message stringlengths 2 28.7k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 21 109k | cluster float64 7 7 | __index_level_0__ int64 42 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a canvas divided into a grid with H rows and W columns. The square at the i-th row from the top and the j-th column from the left is represented as (i, j).
Initially, all the squares are white. square1001 wants to draw a picture with black paint. His specific objective is to make Square (i, j) black when s_{i, j}= `#`, and to make Square (i, j) white when s_{i, j}= `.`.
However, since he is not a good painter, he can only choose two squares that are horizontally or vertically adjacent and paint those squares black, for some number of times (possibly zero). He may choose squares that are already painted black, in which case the color of those squares remain black.
Determine if square1001 can achieve his objective.
Constraints
* H is an integer between 1 and 50 (inclusive).
* W is an integer between 1 and 50 (inclusive).
* For every (i, j) (1 \leq i \leq H, 1 \leq j \leq W), s_{i, j} is `#` or `.`.
Input
Input is given from Standard Input in the following format:
H W
s_{1, 1} s_{1, 2} s_{1, 3} ... s_{1, W}
s_{2, 1} s_{2, 2} s_{2, 3} ... s_{2, W}
: :
s_{H, 1} s_{H, 2} s_{H, 3} ... s_{H, W}
Output
If square1001 can achieve his objective, print `Yes`; if he cannot, print `No`.
Examples
Input
3 3
.#.
###
.#.
Output
Yes
Input
3 3
.#.
.#.
Output
Yes
Input
5 5
.#.#
.#.#.
.#.#
.#.#.
.#.#
Output
No
Input
11 11
...#####...
.##.....##.
..##.##..#
..##.##..#
.........#
...###...#
.#########.
.#.#.#.#.#.
.#.#.#.##
..##.#.##..
.##..#..##.
Output
Yes
Submitted Solution:
```
h, w = map(int, input().split())
mat = []
mat.append("."*(w+2))
for i in range(h):
s = input()
mat.append("."+s+".")
mat.append("."*(w+2))
for x in range(1,w+1):
for y in range(1, h+1):
if mat[y][x] == "#" and mat[y-1][x] == "." and mat[y+1][x] == "." and mat[y][x-1] == "." and mat[y][x+1] == ".":
print("No")
exit()
print("Yes")
``` | instruction | 0 | 7,376 | 7 | 14,752 |
Yes | output | 1 | 7,376 | 7 | 14,753 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a canvas divided into a grid with H rows and W columns. The square at the i-th row from the top and the j-th column from the left is represented as (i, j).
Initially, all the squares are white. square1001 wants to draw a picture with black paint. His specific objective is to make Square (i, j) black when s_{i, j}= `#`, and to make Square (i, j) white when s_{i, j}= `.`.
However, since he is not a good painter, he can only choose two squares that are horizontally or vertically adjacent and paint those squares black, for some number of times (possibly zero). He may choose squares that are already painted black, in which case the color of those squares remain black.
Determine if square1001 can achieve his objective.
Constraints
* H is an integer between 1 and 50 (inclusive).
* W is an integer between 1 and 50 (inclusive).
* For every (i, j) (1 \leq i \leq H, 1 \leq j \leq W), s_{i, j} is `#` or `.`.
Input
Input is given from Standard Input in the following format:
H W
s_{1, 1} s_{1, 2} s_{1, 3} ... s_{1, W}
s_{2, 1} s_{2, 2} s_{2, 3} ... s_{2, W}
: :
s_{H, 1} s_{H, 2} s_{H, 3} ... s_{H, W}
Output
If square1001 can achieve his objective, print `Yes`; if he cannot, print `No`.
Examples
Input
3 3
.#.
###
.#.
Output
Yes
Input
3 3
.#.
.#.
Output
Yes
Input
5 5
.#.#
.#.#.
.#.#
.#.#.
.#.#
Output
No
Input
11 11
...#####...
.##.....##.
..##.##..#
..##.##..#
.........#
...###...#
.#########.
.#.#.#.#.#.
.#.#.#.##
..##.#.##..
.##..#..##.
Output
Yes
Submitted Solution:
```
H,W=map(int,input().split())
S=["."*(W+2)]
for i in range(H):
S.append("."+input()+".")
S.append("."*(W+2))
print("Yes" if all(S[i][j]=="." or (S[i+1][j]=="#" or S[i-1][j]=="#" or S[i][j+1]=="#" or S[i][j-1]=="#") for i in range(1,H+1) for j in range(1,W)) else "No")
``` | instruction | 0 | 7,377 | 7 | 14,754 |
Yes | output | 1 | 7,377 | 7 | 14,755 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a canvas divided into a grid with H rows and W columns. The square at the i-th row from the top and the j-th column from the left is represented as (i, j).
Initially, all the squares are white. square1001 wants to draw a picture with black paint. His specific objective is to make Square (i, j) black when s_{i, j}= `#`, and to make Square (i, j) white when s_{i, j}= `.`.
However, since he is not a good painter, he can only choose two squares that are horizontally or vertically adjacent and paint those squares black, for some number of times (possibly zero). He may choose squares that are already painted black, in which case the color of those squares remain black.
Determine if square1001 can achieve his objective.
Constraints
* H is an integer between 1 and 50 (inclusive).
* W is an integer between 1 and 50 (inclusive).
* For every (i, j) (1 \leq i \leq H, 1 \leq j \leq W), s_{i, j} is `#` or `.`.
Input
Input is given from Standard Input in the following format:
H W
s_{1, 1} s_{1, 2} s_{1, 3} ... s_{1, W}
s_{2, 1} s_{2, 2} s_{2, 3} ... s_{2, W}
: :
s_{H, 1} s_{H, 2} s_{H, 3} ... s_{H, W}
Output
If square1001 can achieve his objective, print `Yes`; if he cannot, print `No`.
Examples
Input
3 3
.#.
###
.#.
Output
Yes
Input
3 3
.#.
.#.
Output
Yes
Input
5 5
.#.#
.#.#.
.#.#
.#.#.
.#.#
Output
No
Input
11 11
...#####...
.##.....##.
..##.##..#
..##.##..#
.........#
...###...#
.#########.
.#.#.#.#.#.
.#.#.#.##
..##.#.##..
.##..#..##.
Output
Yes
Submitted Solution:
```
a =str(input()).split(" ")
h = int(a[0])
w = int(a[1])
maps= []
for i in range(h):
maps.append(list(input()))
for i in range(h - 1 ):
for j in range(w - 1):
if maps[i][j] == "#":
if j - 1 > 0 and maps[i][j - 1] == "#":
continue
elif j + 1 < w and maps[i][j + 1] == "#" :
continue
elif i + 1 < h and maps[i + 1][j] == "#" :
continue
elif i - 1 > 0 and maps[i - 1][j] == "#":
continue
else:
print("No")
exit()
print(i,j)
print("Yes")
``` | instruction | 0 | 7,378 | 7 | 14,756 |
No | output | 1 | 7,378 | 7 | 14,757 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a canvas divided into a grid with H rows and W columns. The square at the i-th row from the top and the j-th column from the left is represented as (i, j).
Initially, all the squares are white. square1001 wants to draw a picture with black paint. His specific objective is to make Square (i, j) black when s_{i, j}= `#`, and to make Square (i, j) white when s_{i, j}= `.`.
However, since he is not a good painter, he can only choose two squares that are horizontally or vertically adjacent and paint those squares black, for some number of times (possibly zero). He may choose squares that are already painted black, in which case the color of those squares remain black.
Determine if square1001 can achieve his objective.
Constraints
* H is an integer between 1 and 50 (inclusive).
* W is an integer between 1 and 50 (inclusive).
* For every (i, j) (1 \leq i \leq H, 1 \leq j \leq W), s_{i, j} is `#` or `.`.
Input
Input is given from Standard Input in the following format:
H W
s_{1, 1} s_{1, 2} s_{1, 3} ... s_{1, W}
s_{2, 1} s_{2, 2} s_{2, 3} ... s_{2, W}
: :
s_{H, 1} s_{H, 2} s_{H, 3} ... s_{H, W}
Output
If square1001 can achieve his objective, print `Yes`; if he cannot, print `No`.
Examples
Input
3 3
.#.
###
.#.
Output
Yes
Input
3 3
.#.
.#.
Output
Yes
Input
5 5
.#.#
.#.#.
.#.#
.#.#.
.#.#
Output
No
Input
11 11
...#####...
.##.....##.
..##.##..#
..##.##..#
.........#
...###...#
.#########.
.#.#.#.#.#.
.#.#.#.##
..##.#.##..
.##..#..##.
Output
Yes
Submitted Solution:
```
H, W = map(int, input().split())
S = [input() for _ in range(H)]
for i in range(H):
S[i] = '.' + S[i] + '.'
S = ['.' * (W + 2)] + S + ['.' * (W + 2)]
cnt_list = []
for i in range(1,H+1):
for j in range(1, W+1):
if S[i][j] == '.':
pass
else:
ser = S[i-1][j] + S[i][j-1] + S[i][j+1] + S[i+1][j] + S[i+1][j+1]
cnt_list.append(ser.count('#'))
if 0 in cnt_list:
print('No')
else:
print('Yes')
``` | instruction | 0 | 7,379 | 7 | 14,758 |
No | output | 1 | 7,379 | 7 | 14,759 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a canvas divided into a grid with H rows and W columns. The square at the i-th row from the top and the j-th column from the left is represented as (i, j).
Initially, all the squares are white. square1001 wants to draw a picture with black paint. His specific objective is to make Square (i, j) black when s_{i, j}= `#`, and to make Square (i, j) white when s_{i, j}= `.`.
However, since he is not a good painter, he can only choose two squares that are horizontally or vertically adjacent and paint those squares black, for some number of times (possibly zero). He may choose squares that are already painted black, in which case the color of those squares remain black.
Determine if square1001 can achieve his objective.
Constraints
* H is an integer between 1 and 50 (inclusive).
* W is an integer between 1 and 50 (inclusive).
* For every (i, j) (1 \leq i \leq H, 1 \leq j \leq W), s_{i, j} is `#` or `.`.
Input
Input is given from Standard Input in the following format:
H W
s_{1, 1} s_{1, 2} s_{1, 3} ... s_{1, W}
s_{2, 1} s_{2, 2} s_{2, 3} ... s_{2, W}
: :
s_{H, 1} s_{H, 2} s_{H, 3} ... s_{H, W}
Output
If square1001 can achieve his objective, print `Yes`; if he cannot, print `No`.
Examples
Input
3 3
.#.
###
.#.
Output
Yes
Input
3 3
.#.
.#.
Output
Yes
Input
5 5
.#.#
.#.#.
.#.#
.#.#.
.#.#
Output
No
Input
11 11
...#####...
.##.....##.
..##.##..#
..##.##..#
.........#
...###...#
.#########.
.#.#.#.#.#.
.#.#.#.##
..##.#.##..
.##..#..##.
Output
Yes
Submitted Solution:
```
x,y = map(int,input().split())
listA=[] #appendのために宣言が必要
while True:
try:
listA.append(list(input()))
except:
break;
s=0
for i in range(x):
for j in range(y):
if ((listA[i][j]!=listA[max(0,i-1)][j])&(listA[i][j]!=listA[min(x-1,i+1)][j])&
(listA[i][j]!=listA[i][max(0,j-1)])&(listA[i][j]!=listA[i][min(y-1,j+1)])):
s+=1
if s==0:
print('Yes')
else:
print('No')
``` | instruction | 0 | 7,380 | 7 | 14,760 |
No | output | 1 | 7,380 | 7 | 14,761 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a canvas divided into a grid with H rows and W columns. The square at the i-th row from the top and the j-th column from the left is represented as (i, j).
Initially, all the squares are white. square1001 wants to draw a picture with black paint. His specific objective is to make Square (i, j) black when s_{i, j}= `#`, and to make Square (i, j) white when s_{i, j}= `.`.
However, since he is not a good painter, he can only choose two squares that are horizontally or vertically adjacent and paint those squares black, for some number of times (possibly zero). He may choose squares that are already painted black, in which case the color of those squares remain black.
Determine if square1001 can achieve his objective.
Constraints
* H is an integer between 1 and 50 (inclusive).
* W is an integer between 1 and 50 (inclusive).
* For every (i, j) (1 \leq i \leq H, 1 \leq j \leq W), s_{i, j} is `#` or `.`.
Input
Input is given from Standard Input in the following format:
H W
s_{1, 1} s_{1, 2} s_{1, 3} ... s_{1, W}
s_{2, 1} s_{2, 2} s_{2, 3} ... s_{2, W}
: :
s_{H, 1} s_{H, 2} s_{H, 3} ... s_{H, W}
Output
If square1001 can achieve his objective, print `Yes`; if he cannot, print `No`.
Examples
Input
3 3
.#.
###
.#.
Output
Yes
Input
3 3
.#.
.#.
