)-solve(v,l-1))
| {
"input": [
"4\n3 4 7\n5 1 2\n5 5 8\n5 1 2 4 3\n4 100 1000\n1 1 1 1\n5 9 13\n2 5 5 1 1\n"
],
"output": [
"\n2\n7\n0\n1\n"
]
} |
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