message stringlengths 2 67k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 463 109k | cluster float64 19 19 | __index_level_0__ int64 926 217k |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
Gildong is now developing a puzzle game. The puzzle consists of n platforms numbered from 1 to n. The player plays the game as a character that can stand on each platform and the goal of the game is to move the character from the 1-st platform to the n-th platform.
The i-th platform is labeled with an integer a_i (0 β€ a_i β€ n-i). When the character is standing on the i-th platform, the player can move the character to any of the j-th platforms where i+1 β€ j β€ i+a_i. If the character is on the i-th platform where a_i=0 and i β n, the player loses the game.
Since Gildong thinks the current game is not hard enough, he wants to make it even harder. He wants to change some (possibly zero) labels to 0 so that there remains exactly one way to win. He wants to modify the game as little as possible, so he's asking you to find the minimum number of platforms that should have their labels changed. Two ways are different if and only if there exists a platform the character gets to in one way but not in the other way.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 β€ t β€ 500).
Each test case contains two lines. The first line of each test case consists of an integer n (2 β€ n β€ 3000) β the number of platforms of the game.
The second line of each test case contains n integers. The i-th integer is a_i (0 β€ a_i β€ n-i) β the integer of the i-th platform.
It is guaranteed that:
* For each test case, there is at least one way to win initially.
* The sum of n in all test cases doesn't exceed 3000.
Output
For each test case, print one integer β the minimum number of different labels that should be changed to 0 so that there remains exactly one way to win.
Example
Input
3
4
1 1 1 0
5
4 3 2 1 0
9
4 1 4 2 1 0 2 1 0
Output
0
3
2
Note
In the first case, the player can only move to the next platform until they get to the 4-th platform. Since there is already only one way to win, the answer is zero.
In the second case, Gildong can change a_2, a_3, and a_4 to 0 so that the game becomes 4 0 0 0 0. Now the only way the player can win is to move directly from the 1-st platform to the 5-th platform.
In the third case, Gildong can change a_2 and a_8 to 0, then the only way to win is to move in the following way: 1 β 3 β 7 β 9. | instruction | 0 | 68,364 | 19 | 136,728 |
Tags: dp
Correct Solution:
```
MAX_N = 3000
inf = MAX_N
def solve():
global inf
n = int(input())
a = [0] + list(map(int, input().split()))
dp = [[0] * (n + 1) for i in range(n + 1)]
for i in range(2, n + 1):
cnt = 0
for j in range(i, n + 1):
dp[i][j] = inf
for j in range(i - 1, 0, -1):
if j + a[j] >= i:
dp[i][j + a[j]] = min(dp[i][j + a[j]], dp[j][i - 1 ] + cnt)
cnt += 1
for j in range(i + 1, n + 1):
dp[i][j] = min(dp[i][j], dp[i][j - 1])
return dp[n][n]
for _ in range(int(input())):print(solve())
``` | output | 1 | 68,364 | 19 | 136,729 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Gildong is now developing a puzzle game. The puzzle consists of n platforms numbered from 1 to n. The player plays the game as a character that can stand on each platform and the goal of the game is to move the character from the 1-st platform to the n-th platform.
The i-th platform is labeled with an integer a_i (0 β€ a_i β€ n-i). When the character is standing on the i-th platform, the player can move the character to any of the j-th platforms where i+1 β€ j β€ i+a_i. If the character is on the i-th platform where a_i=0 and i β n, the player loses the game.
Since Gildong thinks the current game is not hard enough, he wants to make it even harder. He wants to change some (possibly zero) labels to 0 so that there remains exactly one way to win. He wants to modify the game as little as possible, so he's asking you to find the minimum number of platforms that should have their labels changed. Two ways are different if and only if there exists a platform the character gets to in one way but not in the other way.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 β€ t β€ 500).
Each test case contains two lines. The first line of each test case consists of an integer n (2 β€ n β€ 3000) β the number of platforms of the game.
The second line of each test case contains n integers. The i-th integer is a_i (0 β€ a_i β€ n-i) β the integer of the i-th platform.
It is guaranteed that:
* For each test case, there is at least one way to win initially.
* The sum of n in all test cases doesn't exceed 3000.
Output
For each test case, print one integer β the minimum number of different labels that should be changed to 0 so that there remains exactly one way to win.
Example
Input
3
4
1 1 1 0
5
4 3 2 1 0
9
4 1 4 2 1 0 2 1 0
Output
0
3
2
Note
In the first case, the player can only move to the next platform until they get to the 4-th platform. Since there is already only one way to win, the answer is zero.
In the second case, Gildong can change a_2, a_3, and a_4 to 0 so that the game becomes 4 0 0 0 0. Now the only way the player can win is to move directly from the 1-st platform to the 5-th platform.
In the third case, Gildong can change a_2 and a_8 to 0, then the only way to win is to move in the following way: 1 β 3 β 7 β 9. | instruction | 0 | 68,365 | 19 | 136,730 |
Tags: dp
Correct Solution:
```
t = int(input())
for _ in range(t):
n = int(input())
arr = [0] + [int(val) for val in input().split(' ')]
dp = [[0] * (n + 1) for i in range(n + 1)]
for i in range(2, n + 1):
cnt = 0
for j in range(i, n + 1):
dp[i][j] = 3000
for j in range(i - 1, 0, -1):
if j + arr[j] >= i:
dp[i][j + arr[j]] = min(dp[i][j + arr[j]], dp[j][i-1] + cnt)
cnt += 1
for j in range(i + 1, n + 1):
dp[i][j] = min(dp[i][j], dp[i][j-1])
print(dp[n][n])
``` | output | 1 | 68,365 | 19 | 136,731 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Gildong is now developing a puzzle game. The puzzle consists of n platforms numbered from 1 to n. The player plays the game as a character that can stand on each platform and the goal of the game is to move the character from the 1-st platform to the n-th platform.
The i-th platform is labeled with an integer a_i (0 β€ a_i β€ n-i). When the character is standing on the i-th platform, the player can move the character to any of the j-th platforms where i+1 β€ j β€ i+a_i. If the character is on the i-th platform where a_i=0 and i β n, the player loses the game.
Since Gildong thinks the current game is not hard enough, he wants to make it even harder. He wants to change some (possibly zero) labels to 0 so that there remains exactly one way to win. He wants to modify the game as little as possible, so he's asking you to find the minimum number of platforms that should have their labels changed. Two ways are different if and only if there exists a platform the character gets to in one way but not in the other way.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 β€ t β€ 500).
Each test case contains two lines. The first line of each test case consists of an integer n (2 β€ n β€ 3000) β the number of platforms of the game.
The second line of each test case contains n integers. The i-th integer is a_i (0 β€ a_i β€ n-i) β the integer of the i-th platform.
It is guaranteed that:
* For each test case, there is at least one way to win initially.
* The sum of n in all test cases doesn't exceed 3000.
Output
For each test case, print one integer β the minimum number of different labels that should be changed to 0 so that there remains exactly one way to win.
Example
Input
3
4
1 1 1 0
5
4 3 2 1 0
9
4 1 4 2 1 0 2 1 0
Output
0
3
2
Note
In the first case, the player can only move to the next platform until they get to the 4-th platform. Since there is already only one way to win, the answer is zero.
In the second case, Gildong can change a_2, a_3, and a_4 to 0 so that the game becomes 4 0 0 0 0. Now the only way the player can win is to move directly from the 1-st platform to the 5-th platform.
In the third case, Gildong can change a_2 and a_8 to 0, then the only way to win is to move in the following way: 1 β 3 β 7 β 9.
Submitted Solution:
```
for _ in range(int(input())):
n = int(input())
cnt = 0
arr = list(map(int,input().split()))
if(arr[n-1]==0):
cnt += 1
print(len(set(arr))-2)
``` | instruction | 0 | 68,366 | 19 | 136,732 |
No | output | 1 | 68,366 | 19 | 136,733 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Gildong is now developing a puzzle game. The puzzle consists of n platforms numbered from 1 to n. The player plays the game as a character that can stand on each platform and the goal of the game is to move the character from the 1-st platform to the n-th platform.
The i-th platform is labeled with an integer a_i (0 β€ a_i β€ n-i). When the character is standing on the i-th platform, the player can move the character to any of the j-th platforms where i+1 β€ j β€ i+a_i. If the character is on the i-th platform where a_i=0 and i β n, the player loses the game.
Since Gildong thinks the current game is not hard enough, he wants to make it even harder. He wants to change some (possibly zero) labels to 0 so that there remains exactly one way to win. He wants to modify the game as little as possible, so he's asking you to find the minimum number of platforms that should have their labels changed. Two ways are different if and only if there exists a platform the character gets to in one way but not in the other way.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 β€ t β€ 500).
Each test case contains two lines. The first line of each test case consists of an integer n (2 β€ n β€ 3000) β the number of platforms of the game.
The second line of each test case contains n integers. The i-th integer is a_i (0 β€ a_i β€ n-i) β the integer of the i-th platform.
It is guaranteed that:
* For each test case, there is at least one way to win initially.
* The sum of n in all test cases doesn't exceed 3000.
Output
For each test case, print one integer β the minimum number of different labels that should be changed to 0 so that there remains exactly one way to win.
Example
Input
3
4
1 1 1 0
5
4 3 2 1 0
9
4 1 4 2 1 0 2 1 0
Output
0
3
2
Note
In the first case, the player can only move to the next platform until they get to the 4-th platform. Since there is already only one way to win, the answer is zero.
In the second case, Gildong can change a_2, a_3, and a_4 to 0 so that the game becomes 4 0 0 0 0. Now the only way the player can win is to move directly from the 1-st platform to the 5-th platform.
In the third case, Gildong can change a_2 and a_8 to 0, then the only way to win is to move in the following way: 1 β 3 β 7 β 9.
Submitted Solution:
```
def compare(a1,a2,n):
cnt=0
for i in range(n):
if a1[i]!=a2[i]:
cnt+=1
return cnt
def solve(a,index,b):
if index==0:
return
cand=[]
count=0
for i in range(1,index):
if(i+a[i]>=index):
cand.append(i)
count+=1
b_arr=[b]
b_arr[0][cand[0]]=0
for i in range(1,count):
b_arr.append(b)
b_arr[i][cand[i]]=0
if compare(a,b_arr[i])<compare(a,b_arr[i-1]):
b=b_arr[i]
solve(a,index-1,b)
def makeHarder(i,j):
if i==0 or i>j:
dp[i][j]=0
return 0
count=0
cands=[]
for k in range(i):
if a[k]+k>=i and a[k]+k<=j:
cands.append(k)
count+=1
l=count
print(count)
while l>0:
l-=1
dp[i][j]=min(dp[i][j],makeHarder(cands[l],i-1)+count-l-1)
return dp[i][j]
t=int(input())
for i in range(t):
n=int(input())
a=[int(x) for x in input().split(" ")]
dp=[[3000 for x in range(n)]for y in range(n)]
#makeHarder(n-1,n-1)
for i in range(n):
print(dp[i])
``` | instruction | 0 | 68,367 | 19 | 136,734 |
No | output | 1 | 68,367 | 19 | 136,735 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Gildong is now developing a puzzle game. The puzzle consists of n platforms numbered from 1 to n. The player plays the game as a character that can stand on each platform and the goal of the game is to move the character from the 1-st platform to the n-th platform.
The i-th platform is labeled with an integer a_i (0 β€ a_i β€ n-i). When the character is standing on the i-th platform, the player can move the character to any of the j-th platforms where i+1 β€ j β€ i+a_i. If the character is on the i-th platform where a_i=0 and i β n, the player loses the game.
Since Gildong thinks the current game is not hard enough, he wants to make it even harder. He wants to change some (possibly zero) labels to 0 so that there remains exactly one way to win. He wants to modify the game as little as possible, so he's asking you to find the minimum number of platforms that should have their labels changed. Two ways are different if and only if there exists a platform the character gets to in one way but not in the other way.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 β€ t β€ 500).
Each test case contains two lines. The first line of each test case consists of an integer n (2 β€ n β€ 3000) β the number of platforms of the game.
The second line of each test case contains n integers. The i-th integer is a_i (0 β€ a_i β€ n-i) β the integer of the i-th platform.
It is guaranteed that:
* For each test case, there is at least one way to win initially.
* The sum of n in all test cases doesn't exceed 3000.
Output
For each test case, print one integer β the minimum number of different labels that should be changed to 0 so that there remains exactly one way to win.
Example
Input
3
4
1 1 1 0
5
4 3 2 1 0
9
4 1 4 2 1 0 2 1 0
Output
0
3
2
Note
In the first case, the player can only move to the next platform until they get to the 4-th platform. Since there is already only one way to win, the answer is zero.
In the second case, Gildong can change a_2, a_3, and a_4 to 0 so that the game becomes 4 0 0 0 0. Now the only way the player can win is to move directly from the 1-st platform to the 5-th platform.