Output
Yes
Input
5 5
.#.#
.#.#.
.#.#
.#.#.
.#.#
Output
No
Input
11 11
...#####...
.##.....##.
..##.##..#
..##.##..#
.........#
...###...#
.#########.
.#.#.#.#.#.
.#.#.#.##
..##.#.##..
.##..#..##.
Output
Yes
Submitted Solution:
```
import sys
h, w = map(int, input().split())
s = [list(input()) for _ in range(h)]
hantei = [[0 for _ in range(h)] for _ in range(w)]
for i in range(h):
for j in range(w):
if s[i][j] == "#":
if i < h-1:
if s[i+1][j] == "#":
hantei[i][j] = 1
hantei[i+1][j] = 1
if j < w-1:
if s[i][j+1] == "#":
hantei[i][j] = 1
hantei[i][j+1] = 1
for i in range(h):
for j in range(w):
if s[i][j] == "#":
if hantei[i][j] == 0:
print("No")
sys.exit()
print("Yes")
``` | instruction | 0 | 7,381 | 7 | 14,762 |
No | output | 1 | 7,381 | 7 | 14,763 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Evlampiy has found one more cool application to process photos. However the application has certain limitations.
Each photo i has a contrast vi. In order for the processing to be truly of high quality, the application must receive at least k photos with contrasts which differ as little as possible.
Evlampiy already knows the contrast vi for each of his n photos. Now he wants to split the photos into groups, so that each group contains at least k photos. As a result, each photo must belong to exactly one group.
He considers a processing time of the j-th group to be the difference between the maximum and minimum values of vi in the group. Because of multithreading the processing time of a division into groups is the maximum processing time among all groups.
Split n photos into groups in a such way that the processing time of the division is the minimum possible, i.e. that the the maximum processing time over all groups as least as possible.
Input
The first line contains two integers n and k (1 ≤ k ≤ n ≤ 3·105) — number of photos and minimum size of a group.
The second line contains n integers v1, v2, ..., vn (1 ≤ vi ≤ 109), where vi is the contrast of the i-th photo.
Output
Print the minimal processing time of the division into groups.
Examples
Input
5 2
50 110 130 40 120
Output
20
Input
4 1
2 3 4 1
Output
0
Note
In the first example the photos should be split into 2 groups: [40, 50] and [110, 120, 130]. The processing time of the first group is 10, and the processing time of the second group is 20. Maximum among 10 and 20 is 20. It is impossible to split the photos into groups in a such way that the processing time of division is less than 20.
In the second example the photos should be split into four groups, each containing one photo. So the minimal possible processing time of a division is 0. | instruction | 0 | 8,103 | 7 | 16,206 |
Tags: binary search, dp
Correct Solution:
```
def f(m):
global dp, sdp
l = 0
for i in range(n):
while l < n and v[l] < v[i] - m:
l += 1
if l - 1 > i - k:
dp[i] = False
else:
dp[i] = (sdp[i - k + 1] != sdp[l - 1])
sdp[i + 1] = sdp[i] + (1 if dp[i] else 0)
return dp[n - 1]
n, k = map(int, input().split())
dp = [False for i in range(n + 2)]
sdp = [0 for i in range(n + 2)]
dp[-1] = True
sdp[0] = 1
v = list(map(int, input().split()))
v.sort()
le = -1
r = v[-1] - v[0]
while r - le > 1:
m = (r + le) // 2
if f(m):
r = m
else:
le = m
print(r)
``` | output | 1 | 8,103 | 7 | 16,207 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Evlampiy has found one more cool application to process photos. However the application has certain limitations.
Each photo i has a contrast vi. In order for the processing to be truly of high quality, the application must receive at least k photos with contrasts which differ as little as possible.
Evlampiy already knows the contrast vi for each of his n photos. Now he wants to split the photos into groups, so that each group contains at least k photos. As a result, each photo must belong to exactly one group.
He considers a processing time of the j-th group to be the difference between the maximum and minimum values of vi in the group. Because of multithreading the processing time of a division into groups is the maximum processing time among all groups.
Split n photos into groups in a such way that the processing time of the division is the minimum possible, i.e. that the the maximum processing time over all groups as least as possible.
Input
The first line contains two integers n and k (1 ≤ k ≤ n ≤ 3·105) — number of photos and minimum size of a group.
The second line contains n integers v1, v2, ..., vn (1 ≤ vi ≤ 109), where vi is the contrast of the i-th photo.
Output
Print the minimal processing time of the division into groups.
Examples
Input
5 2
50 110 130 40 120
Output
20
Input
4 1
2 3 4 1
Output
0
Note
In the first example the photos should be split into 2 groups: [40, 50] and [110, 120, 130]. The processing time of the first group is 10, and the processing time of the second group is 20. Maximum among 10 and 20 is 20. It is impossible to split the photos into groups in a such way that the processing time of division is less than 20.
In the second example the photos should be split into four groups, each containing one photo. So the minimal possible processing time of a division is 0. | instruction | 0 | 8,104 | 7 | 16,208 |
Tags: binary search, dp
Correct Solution:
```
from bisect import bisect_left, bisect_right
class Result:
def __init__(self, index, value):
self.index = index
self.value = value
class BinarySearch:
def __init__(self):
pass
@staticmethod
def greater_than(num: int, func, size: int = 1):
"""Searches for smallest element greater than num!"""
if isinstance(func, list):
index = bisect_right(func, num)
if index == len(func):
return Result(None, None)
else:
return Result(index, func[index])
else:
alpha, omega = 0, size - 1
if func(omega) <= num:
return Result(None, None)
while alpha < omega:
if func(alpha) > num:
return Result(alpha, func(alpha))
if omega == alpha + 1:
return Result(omega, func(omega))
mid = (alpha + omega) // 2
if func(mid) > num:
omega = mid
else:
alpha = mid
@staticmethod
def less_than(num: int, func, size: int = 1):
"""Searches for largest element less than num!"""
if isinstance(func, list):
index = bisect_left(func, num) - 1
if index == -1:
return Result(None, None)
else:
return Result(index, func[index])
else:
alpha, omega = 0, size - 1
if func(alpha) >= num:
return Result(None, None)
while alpha < omega:
if func(omega) < num:
return Result(omega, func(omega))
if omega == alpha + 1:
return Result(alpha, func(alpha))
mid = (alpha + omega) // 2
if func(mid) < num:
alpha = mid
else:
omega = mid
# ------------------- fast io --------------------
import os
import sys
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)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# ------------------- fast io --------------------
from math import gcd, ceil
def pre(s):
n = len(s)
pi = [0] * n
for i in range(1, n):
j = pi[i - 1]
while j and s[i] != s[j]:
j = pi[j - 1]
if s[i] == s[j]:
j += 1
pi[i] = j
return pi
def prod(a):
ans = 1
for each in a:
ans = (ans * each)
return ans
def lcm(a, b): return a * b // gcd(a, b)
def binary(x, length=16):
y = bin(x)[2:]
return y if len(y) >= length else "0" * (length - len(y)) + y
for _ in range(int(input()) if not True else 1):
# l, r, m = map(int, input().split())
n, k = map(int, input().split())
# c, d = map(int, input().split())
a = sorted(list(map(int, input().split())))
if k == 1:
print(0)
continue
# b = list(map(int, input().split()))
bs = BinarySearch()
a=[0] + a
d = [False]*(n+1)
def check(x):
dp = list(d)
dp[0] = True
if a[k] - a[1] <= x:
dp[k] = True
else:return 0
cur = 1
for i in range(k+1, n+1):
while cur <= n and a[i]-a[cur] > x:
cur += 1
while cur <= n and not dp[cur-1]:
cur += 1
if cur <= i-k+1:
dp[i] = True
return dp[-1]
alpha, omega = 0, 10 ** 9
ans = omega
while alpha < omega:
mid = (alpha + omega) // 2
x = mid
if check(x):
omega = mid
ans = omega
else:
alpha = mid + 1
print(ans)
``` | output | 1 | 8,104 | 7 | 16,209 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Evlampiy has found one more cool application to process photos. However the application has certain limitations.
Each photo i has a contrast vi. In order for the processing to be truly of high quality, the application must receive at least k photos with contrasts which differ as little as possible.
Evlampiy already knows the contrast vi for each of his n photos. Now he wants to split the photos into groups, so that each group contains at least k photos. As a result, each photo must belong to exactly one group.
He considers a processing time of the j-th group to be the difference between the maximum and minimum values of vi in the group. Because of multithreading the processing time of a division into groups is the maximum processing time among all groups.
Split n photos into groups in a such way that the processing time of the division is the minimum possible, i.e. that the the maximum processing time over all groups as least as possible.
Input
The first line contains two integers n and k (1 ≤ k ≤ n ≤ 3·105) — number of photos and minimum size of a group.
The second line contains n integers v1, v2, ..., vn (1 ≤ vi ≤ 109), where vi is the contrast of the i-th photo.
Output
Print the minimal processing time of the division into groups.
Examples
Input
5 2
50 110 130 40 120
Output
20
Input
4 1
2 3 4 1
Output
0
Note
In the first example the photos should be split into 2 groups: [40, 50] and [110, 120, 130]. The processing time of the first group is 10, and the processing time of the second group is 20. Maximum among 10 and 20 is 20. It is impossible to split the photos into groups in a such way that the processing time of division is less than 20.
In the second example the photos should be split into four groups, each containing one photo. So the minimal possible processing time of a division is 0.
Submitted Solution:
```
import sys
def min_cluster(ls, k):
print('min_cluster(', ls, ',', k, ')', file=sys.stderr)
if len(ls) == 0:
return 0
order = sorted(zip(map(lambda x: x[0], ls), list(range(len(ls)))), reverse=True)
print(order, file=sys.stderr)
rs = []
for o in order:
if min(o[1], len(ls) - o[1] - 1) >= k - 1:
# on peut cut
print('cut à', o, file=sys.stderr)
return max(min_cluster(ls[0:o[1]], k), min_cluster(ls[o[1] + 1:], k))
else: # on peut pas cut
r = sum(lse[0] for lse in ls)
rs.append(r)
r = max(rs)
print(' return:', r, file=sys.stderr)
return r
n, k = map(int, input().split())
v = list(sorted(map(int, input().split())))
if k == 1:
print(0)
exit(0)
li = []
for i in range(1, n):
t = v[i] - v[i - 1]
li.append((t, v[i - 1], v[i]))
print(li, file=sys.stderr)
print(min_cluster(li, k))
``` | instruction | 0 | 8,105 | 7 | 16,210 |
No | output | 1 | 8,105 | 7 | 16,211 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Evlampiy has found one more cool application to process photos. However the application has certain limitations.
Each photo i has a contrast vi. In order for the processing to be truly of high quality, the application must receive at least k photos with contrasts which differ as little as possible.
Evlampiy already knows the contrast vi for each of his n photos. Now he wants to split the photos into groups, so that each group contains at least k photos. As a result, each photo must belong to exactly one group.
He considers a processing time of the j-th group to be the difference between the maximum and minimum values of vi in the group. Because of multithreading the processing time of a division into groups is the maximum processing time among all groups.
Split n photos into groups in a such way that the processing time of the division is the minimum possible, i.e. that the the maximum processing time over all groups as least as possible.
Input
The first line contains two integers n and k (1 ≤ k ≤ n ≤ 3·105) — number of photos and minimum size of a group.
The second line contains n integers v1, v2, ..., vn (1 ≤ vi ≤ 109), where vi is the contrast of the i-th photo.
Output
Print the minimal processing time of the division into groups.
Examples
Input
5 2
50 110 130 40 120
Output
20
Input
4 1
2 3 4 1
Output
0
Note
In the first example the photos should be split into 2 groups: [40, 50] and [110, 120, 130]. The processing time of the first group is 10, and the processing time of the second group is 20. Maximum among 10 and 20 is 20. It is impossible to split the photos into groups in a such way that the processing time of division is less than 20.
In the second example the photos should be split into four groups, each containing one photo. So the minimal possible processing time of a division is 0.
Submitted Solution:
```
def f(m):
start = 0
for i in range(n):
if v[i] > v[start] + m:
if i - start >= k:
start = i
else:
return False
#if n - start < k:
# return False
return True
n, k = map(int, input().split())
v = list(map(int, input().split()))
v.sort()
l = -1
r = v[-1] - v[0]
while r - l > 1:
m = (r + l) // 2
if f(m):
r = m
else:
l = m
ans = r
for i in range(r - 100, r):
if f(i):
ans = i
print(ans)
``` | instruction | 0 | 8,106 | 7 | 16,212 |
No | output | 1 | 8,106 | 7 | 16,213 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Evlampiy has found one more cool application to process photos. However the application has certain limitations.
Each photo i has a contrast vi. In order for the processing to be truly of high quality, the application must receive at least k photos with contrasts which differ as little as possible.
Evlampiy already knows the contrast vi for each of his n photos. Now he wants to split the photos into groups, so that each group contains at least k photos. As a result, each photo must belong to exactly one group.
He considers a processing time of the j-th group to be the difference between the maximum and minimum values of vi in the group. Because of multithreading the processing time of a division into groups is the maximum processing time among all groups.
Split n photos into groups in a such way that the processing time of the division is the minimum possible, i.e. that the the maximum processing time over all groups as least as possible.
Input
The first line contains two integers n and k (1 ≤ k ≤ n ≤ 3·105) — number of photos and minimum size of a group.
The second line contains n integers v1, v2, ..., vn (1 ≤ vi ≤ 109), where vi is the contrast of the i-th photo.
Output
Print the minimal processing time of the division into groups.
Examples
Input
5 2
50 110 130 40 120
Output
20
Input
4 1
2 3 4 1
Output
0
Note
In the first example the photos should be split into 2 groups: [40, 50] and [110, 120, 130]. The processing time of the first group is 10, and the processing time of the second group is 20. Maximum among 10 and 20 is 20. It is impossible to split the photos into groups in a such way that the processing time of division is less than 20.
In the second example the photos should be split into four groups, each containing one photo. So the minimal possible processing time of a division is 0.
Submitted Solution:
```
import sys
def min_cluster(ls, k):
print('min_cluster(', ls, ',', k, ')', file=sys.stderr)
if len(ls) == 1:
return ls[0][0]
order = sorted(zip(map(lambda x: x[0], ls), list(range(len(ls)))), reverse=True)
print(order, file=sys.stderr)
for o in order:
if min(o[1], len(ls) - o[1] - 1) >= k - 1:
# on peut cut
print('cut à', o, file=sys.stderr)
return max(min_cluster(ls[0:o[1]], k), min_cluster(ls[o[1] + 1:], k))
else: # on peut pas cut
r = sum(lse[0] for lse in ls)
print(' return:', r, file=sys.stderr)
return r
n, k = map(int, input().split())
v = list(sorted(map(int, input().split())))
if k == 1:
print(0)
exit(0)
li = []
for i in range(1, n):
t = v[i] - v[i - 1]
li.append((t, v[i - 1], v[i]))
print(li, file=sys.stderr)
print(min_cluster(li, k))
``` | instruction | 0 | 8,107 | 7 | 16,214 |
No | output | 1 | 8,107 | 7 | 16,215 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Evlampiy has found one more cool application to process photos. However the application has certain limitations.
Each photo i has a contrast vi. In order for the processing to be truly of high quality, the application must receive at least k photos with contrasts which differ as little as possible.
Evlampiy already knows the contrast vi for each of his n photos. Now he wants to split the photos into groups, so that each group contains at least k photos. As a result, each photo must belong to exactly one group.
He considers a processing time of the j-th group to be the difference between the maximum and minimum values of vi in the group. Because of multithreading the processing time of a division into groups is the maximum processing time among all groups.
Split n photos into groups in a such way that the processing time of the division is the minimum possible, i.e. that the the maximum processing time over all groups as least as possible.
Input
The first line contains two integers n and k (1 ≤ k ≤ n ≤ 3·105) — number of photos and minimum size of a group.
The second line contains n integers v1, v2, ..., vn (1 ≤ vi ≤ 109), where vi is the contrast of the i-th photo.
Output
Print the minimal processing time of the division into groups.
Examples
Input
5 2
50 110 130 40 120
Output
20
Input
4 1
2 3 4 1
Output
0
Note
In the first example the photos should be split into 2 groups: [40, 50] and [110, 120, 130]. The processing time of the first group is 10, and the processing time of the second group is 20. Maximum among 10 and 20 is 20. It is impossible to split the photos into groups in a such way that the processing time of division is less than 20.
In the second example the photos should be split into four groups, each containing one photo. So the minimal possible processing time of a division is 0.
Submitted Solution:
```
def check(mid):
i = 0
c = 0
for j in range(n):
if v[j] - v[i] <= mid:
c += 1
else:
if c < k:
return False
c = 0
i = j
return True
def binSearch(a, b):
left, right = a - 1, b + 1
while right - left > 1:
mid = (left + right) // 2
if check(mid):
right = mid
else:
left = mid
return right
n, k = map(int, input().split())
v = sorted(list(map(int, input().split())))
if k == 1:
print(0)
else:
print(binSearch(0, 10 ** 9))
``` | instruction | 0 | 8,108 | 7 | 16,216 |
No | output | 1 | 8,108 | 7 | 16,217 |
Provide a correct Python 3 solution for this coding contest problem.
D: Two Colors Sort
problem
During the walk, umg found a sequence of length N, P_1, P_2, ..., P_N, which can be made by rearranging 1,2, ..., N.
umg can use mysterious powers to exchange places by choosing two different numbers painted in the same color.
umg wanted to be able to sort the sequence in ascending order by painting R of the numbers in the sequence in red and the remaining N-R in blue.
umg Determine if you can reach your goal.
However, the numbers are so heavy that they cannot be moved without mysterious force. Also, since umg is a genius, you can use mysterious powers any number of times.
Input format
N R
P_1 P_2 ... P_N
Constraint
* 1 \ leq N \ leq 3 \ times 10 ^ 5
* 1 \ leq R \ leq N
* 1 \ leq P_i \ leq N
* P_i \ neq P_j (1 \ leq i <j \ leq N)
* All inputs are integers.
Output format
umg Print `Yes` if you can achieve your goal, otherwise` No` on one line.
Input example 1
3 2
1 3 2
Output example 1
Yes
* You can achieve your goal by painting 1 in blue and 2 and 3 in red.
Input example 2
5 2
1 2 3 4 5
Output example 2
Yes
* They are arranged in ascending order from the beginning.
Input example 3
10 7
3 4 8 5 7 6 2 10 1 9
Output example 3
No
Example
Input
3 2
1 3 2
Output
Yes | instruction | 0 | 8,358 | 7 | 16,716 |
"Correct Solution:
```
N, R = map(int, input().split())
if 2*R > N:
R = N - R
P = [0] + list(map(int, input().split()))
L = []
Lapp = L.append
used = [False]*(N+1)
pre = 0
for i in range(1, N+1):
cnt = 0
while not used[i]:
used[i] = True
cnt += 1
i = P[i]
if cnt:
Lapp(cnt)
table = [0]*(N+1)
for l in L:
table[l] += 1
L = []
Lapp = L.append
for i in range(1, min(R,N//2)+1):
x = table[i]
if not x:
continue
if x == 1:
Lapp(i)
else:
Lapp(i)
p = 2
while 2*p <= x:
table[p*i] += 1
p *= 2
if x == p:
Lapp(i)
table[i*(x - p + 1)] += 1
H = 1
for l in L:
H = H|(H<<l)
if H & 1<<R:
print('Yes')
else:
print('No')
``` | output | 1 | 8,358 | 7 | 16,717 |
Provide a correct Python 3 solution for this coding contest problem.
D: Two Colors Sort
problem
During the walk, umg found a sequence of length N, P_1, P_2, ..., P_N, which can be made by rearranging 1,2, ..., N.
umg can use mysterious powers to exchange places by choosing two different numbers painted in the same color.
umg wanted to be able to sort the sequence in ascending order by painting R of the numbers in the sequence in red and the remaining N-R in blue.
umg Determine if you can reach your goal.
However, the numbers are so heavy that they cannot be moved without mysterious force. Also, since umg is a genius, you can use mysterious powers any number of times.
Input format
N R
P_1 P_2 ... P_N
Constraint
* 1 \ leq N \ leq 3 \ times 10 ^ 5
* 1 \ leq R \ leq N
* 1 \ leq P_i \ leq N
* P_i \ neq P_j (1 \ leq i <j \ leq N)
* All inputs are integers.
Output format
umg Print `Yes` if you can achieve your goal, otherwise` No` on one line.
Input example 1
3 2
1 3 2
Output example 1
Yes
* You can achieve your goal by painting 1 in blue and 2 and 3 in red.
Input example 2
5 2
1 2 3 4 5
Output example 2
Yes
* They are arranged in ascending order from the beginning.
Input example 3
10 7
3 4 8 5 7 6 2 10 1 9
Output example 3
No
Example
Input
3 2
1 3 2
Output
Yes | instruction | 0 | 8,359 | 7 | 16,718 |
"Correct Solution:
```
import sys
N, R = map(int, input().split())
if 2*R > N:
R = N - R
P = [0] + list(map(int, input().split()))
L = []
used = [False]*(N+1)
pre = 0
for i in range(1, N+1):
cnt = 0
while not used[i]:
used[i] = True
cnt += 1
i = P[i]
if cnt:
L.append(cnt)
table = [0]*(N+1)
for l in L:
table[l] += 1
L = []
for i in range(N//2, 0, -1):
x = table[i]
if not x:
continue
if x == 1:
L.append(i)
else:
p = 1
while p+p <= x:
L.append(p*i)
p = p+p
if x - p + 1:
L.append(i*(x - p + 1))
L = [l for l in L if l <= R]
L.sort()
H = 1
for l in L:
H = H|(H<<l)
if H & 1<<R:
print('Yes')
else:
print('No')
``` | output | 1 | 8,359 | 7 | 16,719 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Petya loves counting. He wants to count the number of ways to paint a rectangular checkered board of size n × m (n rows, m columns) in k colors. Besides, the coloring should have the following property: for any vertical line that passes along the grid lines and divides the board in two non-empty parts the number of distinct colors in both these parts should be the same. Help Petya to count these colorings.
Input
The first line contains space-separated integers n, m and k (1 ≤ n, m ≤ 1000, 1 ≤ k ≤ 106) — the board's vertical and horizontal sizes and the number of colors respectively.
Output
Print the answer to the problem. As the answer can be quite a large number, you should print it modulo 109 + 7 (1000000007).
Examples
Input
2 2 1
Output
1
Input
2 2 2
Output
8
Input
3 2 2
Output
40 | instruction | 0 | 8,397 | 7 | 16,794 |
Tags: combinatorics, dp
Correct Solution:
```
M, Mod = 10**3 + 3, 10**9 + 7
n, m, k = [int(x) for x in input().split()]
if m == 0:
print(0)
elif m == 1:
print(pow(k, n, Mod))
else:
C = [[0 for j in range(M)] for i in range(M)]
for i in range(n + 1):
C[i][0] = 1
for j in range(1, i + 1):
C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % Mod
F = [0 for j in range(M)]
for i in range(1, n + 1):
F[i] = pow(i, n, Mod)
for j in range(1, i):
F[i] = (F[i] - F[j] * C[i][j] % Mod + Mod) % Mod
R1, R2 = [1], [1]
for i in range(1, n + 1):
R1.append(R1[i-1] * pow(i, Mod-2, Mod) % Mod)
for i in range(1, 2 * n + 1):
if i > k:
break
R2.append(R2[i - 1] * (k - i + 1)%Mod)
tot = 0
for i in range(n + 1):
x = pow(i,(m-2)*n,Mod)
for j in range(n - i + 1):
if i + 2 * j > k:
break
tot = (tot + R2[2 * j + i] * R1[j] % Mod * R1[j] % Mod * R1[i] % Mod * F[i + j] % Mod * F[i + j] % Mod * x) % Mod;
print(tot)
``` | output | 1 | 8,397 | 7 | 16,795 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Petya loves counting. He wants to count the number of ways to paint a rectangular checkered board of size n × m (n rows, m columns) in k colors. Besides, the coloring should have the following property: for any vertical line that passes along the grid lines and divides the board in two non-empty parts the number of distinct colors in both these parts should be the same. Help Petya to count these colorings.
Input
The first line contains space-separated integers n, m and k (1 ≤ n, m ≤ 1000, 1 ≤ k ≤ 106) — the board's vertical and horizontal sizes and the number of colors respectively.
Output
Print the answer to the problem. As the answer can be quite a large number, you should print it modulo 109 + 7 (1000000007).
Examples
Input
2 2 1
Output
1
Input
2 2 2
Output
8
Input
3 2 2
Output
40 | instruction | 0 | 8,398 | 7 | 16,796 |
Tags: combinatorics, dp
Correct Solution:
```
n, m, k = [int(x) for x in input().split()]
mod = int(10**9 + 7)
if m == 1:
print(pow(k, n, mod))
exit()
ans = 0
fac = [1]
ifac = [1]
tav = [0]
for i in range(1, max(k + 1, n + 1)):
fac.append((fac[i - 1] * i) % mod)
for i in range(1, min(k + 1, n + 1)):
ifac.append((ifac[i - 1] * pow(i, mod - 2, mod)) % mod)
tav.append(pow(i, n * (m - 2), mod))
if m == 2:
tav[0] = 1
dp = [0] * (n + 1)
dp[1] = 1
for i in range(2, n + 1):
for j in range(i, 1, -1):
dp[j] = (j * dp[j] + dp[j - 1]) % mod
for i in range(1, min(k + 1, n + 1)):
for j in range(0, i + 1):
if 2 * i - j > k:
continue
ans = (ans + ((((dp[i] * dp[i] * fac[i] * fac[i]) % mod) * tav[j] * ifac[i - j] * fac[k]) % mod) * pow(fac[k - 2 * i + j], mod - 2, mod) * ifac[i - j] * ifac[j]) % mod
print(ans)
``` | output | 1 | 8,398 | 7 | 16,797 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Petya loves counting. He wants to count the number of ways to paint a rectangular checkered board of size n × m (n rows, m columns) in k colors. Besides, the coloring should have the following property: for any vertical line that passes along the grid lines and divides the board in two non-empty parts the number of distinct colors in both these parts should be the same. Help Petya to count these colorings.