In the third case, Gildong can change a_2 and a_8 to 0, then the only way to win is to move in the following way: 1 β 3 β 7 β 9.
Submitted Solution:
```
import sys
input = sys.stdin.readline
for _ in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
dp = [0] * n
dp[-1] = 1
res = 0
for i in range(n - 2, -1, -1):
for j in range(i + 1, min(n, i + a[i] + 1)): dp[i] += dp[j]
for j in range(min(n - 1, i + a[i]), i, -1):
if dp[i] > 1 and dp[j]:
dp[j] = 0
dp[i] -= 1
res += 1
print(res)
``` | instruction | 0 | 68,368 | 19 | 136,736 |
No | output | 1 | 68,368 | 19 | 136,737 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Gildong is now developing a puzzle game. The puzzle consists of n platforms numbered from 1 to n. The player plays the game as a character that can stand on each platform and the goal of the game is to move the character from the 1-st platform to the n-th platform.
The i-th platform is labeled with an integer a_i (0 β€ a_i β€ n-i). When the character is standing on the i-th platform, the player can move the character to any of the j-th platforms where i+1 β€ j β€ i+a_i. If the character is on the i-th platform where a_i=0 and i β n, the player loses the game.
Since Gildong thinks the current game is not hard enough, he wants to make it even harder. He wants to change some (possibly zero) labels to 0 so that there remains exactly one way to win. He wants to modify the game as little as possible, so he's asking you to find the minimum number of platforms that should have their labels changed. Two ways are different if and only if there exists a platform the character gets to in one way but not in the other way.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 β€ t β€ 500).
Each test case contains two lines. The first line of each test case consists of an integer n (2 β€ n β€ 3000) β the number of platforms of the game.
The second line of each test case contains n integers. The i-th integer is a_i (0 β€ a_i β€ n-i) β the integer of the i-th platform.
It is guaranteed that:
* For each test case, there is at least one way to win initially.
* The sum of n in all test cases doesn't exceed 3000.
Output
For each test case, print one integer β the minimum number of different labels that should be changed to 0 so that there remains exactly one way to win.
Example
Input
3
4
1 1 1 0
5
4 3 2 1 0
9
4 1 4 2 1 0 2 1 0
Output
0
3
2
Note
In the first case, the player can only move to the next platform until they get to the 4-th platform. Since there is already only one way to win, the answer is zero.
In the second case, Gildong can change a_2, a_3, and a_4 to 0 so that the game becomes 4 0 0 0 0. Now the only way the player can win is to move directly from the 1-st platform to the 5-th platform.
In the third case, Gildong can change a_2 and a_8 to 0, then the only way to win is to move in the following way: 1 β 3 β 7 β 9.
Submitted Solution:
```
for _ in range(int(input())):
n = int(input())
cnt = 0
arr = list(map(int,input().split()))
if(arr[n-1]==0):
cnt += 1
print(len(set(arr))-1-cnt)
``` | instruction | 0 | 68,369 | 19 | 136,738 |
No | output | 1 | 68,369 | 19 | 136,739 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nezzar designs a brand new game "Hidden Permutations" and shares it with his best friend, Nanako.
At the beginning of the game, Nanako and Nezzar both know integers n and m. The game goes in the following way:
* Firstly, Nezzar hides two permutations p_1,p_2,β¦,p_n and q_1,q_2,β¦,q_n of integers from 1 to n, and Nanako secretly selects m unordered pairs (l_1,r_1),(l_2,r_2),β¦,(l_m,r_m);
* After that, Nanako sends his chosen pairs to Nezzar;
* On receiving those m unordered pairs, Nezzar checks if there exists 1 β€ i β€ m, such that (p_{l_i}-p_{r_i}) and (q_{l_i}-q_{r_i}) have different signs. If so, Nezzar instantly loses the game and gets a score of -1. Otherwise, the score Nezzar gets is equal to the number of indices 1 β€ i β€ n such that p_i β q_i.
However, Nezzar accidentally knows Nanako's unordered pairs and decides to take advantage of them. Please help Nezzar find out two permutations p and q such that the score is maximized.
Input
The first line contains a single integer t (1 β€ t β€ 5 β
10^5) β the number of test cases.
The first line of each test case contains two integers n,m (1 β€ n β€ 5 β
10^5, 0 β€ m β€ min((n(n-1))/(2),5 β
10^5)).
Then m lines follow, i-th of them contains two integers l_i,r_i (1 β€ l_i,r_i β€ n, l_i β r_i), describing the i-th unordered pair Nanako chooses. It is guaranteed that all m unordered pairs are distinct.
It is guaranteed that the sum of n for all test cases does not exceed 5 β
10^5, and the sum of m for all test cases does not exceed 5β
10^5.
Output
For each test case, print two permutations p_1,p_2,β¦,p_n and q_1,q_2,β¦,q_n such that the score Nezzar gets is maximized.
Example
Input
3
4 2
1 2
3 4
6 4
1 2
1 3
3 5
3 6
2 1
1 2
Output
1 2 3 4
3 4 1 2
2 3 4 1 6 5
1 4 3 2 5 6
1 2
1 2
Note
For first test case, for each pair given by Nanako:
* for the first pair (1,2): p_1 - p_2 = 1 - 2 = -1, q_1 - q_2 = 3 - 4 = -1, they have the same sign;
* for the second pair (3,4): p_3 - p_4 = 3 - 4 = -1, q_3 - q_4 = 1 - 2 = -1, they have the same sign.
As Nezzar does not lose instantly, Nezzar gains the score of 4 as p_i β q_i for all 1 β€ i β€ 4. Obviously, it is the maximum possible score Nezzar can get.
Submitted Solution:
```
#!/usr/bin/env python
import os
import sys
from io import BytesIO, IOBase
def connected_components(n, graph):
components, visited = [], [False] * n
def dfs(start):
component, stack = [], [start]
while stack:
start = stack[-1]
if visited[start]:
stack.pop()
continue
else:
visited[start] = True
component.append(start)
for i in graph[start]:
if not visited[i]:
stack.append(i)
return component
for i in range(n):
if not visited[i]:
components.append(dfs(i))
return components
def main():
for _ in range(int(input())):
n,m = map(int,input().split())
graph = []
for _ in range(n):
graph.append([])
for _ in range(m):
a,b = map(int,input().split())
graph[a-1].append(b-1)
graph[b-1].append(a-1)
components = connected_components(n, graph)
if len(components) == 1:
print(" ".join(map(str,list(range(1,n + 1)))))
print(" ".join(map(str,list(range(1,n + 1)))))
else:
ans1 = [0] * n
ans2 = [0] * n
curAns1 = 1
curAns2 = 1
for i in range(len(components)):
for j in components[i]:
ans1[j] = curAns1
curAns1 += 1
if i != 0:
for j in components[i]:
ans2[j] = curAns2
curAns2 += 1
for j in components[0]:
ans2[j] = curAns2
curAns2 += 1
print(" ".join(map(str,ans1)))
print(" ".join(map(str,ans2)))
# 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")
# endregion
if __name__ == "__main__":
main()
``` | instruction | 0 | 68,370 | 19 | 136,740 |
No | output | 1 | 68,370 | 19 | 136,741 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nezzar designs a brand new game "Hidden Permutations" and shares it with his best friend, Nanako.
At the beginning of the game, Nanako and Nezzar both know integers n and m. The game goes in the following way:
* Firstly, Nezzar hides two permutations p_1,p_2,β¦,p_n and q_1,q_2,β¦,q_n of integers from 1 to n, and Nanako secretly selects m unordered pairs (l_1,r_1),(l_2,r_2),β¦,(l_m,r_m);
* After that, Nanako sends his chosen pairs to Nezzar;
* On receiving those m unordered pairs, Nezzar checks if there exists 1 β€ i β€ m, such that (p_{l_i}-p_{r_i}) and (q_{l_i}-q_{r_i}) have different signs. If so, Nezzar instantly loses the game and gets a score of -1. Otherwise, the score Nezzar gets is equal to the number of indices 1 β€ i β€ n such that p_i β q_i.
However, Nezzar accidentally knows Nanako's unordered pairs and decides to take advantage of them. Please help Nezzar find out two permutations p and q such that the score is maximized.
Input
The first line contains a single integer t (1 β€ t β€ 5 β
10^5) β the number of test cases.
The first line of each test case contains two integers n,m (1 β€ n β€ 5 β
10^5, 0 β€ m β€ min((n(n-1))/(2),5 β
10^5)).
Then m lines follow, i-th of them contains two integers l_i,r_i (1 β€ l_i,r_i β€ n, l_i β r_i), describing the i-th unordered pair Nanako chooses. It is guaranteed that all m unordered pairs are distinct.
It is guaranteed that the sum of n for all test cases does not exceed 5 β
10^5, and the sum of m for all test cases does not exceed 5β
10^5.
Output
For each test case, print two permutations p_1,p_2,β¦,p_n and q_1,q_2,β¦,q_n such that the score Nezzar gets is maximized.
Example
Input
3
4 2
1 2
3 4
6 4
1 2
1 3
3 5
3 6
2 1
1 2
Output
1 2 3 4
3 4 1 2
2 3 4 1 6 5
1 4 3 2 5 6
1 2
1 2
Note
For first test case, for each pair given by Nanako:
* for the first pair (1,2): p_1 - p_2 = 1 - 2 = -1, q_1 - q_2 = 3 - 4 = -1, they have the same sign;
* for the second pair (3,4): p_3 - p_4 = 3 - 4 = -1, q_3 - q_4 = 1 - 2 = -1, they have the same sign.
As Nezzar does not lose instantly, Nezzar gains the score of 4 as p_i β q_i for all 1 β€ i β€ 4. Obviously, it is the maximum possible score Nezzar can get.
Submitted Solution:
```
#!/usr/bin/env python
import os
import sys
from io import BytesIO, IOBase
def connected_components(n, graph):
components, visited = [], [False] * n
def dfs(start):
component, stack = [], [start]
while stack:
start = stack[-1]
if visited[start]:
stack.pop()
continue
else:
visited[start] = True
component.append(start)
for i in graph[start]:
if not visited[i]:
stack.append(i)
return component
for i in range(n):
if not visited[i]:
components.append(dfs(i))
return components
def main():
for _ in range(int(input())):
n,m = map(int,input().split())
graph = []
for _ in range(n):
graph.append([])
for _ in range(m):
a,b = map(int,input().split())
graph[a-1].append(b-1)
graph[b-1].append(a-1)
components = connected_components(n, graph)
if len(components) == 1:
ans1 = [0] * n
ans2 = [0] * n
curAns = 1
for i in range(n):
if len(graph[i]) == n - 1:
ans1[i] = curAns
ans2[i] = curAns
curAns += 1
graph[i] = []
invComponents = []
invComponentsLoc = [-1] * n
for i in range(n):
if ans1[i] != 0 or invComponentsLoc[i] != -1:
continue
graph[i].append(i)
mexList = [0] * len(graph[i])
for j in graph[i]:
if j < len(graph[i]):
mexList[j] = 1
mex = len(graph[i])
for k in range(len(mexList)):
if mexList[k] == 0:
mex = k
break
if invComponentsLoc[mex] != -1:
invComponents[invComponentsLoc[mex]].append(i)
invComponentsLoc[i] = invComponentsLoc[mex]
else:
invComponents.append([i, mex])
invComponentsLoc[i] = len(invComponents) - 1
invComponentsLoc[mex] = len(invComponents) - 1
for l in range(len(invComponents)):
ans1[invComponents[l][0]] = curAns
ans2[invComponents[l][0]] = curAns + 1
ans1[invComponents[l][1]] = curAns + 1
ans2[invComponents[l][1]] = curAns
curAns += 2
for k in range(2, len(invComponents[l])):
ans1[invComponents[l][k]] = curAns
ans2[invComponents[l][k]] = curAns
if ans1[invComponents[l][k - 1]] == curAns - 1:
ans1[invComponents[l][k]] -= 1
ans1[invComponents[l][k - 1]] += 1
else:
ans2[invComponents[l][k]] -= 1
ans2[invComponents[l][k - 1]] += 1
curAns += 1
print(" ".join(map(str,ans1)))
print(" ".join(map(str,ans2)))
else:
ans1 = [0] * n
ans2 = [0] * n
curAns1 = 1
curAns2 = 1
for i in range(len(components)):
for j in components[i]:
ans1[j] = curAns1
curAns1 += 1
if i != 0:
for j in components[i]:
ans2[j] = curAns2
curAns2 += 1
for j in components[0]:
ans2[j] = curAns2
curAns2 += 1
print(" ".join(map(str,ans1)))
print(" ".join(map(str,ans2)))
# 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")
# endregion
if __name__ == "__main__":
main()
``` | instruction | 0 | 68,371 | 19 | 136,742 |
No | output | 1 | 68,371 | 19 | 136,743 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nezzar designs a brand new game "Hidden Permutations" and shares it with his best friend, Nanako.