Input
The first line contains space-separated integers n, m and k (1 ≤ n, m ≤ 1000, 1 ≤ k ≤ 106) — the board's vertical and horizontal sizes and the number of colors respectively.
Output
Print the answer to the problem. As the answer can be quite a large number, you should print it modulo 109 + 7 (1000000007).
Examples
Input
2 2 1
Output
1
Input
2 2 2
Output
8
Input
3 2 2
Output
40
Submitted Solution:
```
from math import factorial
MOD = 1000000007
n,m,k = map(int,input().split())
if m==1:
print(k**m%MOD)
exit()
def Comb(n,k): #Hallar las combinaciones de n en k
comb= factorial(n)/factorial(k)*factorial(n-k)
return int(comb)%MOD
def ColorearCeldas(c): #Cantidad de formas de rellenar n celdas con exactamente c colores
resp = c**n #Toda las posibles formas de distribuir c colores en n celdas
for i in range(1,c):
resp-= Comb(c,i)*ColorearCeldas(i) #Le restamos la cantidad de formas en que coloreamos con menos de c colores, multiplicado por la cantidad de formas en que podemos escoger los i colores
return resp%MOD
resp = 0
for y in range(1,min(k,2*n)+1): #Iteramos por todos los posibles colores usados para colorear el tablero entero
for x in range(y//2+1): #Iterar por todos los posibles colores unicos en la primera columna y consecuentemente los posibles colores de la ultoma columna
cc= y-2*x #Colores comunes
pc=Comb(k,y) #Escogemos los colores utilizados
pc*=Comb(y,x) #Formas de escoger del conjunto total los colores únicos para la columna 1
pc*=Comb(y-x,x) #Formas de elegir del resto los colores únicos para la última columna
po=ColorearCeldas(x+cc)**2 #Formas de organizar los colores elegidos en la primera y ultima columna
po*=cc**(n*(m-2)) #Formas de organizar los colores elegidos en las columnas del medio
resp= (resp+pc*po)%MOD
print(resp%MOD)
``` | instruction | 0 | 8,399 | 7 | 16,798 |
No | output | 1 | 8,399 | 7 | 16,799 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Petya loves counting. He wants to count the number of ways to paint a rectangular checkered board of size n × m (n rows, m columns) in k colors. Besides, the coloring should have the following property: for any vertical line that passes along the grid lines and divides the board in two non-empty parts the number of distinct colors in both these parts should be the same. Help Petya to count these colorings.
Input
The first line contains space-separated integers n, m and k (1 ≤ n, m ≤ 1000, 1 ≤ k ≤ 106) — the board's vertical and horizontal sizes and the number of colors respectively.
Output
Print the answer to the problem. As the answer can be quite a large number, you should print it modulo 109 + 7 (1000000007).
Examples
Input
2 2 1
Output
1
Input
2 2 2
Output
8
Input
3 2 2
Output
40
Submitted Solution:
```
def inv(x):
return pow(x, mod - 2, mod)
def C(n, k):
if (k > n):
return 0
return f[n] * inv(f[k]) % mod * inv(f[n - k]) % mod
def calc(x, y):
if not y:
return pow(x, n, mod) - pow(x - 1, n, mod)
return (pow(x, n, mod) * C(k, y) - pow(x - 1, n, mod) * C(k, y - 1)) % mod
n, m, k = map(int, input().split())
mod = 1000000007
f = [1]
for i in range(1, 1000001):
f.append(f[-1] * i % mod)
ans = 0
if m == 1:
print(pow(k, n, mod))
exit()
for x in range(1, min(k, n) + 1):
if m == 2:
ans += calc(x, x) * calc(x, x)
ans %= mod
else:
for y in range(0, x + 1):
ans += calc(x, x) * calc(x, x - y) % mod * pow(y, n * (m - 2), mod)
ans %= mod
print(ans)
``` | instruction | 0 | 8,400 | 7 | 16,800 |
No | output | 1 | 8,400 | 7 | 16,801 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Petya loves counting. He wants to count the number of ways to paint a rectangular checkered board of size n × m (n rows, m columns) in k colors. Besides, the coloring should have the following property: for any vertical line that passes along the grid lines and divides the board in two non-empty parts the number of distinct colors in both these parts should be the same. Help Petya to count these colorings.
Input
The first line contains space-separated integers n, m and k (1 ≤ n, m ≤ 1000, 1 ≤ k ≤ 106) — the board's vertical and horizontal sizes and the number of colors respectively.
Output
Print the answer to the problem. As the answer can be quite a large number, you should print it modulo 109 + 7 (1000000007).
Examples
Input
2 2 1
Output
1
Input
2 2 2
Output
8
Input
3 2 2
Output
40
Submitted Solution:
```
#!/usr/bin/env python3
import math
MODULO = 1000000007
def nr_choices(total, to_pick):
# n choose k function
return int(math.factorial(total) / math.factorial(to_pick) * math.factorial(total-to_pick))
(n,m,k) = map(int, input().split())
prev_combinations = k
answer = k**m % MODULO
for nr_colors in range(2, k+1):
new_combinations = ((nr_colors**n) - prev_combinations) % MODULO
prev_combinations = new_combinations
answer += nr_choices(k, nr_colors) * new_combinations**m
print(answer % MODULO)
``` | instruction | 0 | 8,401 | 7 | 16,802 |
No | output | 1 | 8,401 | 7 | 16,803 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Petya loves counting. He wants to count the number of ways to paint a rectangular checkered board of size n × m (n rows, m columns) in k colors. Besides, the coloring should have the following property: for any vertical line that passes along the grid lines and divides the board in two non-empty parts the number of distinct colors in both these parts should be the same. Help Petya to count these colorings.
Input
The first line contains space-separated integers n, m and k (1 ≤ n, m ≤ 1000, 1 ≤ k ≤ 106) — the board's vertical and horizontal sizes and the number of colors respectively.
Output
Print the answer to the problem. As the answer can be quite a large number, you should print it modulo 109 + 7 (1000000007).
Examples
Input
2 2 1
Output
1
Input
2 2 2
Output
8
Input
3 2 2
Output
40
Submitted Solution:
```
def inv(x):
return pow(x, mod - 2, mod)
def C(n, k):
if (k > n):
return 0
return f[n] * inv(f[k]) % mod * inv(f[n - k]) % mod
def calc(x):
return (pow(x, n, mod) * C(k, x) - pow(x - 1, n, mod) * C(k, x - 1)) % mod
n, m, k = map(int, input().split())
mod = 1000000007
f = [1]
for i in range(1, 1000001):
f.append(f[-1] * i % mod)
ans = 0
for x in range(1, k + 1):
if (2 * x >= k):
ans += calc(x) * calc(x) % mod * pow(max(1, n * (m - 2)), 2 * x - k, mod) % mod
print(ans)
``` | instruction | 0 | 8,402 | 7 | 16,804 |
No | output | 1 | 8,402 | 7 | 16,805 |
Provide a correct Python 3 solution for this coding contest problem.
Balph is learning to play a game called Buma. In this game, he is given a row of colored balls. He has to choose the color of one new ball and the place to insert it (between two balls, or to the left of all the balls, or to the right of all the balls).
When the ball is inserted the following happens repeatedly: if some segment of balls of the same color became longer as a result of a previous action and its length became at least 3, then all the balls of this segment are eliminated.
Consider, for example, a row of balls 'AAABBBWWBB'. Suppose Balph chooses a ball of color 'W' and the place to insert it after the sixth ball, i. e. to the left of the two 'W's. After Balph inserts this ball, the balls of color 'W' are eliminated, since this segment was made longer and has length 3 now, so the row becomes 'AAABBBBB'. The balls of color 'B' are eliminated now, because the segment of balls of color 'B' became longer and has length 5 now. Thus, the row becomes 'AAA'. However, none of the balls are eliminated now, because there is no elongated segment.
Help Balph count the number of possible ways to choose a color of a new ball and a place to insert it that leads to the elimination of all the balls.
Input
The only line contains a non-empty string of uppercase English letters of length at most 3 ⋅ 10^5. Each letter represents a ball with the corresponding color.
Output
Output the number of ways to choose a color and a position of a new ball in order to eliminate all the balls.
Examples
Input
BBWWBB
Output
3
Input
BWWB
Output
0
Input
BBWBB
Output
0
Input
OOOWWW
Output
0
Input
WWWOOOOOOWWW
Output
7 | instruction | 0 | 8,462 | 7 | 16,924 |
"Correct Solution:
```
s=list(input())
i=0
l=len(s)-1
t=0
while i<=l :
a=s[i]
c=0
if s[i]!=s[l]:
print(0)
exit()
while i<=l and s[i]==a:
c+=1
i+=1
while i<=l and s[l]==a:
c+=1
l-=1
if c<=2 and i<l:
print(0)
exit()
if c>1:
print(c+1)
else:
print(0)
``` | output | 1 | 8,462 | 7 | 16,925 |
Provide a correct Python 3 solution for this coding contest problem.
Balph is learning to play a game called Buma. In this game, he is given a row of colored balls. He has to choose the color of one new ball and the place to insert it (between two balls, or to the left of all the balls, or to the right of all the balls).
When the ball is inserted the following happens repeatedly: if some segment of balls of the same color became longer as a result of a previous action and its length became at least 3, then all the balls of this segment are eliminated.
Consider, for example, a row of balls 'AAABBBWWBB'. Suppose Balph chooses a ball of color 'W' and the place to insert it after the sixth ball, i. e. to the left of the two 'W's. After Balph inserts this ball, the balls of color 'W' are eliminated, since this segment was made longer and has length 3 now, so the row becomes 'AAABBBBB'. The balls of color 'B' are eliminated now, because the segment of balls of color 'B' became longer and has length 5 now. Thus, the row becomes 'AAA'. However, none of the balls are eliminated now, because there is no elongated segment.
Help Balph count the number of possible ways to choose a color of a new ball and a place to insert it that leads to the elimination of all the balls.
Input
The only line contains a non-empty string of uppercase English letters of length at most 3 ⋅ 10^5. Each letter represents a ball with the corresponding color.
Output
Output the number of ways to choose a color and a position of a new ball in order to eliminate all the balls.
Examples
Input
BBWWBB
Output
3
Input
BWWB
Output
0
Input
BBWBB
Output
0
Input
OOOWWW
Output
0
Input
WWWOOOOOOWWW
Output
7 | instruction | 0 | 8,463 | 7 | 16,926 |
"Correct Solution:
```
from collections import defaultdict as dd
s=input()
l=[s[0]]
n=s[0]
d=dd(int)
d[s[0]]+=1
cou=1
li=[]
for i in range(1,len(s)):
d[s[i]]+=1
if(s[i]==n):
cou+=1
if(s[i]!=n):
l.append(s[i])
n=s[i]
li.append(cou)
cou=1
li.append(cou)
#print(l,d)
if(l==l[::-1]):
mid=len(l)//2
lol=0
for i in range(len(li)):
j=len(li)-i-1
if(i!=j and li[i]+li[j]<=2):
lol=1
break
if(i==j and li[i]<2):
lol=1
break
if(lol):
print(0)
else:
print(li[mid]+1)
else:
print(0)
``` | output | 1 | 8,463 | 7 | 16,927 |
Provide a correct Python 3 solution for this coding contest problem.
Balph is learning to play a game called Buma. In this game, he is given a row of colored balls. He has to choose the color of one new ball and the place to insert it (between two balls, or to the left of all the balls, or to the right of all the balls).
When the ball is inserted the following happens repeatedly: if some segment of balls of the same color became longer as a result of a previous action and its length became at least 3, then all the balls of this segment are eliminated.
Consider, for example, a row of balls 'AAABBBWWBB'. Suppose Balph chooses a ball of color 'W' and the place to insert it after the sixth ball, i. e. to the left of the two 'W's. After Balph inserts this ball, the balls of color 'W' are eliminated, since this segment was made longer and has length 3 now, so the row becomes 'AAABBBBB'. The balls of color 'B' are eliminated now, because the segment of balls of color 'B' became longer and has length 5 now. Thus, the row becomes 'AAA'. However, none of the balls are eliminated now, because there is no elongated segment.
Help Balph count the number of possible ways to choose a color of a new ball and a place to insert it that leads to the elimination of all the balls.
Input
The only line contains a non-empty string of uppercase English letters of length at most 3 ⋅ 10^5. Each letter represents a ball with the corresponding color.
Output
Output the number of ways to choose a color and a position of a new ball in order to eliminate all the balls.
Examples
Input
BBWWBB
Output
3
Input
BWWB
Output
0
Input
BBWBB
Output
0
Input
OOOWWW
Output
0
Input
WWWOOOOOOWWW
Output
7 | instruction | 0 | 8,464 | 7 | 16,928 |
"Correct Solution:
```
#!/usr/bin/env python3
from functools import lru_cache, reduce
from sys import setrecursionlimit
import math
ran = []
def success(i):
if ran[i][1]+1<3:
return False
l = i
r = i
while l>=0 and r<len(ran):
if ran[l][0] != ran[r][0] or ran[l][1] + ran[r][1] < 3:
return False
l -= 1
r += 1
if (l<0) ^ (r>=len(ran)):
return False
return True
if __name__ == "__main__":
for v in input().strip():
if len(ran)>0:
if ran[-1][0] == v:
ran[-1][1] += 1
else:
ran.append([v,1])
else:
ran.append([v,1])
res = 0
#for i in range(len(ran)):
if len(ran) % 2 != 0:
i = len(ran) // 2
if(success(i)):
res=ran[i][1]+1
print(res)
``` | output | 1 | 8,464 | 7 | 16,929 |
Provide a correct Python 3 solution for this coding contest problem.
Balph is learning to play a game called Buma. In this game, he is given a row of colored balls. He has to choose the color of one new ball and the place to insert it (between two balls, or to the left of all the balls, or to the right of all the balls).
When the ball is inserted the following happens repeatedly: if some segment of balls of the same color became longer as a result of a previous action and its length became at least 3, then all the balls of this segment are eliminated.
Consider, for example, a row of balls 'AAABBBWWBB'. Suppose Balph chooses a ball of color 'W' and the place to insert it after the sixth ball, i. e. to the left of the two 'W's. After Balph inserts this ball, the balls of color 'W' are eliminated, since this segment was made longer and has length 3 now, so the row becomes 'AAABBBBB'. The balls of color 'B' are eliminated now, because the segment of balls of color 'B' became longer and has length 5 now. Thus, the row becomes 'AAA'. However, none of the balls are eliminated now, because there is no elongated segment.
Help Balph count the number of possible ways to choose a color of a new ball and a place to insert it that leads to the elimination of all the balls.
Input
The only line contains a non-empty string of uppercase English letters of length at most 3 ⋅ 10^5. Each letter represents a ball with the corresponding color.
Output
Output the number of ways to choose a color and a position of a new ball in order to eliminate all the balls.
Examples
Input
BBWWBB
Output
3
Input
BWWB
Output
0
Input
BBWBB
Output
0
Input
OOOWWW
Output
0
Input
WWWOOOOOOWWW
Output
7 | instruction | 0 | 8,465 | 7 | 16,930 |
"Correct Solution:
```
s = list(input())
p0 = 0
p1 = len(s)-1
if(s[p0]!= s[p1]):
print(0)
else:
possible = 1
curr_len = 2
while p1 > p0:
if s[p0+1] == s[p0]:
curr_len += 1
p0+=1
elif s[p1-1] == s[p1]:
curr_len += 1
p1-=1
else:
if(curr_len<3 or s[p0+1]!=s[p1-1]):
possible = 0
break
else:
p0+=1
p1-=1
curr_len = 2
if not possible:
print(0)
else:
if p1==p0:
if(curr_len>2):
print(curr_len)
else:
print(0)
else:
print(0)
``` | output | 1 | 8,465 | 7 | 16,931 |
Provide a correct Python 3 solution for this coding contest problem.
Balph is learning to play a game called Buma. In this game, he is given a row of colored balls. He has to choose the color of one new ball and the place to insert it (between two balls, or to the left of all the balls, or to the right of all the balls).
When the ball is inserted the following happens repeatedly: if some segment of balls of the same color became longer as a result of a previous action and its length became at least 3, then all the balls of this segment are eliminated.
Consider, for example, a row of balls 'AAABBBWWBB'. Suppose Balph chooses a ball of color 'W' and the place to insert it after the sixth ball, i. e. to the left of the two 'W's. After Balph inserts this ball, the balls of color 'W' are eliminated, since this segment was made longer and has length 3 now, so the row becomes 'AAABBBBB'. The balls of color 'B' are eliminated now, because the segment of balls of color 'B' became longer and has length 5 now. Thus, the row becomes 'AAA'. However, none of the balls are eliminated now, because there is no elongated segment.
Help Balph count the number of possible ways to choose a color of a new ball and a place to insert it that leads to the elimination of all the balls.
Input
The only line contains a non-empty string of uppercase English letters of length at most 3 ⋅ 10^5. Each letter represents a ball with the corresponding color.
Output
Output the number of ways to choose a color and a position of a new ball in order to eliminate all the balls.
Examples
Input
BBWWBB
Output
3
Input
BWWB
Output
0
Input
BBWBB
Output
0
Input
OOOWWW
Output
0
Input
WWWOOOOOOWWW
Output
7 | instruction | 0 | 8,466 | 7 | 16,932 |
"Correct Solution:
```
from itertools import groupby
s = input()
lst = []
grp_i = []
for i, grp in groupby(s):
grp_i.append(i)
lst.append(list(grp))
if len(grp_i) % 2 == 0 or grp_i != grp_i[::-1]:
print(0)
else:
mid = grp_i[len(grp_i)//2]
nm = len(grp_i)//2
l = nm -1
r = nm + 1
ln = len(grp_i)
while l >= 0 and r < ln:
if len(lst[l]) + len(lst[r]) < 3:
print(0)
exit()
l -= 1
r += 1
if len(lst[len(grp_i)//2]) < 2:
print(0)
else:
ans = len(lst[len(grp_i)//2])+1
print(ans)
``` | output | 1 | 8,466 | 7 | 16,933 |
Provide a correct Python 3 solution for this coding contest problem.
Balph is learning to play a game called Buma. In this game, he is given a row of colored balls. He has to choose the color of one new ball and the place to insert it (between two balls, or to the left of all the balls, or to the right of all the balls).
When the ball is inserted the following happens repeatedly: if some segment of balls of the same color became longer as a result of a previous action and its length became at least 3, then all the balls of this segment are eliminated.
Consider, for example, a row of balls 'AAABBBWWBB'. Suppose Balph chooses a ball of color 'W' and the place to insert it after the sixth ball, i. e. to the left of the two 'W's. After Balph inserts this ball, the balls of color 'W' are eliminated, since this segment was made longer and has length 3 now, so the row becomes 'AAABBBBB'. The balls of color 'B' are eliminated now, because the segment of balls of color 'B' became longer and has length 5 now. Thus, the row becomes 'AAA'. However, none of the balls are eliminated now, because there is no elongated segment.
Help Balph count the number of possible ways to choose a color of a new ball and a place to insert it that leads to the elimination of all the balls.
Input
The only line contains a non-empty string of uppercase English letters of length at most 3 ⋅ 10^5. Each letter represents a ball with the corresponding color.
Output
Output the number of ways to choose a color and a position of a new ball in order to eliminate all the balls.
Examples
Input
BBWWBB
Output
3
Input
BWWB
Output
0
Input
BBWBB
Output
0
Input
OOOWWW
Output
0
Input
WWWOOOOOOWWW
Output
7 | instruction | 0 | 8,467 | 7 | 16,934 |
"Correct Solution:
```
S = input().strip()
P = []
j = 0
for i in range(1, len(S)+1):
if i == len(S) or S[i] != S[j]:
P.append((S[j], i-j))
j = i
l = 0
r = len(P)-1
while l < r:
if P[l][0] != P[r][0] or P[l][1] + P[r][1] < 3:
break
l += 1
r -= 1
if l == r and P[l][1] >= 2:
print(P[l][1]+1)
else:
print(0)
``` | output | 1 | 8,467 | 7 | 16,935 |
Provide a correct Python 3 solution for this coding contest problem.
Balph is learning to play a game called Buma. In this game, he is given a row of colored balls. He has to choose the color of one new ball and the place to insert it (between two balls, or to the left of all the balls, or to the right of all the balls).
When the ball is inserted the following happens repeatedly: if some segment of balls of the same color became longer as a result of a previous action and its length became at least 3, then all the balls of this segment are eliminated.
Consider, for example, a row of balls 'AAABBBWWBB'. Suppose Balph chooses a ball of color 'W' and the place to insert it after the sixth ball, i. e. to the left of the two 'W's. After Balph inserts this ball, the balls of color 'W' are eliminated, since this segment was made longer and has length 3 now, so the row becomes 'AAABBBBB'. The balls of color 'B' are eliminated now, because the segment of balls of color 'B' became longer and has length 5 now. Thus, the row becomes 'AAA'. However, none of the balls are eliminated now, because there is no elongated segment.
Help Balph count the number of possible ways to choose a color of a new ball and a place to insert it that leads to the elimination of all the balls.
Input
The only line contains a non-empty string of uppercase English letters of length at most 3 ⋅ 10^5. Each letter represents a ball with the corresponding color.
Output
Output the number of ways to choose a color and a position of a new ball in order to eliminate all the balls.
Examples
Input
BBWWBB
Output
3
Input
BWWB
Output
0
Input
BBWBB
Output
0
Input
OOOWWW
Output
0
Input
WWWOOOOOOWWW
Output
7 | instruction | 0 | 8,468 | 7 | 16,936 |
"Correct Solution:
```
def main():
freq = []
symb = []
s = input()
symb.append(s[0]);
freq.append(1);
for ch in s[1:]:
if ch == symb[-1]:
freq[-1] += 1
continue
symb.append(ch)
freq.append(1)
if (len(freq) % 2 == 0) or (freq[len(freq) // 2] == 1):
print(0)
return
for i in range(len(freq) // 2):
if (symb[i] != symb[-i-1]) or (freq[i] + freq[-i-1] < 3):
print(0)
return
print(freq[len(freq) // 2] + 1)
if __name__ == '__main__':
main()
``` | output | 1 | 8,468 | 7 | 16,937 |
Provide a correct Python 3 solution for this coding contest problem.
Balph is learning to play a game called Buma. In this game, he is given a row of colored balls. He has to choose the color of one new ball and the place to insert it (between two balls, or to the left of all the balls, or to the right of all the balls).
When the ball is inserted the following happens repeatedly: if some segment of balls of the same color became longer as a result of a previous action and its length became at least 3, then all the balls of this segment are eliminated.
Consider, for example, a row of balls 'AAABBBWWBB'. Suppose Balph chooses a ball of color 'W' and the place to insert it after the sixth ball, i. e. to the left of the two 'W's. After Balph inserts this ball, the balls of color 'W' are eliminated, since this segment was made longer and has length 3 now, so the row becomes 'AAABBBBB'. The balls of color 'B' are eliminated now, because the segment of balls of color 'B' became longer and has length 5 now. Thus, the row becomes 'AAA'. However, none of the balls are eliminated now, because there is no elongated segment.
Help Balph count the number of possible ways to choose a color of a new ball and a place to insert it that leads to the elimination of all the balls.
Input
The only line contains a non-empty string of uppercase English letters of length at most 3 ⋅ 10^5. Each letter represents a ball with the corresponding color.
Output
Output the number of ways to choose a color and a position of a new ball in order to eliminate all the balls.
Examples
Input
BBWWBB
Output
3
Input
BWWB
Output
0
Input
BBWBB
Output
0
Input
OOOWWW
Output
0
Input
WWWOOOOOOWWW
Output
7 | instruction | 0 | 8,469 | 7 | 16,938 |
"Correct Solution:
```
#!/usr/bin/env python3
import sys
def main_proc():
res = 0
data = sys.stdin.buffer.read()
data = data.strip()
s = 0; e = len(data) - 1
while True:
cnt = 0
if data[s] != data[e]: break
t = s
while data[s] == data[t] and t != e: t += 1; cnt += 1
if t == e:
res = cnt + 2 if cnt >= 1 else res
break
else:
s = t
t = e
while data[t] == data[e]: cnt += 1; t -= 1
if cnt < 3: break
e = t
print(res)
if __name__ == '__main__':
main_proc()
``` | output | 1 | 8,469 | 7 | 16,939 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Balph is learning to play a game called Buma. In this game, he is given a row of colored balls. He has to choose the color of one new ball and the place to insert it (between two balls, or to the left of all the balls, or to the right of all the balls).
When the ball is inserted the following happens repeatedly: if some segment of balls of the same color became longer as a result of a previous action and its length became at least 3, then all the balls of this segment are eliminated.