At the beginning of the game, Nanako and Nezzar both know integers n and m. The game goes in the following way:
* Firstly, Nezzar hides two permutations p_1,p_2,β¦,p_n and q_1,q_2,β¦,q_n of integers from 1 to n, and Nanako secretly selects m unordered pairs (l_1,r_1),(l_2,r_2),β¦,(l_m,r_m);
* After that, Nanako sends his chosen pairs to Nezzar;
* On receiving those m unordered pairs, Nezzar checks if there exists 1 β€ i β€ m, such that (p_{l_i}-p_{r_i}) and (q_{l_i}-q_{r_i}) have different signs. If so, Nezzar instantly loses the game and gets a score of -1. Otherwise, the score Nezzar gets is equal to the number of indices 1 β€ i β€ n such that p_i β q_i.
However, Nezzar accidentally knows Nanako's unordered pairs and decides to take advantage of them. Please help Nezzar find out two permutations p and q such that the score is maximized.
Input
The first line contains a single integer t (1 β€ t β€ 5 β
10^5) β the number of test cases.
The first line of each test case contains two integers n,m (1 β€ n β€ 5 β
10^5, 0 β€ m β€ min((n(n-1))/(2),5 β
10^5)).
Then m lines follow, i-th of them contains two integers l_i,r_i (1 β€ l_i,r_i β€ n, l_i β r_i), describing the i-th unordered pair Nanako chooses. It is guaranteed that all m unordered pairs are distinct.
It is guaranteed that the sum of n for all test cases does not exceed 5 β
10^5, and the sum of m for all test cases does not exceed 5β
10^5.
Output
For each test case, print two permutations p_1,p_2,β¦,p_n and q_1,q_2,β¦,q_n such that the score Nezzar gets is maximized.
Example
Input
3
4 2
1 2
3 4
6 4
1 2
1 3
3 5
3 6
2 1
1 2
Output
1 2 3 4
3 4 1 2
2 3 4 1 6 5
1 4 3 2 5 6
1 2
1 2
Note
For first test case, for each pair given by Nanako:
* for the first pair (1,2): p_1 - p_2 = 1 - 2 = -1, q_1 - q_2 = 3 - 4 = -1, they have the same sign;
* for the second pair (3,4): p_3 - p_4 = 3 - 4 = -1, q_3 - q_4 = 1 - 2 = -1, they have the same sign.
As Nezzar does not lose instantly, Nezzar gains the score of 4 as p_i β q_i for all 1 β€ i β€ 4. Obviously, it is the maximum possible score Nezzar can get.
Submitted Solution:
```
#!/usr/bin/env python
import os
import sys
from io import BytesIO, IOBase
def connected_components(n, graph):
components, visited = [], [False] * n
def dfs(start):
component, stack = [], [start]
while stack:
start = stack[-1]
if visited[start]:
stack.pop()
continue
else:
visited[start] = True
component.append(start)
for i in graph[start]:
if not visited[i]:
stack.append(i)
return component
for i in range(n):
if not visited[i]:
components.append(dfs(i))
return components
def main():
for _ in range(int(input())):
n,m = map(int,input().split())
graph = []
for _ in range(n):
graph.append([])
for _ in range(m):
a,b = map(int,input().split())
graph[a-1].append(b-1)
graph[b-1].append(a-1)
components = connected_components(n, graph)
if len(components) == 1:
ans1 = [0] * n
ans2 = [0] * n
curAns = 1
for i in range(n):
if len(graph[i]) == n - 1:
ans1[i] = curAns
ans2[i] = curAns
curAns += 1
graph[i] = []
for i in range(n):
newArray = []
for j in graph[i]:
if ans1[j] == 0:
newArray.append(j)
graph[i] = newArray
invComponents = []
invComponentsLoc = [-1] * n
for i in range(n):
if ans1[i] != 0 or invComponentsLoc[i] != -1:
continue
graph[i].append(i)
mexList = [0] * len(graph[i])
for j in graph[i]:
if j < len(graph[i]):
mexList[j] = 1
mex = len(graph[i])
for k in range(len(mexList)):
if mexList[k] == 0:
mex = k
break
if invComponentsLoc[mex] != -1:
invComponents[invComponentsLoc[mex]].append(i)
invComponentsLoc[i] = invComponentsLoc[mex]
else:
invComponents.append([i, mex])
invComponentsLoc[i] = len(invComponents) - 1
invComponentsLoc[mex] = len(invComponents) - 1
for l in range(len(invComponents)):
ans1[invComponents[l][0]] = curAns
ans2[invComponents[l][0]] = curAns + 1
ans1[invComponents[l][1]] = curAns + 1
ans2[invComponents[l][1]] = curAns
curAns += 2
for k in range(2, len(invComponents[l])):
ans1[invComponents[l][k]] = curAns
ans2[invComponents[l][k]] = curAns
if ans1[invComponents[l][k - 1]] == curAns - 1:
ans1[invComponents[l][k]] -= 1
ans1[invComponents[l][k - 1]] += 1
else:
ans2[invComponents[l][k]] -= 1
ans2[invComponents[l][k - 1]] += 1
curAns += 1
print(" ".join(map(str,ans1)))
print(" ".join(map(str,ans2)))
else:
ans1 = [0] * n
ans2 = [0] * n
curAns1 = 1
curAns2 = 1
for i in range(len(components)):
for j in components[i]:
ans1[j] = curAns1
curAns1 += 1
if i != 0:
for j in components[i]:
ans2[j] = curAns2
curAns2 += 1
for j in components[0]:
ans2[j] = curAns2
curAns2 += 1
print(" ".join(map(str,ans1)))
print(" ".join(map(str,ans2)))
# 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")
# endregion
if __name__ == "__main__":
main()
``` | instruction | 0 | 68,372 | 19 | 136,744 |
No | output | 1 | 68,372 | 19 | 136,745 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nezzar designs a brand new game "Hidden Permutations" and shares it with his best friend, Nanako.
At the beginning of the game, Nanako and Nezzar both know integers n and m. The game goes in the following way:
* Firstly, Nezzar hides two permutations p_1,p_2,β¦,p_n and q_1,q_2,β¦,q_n of integers from 1 to n, and Nanako secretly selects m unordered pairs (l_1,r_1),(l_2,r_2),β¦,(l_m,r_m);
* After that, Nanako sends his chosen pairs to Nezzar;
* On receiving those m unordered pairs, Nezzar checks if there exists 1 β€ i β€ m, such that (p_{l_i}-p_{r_i}) and (q_{l_i}-q_{r_i}) have different signs. If so, Nezzar instantly loses the game and gets a score of -1. Otherwise, the score Nezzar gets is equal to the number of indices 1 β€ i β€ n such that p_i β q_i.
However, Nezzar accidentally knows Nanako's unordered pairs and decides to take advantage of them. Please help Nezzar find out two permutations p and q such that the score is maximized.
Input
The first line contains a single integer t (1 β€ t β€ 5 β
10^5) β the number of test cases.
The first line of each test case contains two integers n,m (1 β€ n β€ 5 β
10^5, 0 β€ m β€ min((n(n-1))/(2),5 β
10^5)).
Then m lines follow, i-th of them contains two integers l_i,r_i (1 β€ l_i,r_i β€ n, l_i β r_i), describing the i-th unordered pair Nanako chooses. It is guaranteed that all m unordered pairs are distinct.
It is guaranteed that the sum of n for all test cases does not exceed 5 β
10^5, and the sum of m for all test cases does not exceed 5β
10^5.
Output
For each test case, print two permutations p_1,p_2,β¦,p_n and q_1,q_2,β¦,q_n such that the score Nezzar gets is maximized.
Example
Input
3
4 2
1 2
3 4
6 4
1 2
1 3
3 5
3 6
2 1
1 2
Output
1 2 3 4
3 4 1 2
2 3 4 1 6 5
1 4 3 2 5 6
1 2
1 2
Note
For first test case, for each pair given by Nanako:
* for the first pair (1,2): p_1 - p_2 = 1 - 2 = -1, q_1 - q_2 = 3 - 4 = -1, they have the same sign;
* for the second pair (3,4): p_3 - p_4 = 3 - 4 = -1, q_3 - q_4 = 1 - 2 = -1, they have the same sign.
As Nezzar does not lose instantly, Nezzar gains the score of 4 as p_i β q_i for all 1 β€ i β€ 4. Obviously, it is the maximum possible score Nezzar can get.
Submitted Solution:
```
import bisect
from itertools import accumulate
import os
import sys
import math
from decimal import *
from io import BytesIO, IOBase
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)
def input(): return sys.stdin.readline().rstrip("\r\n")
def isPrime(n) :
if (n <= 1) : return False
if (n <= 3) : return True
if (n % 2 == 0 or n % 3 == 0) : return False
i = 5
while(i * i <= n) :
if (n % i == 0 or n % (i + 2) == 0) :
return False
i = i + 6
return True
def SieveOfEratosthenes(n):
prime=[]
primes = [True for i in range(n+1)]
p = 2
while (p * p <= n):
if (primes[p] == True):
prime.append(p)
for i in range(p * p, n+1, p):
primes[i] = False
p += 1
return prime
def primefactors(n):
fac=[]
while(n%2==0):
fac.append(2)
n=n//2
for i in range(3,int(math.sqrt(n))+2):
while(n%i==0):
fac.append(i)
n=n//i
if n>1:
fac.append(n)
return fac
def factors(n):
fac=set()
fac.add(1)
fac.add(n)
for i in range(2,int(math.sqrt(n))+1):
if n%i==0:
fac.add(i)
fac.add(n//i)
return list(fac)
#-------------------------------------------------------
for _ in range(int(input())):
n,k=map(int,input().split())
a=sorted(list(map(int,input().split())))
g=0
for i in range(1,len(a)):
g=math.gcd(g,a[i]-a[0])
if (k-a[0])%g==0:
print("YES")
else:
print("NO")
``` | instruction | 0 | 68,373 | 19 | 136,746 |
No | output | 1 | 68,373 | 19 | 136,747 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are playing a video game and you have just reached the bonus level, where the only possible goal is to score as many points as possible. Being a perfectionist, you've decided that you won't leave this level until you've gained the maximum possible number of points there.
The bonus level consists of n small platforms placed in a line and numbered from 1 to n from left to right and (n - 1) bridges connecting adjacent platforms. The bridges between the platforms are very fragile, and for each bridge the number of times one can pass this bridge from one of its ends to the other before it collapses forever is known in advance.
The player's actions are as follows. First, he selects one of the platforms to be the starting position for his hero. After that the player can freely move the hero across the platforms moving by the undestroyed bridges. As soon as the hero finds himself on a platform with no undestroyed bridge attached to it, the level is automatically ended. The number of points scored by the player at the end of the level is calculated as the number of transitions made by the hero between the platforms. Note that if the hero started moving by a certain bridge, he has to continue moving in the same direction until he is on a platform.
Find how many points you need to score to be sure that nobody will beat your record, and move to the next level with a quiet heart.
Input
The first line contains a single integer n (2 β€ n β€ 105) β the number of platforms on the bonus level. The second line contains (n - 1) integers ai (1 β€ ai β€ 109, 1 β€ i < n) β the number of transitions from one end to the other that the bridge between platforms i and i + 1 can bear.
Output
Print a single integer β the maximum number of points a player can get on the bonus level.
Please, do not use the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5
2 1 2 1
Output
5
Note
One possibility of getting 5 points in the sample is starting from platform 3 and consequently moving to platforms 4, 3, 2, 1 and 2. After that the only undestroyed bridge is the bridge between platforms 4 and 5, but this bridge is too far from platform 2 where the hero is located now.
Submitted Solution:
```
print("Hello")
print("Hello")
print("Hello")
print("Hello")
print("Hello")
print("Hello")
print("Hello")
``` | instruction | 0 | 68,407 | 19 | 136,814 |
No | output | 1 | 68,407 | 19 | 136,815 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Appleman has n cards. Each card has an uppercase letter written on it. Toastman must choose k cards from Appleman's cards. Then Appleman should give Toastman some coins depending on the chosen cards. Formally, for each Toastman's card i you should calculate how much Toastman's cards have the letter equal to letter on ith, then sum up all these quantities, such a number of coins Appleman should give to Toastman.
Given the description of Appleman's cards. What is the maximum number of coins Toastman can get?
Input
The first line contains two integers n and k (1 β€ k β€ n β€ 105). The next line contains n uppercase letters without spaces β the i-th letter describes the i-th card of the Appleman.
Output
Print a single integer β the answer to the problem.