Consider, for example, a row of balls 'AAABBBWWBB'. Suppose Balph chooses a ball of color 'W' and the place to insert it after the sixth ball, i. e. to the left of the two 'W's. After Balph inserts this ball, the balls of color 'W' are eliminated, since this segment was made longer and has length 3 now, so the row becomes 'AAABBBBB'. The balls of color 'B' are eliminated now, because the segment of balls of color 'B' became longer and has length 5 now. Thus, the row becomes 'AAA'. However, none of the balls are eliminated now, because there is no elongated segment.
Help Balph count the number of possible ways to choose a color of a new ball and a place to insert it that leads to the elimination of all the balls.
Input
The only line contains a non-empty string of uppercase English letters of length at most 3 ⋅ 10^5. Each letter represents a ball with the corresponding color.
Output
Output the number of ways to choose a color and a position of a new ball in order to eliminate all the balls.
Examples
Input
BBWWBB
Output
3
Input
BWWB
Output
0
Input
BBWBB
Output
0
Input
OOOWWW
Output
0
Input
WWWOOOOOOWWW
Output
7
Submitted Solution:
```
def main():
s=input()
# first check if it's possible...
n=len(s)
l=0
r=n-1
ok=True
while True:
if s[l]!=s[r]:
ok=False
break
c=s[l]
lCnt=0
rCnt=0
while s[l]==c and l<r:
l+=1
lCnt+=1
while s[r]==c and r>l:
r-=1
rCnt+=1
if l==r: # possible
break
if lCnt+rCnt<3:
ok=False
break
if ok:
c=s[l]
while l-1>=0 and s[l-1]==c:
l-=1
while r+1<n and s[r+1]==c:
r+=1
if l==r: # only 1 ball... cannot
print(0)
else:
print(r-l+2)
else:
print(0)
return
import sys
# input=sys.stdin.buffer.readline #FOR READING PURE INTEGER INPUTS (space separation ok)
input=lambda: sys.stdin.readline().rstrip("\r\n") #FOR READING STRING/TEXT INPUTS.
def oneLineArrayPrint(arr):
print(' '.join([str(x) for x in arr]))
def multiLineArrayPrint(arr):
print('\n'.join([str(x) for x in arr]))
def multiLineArrayOfArraysPrint(arr):
print('\n'.join([' '.join([str(x) for x in y]) for y in arr]))
def readIntArr():
return [int(x) for x in input().split()]
# def readFloatArr():
# return [float(x) for x in input().split()]
def makeArr(defaultVal,dimensionArr): # eg. makeArr(0,[n,m])
dv=defaultVal;da=dimensionArr
if len(da)==1:return [dv for _ in range(da[0])]
else:return [makeArr(dv,da[1:]) for _ in range(da[0])]
def queryInteractive(x,y):
print('? {} {}'.format(x,y))
sys.stdout.flush()
return int(input())
def answerInteractive(ans):
print('! {}'.format(ans))
sys.stdout.flush()
inf=float('inf')
MOD=10**9+7
for _abc in range(1):
main()
``` | instruction | 0 | 8,470 | 7 | 16,940 |
Yes | output | 1 | 8,470 | 7 | 16,941 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Balph is learning to play a game called Buma. In this game, he is given a row of colored balls. He has to choose the color of one new ball and the place to insert it (between two balls, or to the left of all the balls, or to the right of all the balls).
When the ball is inserted the following happens repeatedly: if some segment of balls of the same color became longer as a result of a previous action and its length became at least 3, then all the balls of this segment are eliminated.
Consider, for example, a row of balls 'AAABBBWWBB'. Suppose Balph chooses a ball of color 'W' and the place to insert it after the sixth ball, i. e. to the left of the two 'W's. After Balph inserts this ball, the balls of color 'W' are eliminated, since this segment was made longer and has length 3 now, so the row becomes 'AAABBBBB'. The balls of color 'B' are eliminated now, because the segment of balls of color 'B' became longer and has length 5 now. Thus, the row becomes 'AAA'. However, none of the balls are eliminated now, because there is no elongated segment.
Help Balph count the number of possible ways to choose a color of a new ball and a place to insert it that leads to the elimination of all the balls.
Input
The only line contains a non-empty string of uppercase English letters of length at most 3 ⋅ 10^5. Each letter represents a ball with the corresponding color.
Output
Output the number of ways to choose a color and a position of a new ball in order to eliminate all the balls.
Examples
Input
BBWWBB
Output
3
Input
BWWB
Output
0
Input
BBWBB
Output
0
Input
OOOWWW
Output
0
Input
WWWOOOOOOWWW
Output
7
Submitted Solution:
```
s = input()
j = len(s)-1
i = 0
while i<=j:
c = 0
p = s[i]
if p!=s[j]:
exit(print(0))
while i<=j and s[i]==p:
i+=1
c+=1
while i<=j and s[j]==p:
j-=1
c+=1
if c<=2 and i<j:
exit(print(0))
if c>1:
print(c+1)
else:
print(0)
``` | instruction | 0 | 8,471 | 7 | 16,942 |
Yes | output | 1 | 8,471 | 7 | 16,943 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Balph is learning to play a game called Buma. In this game, he is given a row of colored balls. He has to choose the color of one new ball and the place to insert it (between two balls, or to the left of all the balls, or to the right of all the balls).
When the ball is inserted the following happens repeatedly: if some segment of balls of the same color became longer as a result of a previous action and its length became at least 3, then all the balls of this segment are eliminated.
Consider, for example, a row of balls 'AAABBBWWBB'. Suppose Balph chooses a ball of color 'W' and the place to insert it after the sixth ball, i. e. to the left of the two 'W's. After Balph inserts this ball, the balls of color 'W' are eliminated, since this segment was made longer and has length 3 now, so the row becomes 'AAABBBBB'. The balls of color 'B' are eliminated now, because the segment of balls of color 'B' became longer and has length 5 now. Thus, the row becomes 'AAA'. However, none of the balls are eliminated now, because there is no elongated segment.
Help Balph count the number of possible ways to choose a color of a new ball and a place to insert it that leads to the elimination of all the balls.
Input
The only line contains a non-empty string of uppercase English letters of length at most 3 ⋅ 10^5. Each letter represents a ball with the corresponding color.
Output
Output the number of ways to choose a color and a position of a new ball in order to eliminate all the balls.
Examples
Input
BBWWBB
Output
3
Input
BWWB
Output
0
Input
BBWBB
Output
0
Input
OOOWWW
Output
0
Input
WWWOOOOOOWWW
Output
7
Submitted Solution:
```
s = input()
lista = []
aux = s[0]
cont = 0
for i in s:
if i == aux:
cont += 1
else:
lista.append([aux, cont])
cont = 1
aux = i
lista.append([aux, cont])
if len(lista)%2 == 0:
print(0)
else:
flag = False
while len(lista) > 1:
a = lista.pop(0)
b = lista.pop(len(lista) - 1)
if a[0] != b[0]:
flag = True
break
else:
if a[1] + b[1] < 3:
flag = True
break
if flag:
print(0)
else:
if lista[0][1] == 1:
print(0)
else:
print(lista[0][1] + 1)
``` | instruction | 0 | 8,472 | 7 | 16,944 |
Yes | output | 1 | 8,472 | 7 | 16,945 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Balph is learning to play a game called Buma. In this game, he is given a row of colored balls. He has to choose the color of one new ball and the place to insert it (between two balls, or to the left of all the balls, or to the right of all the balls).
When the ball is inserted the following happens repeatedly: if some segment of balls of the same color became longer as a result of a previous action and its length became at least 3, then all the balls of this segment are eliminated.
Consider, for example, a row of balls 'AAABBBWWBB'. Suppose Balph chooses a ball of color 'W' and the place to insert it after the sixth ball, i. e. to the left of the two 'W's. After Balph inserts this ball, the balls of color 'W' are eliminated, since this segment was made longer and has length 3 now, so the row becomes 'AAABBBBB'. The balls of color 'B' are eliminated now, because the segment of balls of color 'B' became longer and has length 5 now. Thus, the row becomes 'AAA'. However, none of the balls are eliminated now, because there is no elongated segment.
Help Balph count the number of possible ways to choose a color of a new ball and a place to insert it that leads to the elimination of all the balls.
Input
The only line contains a non-empty string of uppercase English letters of length at most 3 ⋅ 10^5. Each letter represents a ball with the corresponding color.
Output
Output the number of ways to choose a color and a position of a new ball in order to eliminate all the balls.
Examples
Input
BBWWBB
Output
3
Input
BWWB
Output
0
Input
BBWBB
Output
0
Input
OOOWWW
Output
0
Input
WWWOOOOOOWWW
Output
7
Submitted Solution:
```
from collections import deque
from collections import OrderedDict
import math
import sys
import os
from io import BytesIO
import threading
import bisect
import operator
import heapq
#sys.stdin = open("F:\PY\\test.txt", "r")
input = lambda: sys.stdin.readline().rstrip("\r\n")
s = input()
l = len(s)
dp = [s[0]]
dpC = [1]
idx = 0
for i in range(1,l):
if s[i]==s[i-1]:
dpC[idx]+=1
else:
idx+=1
dp.append(s[i])
dpC.append(1)
if len(dp)%2==0 or dpC[len(dpC)//2]<2:
print(0)
else:
for i in range(0,len(dp)//2):
#print(i, len(dp)-1-i)
if dpC[i]+dpC[len(dp)-1-i]<3 or dp[i]!=dp[len(dp)-1-i]:
print(0)
break
else:
print(dpC[len(dpC)//2]+1)
``` | instruction | 0 | 8,473 | 7 | 16,946 |
Yes | output | 1 | 8,473 | 7 | 16,947 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Balph is learning to play a game called Buma. In this game, he is given a row of colored balls. He has to choose the color of one new ball and the place to insert it (between two balls, or to the left of all the balls, or to the right of all the balls).
When the ball is inserted the following happens repeatedly: if some segment of balls of the same color became longer as a result of a previous action and its length became at least 3, then all the balls of this segment are eliminated.
Consider, for example, a row of balls 'AAABBBWWBB'. Suppose Balph chooses a ball of color 'W' and the place to insert it after the sixth ball, i. e. to the left of the two 'W's. After Balph inserts this ball, the balls of color 'W' are eliminated, since this segment was made longer and has length 3 now, so the row becomes 'AAABBBBB'. The balls of color 'B' are eliminated now, because the segment of balls of color 'B' became longer and has length 5 now. Thus, the row becomes 'AAA'. However, none of the balls are eliminated now, because there is no elongated segment.
Help Balph count the number of possible ways to choose a color of a new ball and a place to insert it that leads to the elimination of all the balls.
Input
The only line contains a non-empty string of uppercase English letters of length at most 3 ⋅ 10^5. Each letter represents a ball with the corresponding color.
Output
Output the number of ways to choose a color and a position of a new ball in order to eliminate all the balls.
Examples
Input
BBWWBB
Output
3
Input
BWWB
Output
0
Input
BBWBB
Output
0
Input
OOOWWW
Output
0
Input
WWWOOOOOOWWW
Output
7
Submitted Solution:
```
s = input()
n = len(s)
a = ''
b = []
t = 1
for i in range(n-1):
if s[i] != s[i+1]:
a += s[i]
b.append(t)
t = 1
else:
t += 1
a += s[n-1]
b.append(t)
k = len(a)
if s == 'BWWB' or s == 'BBWBB' or s == 'OOOWWW' or 'ABCDEF':
print(0)
else:
if k % 2 == 0:
print(1)
else:
if k > 1:
for i in range((k - 1) // 2):
if a[i] != a[k - 1 - i] or b[i] + b[k - 1 - i] < 3:
print(2)
break
elif i == ((k - 1) // 2 - 1) and b[(k + 1) // 2 - 1] >= 2:
print(b[(k + 1) // 2 - 1] + 1)
break
else:
print(3)
break
elif b[0] >= 2:
print(b[0] + 1)
else:
print(4)
``` | instruction | 0 | 8,474 | 7 | 16,948 |
No | output | 1 | 8,474 | 7 | 16,949 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Balph is learning to play a game called Buma. In this game, he is given a row of colored balls. He has to choose the color of one new ball and the place to insert it (between two balls, or to the left of all the balls, or to the right of all the balls).
When the ball is inserted the following happens repeatedly: if some segment of balls of the same color became longer as a result of a previous action and its length became at least 3, then all the balls of this segment are eliminated.
Consider, for example, a row of balls 'AAABBBWWBB'. Suppose Balph chooses a ball of color 'W' and the place to insert it after the sixth ball, i. e. to the left of the two 'W's. After Balph inserts this ball, the balls of color 'W' are eliminated, since this segment was made longer and has length 3 now, so the row becomes 'AAABBBBB'. The balls of color 'B' are eliminated now, because the segment of balls of color 'B' became longer and has length 5 now. Thus, the row becomes 'AAA'. However, none of the balls are eliminated now, because there is no elongated segment.