Examples
Input
15 10
DZFDFZDFDDDDDDF
Output
82
Input
6 4
YJSNPI
Output
4
Note
In the first test example Toastman can choose nine cards with letter D and one additional card with any letter. For each card with D he will get 9 coins and for the additional card he will get 1 coin. | instruction | 0 | 68,495 | 19 | 136,990 |
Tags: greedy
Correct Solution:
```
n,k=map(int,input().split())
m=26
d=[0]*m
c=input()
for e in c:
d[ord(e)-65]+=1
d.sort(reverse=True)
ans=0
res=0
for i in range(m):
if ans>=k:
break
if ans+d[i]<=k:
ans+=d[i]
res+=d[i]*d[i]
else:
res+=(k-ans)*(k-ans)
ans=k
print(res)
``` | output | 1 | 68,495 | 19 | 136,991 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Appleman has n cards. Each card has an uppercase letter written on it. Toastman must choose k cards from Appleman's cards. Then Appleman should give Toastman some coins depending on the chosen cards. Formally, for each Toastman's card i you should calculate how much Toastman's cards have the letter equal to letter on ith, then sum up all these quantities, such a number of coins Appleman should give to Toastman.
Given the description of Appleman's cards. What is the maximum number of coins Toastman can get?
Input
The first line contains two integers n and k (1 β€ k β€ n β€ 105). The next line contains n uppercase letters without spaces β the i-th letter describes the i-th card of the Appleman.
Output
Print a single integer β the answer to the problem.
Examples
Input
15 10
DZFDFZDFDDDDDDF
Output
82
Input
6 4
YJSNPI
Output
4
Note
In the first test example Toastman can choose nine cards with letter D and one additional card with any letter. For each card with D he will get 9 coins and for the additional card he will get 1 coin. | instruction | 0 | 68,496 | 19 | 136,992 |
Tags: greedy
Correct Solution:
```
from collections import Counter
def solve(n,k,ar):
dict = Counter(ar)
dict = sorted(dict.values())
dict.reverse()
ans = []
cnt = 0
for i in range(len(dict)):
if k <= 0: break
if dict[i] <= k:
ans.append(dict[i])
k -= dict[i]
else:
ans.append(k)
k = 0
break
for i in range(len(ans)):
cnt += ans[i]**2
print(cnt)
if __name__ == '__main__':
n,k = map(int,input().split())
ar = list(input())
solve(n,k,ar)
``` | output | 1 | 68,496 | 19 | 136,993 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Appleman has n cards. Each card has an uppercase letter written on it. Toastman must choose k cards from Appleman's cards. Then Appleman should give Toastman some coins depending on the chosen cards. Formally, for each Toastman's card i you should calculate how much Toastman's cards have the letter equal to letter on ith, then sum up all these quantities, such a number of coins Appleman should give to Toastman.
Given the description of Appleman's cards. What is the maximum number of coins Toastman can get?
Input
The first line contains two integers n and k (1 β€ k β€ n β€ 105). The next line contains n uppercase letters without spaces β the i-th letter describes the i-th card of the Appleman.
Output
Print a single integer β the answer to the problem.
Examples
Input
15 10
DZFDFZDFDDDDDDF
Output
82
Input
6 4
YJSNPI
Output
4
Note
In the first test example Toastman can choose nine cards with letter D and one additional card with any letter. For each card with D he will get 9 coins and for the additional card he will get 1 coin. | instruction | 0 | 68,497 | 19 | 136,994 |
Tags: greedy
Correct Solution:
```
##__________________________________________________________________
##
## Author: GogolGrind
##__________________________________________________________________
from sys import *
from math import *
def main ():
n,k = list(map(int,input().split()))
s = input()
l = [0]*(26)
for c in s:
f = lambda x: ord(x)-ord('A')
l[f(c)] += 1
l = sorted(l, reverse = True)
ans = 0
for i in l:
count = min(k,i)
ans += count*count
k -= count
if k == 0:
break
print(ans)
if __name__ == '__main__':
main()
``` | output | 1 | 68,497 | 19 | 136,995 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Appleman has n cards. Each card has an uppercase letter written on it. Toastman must choose k cards from Appleman's cards. Then Appleman should give Toastman some coins depending on the chosen cards. Formally, for each Toastman's card i you should calculate how much Toastman's cards have the letter equal to letter on ith, then sum up all these quantities, such a number of coins Appleman should give to Toastman.
Given the description of Appleman's cards. What is the maximum number of coins Toastman can get?
Input
The first line contains two integers n and k (1 β€ k β€ n β€ 105). The next line contains n uppercase letters without spaces β the i-th letter describes the i-th card of the Appleman.
Output
Print a single integer β the answer to the problem.
Examples
Input
15 10
DZFDFZDFDDDDDDF
Output
82
Input
6 4
YJSNPI
Output
4
Note
In the first test example Toastman can choose nine cards with letter D and one additional card with any letter. For each card with D he will get 9 coins and for the additional card he will get 1 coin. | instruction | 0 | 68,498 | 19 | 136,996 |
Tags: greedy
Correct Solution:
```
n,k = map(int,input().split())
d,r = {},0
for c in input(): d[c] = d.get(c, 0)+1
for v in sorted(d.values())[::-1]:
x = min(k,v)
r += x*x
k -= x
print(r)
# Made By Mostafa_Khaled
``` | output | 1 | 68,498 | 19 | 136,997 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Appleman has n cards. Each card has an uppercase letter written on it. Toastman must choose k cards from Appleman's cards. Then Appleman should give Toastman some coins depending on the chosen cards. Formally, for each Toastman's card i you should calculate how much Toastman's cards have the letter equal to letter on ith, then sum up all these quantities, such a number of coins Appleman should give to Toastman.
Given the description of Appleman's cards. What is the maximum number of coins Toastman can get?
Input
The first line contains two integers n and k (1 β€ k β€ n β€ 105). The next line contains n uppercase letters without spaces β the i-th letter describes the i-th card of the Appleman.
Output
Print a single integer β the answer to the problem.
Examples
Input
15 10
DZFDFZDFDDDDDDF
Output
82
Input
6 4
YJSNPI
Output
4
Note
In the first test example Toastman can choose nine cards with letter D and one additional card with any letter. For each card with D he will get 9 coins and for the additional card he will get 1 coin. | instruction | 0 | 68,499 | 19 | 136,998 |
Tags: greedy
Correct Solution:
```
a= input().split(" ")
#print(a)
n=int(a[0])
k=int(a[1])
#print(n,k)
q=input()
w=set(q)
e=list(w)
l=len(e)
i=0
t=[]
#print('q',q)
#print('e',e)
while i<l:
r=e[i]
y=q.count(r)
t.append(y)
i+=1
#print('t',t)
t.sort()
t.reverse()
#print(t)
c=list(set(t))
c.sort()
c.reverse()
#print(c)
v=len(t)
b=0
m=0
bn=0
while b<v:
nj=t[b]
#print('.')
#print('nj',nj)
if nj<k:
#print('yes')
m=m+nj*nj
k=k-nj
#print(k,m)
elif k==0:
#print('ll')
break
elif k<nj:
#print('kk')
m=m+k*k
break
elif k==nj:
m=m+nj*nj
k=k-nj
b+=1
print(m)
##x=k-z
##print(z*z+x)
``` | output | 1 | 68,499 | 19 | 136,999 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Appleman has n cards. Each card has an uppercase letter written on it. Toastman must choose k cards from Appleman's cards. Then Appleman should give Toastman some coins depending on the chosen cards. Formally, for each Toastman's card i you should calculate how much Toastman's cards have the letter equal to letter on ith, then sum up all these quantities, such a number of coins Appleman should give to Toastman.
Given the description of Appleman's cards. What is the maximum number of coins Toastman can get?
Input
The first line contains two integers n and k (1 β€ k β€ n β€ 105). The next line contains n uppercase letters without spaces β the i-th letter describes the i-th card of the Appleman.
Output
Print a single integer β the answer to the problem.
Examples
Input
15 10
DZFDFZDFDDDDDDF
Output
82
Input
6 4
YJSNPI
Output
4
Note
In the first test example Toastman can choose nine cards with letter D and one additional card with any letter. For each card with D he will get 9 coins and for the additional card he will get 1 coin. | instruction | 0 | 68,500 | 19 | 137,000 |
Tags: greedy
Correct Solution:
```
# -*- coding: utf-8 -*-
"""
Created on Sun Jul 26 03:54:17 2020
@author: Samiul2651
"""
x = input().split(" ")
a = int(x[0])
b = int(x[1])
n = input()
list_1 = list(n)
list_1 = sorted(list_1)
#print(list_1)
i = 0
length_1 = len(list_1)
list_2 = []
while i < length_1:
g = list_1[i]
h = list_2.count(g)
if h == 0:
list_2.append(g)
i += 1
#print(list_2)
length_2 = len(list_2)
list_3 = []
i = 0
while i < length_2:
g = list_2[i]
h = list_1.count(g)
list_3.append(h)
i += 1
list_3 = sorted(list_3)
list_3.reverse()
#print(list_3)
coins = 0
i = 0
while i < len(list_3):
#print(coins)
j = list_3[i]
if b >= j:
b = b - j
coins += j*j
else:
l = b
b = b - l
coins += l*l
i += 1
if b <= 0:
break
print(coins)
``` | output | 1 | 68,500 | 19 | 137,001 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Appleman has n cards. Each card has an uppercase letter written on it. Toastman must choose k cards from Appleman's cards. Then Appleman should give Toastman some coins depending on the chosen cards. Formally, for each Toastman's card i you should calculate how much Toastman's cards have the letter equal to letter on ith, then sum up all these quantities, such a number of coins Appleman should give to Toastman.
Given the description of Appleman's cards. What is the maximum number of coins Toastman can get?
Input
The first line contains two integers n and k (1 β€ k β€ n β€ 105). The next line contains n uppercase letters without spaces β the i-th letter describes the i-th card of the Appleman.
Output
Print a single integer β the answer to the problem.
Examples
Input
15 10
DZFDFZDFDDDDDDF
Output
82
Input
6 4
YJSNPI
Output
4
Note
In the first test example Toastman can choose nine cards with letter D and one additional card with any letter. For each card with D he will get 9 coins and for the additional card he will get 1 coin. | instruction | 0 | 68,501 | 19 | 137,002 |
Tags: greedy
Correct Solution:
```
from sys import stdin
A = list(map(int,stdin.readline().split()))
N = A[0]
K = A[1]
B = stdin.readline()
D=dict()
S=set()
T=list()
for t in range(0,len(B)-1):
l=len(S)
S.add(B[t])
if l==len(S):
D[B[t]]+=1
T.remove(D[B[t]]-1)
T.append(D[B[t]])
else:
D[B[t]]=1
T.append(1)
T.sort(reverse=True)
count=0
ans=0
while K>0:
ans+=(min(K,T[count])**2)
K-=T[count]
count+=1
print(ans)
``` | output | 1 | 68,501 | 19 | 137,003 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Appleman has n cards. Each card has an uppercase letter written on it. Toastman must choose k cards from Appleman's cards. Then Appleman should give Toastman some coins depending on the chosen cards. Formally, for each Toastman's card i you should calculate how much Toastman's cards have the letter equal to letter on ith, then sum up all these quantities, such a number of coins Appleman should give to Toastman.
Given the description of Appleman's cards. What is the maximum number of coins Toastman can get?
Input
The first line contains two integers n and k (1 β€ k β€ n β€ 105). The next line contains n uppercase letters without spaces β the i-th letter describes the i-th card of the Appleman.
Output
Print a single integer β the answer to the problem.
Examples
Input
15 10
DZFDFZDFDDDDDDF
Output
82
Input
6 4
YJSNPI
Output
4
Note
In the first test example Toastman can choose nine cards with letter D and one additional card with any letter. For each card with D he will get 9 coins and for the additional card he will get 1 coin. | instruction | 0 | 68,502 | 19 | 137,004 |
Tags: greedy
Correct Solution:
```
n, k = list(map(int ,input().split()))
s = input()
d = {}
for char in s:
d[char] = d.get(char, 0) + 1
L = sorted([[key, d[key]] for key in d], key=lambda x:x[1], reverse=True)
ans, i = 0, 0
while(k > 0):
if k >= L[i][1]:
ans += L[i][1] ** 2
k -= L[i][1]
else:
ans += k ** 2
k = 0
i += 1
print(ans)
``` | output | 1 | 68,502 | 19 | 137,005 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Appleman has n cards. Each card has an uppercase letter written on it. Toastman must choose k cards from Appleman's cards. Then Appleman should give Toastman some coins depending on the chosen cards. Formally, for each Toastman's card i you should calculate how much Toastman's cards have the letter equal to letter on ith, then sum up all these quantities, such a number of coins Appleman should give to Toastman.
Given the description of Appleman's cards. What is the maximum number of coins Toastman can get?
Input
The first line contains two integers n and k (1 β€ k β€ n β€ 105). The next line contains n uppercase letters without spaces β the i-th letter describes the i-th card of the Appleman.
Output
Print a single integer β the answer to the problem.
Examples
Input
15 10
DZFDFZDFDDDDDDF
Output
82
Input
6 4
YJSNPI
Output
4
Note
In the first test example Toastman can choose nine cards with letter D and one additional card with any letter. For each card with D he will get 9 coins and for the additional card he will get 1 coin.
Submitted Solution:
```
n,k=[int(x) for x in input().split()]
s=input()
dict={}
ans=[]
for i in range(n):
if s[i] not in dict:
dict[s[i]]=1
else:
dict[s[i]]+=1
for key in dict:
ans.append(dict[key])
ans.sort()
res=0
while True:
val=ans.pop()
if k>val:
k-=val
res+=val**2
else:
res+=k**2
break
print(res)
``` | instruction | 0 | 68,503 | 19 | 137,006 |
Yes | output | 1 | 68,503 | 19 | 137,007 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Appleman has n cards. Each card has an uppercase letter written on it. Toastman must choose k cards from Appleman's cards. Then Appleman should give Toastman some coins depending on the chosen cards. Formally, for each Toastman's card i you should calculate how much Toastman's cards have the letter equal to letter on ith, then sum up all these quantities, such a number of coins Appleman should give to Toastman.
Given the description of Appleman's cards. What is the maximum number of coins Toastman can get?
Input
The first line contains two integers n and k (1 β€ k β€ n β€ 105). The next line contains n uppercase letters without spaces β the i-th letter describes the i-th card of the Appleman.
Output
Print a single integer β the answer to the problem.
Examples
Input
15 10
DZFDFZDFDDDDDDF
Output
82
Input
6 4
YJSNPI
Output
4
Note
In the first test example Toastman can choose nine cards with letter D and one additional card with any letter. For each card with D he will get 9 coins and for the additional card he will get 1 coin.
Submitted Solution:
```
if __name__ == '__main__':
n,k = input().split()
n =int(n)
k = int(k)
arr = input()
arr = list(arr)
count_arr = []
check_arr = []
for i in range(0,n):
if(arr[i] not in check_arr):
count_arr.append(arr.count(arr[i]))
check_arr.append(arr[i])
# print(check_arr)
# print(count_arr)
i =0
tally = 0
while(i<k):
m = max(count_arr)
count_arr.remove(m)
if(i<k):
if((i+m)<k):
i =i + m
tally+=m*m
else:
tally+=(k-i)*(k-i)
break
else:
break
print(tally)
# print(check_arr)
# print(count_arr)
``` | instruction | 0 | 68,505 | 19 | 137,010 |
Yes | output | 1 | 68,505 | 19 | 137,011 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Appleman has n cards. Each card has an uppercase letter written on it. Toastman must choose k cards from Appleman's cards. Then Appleman should give Toastman some coins depending on the chosen cards. Formally, for each Toastman's card i you should calculate how much Toastman's cards have the letter equal to letter on ith, then sum up all these quantities, such a number of coins Appleman should give to Toastman.
Given the description of Appleman's cards. What is the maximum number of coins Toastman can get?
Input
The first line contains two integers n and k (1 β€ k β€ n β€ 105). The next line contains n uppercase letters without spaces β the i-th letter describes the i-th card of the Appleman.
Output
Print a single integer β the answer to the problem.
Examples
Input
15 10
DZFDFZDFDDDDDDF
Output
82
Input
6 4
YJSNPI
Output
4
Note
In the first test example Toastman can choose nine cards with letter D and one additional card with any letter. For each card with D he will get 9 coins and for the additional card he will get 1 coin.
Submitted Solution:
```
n,k=map(int,input().split())
s=input()
s=list(s)
dic={}
d=[]
ans=0
for i in range(n):
if s[i] not in dic:
dic[s[i]]=1
else:
dic[s[i]]+=1
for i in s:
d.append((dic[i],i))
d=set(d)
d=list(d)
d.sort(reverse=True)
t=0
for i in d:
if i[0]>k-t:
ans=ans+(k-t)**2
break
else:
ans=ans+i[0]**2
t+=i[0]
print(ans)
``` | instruction | 0 | 68,506 | 19 | 137,012 |
Yes | output | 1 | 68,506 | 19 | 137,013 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Appleman has n cards. Each card has an uppercase letter written on it. Toastman must choose k cards from Appleman's cards. Then Appleman should give Toastman some coins depending on the chosen cards. Formally, for each Toastman's card i you should calculate how much Toastman's cards have the letter equal to letter on ith, then sum up all these quantities, such a number of coins Appleman should give to Toastman.
Given the description of Appleman's cards. What is the maximum number of coins Toastman can get?
Input
The first line contains two integers n and k (1 β€ k β€ n β€ 105). The next line contains n uppercase letters without spaces β the i-th letter describes the i-th card of the Appleman.
Output
Print a single integer β the answer to the problem.
Examples
Input
15 10
DZFDFZDFDDDDDDF
Output
82
Input
6 4
YJSNPI
Output
4
Note
In the first test example Toastman can choose nine cards with letter D and one additional card with any letter. For each card with D he will get 9 coins and for the additional card he will get 1 coin.
Submitted Solution:
```
n, k = input().split()
n, k = int(n), int(k)
cards = list(input())
cards.sort()
combo = 1
for i in range(1, k):
if cards[i] == cards[i - 1]:
combo += 1
lfto = k - combo
print(combo*combo + lfto)
``` | instruction | 0 | 68,507 | 19 | 137,014 |
No | output | 1 | 68,507 | 19 | 137,015 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Appleman has n cards. Each card has an uppercase letter written on it. Toastman must choose k cards from Appleman's cards. Then Appleman should give Toastman some coins depending on the chosen cards. Formally, for each Toastman's card i you should calculate how much Toastman's cards have the letter equal to letter on ith, then sum up all these quantities, such a number of coins Appleman should give to Toastman.
Given the description of Appleman's cards. What is the maximum number of coins Toastman can get?
Input
The first line contains two integers n and k (1 β€ k β€ n β€ 105). The next line contains n uppercase letters without spaces β the i-th letter describes the i-th card of the Appleman.
Output
Print a single integer β the answer to the problem.
Examples
Input
15 10
DZFDFZDFDDDDDDF
Output
82
Input
6 4
YJSNPI
Output
4
Note
In the first test example Toastman can choose nine cards with letter D and one additional card with any letter. For each card with D he will get 9 coins and for the additional card he will get 1 coin.
Submitted Solution:
```
import collections
n, k = map(int, input().split())
a = input()
counter = dict(collections.Counter(a))
counter = sorted(list(counter.values()), reverse=True)
ans, i = 0, 0
while i < k and i + counter[0] < k:
ans += counter[0]*counter[0]
i += counter[0]
k -= i
if k != i and len(counter) > 0:
i -= k
ans += i*i
print(ans)
``` | instruction | 0 | 68,508 | 19 | 137,016 |
No | output | 1 | 68,508 | 19 | 137,017 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Appleman has n cards. Each card has an uppercase letter written on it. Toastman must choose k cards from Appleman's cards. Then Appleman should give Toastman some coins depending on the chosen cards. Formally, for each Toastman's card i you should calculate how much Toastman's cards have the letter equal to letter on ith, then sum up all these quantities, such a number of coins Appleman should give to Toastman.
Given the description of Appleman's cards. What is the maximum number of coins Toastman can get?
Input
The first line contains two integers n and k (1 β€ k β€ n β€ 105). The next line contains n uppercase letters without spaces β the i-th letter describes the i-th card of the Appleman.
Output
Print a single integer β the answer to the problem.
Examples
Input
15 10
DZFDFZDFDDDDDDF
Output
82
Input
6 4
YJSNPI
Output
4
Note
In the first test example Toastman can choose nine cards with letter D and one additional card with any letter. For each card with D he will get 9 coins and for the additional card he will get 1 coin.
Submitted Solution:
```
n, k = map(int, input().split())
s = input()
A = [0 for i in range(26)]
for i in s:
A[ord(i) - 65] += 1
A.sort()
i = 25
taked = 0
ans = 0
tk = k
for j in range(k):
A[i] -= 1
taked += 1
if A[i] == 0:
i -= 1
ans += taked * taked
tk -= taked
taked = 0
if tk == 1:
ans += 1
print(ans)
``` | instruction | 0 | 68,509 | 19 | 137,018 |
No | output | 1 | 68,509 | 19 | 137,019 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Appleman has n cards. Each card has an uppercase letter written on it. Toastman must choose k cards from Appleman's cards. Then Appleman should give Toastman some coins depending on the chosen cards. Formally, for each Toastman's card i you should calculate how much Toastman's cards have the letter equal to letter on ith, then sum up all these quantities, such a number of coins Appleman should give to Toastman.
Given the description of Appleman's cards. What is the maximum number of coins Toastman can get?
Input
The first line contains two integers n and k (1 β€ k β€ n β€ 105). The next line contains n uppercase letters without spaces β the i-th letter describes the i-th card of the Appleman.
Output
Print a single integer β the answer to the problem.
Examples
Input
15 10
DZFDFZDFDDDDDDF
Output
82
Input
6 4
YJSNPI
Output
4
Note
In the first test example Toastman can choose nine cards with letter D and one additional card with any letter. For each card with D he will get 9 coins and for the additional card he will get 1 coin.
Submitted Solution:
```
a= input().split(" ")
#print(a)
n=int(a[0])
k=int(a[1])
#print(n,k)
q=input()
w=set(q)
e=list(w)
l=len(e)
i=0
t=[]
#print('q',q)
#print('e',e)
while i<l:
r=e[i]
y=q.count(r)
t.append(y)
i+=1
#print('t',t)
c=list(set(t))
c.sort()
c.reverse()
#print(c)
v=len(c)
b=0
m=0
bn=0
if v>1:
while b<v:
n1=c[b]
m=m+n1*n1
#print('n1',n1)
#print('k',k)
if k>=n1:
k=k-n1
elif k==0:
break
else:
bn=k
break
b+=1
else:
m=k
print(m+bn)
##x=k-z
##print(z*z+x)
``` | instruction | 0 | 68,510 | 19 | 137,020 |
No | output | 1 | 68,510 | 19 | 137,021 |
Provide tags and a correct Python 3 solution for this coding contest problem.
n people are standing in a line to play table tennis. At first, the first two players in the line play a game. Then the loser goes to the end of the line, and the winner plays with the next person from the line, and so on. They play until someone wins k games in a row. This player becomes the winner.
For each of the participants, you know the power to play table tennis, and for all players these values are different. In a game the player with greater power always wins. Determine who will be the winner.
Input
The first line contains two integers: n and k (2 β€ n β€ 500, 2 β€ k β€ 1012) β the number of people and the number of wins.
The second line contains n integers a1, a2, ..., an (1 β€ ai β€ n) β powers of the player. It's guaranteed that this line contains a valid permutation, i.e. all ai are distinct.
Output
Output a single integer β power of the winner.
Examples
Input
2 2
1 2
Output
2
Input
4 2
3 1 2 4
Output
3
Input
6 2
6 5 3 1 2 4
Output
6
Input
2 10000000000
2 1
Output
2
Note
Games in the second sample:
3 plays with 1. 3 wins. 1 goes to the end of the line.
3 plays with 2. 3 wins. He wins twice in a row. He becomes the winner. | instruction | 0 | 69,540 | 19 | 139,080 |
Tags: data structures, implementation
Correct Solution:
```
inp = list(map(int, input().split()))
players = list(map(int, input().split()))
n = inp[0]
k = inp[1]
if (n == 2):
print(max(players[0], players[1]))
else:
flag = False
indice_prox = 2
maior = max(players)
num_vitorias = 0
prev_vencedor = max(players[0], players[1])
num_vitorias += 1
if(prev_vencedor == maior):
print(prev_vencedor)
else:
while(num_vitorias < k):
prox = players[indice_prox]
if(prox == maior):
flag = True
print(maior)
break
vencedor = max(prev_vencedor, prox)
if(vencedor == prev_vencedor):
num_vitorias += 1
else:
num_vitorias = 1
prev_vencedor = vencedor
indice_prox += 1
if(flag == False):
print(prev_vencedor)
``` | output | 1 | 69,540 | 19 | 139,081 |
Provide tags and a correct Python 3 solution for this coding contest problem.
n people are standing in a line to play table tennis. At first, the first two players in the line play a game. Then the loser goes to the end of the line, and the winner plays with the next person from the line, and so on. They play until someone wins k games in a row. This player becomes the winner.
For each of the participants, you know the power to play table tennis, and for all players these values are different. In a game the player with greater power always wins. Determine who will be the winner.
Input
The first line contains two integers: n and k (2 β€ n β€ 500, 2 β€ k β€ 1012) β the number of people and the number of wins.
The second line contains n integers a1, a2, ..., an (1 β€ ai β€ n) β powers of the player. It's guaranteed that this line contains a valid permutation, i.e. all ai are distinct.
Output
Output a single integer β power of the winner.
Examples
Input
2 2
1 2
Output
2
Input
4 2
3 1 2 4
Output
3
Input
6 2
6 5 3 1 2 4
Output
6
Input
2 10000000000
2 1
Output
2
Note
Games in the second sample:
3 plays with 1. 3 wins. 1 goes to the end of the line.