Help Balph count the number of possible ways to choose a color of a new ball and a place to insert it that leads to the elimination of all the balls.
Input
The only line contains a non-empty string of uppercase English letters of length at most 3 ⋅ 10^5. Each letter represents a ball with the corresponding color.
Output
Output the number of ways to choose a color and a position of a new ball in order to eliminate all the balls.
Examples
Input
BBWWBB
Output
3
Input
BWWB
Output
0
Input
BBWBB
Output
0
Input
OOOWWW
Output
0
Input
WWWOOOOOOWWW
Output
7
Submitted Solution:
```
def build_ball_dictionary(ball_arrangement):
"""
Helper function for 'compute_ways_2_eliminate_all_balls' that builds a
dictionary that maps the starting position of a ball to a list that
contains the color of the ball as well as 1 + the number of balls of the
same color that follow it.
This function only needs to be called once.
Input:
ball_arrangement : <str>
# Contains arrangement of original row of balls
# Output :
ball_dictionary : < <int> : <str, int> >
Dictionary that maps the starting position of a ball to a list that
contains the color of the ball as well as 1 + the number of balls of the
same color that follow consecutively.
num_unique_colors : int
Represents the number of unique colors found in the initial set of balls
"""
# Initialize dictionary object
ball_dictionary = {}
# Initialize list object
unique_ball_colors_list = []
# Initialize counter that points to a location in the string representing
# the initial set of balls
index = 0
# Get number of balls at game's outset
n_balls = len(ball_arrangement)
while index < n_balls:
# Get the color of the ball at location 'index' in the
# ball_arrangement variable
current_ball_color = ball_arrangement[index]
# Reset length of current ball segment that contains all balls of
# the same color
current_segment_length = 1
# If ball color has not yet been encountered, add to list tracking
# unique ball colors
if current_ball_color not in unique_ball_colors_list:
unique_ball_colors_list.append(current_ball_color)
# Continue extending search towards the end of the original string
# until a new ball color is found. However, if we are at the end of
# the string, then there is no need to extend search (unless we want
# an 'index out of bounds' exception error)
while index < n_balls and index+current_segment_length < n_balls \
and ball_arrangement[index + current_segment_length] == \
current_ball_color:
current_segment_length += 1
ball_dictionary[index] = [current_ball_color, current_segment_length]
index += current_segment_length
# Get number of unique colors in row of balls
if "X"*int(511) in ball_arrangement:
print(unique_ball_colors_list)
num_unique_colors = len(unique_ball_colors_list)
return ball_dictionary, num_unique_colors
def compute_ways_2_eliminate_all_balls(ball_arrangement):
"""
Balph is learning to play a game called Buma. In this game, he is given
a row of colored balls. He has to choose the color of one new ball as
well as the place to insert it (between two balls, or to the left
of all the balls, or to the right of all the balls).
When the ball is inserted the following happens repeatedly: if some
segment of balls of the same color became longer as a result of a
previous action, and the length of this segment becomes at least 3,
then all the balls of this segment are eliminated.
Consider, for example, a row of balls 'AAABBBWWBB'. Suppose Balph
chooses a ball of color 'W' and the places it after the
sixth ball, i. e. to the left of the two 'W's. After Balph inserts
this ball, the balls of color 'W' are eliminated, since this segment
was made longer and now has length 3. The row then becomes 'AAABBBBB'.
The balls of color 'B' are now eliminated, because the segment of balls
of color 'B' became longer, and have length 5 now. Thus, the row becomes
'AAA'. However, balls of color 'A' remain, because they have not become
elongated (i.e. the number of consecutive A's did not change).
Help Balph count the number of possible ways to choose a color of a
new ball, and a place to insert it such that the insertion leads to the
elimination of all balls.
Input :
colored_balls_arrangement : <str>
Contains a non-empty string of uppercase English letters of length at
most 3*10^5. Each letter represents a ball with the corresponding color.
Output :
Output the number of ways to choose a color and a position of a new ball
in order to eliminate all the balls.
"""
# Condition 1:
# In order for any segment of colored balls to be eliminated, a ball of
# the same color must be placed next to a segment that contains at least
# two balls of that color. Ball can be inserted anywhere next to that
# segment.
# Create a dictionary from 'ball_arrangement'
ball_segments_map, n_unique_colors = build_ball_dictionary(ball_arrangement)
# Since the keys of 'ball_dict' refer to starting positions of a distinct
# ball segment, we can use a list of sorted keys to look through the
# collection of balls in an ordered manner.
sorted_ball_segments_by_pos = sorted(ball_segments_map.keys())
# There are several situations where it is quick and easy to determine
# that a total elimination of the original row of balls is not possible.
# This is the case when:
# 1) There are an even number of distinct ball segments in the original
# row of balls. Can be determined by the number of key entries in the
# ball segments map/dictionary
# 2) The total number of balls is fewer than 1 less than three times
# the number of uniquely colored ball colors
# 3) The ball segments at each end of the original row of balls do not
# share the same color
# 4) The ball segments at each end share the same color, but their sum
# does not exceed 2
# Condition 1 for no solution
if len(sorted_ball_segments_by_pos) // 2 == 0:
print(0)
return 0
# Condition 2 for no solution
if len(ball_arrangement) < 3 * n_unique_colors - 1:
print(0)
return 0
first_ball_segment_pos = sorted_ball_segments_by_pos[0]
last_ball_segment_pos = sorted_ball_segments_by_pos[-1]
first_ball_segment_color = ball_segments_map[first_ball_segment_pos][0]
last_ball_segment_color = ball_segments_map[last_ball_segment_pos][0]
# Condition 3 for no solution
if first_ball_segment_color != last_ball_segment_color:
print(0)
return 0
first_ball_segment_length = ball_segments_map[first_ball_segment_pos][1]
last_ball_segment_length = ball_segments_map[last_ball_segment_pos][1]
# Condition 4 for no solution
if first_ball_segment_length + last_ball_segment_length < 3:
print(0)
return 0
# The key to determining if all balls can be eliminated is by starting from
# the ball segment that has an equal number of ball segments on either
# side of it. If there is no complete collapse of balls by inserting a ball
# into the middle segment, then no solution exists.
# Find middle index of the sorted key list
middle_ball_segment_index = len(sorted_ball_segments_by_pos)//2
# Get the starting position of the current ball segment<key> from the
# list of sorted ball segment positions<list of keys> using the
# current ball segment index<int>
middle_ball_segment_pos = \
sorted_ball_segments_by_pos[middle_ball_segment_index]
# Determine if the length of the ball segment (i.e. the 2nd element of
# the list) associated with the middle ball segment is greater than
# or equal to 2.
min_initial_length = 2
if ball_segments_map[middle_ball_segment_pos][1] >= min_initial_length:
# If length requirement is satisfied, determine whether the ball
# segments equidistant from the the middle ball segment share the
# same color
count_left_segments = 1
count_right_segments = 1
# Continue process outwards until either:
# 1) one end of the ball segment dictionary has been reached or
# 2) surrounding ball segments do not share the same ball color or
# 3) the sum of the outer two surrounding ball segments is not of
# sufficient length.
# Ensure that the bounds of the sorted key list are not exceeded
while middle_ball_segment_index - count_left_segments >= 0:
left_segment_index = middle_ball_segment_index - \
count_left_segments
right_segment_index = middle_ball_segment_index + \
count_right_segments
left_ball_segment_pos = \
sorted_ball_segments_by_pos[left_segment_index]
right_ball_segment_pos =\
sorted_ball_segments_by_pos[right_segment_index]
if ball_segments_map[left_ball_segment_pos][0] == \
ball_segments_map[right_ball_segment_pos][0]:
# If the two ball segments equidistant from the middle
# ball segment are of the same color, check if the sum of
# the lengths of those two ball segments is greater than 3.
left_segment_length = ball_segments_map[
left_ball_segment_pos][1]
right_segment_length = ball_segments_map[
right_ball_segment_pos][1]
# If ball segments equidistant from middle ball segment
# satisfy all constraints, continue search on the next two
# ball segments outwards from the middle.
if left_segment_length + right_segment_index >= 3:
count_left_segments += 1
count_right_segments += 1
# If the ball segments equidistant from the middle ball
# segment share the same color but, when joined, do not
# become elongated enough to exceed a length of three,
# end search, print zero, and return no solution.
else:
print(0)
return 0
# If ball segments equidistant from middle ball segment do not
# share the same color, end search, print zero, and return no
# solution
else:
print(0)
return 0
# if end of while loop is reached, then all balls have been
# successfully eliminated. Number of solutions is just the length of
# the middle ball segment + 1
num_solutions = ball_segments_map[middle_ball_segment_pos][1]+1
print(num_solutions)
# if middle ball segment does not have at least two balls of the same
# color, then end search, print 0, and return no solution
else:
print(0)
return 0
if __name__ == "__main__":
new_ball_arrangement = input()
compute_ways_2_eliminate_all_balls(new_ball_arrangement)
# Examples
# inputCopy
# BBWWBB
# outputCopy
# 3
# inputCopy
# BWWB
# outputCopy
# 0
# inputCopy
# BBWBB
# outputCopy
# 0
# inputCopy
# OOOWWW
# outputCopy
# 0
# inputCopy
# WWWOOOOOOWWW
# outputCopy
# 7
``` | instruction | 0 | 8,475 | 7 | 16,950 |
No | output | 1 | 8,475 | 7 | 16,951 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Balph is learning to play a game called Buma. In this game, he is given a row of colored balls. He has to choose the color of one new ball and the place to insert it (between two balls, or to the left of all the balls, or to the right of all the balls).
When the ball is inserted the following happens repeatedly: if some segment of balls of the same color became longer as a result of a previous action and its length became at least 3, then all the balls of this segment are eliminated.
Consider, for example, a row of balls 'AAABBBWWBB'. Suppose Balph chooses a ball of color 'W' and the place to insert it after the sixth ball, i. e. to the left of the two 'W's. After Balph inserts this ball, the balls of color 'W' are eliminated, since this segment was made longer and has length 3 now, so the row becomes 'AAABBBBB'. The balls of color 'B' are eliminated now, because the segment of balls of color 'B' became longer and has length 5 now. Thus, the row becomes 'AAA'. However, none of the balls are eliminated now, because there is no elongated segment.
Help Balph count the number of possible ways to choose a color of a new ball and a place to insert it that leads to the elimination of all the balls.
Input
The only line contains a non-empty string of uppercase English letters of length at most 3 ⋅ 10^5. Each letter represents a ball with the corresponding color.
Output
Output the number of ways to choose a color and a position of a new ball in order to eliminate all the balls.
Examples
Input
BBWWBB
Output
3
Input
BWWB
Output
0
Input
BBWBB
Output
0
Input
OOOWWW
Output
0
Input
WWWOOOOOOWWW
Output
7
Submitted Solution:
```
from collections import defaultdict as dd
s=input()
l=[s[0]]
n=s[0]
d=dd(int)
d[s[0]]+=1
cou=1
for i in range(1,len(s)):
d[s[i]]+=1
if(s[i]!=n):
l.append(s[i])
n=s[i]
if(l==l[::-1]):
mid=len(l)//2
lol=0
for i in d:
if(d[i]<=2 and i!=l[mid]):
lol=1
if(lol):
print(0)
else:
print(d[l[mid]]+1)
else:
print(0)
``` | instruction | 0 | 8,476 | 7 | 16,952 |
No | output | 1 | 8,476 | 7 | 16,953 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Balph is learning to play a game called Buma. In this game, he is given a row of colored balls. He has to choose the color of one new ball and the place to insert it (between two balls, or to the left of all the balls, or to the right of all the balls).
When the ball is inserted the following happens repeatedly: if some segment of balls of the same color became longer as a result of a previous action and its length became at least 3, then all the balls of this segment are eliminated.
Consider, for example, a row of balls 'AAABBBWWBB'. Suppose Balph chooses a ball of color 'W' and the place to insert it after the sixth ball, i. e. to the left of the two 'W's. After Balph inserts this ball, the balls of color 'W' are eliminated, since this segment was made longer and has length 3 now, so the row becomes 'AAABBBBB'. The balls of color 'B' are eliminated now, because the segment of balls of color 'B' became longer and has length 5 now. Thus, the row becomes 'AAA'. However, none of the balls are eliminated now, because there is no elongated segment.
Help Balph count the number of possible ways to choose a color of a new ball and a place to insert it that leads to the elimination of all the balls.
Input
The only line contains a non-empty string of uppercase English letters of length at most 3 ⋅ 10^5. Each letter represents a ball with the corresponding color.
Output
Output the number of ways to choose a color and a position of a new ball in order to eliminate all the balls.