3 plays with 2. 3 wins. He wins twice in a row. He becomes the winner. | instruction | 0 | 69,541 | 19 | 139,082 |
Tags: data structures, implementation
Correct Solution:
```
n, k = map(int, input().split())
a = list(map(int, input().split()))
high = max(a.pop(0), a.pop(0))
try:
if n == 2:
raise IndexError
while True:
for i in range(k-1):
if a[i] > high:
high = a[i]
a = a[i+1::]
break
elif i+2 == k:
raise IndexError
except IndexError:
print(high)
``` | output | 1 | 69,541 | 19 | 139,083 |
Provide tags and a correct Python 3 solution for this coding contest problem.
n people are standing in a line to play table tennis. At first, the first two players in the line play a game. Then the loser goes to the end of the line, and the winner plays with the next person from the line, and so on. They play until someone wins k games in a row. This player becomes the winner.
For each of the participants, you know the power to play table tennis, and for all players these values are different. In a game the player with greater power always wins. Determine who will be the winner.
Input
The first line contains two integers: n and k (2 β€ n β€ 500, 2 β€ k β€ 1012) β the number of people and the number of wins.
The second line contains n integers a1, a2, ..., an (1 β€ ai β€ n) β powers of the player. It's guaranteed that this line contains a valid permutation, i.e. all ai are distinct.
Output
Output a single integer β power of the winner.
Examples
Input
2 2
1 2
Output
2
Input
4 2
3 1 2 4
Output
3
Input
6 2
6 5 3 1 2 4
Output
6
Input
2 10000000000
2 1
Output
2
Note
Games in the second sample:
3 plays with 1. 3 wins. 1 goes to the end of the line.
3 plays with 2. 3 wins. He wins twice in a row. He becomes the winner. | instruction | 0 | 69,542 | 19 | 139,084 |
Tags: data structures, implementation
Correct Solution:
```
n,k=map(int,input().split())
a=list(map(int,input().split()))
if k>=n:
print(max(a))
else:
for i in range(k-1):
a.append(a[i])
if max(a[0:k+1])==a[0]:
print(a[0])
else:
i=a.index(max(a[0:k+1]))
while True:
ma=max(a[i:i+k])
if ma==a[i]:
print(ma)
break
else:
i=a.index(ma)
``` | output | 1 | 69,542 | 19 | 139,085 |
Provide tags and a correct Python 3 solution for this coding contest problem.
n people are standing in a line to play table tennis. At first, the first two players in the line play a game. Then the loser goes to the end of the line, and the winner plays with the next person from the line, and so on. They play until someone wins k games in a row. This player becomes the winner.
For each of the participants, you know the power to play table tennis, and for all players these values are different. In a game the player with greater power always wins. Determine who will be the winner.
Input
The first line contains two integers: n and k (2 β€ n β€ 500, 2 β€ k β€ 1012) β the number of people and the number of wins.
The second line contains n integers a1, a2, ..., an (1 β€ ai β€ n) β powers of the player. It's guaranteed that this line contains a valid permutation, i.e. all ai are distinct.
Output
Output a single integer β power of the winner.
Examples
Input
2 2
1 2
Output
2
Input
4 2
3 1 2 4
Output
3
Input
6 2
6 5 3 1 2 4
Output
6
Input
2 10000000000
2 1
Output
2
Note
Games in the second sample:
3 plays with 1. 3 wins. 1 goes to the end of the line.
3 plays with 2. 3 wins. He wins twice in a row. He becomes the winner. | instruction | 0 | 69,543 | 19 | 139,086 |
Tags: data structures, implementation
Correct Solution:
```
import io, os
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
n, k = map(int, input().split())
a = list(map(int, input().split()))
t = a[0]
temp = 0
if k >= n-1:
print(max(a))
else:
while temp != k:
x, y = a[0], a[1]
if x > y:
a.append(y)
del a[1]
else:
a.append(x)
del a[0]
if t == a[0]:
temp += 1
else:
t = a[0]
temp = 1
print(a[0])
``` | output | 1 | 69,543 | 19 | 139,087 |
Provide tags and a correct Python 3 solution for this coding contest problem.
n people are standing in a line to play table tennis. At first, the first two players in the line play a game. Then the loser goes to the end of the line, and the winner plays with the next person from the line, and so on. They play until someone wins k games in a row. This player becomes the winner.
For each of the participants, you know the power to play table tennis, and for all players these values are different. In a game the player with greater power always wins. Determine who will be the winner.
Input
The first line contains two integers: n and k (2 β€ n β€ 500, 2 β€ k β€ 1012) β the number of people and the number of wins.
The second line contains n integers a1, a2, ..., an (1 β€ ai β€ n) β powers of the player. It's guaranteed that this line contains a valid permutation, i.e. all ai are distinct.
Output
Output a single integer β power of the winner.
Examples
Input
2 2
1 2
Output
2
Input
4 2
3 1 2 4
Output
3
Input
6 2
6 5 3 1 2 4
Output
6
Input
2 10000000000
2 1
Output
2
Note
Games in the second sample:
3 plays with 1. 3 wins. 1 goes to the end of the line.
3 plays with 2. 3 wins. He wins twice in a row. He becomes the winner. | instruction | 0 | 69,544 | 19 | 139,088 |
Tags: data structures, implementation
Correct Solution:
```
IL = lambda: list(map(int, input().split()))
I = lambda: int(input())
n, k = IL()
a = IL()
ans = 0
score = 0
for i in range(n-1):
if a[0] > a[1]:
score += 1
else:
score = 1
if score == k:
ans = a[0]
break
p1, p2 = a[:2]
a.pop(0)
a[0] = max(p1, p2)
a.append(min(p1, p2))
if ans==0:
ans = max(a)
print(ans)
``` | output | 1 | 69,544 | 19 | 139,089 |
Provide tags and a correct Python 3 solution for this coding contest problem.
n people are standing in a line to play table tennis. At first, the first two players in the line play a game. Then the loser goes to the end of the line, and the winner plays with the next person from the line, and so on. They play until someone wins k games in a row. This player becomes the winner.
For each of the participants, you know the power to play table tennis, and for all players these values are different. In a game the player with greater power always wins. Determine who will be the winner.
Input
The first line contains two integers: n and k (2 β€ n β€ 500, 2 β€ k β€ 1012) β the number of people and the number of wins.
The second line contains n integers a1, a2, ..., an (1 β€ ai β€ n) β powers of the player. It's guaranteed that this line contains a valid permutation, i.e. all ai are distinct.
Output
Output a single integer β power of the winner.
Examples
Input
2 2
1 2
Output
2
Input
4 2
3 1 2 4
Output
3
Input
6 2
6 5 3 1 2 4
Output
6
Input
2 10000000000
2 1
Output
2
Note
Games in the second sample:
3 plays with 1. 3 wins. 1 goes to the end of the line.
3 plays with 2. 3 wins. He wins twice in a row. He becomes the winner. | instruction | 0 | 69,545 | 19 | 139,090 |
Tags: data structures, implementation
Correct Solution:
```
n,k=map(int,input().split())
l=list(map(int,input().split()))
if k<=len(l):
pos = l.index(max(l))
h= l[:pos]
m=0
i=0
j=1
while i<pos and j<pos and m<k:
if h[i]>h[j]:
m+=1
j+=1
else :
i=j
j+=1
m=1
if m>=k:
print(h[i])
else :
print(max(l))
else :
print(max(l))
``` | output | 1 | 69,545 | 19 | 139,091 |
Provide tags and a correct Python 3 solution for this coding contest problem.
n people are standing in a line to play table tennis. At first, the first two players in the line play a game. Then the loser goes to the end of the line, and the winner plays with the next person from the line, and so on. They play until someone wins k games in a row. This player becomes the winner.
For each of the participants, you know the power to play table tennis, and for all players these values are different. In a game the player with greater power always wins. Determine who will be the winner.
Input
The first line contains two integers: n and k (2 β€ n β€ 500, 2 β€ k β€ 1012) β the number of people and the number of wins.
The second line contains n integers a1, a2, ..., an (1 β€ ai β€ n) β powers of the player. It's guaranteed that this line contains a valid permutation, i.e. all ai are distinct.
Output
Output a single integer β power of the winner.
Examples
Input
2 2
1 2
Output
2
Input
4 2
3 1 2 4
Output
3
Input
6 2
6 5 3 1 2 4
Output
6
Input
2 10000000000
2 1
Output
2
Note
Games in the second sample:
3 plays with 1. 3 wins. 1 goes to the end of the line.
3 plays with 2. 3 wins. He wins twice in a row. He becomes the winner. | instruction | 0 | 69,546 | 19 | 139,092 |
Tags: data structures, implementation
Correct Solution:
```
n,k=map(int,input().split())
t=list(map(int,input().split()))
j=0
maxx=0
a=1
for i in t:
if i>maxx:
maxx=i
h=1
if a==1:
h=h-1
a=0
else:
h+=1
if h==k:
break
print(maxx)
``` | output | 1 | 69,546 | 19 | 139,093 |
Provide tags and a correct Python 3 solution for this coding contest problem.
n people are standing in a line to play table tennis. At first, the first two players in the line play a game. Then the loser goes to the end of the line, and the winner plays with the next person from the line, and so on. They play until someone wins k games in a row. This player becomes the winner.
For each of the participants, you know the power to play table tennis, and for all players these values are different. In a game the player with greater power always wins. Determine who will be the winner.
Input
The first line contains two integers: n and k (2 β€ n β€ 500, 2 β€ k β€ 1012) β the number of people and the number of wins.
The second line contains n integers a1, a2, ..., an (1 β€ ai β€ n) β powers of the player. It's guaranteed that this line contains a valid permutation, i.e. all ai are distinct.
Output
Output a single integer β power of the winner.
Examples
Input
2 2
1 2
Output
2
Input
4 2
3 1 2 4
Output
3
Input
6 2
6 5 3 1 2 4
Output
6
Input
2 10000000000
2 1
Output
2
Note
Games in the second sample:
3 plays with 1. 3 wins. 1 goes to the end of the line.
3 plays with 2. 3 wins. He wins twice in a row. He becomes the winner. | instruction | 0 | 69,547 | 19 | 139,094 |
Tags: data structures, implementation
Correct Solution:
```
from sys import stdin,stdout
n,k = map(int,stdin.readline().split())
p = list(map(int,stdin.readline().split()))
if k >= (n-1):
stdout.write(str(max(p)))
else:
n1 = p[0]
del p[0]
c = 0
for item in p:
if item > n1:
n1 = item
c = 1
else:
c = c+1
if c == k:
break
stdout.write(str(n1))
``` | output | 1 | 69,547 | 19 | 139,095 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
n people are standing in a line to play table tennis. At first, the first two players in the line play a game. Then the loser goes to the end of the line, and the winner plays with the next person from the line, and so on. They play until someone wins k games in a row. This player becomes the winner.
For each of the participants, you know the power to play table tennis, and for all players these values are different. In a game the player with greater power always wins. Determine who will be the winner.
Input
The first line contains two integers: n and k (2 β€ n β€ 500, 2 β€ k β€ 1012) β the number of people and the number of wins.
The second line contains n integers a1, a2, ..., an (1 β€ ai β€ n) β powers of the player. It's guaranteed that this line contains a valid permutation, i.e. all ai are distinct.
Output
Output a single integer β power of the winner.
Examples
Input
2 2
1 2
Output
2
Input
4 2
3 1 2 4
Output
3
Input
6 2
6 5 3 1 2 4
Output
6
Input
2 10000000000
2 1
Output
2
Note
Games in the second sample:
3 plays with 1. 3 wins. 1 goes to the end of the line.
3 plays with 2. 3 wins. He wins twice in a row. He becomes the winner.
Submitted Solution:
```
from math import*
n, k = [int(i) for i in input().split()]
gg = [int(i) for i in input().split()]
cnt = 0
comper = gg[0]
for i in range(1, len(gg)):
if cnt >= k:
print(comper)
break
if comper > gg[i]:
cnt += 1
else:
comper = gg[i]
cnt = 1
else:
print(max(gg))
``` | instruction | 0 | 69,548 | 19 | 139,096 |
Yes | output | 1 | 69,548 | 19 | 139,097 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
n people are standing in a line to play table tennis. At first, the first two players in the line play a game. Then the loser goes to the end of the line, and the winner plays with the next person from the line, and so on. They play until someone wins k games in a row. This player becomes the winner.
For each of the participants, you know the power to play table tennis, and for all players these values are different. In a game the player with greater power always wins. Determine who will be the winner.
Input
The first line contains two integers: n and k (2 β€ n β€ 500, 2 β€ k β€ 1012) β the number of people and the number of wins.
The second line contains n integers a1, a2, ..., an (1 β€ ai β€ n) β powers of the player. It's guaranteed that this line contains a valid permutation, i.e. all ai are distinct.
Output
Output a single integer β power of the winner.
Examples
Input
2 2
1 2
Output
2
Input
4 2
3 1 2 4
Output
3
Input
6 2
6 5 3 1 2 4
Output
6
Input
2 10000000000
2 1
Output
2
Note
Games in the second sample:
3 plays with 1. 3 wins. 1 goes to the end of the line.
3 plays with 2. 3 wins. He wins twice in a row. He becomes the winner.
Submitted Solution:
```
# n=int(input())
# n,k=map(int,input().split())
# arr=list(map(int,input().split()))
#ls=list(map(int,input().split()))
#for i in range(m):
# for _ in range(int(input())):
#from collections import Counter
#from fractions import Fraction
#s=iter(input())
from collections import deque
n,k=map(int,input().split())
arr=list(map(int,input().split()))
'''if n==2:
print(max(arr))
exit()'''
for i in range(n):
var=i+k
if i==0:
if var>=n:
if arr[i]>=max(arr[i:n]) and arr[i]>=max(arr[:(k-(n-i-1))+1]):
print(arr[i])
break
else:
if arr[i] >= max(arr[i:i+k+1]):
print(arr[i])
break
else:
#print(i)
if var>=n:
if arr[i]>=max(arr[i-1:n]) and arr[i]>=max(arr[:(k-(n-i-1))]):
print(arr[i])
break
else:
if arr[i]>=max(arr[i-1:i+k]):
print(arr[i])
break
``` | instruction | 0 | 69,549 | 19 | 139,098 |
Yes | output | 1 | 69,549 | 19 | 139,099 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
n people are standing in a line to play table tennis. At first, the first two players in the line play a game. Then the loser goes to the end of the line, and the winner plays with the next person from the line, and so on. They play until someone wins k games in a row. This player becomes the winner.
For each of the participants, you know the power to play table tennis, and for all players these values are different. In a game the player with greater power always wins. Determine who will be the winner.
Input
The first line contains two integers: n and k (2 β€ n β€ 500, 2 β€ k β€ 1012) β the number of people and the number of wins.
The second line contains n integers a1, a2, ..., an (1 β€ ai β€ n) β powers of the player. It's guaranteed that this line contains a valid permutation, i.e. all ai are distinct.
Output
Output a single integer β power of the winner.
Examples
Input
2 2
1 2
Output
2
Input
4 2
3 1 2 4
Output
3
Input
6 2
6 5 3 1 2 4
Output
6
Input
2 10000000000
2 1
Output
2
Note
Games in the second sample:
3 plays with 1. 3 wins. 1 goes to the end of the line.
3 plays with 2. 3 wins. He wins twice in a row. He becomes the winner.
Submitted Solution:
```
z,zz=input,lambda:list(map(int,z().split()))
fast=lambda:stdin.readline().strip()
zzz=lambda:[int(i) for i in fast().split()]
szz,graph,mod,szzz=lambda:sorted(zz()),{},10**9+7,lambda:sorted(zzz())
from string import *
from re import *
from collections import *
from queue import *
from sys import *
from collections import *
from math import *
from heapq import *
from itertools import *
from bisect import *
from collections import Counter as cc
from math import factorial as f
from bisect import bisect as bs
from bisect import bisect_left as bsl
from itertools import accumulate as ac
def lcd(xnum1,xnum2):return (xnum1*xnum2//gcd(xnum1,xnum2))
def prime(x):
p=ceil(x**.5)+1
for i in range(2,p):
if (x%i==0 and x!=2) or x==0:return 0
return 1
def dfs(u,visit,graph):
visit[u]=1
for i in graph[u]:
if not visit[i]:
dfs(i,visit,graph)
###########################---Test-Case---#################################
"""
"""
###########################---START-CODING---##############################
quee=deque()
n,k=zzz()
quee.extend(zzz())
cnt=0
maxper=max(list(quee))
last_per=-1
while True:
u=quee.popleft()
v=quee.popleft()
loser=min(u,v)
win=max(u,v)
if win==last_per:
cnt+=1
else:
cnt=1
last_per=win
if cnt==k:
break
if win==maxper:
break
quee.append(loser)
quee.appendleft(win)
print(last_per)
``` | instruction | 0 | 69,550 | 19 | 139,100 |
Yes | output | 1 | 69,550 | 19 | 139,101 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
n people are standing in a line to play table tennis. At first, the first two players in the line play a game. Then the loser goes to the end of the line, and the winner plays with the next person from the line, and so on. They play until someone wins k games in a row. This player becomes the winner.
For each of the participants, you know the power to play table tennis, and for all players these values are different. In a game the player with greater power always wins. Determine who will be the winner.
Input
The first line contains two integers: n and k (2 β€ n β€ 500, 2 β€ k β€ 1012) β the number of people and the number of wins.
The second line contains n integers a1, a2, ..., an (1 β€ ai β€ n) β powers of the player. It's guaranteed that this line contains a valid permutation, i.e. all ai are distinct.
Output
Output a single integer β power of the winner.
Examples
Input
2 2
1 2
Output
2
Input
4 2
3 1 2 4
Output
3
Input
6 2
6 5 3 1 2 4
Output
6
Input
2 10000000000
2 1
Output
2
Note
Games in the second sample:
3 plays with 1. 3 wins. 1 goes to the end of the line.
3 plays with 2. 3 wins. He wins twice in a row. He becomes the winner.
Submitted Solution:
```
from collections import deque
p, k = list(map(int, input().split()))
players = deque(list(map(int, input().split())))
wins = {}
for i in players:
wins[i] = 0
max_p = max(players)
index_max = players.index(max_p)
if index_max <= k:
print(max_p)
else:
while True:
if players[0] > players[1]:
aux = players[0]
players[0] = players[1]
players[1] = aux
first = players.popleft()
players.append(first)
wins[players[0]] += 1
wins[players[1]] = 0
else:
wins[players[0]] = 0
wins[players[1]] += 1
aux = players.popleft()
players.append(aux)
if wins[players[0]] == k:
print(players[0])
break
``` | instruction | 0 | 69,551 | 19 | 139,102 |
Yes | output | 1 | 69,551 | 19 | 139,103 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
n people are standing in a line to play table tennis. At first, the first two players in the line play a game. Then the loser goes to the end of the line, and the winner plays with the next person from the line, and so on. They play until someone wins k games in a row. This player becomes the winner.
For each of the participants, you know the power to play table tennis, and for all players these values are different. In a game the player with greater power always wins. Determine who will be the winner.
Input
The first line contains two integers: n and k (2 β€ n β€ 500, 2 β€ k β€ 1012) β the number of people and the number of wins.
The second line contains n integers a1, a2, ..., an (1 β€ ai β€ n) β powers of the player. It's guaranteed that this line contains a valid permutation, i.e. all ai are distinct.
Output
Output a single integer β power of the winner.
Examples
Input
2 2
1 2
Output
2
Input
4 2
3 1 2 4
Output
3
Input
6 2
6 5 3 1 2 4
Output
6
Input
2 10000000000
2 1
Output
2
Note
Games in the second sample:
3 plays with 1. 3 wins. 1 goes to the end of the line.
3 plays with 2. 3 wins. He wins twice in a row. He becomes the winner.
Submitted Solution:
```
from collections import deque
players=[]
p1=0
p2=0
wins=0
isP1Stronk=True
n,k=input().split()
players=input().split()
p1=players[0]
players.pop(0)
p2=players[0]
power=p1
while wins<int(k):
if len(players)==1 and p1>p2:
break
elif len(players)==1 and p2>p1:
power=p2
break
if p1>p2:
players.append(p2)
players.pop(0)
p2=players[0]
wins+=1
else:
wins=1
power=p2
players.append(p1)
p1=p2
players.pop(0)
p2=players[0]
print(power)
``` | instruction | 0 | 69,552 | 19 | 139,104 |
No | output | 1 | 69,552 | 19 | 139,105 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
n people are standing in a line to play table tennis. At first, the first two players in the line play a game. Then the loser goes to the end of the line, and the winner plays with the next person from the line, and so on. They play until someone wins k games in a row. This player becomes the winner.
For each of the participants, you know the power to play table tennis, and for all players these values are different. In a game the player with greater power always wins. Determine who will be the winner.
Input
The first line contains two integers: n and k (2 β€ n β€ 500, 2 β€ k β€ 1012) β the number of people and the number of wins.
The second line contains n integers a1, a2, ..., an (1 β€ ai β€ n) β powers of the player. It's guaranteed that this line contains a valid permutation, i.e. all ai are distinct.
Output
Output a single integer β power of the winner.
Examples
Input
2 2
1 2
Output
2
Input
4 2
3 1 2 4
Output
3
Input
6 2
6 5 3 1 2 4
Output
6
Input
2 10000000000
2 1
Output
2
Note
Games in the second sample:
3 plays with 1. 3 wins. 1 goes to the end of the line.
3 plays with 2. 3 wins. He wins twice in a row. He becomes the winner.
Submitted Solution:
```
nk = input().split()
n = int(nk[0])
k = int(nk[1])
players = input().split()
if k >= n:
k = n
x = 0
while 2 <= n <= 500 and 2 <= k <= 1000000000000:
if len(players) > n or len(players) < 1:
break
A = int(players[0])
for i in range(1, len(players)):
if x == k:
break
if A > int(players[i]):
x += 1
elif A < int(players[i]):
print(players[i])
players[0] = players[i]
players.remove(players[i])
players.append(str(A))
print(players)
x = 0
break
if x == k:
print(A)
break
``` | instruction | 0 | 69,553 | 19 | 139,106 |
No | output | 1 | 69,553 | 19 | 139,107 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
n people are standing in a line to play table tennis. At first, the first two players in the line play a game. Then the loser goes to the end of the line, and the winner plays with the next person from the line, and so on. They play until someone wins k games in a row. This player becomes the winner.
For each of the participants, you know the power to play table tennis, and for all players these values are different. In a game the player with greater power always wins. Determine who will be the winner.
Input
The first line contains two integers: n and k (2 β€ n β€ 500, 2 β€ k β€ 1012) β the number of people and the number of wins.
The second line contains n integers a1, a2, ..., an (1 β€ ai β€ n) β powers of the player. It's guaranteed that this line contains a valid permutation, i.e. all ai are distinct.
Output
Output a single integer β power of the winner.
Examples
Input
2 2
1 2
Output
2
Input
4 2
3 1 2 4
Output
3
Input
6 2
6 5 3 1 2 4
Output
6
Input
2 10000000000
2 1
Output
2
Note
Games in the second sample:
3 plays with 1. 3 wins. 1 goes to the end of the line.
3 plays with 2. 3 wins. He wins twice in a row. He becomes the winner.
Submitted Solution:
```
inp = list(map(int, input().split()))
players = list(map(int, input().split()))
n = inp[0]
k = inp[1]
if (n == 2):
print(max(players[0], players[1]))
else:
flag = False
indice_prox = 2
maior = max(players)
num_vitorias = 0
prev_vencedor = max(players[0], players[1])
num_vitorias += 1
if(prev_vencedor == maior):
flag = True
print(prev_vencedor)
else:
while(num_vitorias < k):
prox = players[indice_prox]
if(prox == maior):
flag = True
print(maior)
break
vencedor = max(prev_vencedor, prox)
if(vencedor == prev_vencedor):
num_vitorias += 1
else:
prev_vencedor = vencedor
indice_prox += 1
if(flag == False):
print(prev_vencedor)
``` | instruction | 0 | 69,554 | 19 | 139,108 |
No | output | 1 | 69,554 | 19 | 139,109 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
n people are standing in a line to play table tennis. At first, the first two players in the line play a game. Then the loser goes to the end of the line, and the winner plays with the next person from the line, and so on. They play until someone wins k games in a row. This player becomes the winner.
For each of the participants, you know the power to play table tennis, and for all players these values are different. In a game the player with greater power always wins. Determine who will be the winner.
Input
The first line contains two integers: n and k (2 β€ n β€ 500, 2 β€ k β€ 1012) β the number of people and the number of wins.
The second line contains n integers a1, a2, ..., an (1 β€ ai β€ n) β powers of the player. It's guaranteed that this line contains a valid permutation, i.e. all ai are distinct.
Output
Output a single integer β power of the winner.
Examples
Input
2 2
1 2
Output
2
Input
4 2
3 1 2 4
Output
3
Input
6 2
6 5 3 1 2 4
Output
6
Input
2 10000000000
2 1
Output
2
Note
Games in the second sample:
3 plays with 1. 3 wins. 1 goes to the end of the line.
3 plays with 2. 3 wins. He wins twice in a row. He becomes the winner.
Submitted Solution:
```
firstGiven = input().split()
testCase = int(firstGiven[0])
winCondition = int(firstGiven[1])
players = input().split()
activePlayerOne = int(players[0])
activePlayerTwo = int(players[1])
playerOneWins = 0
playerTwoWins = 0
if len(players) == 2:
if activePlayerOne > activePlayerTwo:
playerOneWins = winCondition
else:
playerTwoWins = winCondition
while playerOneWins != winCondition and playerTwoWins != winCondition:
print(players)
if activePlayerOne > activePlayerTwo:
players.append(str(activePlayerTwo))
players.remove(str(activePlayerTwo))
activePlayerTwo = int(players[1])
playerTwoWins = 0
playerOneWins += 1
else:
players.append(str(activePlayerOne))
players.remove(str(activePlayerOne))
activePlayerOne = int(players[1])
playerOneWins = 0
playerTwoWins += 1
if playerOneWins == winCondition:
print(activePlayerOne)
elif playerTwoWins == winCondition:
print(activePlayerTwo)
``` | instruction | 0 | 69,555 | 19 | 139,110 |
No | output | 1 | 69,555 | 19 | 139,111 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice and Bob are playing a game with n piles of stones. It is guaranteed that n is an even number. The i-th pile has a_i stones.
Alice and Bob will play a game alternating turns with Alice going first.
On a player's turn, they must choose exactly n/2 nonempty piles and independently remove a positive number of stones from each of the chosen piles. They can remove a different number of stones from the piles in a single turn. The first player unable to make a move loses (when there are less than n/2 nonempty piles).
Given the starting configuration, determine who will win the game.
Input
The first line contains one integer n (2 β€ n β€ 50) β the number of piles. It is guaranteed that n is an even number.
The second line contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ 50) β the number of stones in the piles.
Output
Print a single string "Alice" if Alice wins; otherwise, print "Bob" (without double quotes).
Examples
Input
2
8 8
Output
Bob
Input
4
3 1 4 1
Output
Alice
Note
In the first example, each player can only remove stones from one pile (2/2=1). Alice loses, since Bob can copy whatever Alice does on the other pile, so Alice will run out of moves first.
In the second example, Alice can remove 2 stones from the first pile and 3 stones from the third pile on her first move to guarantee a win. | instruction | 0 | 69,894 | 19 | 139,788 |
Tags: games
Correct Solution:
```
n = int(input())
L = [int(i) for i in input().split()]
s = 0
m = L[0]
for i in L:
if i == 1:
s += 1
if i < m:
m = i
if s > n // 2:
print('Bob')
elif s <= n // 2 and s > 0:
print('Alice')
elif s == 0:
ss = 0
for j in L:
if j == m:
ss += 1
if ss <= n // 2:
print('Alice')
else:
print('Bob')
``` | output | 1 | 69,894 | 19 | 139,789 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice and Bob are playing a game with n piles of stones. It is guaranteed that n is an even number. The i-th pile has a_i stones.
Alice and Bob will play a game alternating turns with Alice going first.
On a player's turn, they must choose exactly n/2 nonempty piles and independently remove a positive number of stones from each of the chosen piles. They can remove a different number of stones from the piles in a single turn. The first player unable to make a move loses (when there are less than n/2 nonempty piles).
Given the starting configuration, determine who will win the game.
Input
The first line contains one integer n (2 β€ n β€ 50) β the number of piles. It is guaranteed that n is an even number.
The second line contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ 50) β the number of stones in the piles.
Output
Print a single string "Alice" if Alice wins; otherwise, print "Bob" (without double quotes).
Examples
Input
2
8 8
Output
Bob
Input
4
3 1 4 1
Output
Alice
Note
In the first example, each player can only remove stones from one pile (2/2=1). Alice loses, since Bob can copy whatever Alice does on the other pile, so Alice will run out of moves first.
In the second example, Alice can remove 2 stones from the first pile and 3 stones from the third pile on her first move to guarantee a win. | instruction | 0 | 69,895 | 19 | 139,790 |
Tags: games
Correct Solution:
```
n=int(input())
a=list(map(int,input().split()))
b=[0]*51
min_a=100
for i in a:
b[i]+=1
min_a=min(min_a,i)
if b[min_a]>n//2 :
print("Bob")
else:
print("Alice")
``` | output | 1 | 69,895 | 19 | 139,791 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice and Bob are playing a game with n piles of stones. It is guaranteed that n is an even number. The i-th pile has a_i stones.
Alice and Bob will play a game alternating turns with Alice going first.
On a player's turn, they must choose exactly n/2 nonempty piles and independently remove a positive number of stones from each of the chosen piles. They can remove a different number of stones from the piles in a single turn. The first player unable to make a move loses (when there are less than n/2 nonempty piles).
Given the starting configuration, determine who will win the game.
Input
The first line contains one integer n (2 β€ n β€ 50) β the number of piles. It is guaranteed that n is an even number.
The second line contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ 50) β the number of stones in the piles.
Output
Print a single string "Alice" if Alice wins; otherwise, print "Bob" (without double quotes).
Examples
Input
2
8 8
Output
Bob
Input
4
3 1 4 1
Output
Alice
Note
In the first example, each player can only remove stones from one pile (2/2=1). Alice loses, since Bob can copy whatever Alice does on the other pile, so Alice will run out of moves first.
In the second example, Alice can remove 2 stones from the first pile and 3 stones from the third pile on her first move to guarantee a win. | instruction | 0 | 69,896 | 19 | 139,792 |
Tags: games
Correct Solution:
```
n = int(input())
A = list(map(int, input().split()))
m = min(A)
cnt = 0
for a in A:
if a == m:
cnt += 1
if cnt > n//2:
print('Bob')
else:
print('Alice')
``` | output | 1 | 69,896 | 19 | 139,793 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice and Bob are playing a game with n piles of stones. It is guaranteed that n is an even number. The i-th pile has a_i stones.
Alice and Bob will play a game alternating turns with Alice going first.
On a player's turn, they must choose exactly n/2 nonempty piles and independently remove a positive number of stones from each of the chosen piles. They can remove a different number of stones from the piles in a single turn. The first player unable to make a move loses (when there are less than n/2 nonempty piles).
Given the starting configuration, determine who will win the game.
Input
The first line contains one integer n (2 β€ n β€ 50) β the number of piles. It is guaranteed that n is an even number.
The second line contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ 50) β the number of stones in the piles.
Output
Print a single string "Alice" if Alice wins; otherwise, print "Bob" (without double quotes).
Examples
Input
2
8 8
Output
Bob
Input
4
3 1 4 1
Output
Alice
Note
In the first example, each player can only remove stones from one pile (2/2=1). Alice loses, since Bob can copy whatever Alice does on the other pile, so Alice will run out of moves first.
In the second example, Alice can remove 2 stones from the first pile and 3 stones from the third pile on her first move to guarantee a win. | instruction | 0 | 69,897 | 19 | 139,794 |
Tags: games
Correct Solution:
```
if __name__ == '__main__':
stone_piles_nr = int(input())
pile_idx___stones_nr = [int(x) for x in input().split()]
min_stones_nr = min(pile_idx___stones_nr)
min_count = pile_idx___stones_nr.count(min_stones_nr)
if min_count > stone_piles_nr // 2:
print('Bob')
else:
print('Alice')
``` | output | 1 | 69,897 | 19 | 139,795 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice and Bob are playing a game with n piles of stones. It is guaranteed that n is an even number. The i-th pile has a_i stones.
Alice and Bob will play a game alternating turns with Alice going first.
On a player's turn, they must choose exactly n/2 nonempty piles and independently remove a positive number of stones from each of the chosen piles. They can remove a different number of stones from the piles in a single turn. The first player unable to make a move loses (when there are less than n/2 nonempty piles).
Given the starting configuration, determine who will win the game.
Input
The first line contains one integer n (2 β€ n β€ 50) β the number of piles. It is guaranteed that n is an even number.
The second line contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ 50) β the number of stones in the piles.
Output
Print a single string "Alice" if Alice wins; otherwise, print "Bob" (without double quotes).
Examples
Input
2
8 8
Output
Bob
Input
4
3 1 4 1
Output
Alice
Note
In the first example, each player can only remove stones from one pile (2/2=1). Alice loses, since Bob can copy whatever Alice does on the other pile, so Alice will run out of moves first.
In the second example, Alice can remove 2 stones from the first pile and 3 stones from the third pile on her first move to guarantee a win. | instruction | 0 | 69,898 | 19 | 139,796 |
Tags: games
Correct Solution:
```
i=input
i()
s=list(map(int,i().split()))
print("Bob"if s.count(min(s))>len(s)/2 else"Alice")
``` | output | 1 | 69,898 | 19 | 139,797 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice and Bob are playing a game with n piles of stones. It is guaranteed that n is an even number. The i-th pile has a_i stones.
Alice and Bob will play a game alternating turns with Alice going first.
On a player's turn, they must choose exactly n/2 nonempty piles and independently remove a positive number of stones from each of the chosen piles. They can remove a different number of stones from the piles in a single turn. The first player unable to make a move loses (when there are less than n/2 nonempty piles).
Given the starting configuration, determine who will win the game.
Input
The first line contains one integer n (2 β€ n β€ 50) β the number of piles. It is guaranteed that n is an even number.
The second line contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ 50) β the number of stones in the piles.
Output
Print a single string "Alice" if Alice wins; otherwise, print "Bob" (without double quotes).
Examples
Input
2
8 8
Output
Bob
Input
4
3 1 4 1
Output
Alice
Note
In the first example, each player can only remove stones from one pile (2/2=1). Alice loses, since Bob can copy whatever Alice does on the other pile, so Alice will run out of moves first.
In the second example, Alice can remove 2 stones from the first pile and 3 stones from the third pile on her first move to guarantee a win. | instruction | 0 | 69,899 | 19 | 139,798 |
Tags: games
Correct Solution:
```
kk=lambda:map(int,input().split())
ll=lambda:list(kk())
n,s=int(input()),ll()
print("Bob" if s.count(min(s)) > n/2 else "Alice")
``` | output | 1 | 69,899 | 19 | 139,799 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice and Bob are playing a game with n piles of stones. It is guaranteed that n is an even number. The i-th pile has a_i stones.
Alice and Bob will play a game alternating turns with Alice going first.
On a player's turn, they must choose exactly n/2 nonempty piles and independently remove a positive number of stones from each of the chosen piles. They can remove a different number of stones from the piles in a single turn. The first player unable to make a move loses (when there are less than n/2 nonempty piles).
Given the starting configuration, determine who will win the game.
Input
The first line contains one integer n (2 β€ n β€ 50) β the number of piles. It is guaranteed that n is an even number.
The second line contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ 50) β the number of stones in the piles.
Output
Print a single string "Alice" if Alice wins; otherwise, print "Bob" (without double quotes).
Examples
Input
2
8 8
Output
Bob
Input
4
3 1 4 1
Output
Alice
Note
In the first example, each player can only remove stones from one pile (2/2=1). Alice loses, since Bob can copy whatever Alice does on the other pile, so Alice will run out of moves first.
In the second example, Alice can remove 2 stones from the first pile and 3 stones from the third pile on her first move to guarantee a win. | instruction | 0 | 69,900 | 19 | 139,800 |
Tags: games
Correct Solution:
```
from collections import Counter as C
n = int(input())
c = C(map(int, input().split()))
m = c[min(c.keys())]
print("Alice" if m <= n / 2 else "Bob")
``` | output | 1 | 69,900 | 19 | 139,801 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice and Bob are playing a game with n piles of stones. It is guaranteed that n is an even number. The i-th pile has a_i stones.
Alice and Bob will play a game alternating turns with Alice going first.
On a player's turn, they must choose exactly n/2 nonempty piles and independently remove a positive number of stones from each of the chosen piles. They can remove a different number of stones from the piles in a single turn. The first player unable to make a move loses (when there are less than n/2 nonempty piles).
Given the starting configuration, determine who will win the game.
Input
The first line contains one integer n (2 β€ n β€ 50) β the number of piles. It is guaranteed that n is an even number.
The second line contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ 50) β the number of stones in the piles.
Output
Print a single string "Alice" if Alice wins; otherwise, print "Bob" (without double quotes).
Examples
Input
2
8 8
Output
Bob
Input
4
3 1 4 1
Output
Alice
Note
In the first example, each player can only remove stones from one pile (2/2=1). Alice loses, since Bob can copy whatever Alice does on the other pile, so Alice will run out of moves first.
In the second example, Alice can remove 2 stones from the first pile and 3 stones from the third pile on her first move to guarantee a win. | instruction | 0 | 69,901 | 19 | 139,802 |
Tags: games
Correct Solution:
```
# AC
import sys
class Main:
def __init__(self):
self.buff = None
self.index = 0
def next(self):
if self.buff is None or self.index == len(self.buff):
self.buff = self.next_line()
self.index = 0
val = self.buff[self.index]
self.index += 1
return val
def next_line(self):
return sys.stdin.readline().split()
def next_ints(self):
return [int(x) for x in sys.stdin.readline().split()]
def next_int(self):
return int(self.next())
def solve(self):
n = self.next_int()
x = self.next_ints()
c = x.count(min(x))
print('Alice' if c <= n // 2 else 'Bob')
if __name__ == '__main__':
Main().solve()
``` | output | 1 | 69,901 | 19 | 139,803 |
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