Examples
Input
BBWWBB
Output
3
Input
BWWB
Output
0
Input
BBWBB
Output
0
Input
OOOWWW
Output
0
Input
WWWOOOOOOWWW
Output
7
Submitted Solution:
```
from collections import defaultdict as dd
s=input()
l=[s[0]]
n=s[0]
d=dd(int)
d[s[0]]+=1
cou=1
li=[]
for i in range(1,len(s)):
d[s[i]]+=1
if(s[i]==n):
cou+=1
if(s[i]!=n):
l.append(s[i])
n=s[i]
li.append(cou)
cou=1
li.append(cou)
#print(l,d)
if(l==l[::-1]):
mid=len(l)//2
lol=0
for i in d:
if(d[i]<=2 and i!=l[mid]) or (d[i]<2 and i==l[mid]):
lol=1
if(lol):
print(0)
else:
print(li[mid]+1)
else:
print(0)
``` | instruction | 0 | 8,477 | 7 | 16,954 |
No | output | 1 | 8,477 | 7 | 16,955 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive task.
We have 2N balls arranged in a row, numbered 1, 2, 3, ..., 2N from left to right, where N is an odd number. Among them, there are N red balls and N blue balls.
While blindfolded, you are challenged to guess the color of every ball correctly, by asking at most 210 questions of the following form:
* You choose any N of the 2N balls and ask whether there are more red balls than blue balls or not among those N balls.
Now, let us begin.
Constraints
* 1 \leq N \leq 99
* N is an odd number.
* * *
Example
Input
Output
Submitted Solution:
```
from sys import exit
import time
def q(l):
print("?",end=" ")
for i in l:
print(i,end=" ")
print("")
ret=input()
if ret=="-1":
exit(0)
return ret=="Red"
n=int(input())
res=[None for i in range(n*2+1)]
if n==1:
if q([1]):
res=[None,"R","B"]
else:
res=[None,"B","R"]
else:
a=[i for i in range(1,(n+1)//2)]
b=[i for i in range((n+1)//2,n+1)]
c=[i for i in range(n+1,(n+1)//2+n+1)]
d=[i for i in range((n+1)//2+n+1,n*2+1)]
ab=q(a+b)
ac=q(a+c)
if ab and ac:
#a red d blue
toadd=b
rr=a
bb=d
elif ab and not ac:
#b red c blue
toadd=a
rr=b
bb=c
elif not ab and ac:
#c red b blue
toadd=a
rr=c
bb=b
elif not ab and not ac:
#d red a blue
toadd=c
rr=d
bb=a
else:
pass
if n<=50:
for en in range(1,len(rr)):
if q(toadd+rr[:en]+bb[en:]):
arr=toadd+rr[:en-1]+bb[en:]
break
else:
blim=0
rlim=len(rr)
while rlim-blim>1:
en=(blim+rlim)//2
if q(toadd+rr[:en]+bb[en:]):
rlim=en
else:
blim=en
arr=toadd+rr[:blim]+bb[blim+1:]
for i in range(1,2*n+1):
if i in arr:
continue
if q(arr+[i]):
res[i]="R"
else:
res[i]="B"
half=[]
rcnt=0
bcnt=0
for i in range(1,2*n+1):
if res[i]=="R":
if rcnt<(n-1)//2:
half.append(i)
rcnt+=1
elif res[i]=="B":
if bcnt<(n-1)//2:
half.append(i)
bcnt+=1
for i in range(1,2*n+1):
if res[i]==None:
ret=q(half+[i])
if ret:
res[i]="R"
else:
res[i]="B"
print("!",end=" ")
for c in res[1:]:
if c==None:
time.sleep(0.5)
break
print(c,end="")
print("",flush=True)
``` | instruction | 0 | 8,872 | 7 | 17,744 |
No | output | 1 | 8,872 | 7 | 17,745 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive task.
We have 2N balls arranged in a row, numbered 1, 2, 3, ..., 2N from left to right, where N is an odd number. Among them, there are N red balls and N blue balls.
While blindfolded, you are challenged to guess the color of every ball correctly, by asking at most 210 questions of the following form:
* You choose any N of the 2N balls and ask whether there are more red balls than blue balls or not among those N balls.
Now, let us begin.
Constraints
* 1 \leq N \leq 99
* N is an odd number.
* * *
Example
Input
Output
Submitted Solution:
```
import sys
def input():
a = sys.stdin.readline()
if a[0] == "-":
return 0/0
return a[0]
N = int(input())
ans = [" "] * (2*N)
def nibu(R,B):
if abs(R-B)==1:
return min(R,B),max(R,B)+N-1
c = (R+B)//2
print("?",*range(c+1,c+N+1))
a = input()
if a == 'R':
return nibu(c,B)
else:
return nibu(R,c)
print("?",*range(1,N+1))
if input() == "R":
R,B = nibu(0,N)
else:
R,B = nibu(N,0)
neutral1 = range(R+2,B+1)
neutral2l = range(1,R+1)
neutral2r = range(B+2,2*N+1)
for i in range(R+1):
print("?",*neutral1,i+1)
ans[i] = input()
for i in range(R+2,B+1):
print("?",*neutral2l,*neutral2r,i)
ans[i-1] = input()
for i in range(B+1,2*N+1):
print("?",*neutral1,i)
ans[i-1] = input()
print("!","".join(ans))
``` | instruction | 0 | 8,873 | 7 | 17,746 |
No | output | 1 | 8,873 | 7 | 17,747 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive task.
We have 2N balls arranged in a row, numbered 1, 2, 3, ..., 2N from left to right, where N is an odd number. Among them, there are N red balls and N blue balls.
While blindfolded, you are challenged to guess the color of every ball correctly, by asking at most 210 questions of the following form:
* You choose any N of the 2N balls and ask whether there are more red balls than blue balls or not among those N balls.
Now, let us begin.
Constraints
* 1 \leq N \leq 99
* N is an odd number.
* * *
Example
Input
Output
Submitted Solution:
```
import math
import sys
def make_divisors(n):
divisors = []
for i in range(1, int(n**0.5)+1):
if n % i == 0:
divisors.append(i)
if i != n // i:
divisors.append(n//i)
# divisors.sort()
return divisors
#a = list(map(int, input().split()))
n = int(input())
left = [i+1 for i in range(n)]
right = [i+n+1 for i in range(n)]
print('!',' '.join(map(str,left)))
``` | instruction | 0 | 8,874 | 7 | 17,748 |
No | output | 1 | 8,874 | 7 | 17,749 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive task.
We have 2N balls arranged in a row, numbered 1, 2, 3, ..., 2N from left to right, where N is an odd number. Among them, there are N red balls and N blue balls.
While blindfolded, you are challenged to guess the color of every ball correctly, by asking at most 210 questions of the following form:
* You choose any N of the 2N balls and ask whether there are more red balls than blue balls or not among those N balls.
Now, let us begin.
Constraints
* 1 \leq N \leq 99
* N is an odd number.
* * *
Example
Input
Output
Submitted Solution:
```
from itertools import chain
def query(indices):
print('? '+' '.join(map(lambda x: str(x+1),indices)), flush=True)
r = input()
if r == '-1':
exit()
return r == 'Red'
def answer(colors):
print('! '+''.join(map(lambda x: 'R' if x else 'B', colors)), flush=True)
def bisect(ng, ok, judge):
while abs(ng-ok) > 1:
m = (ng+ok)//2
if judge(m):
ok = m
else:
ng = m
return ok
from random import shuffle
def solve(N):
# detect change
check = lambda i: query(range(i,i+N)) if i <= N else query(chain(range(i-N),range(i,2*N)))
L = list(range(2*N))
shuffle(L)
x = L[-1]
orig = check(x)
for y in reversed(L[:-1]):
if orig != check(y):
break
if y < x:
x -= N*2
while abs(x-y) > 1:
mid = (x+y)//2
r = check((mid)%(2*N))
if r == orig:
x = mid
else:
y = mid
x %= 2*N
res = [None]*(2*N)
a,b = x, (x+N)%(2*N)
res[a] = orig
res[b] = not orig
# [i+1, ... , i+N] is even
n1 = list(range(a+1,b)) if a < b else list(chain(range(a+1,2*N),range(0,b)))
# [0...i-1] + [i+N+2, ... 2*N] is even
n2 = list(range(b+1,a)) if b < a else list(chain(range(b+1,2*N),range(0,a)))
for i in n1:
res[i] = query(n2+[i])
for i in n2:
res[i] = query(n1+[i])
answer(res)
if __name__ == '__main__':
N = int(input())
solve(N)
``` | instruction | 0 | 8,875 | 7 | 17,750 |
No | output | 1 | 8,875 | 7 | 17,751 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n boxes with colored balls on the table. Colors are numbered from 1 to n. i-th box contains ai balls, all of which have color i. You have to write a program that will divide all balls into sets such that:
* each ball belongs to exactly one of the sets,
* there are no empty sets,
* there is no set containing two (or more) balls of different colors (each set contains only balls of one color),
* there are no two sets such that the difference between their sizes is greater than 1.
Print the minimum possible number of sets.
Input
The first line contains one integer number n (1 ≤ n ≤ 500).
The second line contains n integer numbers a1, a2, ... , an (1 ≤ ai ≤ 109).
Output
Print one integer number — the minimum possible number of sets.
Examples
Input
3
4 7 8
Output
5
Input
2
2 7
Output
4
Note
In the first example the balls can be divided into sets like that: one set with 4 balls of the first color, two sets with 3 and 4 balls, respectively, of the second color, and two sets with 4 balls of the third color. | instruction | 0 | 9,565 | 7 | 19,130 |
Tags: greedy, math, number theory
Correct Solution:
```
#!/usr/bin/python3
import sys
n = int(sys.stdin.readline())
a = [int(x) for x in sys.stdin.readline().split()]
def is_valid(s, k):
# Ali lahko zapakiramo s steklenic v škatle velikosti k in k + 1?
# s = x * k + y * (k + 1) = (x + y) * k + y, kjer je 0 <= y < k
y = s % k
return s // k >= y
def all_valid(a, k):
for s in a:
if not is_valid(s, k):
return False
return True
best_sol = 1
k = 1
while k*k <= a[0]:
if all_valid(a, k):
best_sol = max(best_sol, k)
k += 1
# t je skupno število škatel.
t = 1
while t*t <= a[0]:
k = a[0] // t
if all_valid(a, k):
best_sol = max(best_sol, k)
if a[0] % t == 0 and k > 1:
if all_valid(a, k - 1):
best_sol = max(best_sol, k - 1)
t += 1
# print(best_sol, best_sol + 1)
print(sum(s // (best_sol + 1) + (0 if s % (best_sol + 1) == 0 else 1) for s in a))
``` | output | 1 | 9,565 | 7 | 19,131 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n boxes with colored balls on the table. Colors are numbered from 1 to n. i-th box contains ai balls, all of which have color i. You have to write a program that will divide all balls into sets such that:
* each ball belongs to exactly one of the sets,
* there are no empty sets,
* there is no set containing two (or more) balls of different colors (each set contains only balls of one color),
* there are no two sets such that the difference between their sizes is greater than 1.
Print the minimum possible number of sets.
Input
The first line contains one integer number n (1 ≤ n ≤ 500).
The second line contains n integer numbers a1, a2, ... , an (1 ≤ ai ≤ 109).
Output
Print one integer number — the minimum possible number of sets.
Examples
Input
3
4 7 8
Output
5
Input
2
2 7
Output
4
Note
In the first example the balls can be divided into sets like that: one set with 4 balls of the first color, two sets with 3 and 4 balls, respectively, of the second color, and two sets with 4 balls of the third color. | instruction | 0 | 9,566 | 7 | 19,132 |
Tags: greedy, math, number theory
Correct Solution:
```
# -*- coding: utf-8 -*-
"""
Created on Sun Apr 2 22:42:34 2017
@author: Sean38
"""
n = int(input().rstrip())
s = input()
a = [int(ch) for ch in s.split()]
a = a[0:n]
a.sort()
def check_num(p, i):
# i = ap + b(p+1)
# min(a+b) <=> max(b)
# b(p+1) <= i
# b == i (mod p)
max_b = (i // (p + 1))
b = i % p + ((max_b - i % p) // p) * p
# cur = a + b
cur = (i - b) // p
#print(cur - b, b, p)
if b < 0:
return None
return cur
def sets_num(p):
total = 0
for i in a:
if check_num(p, i):
total += check_num(p, i)
else:
return None
return total
for div_sets in range(1, a[0] + 1):
p, q = divmod(a[0], div_sets)
if (q == 0):
if sets_num(p):
print(sets_num(p))
break
if (p > 0) and sets_num(p - 1):
print(sets_num(p - 1))
break
else:
if sets_num(p):
print(sets_num(p))
break
``` | output | 1 | 9,566 | 7 | 19,133 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n boxes with colored balls on the table. Colors are numbered from 1 to n. i-th box contains ai balls, all of which have color i. You have to write a program that will divide all balls into sets such that:
* each ball belongs to exactly one of the sets,
* there are no empty sets,
* there is no set containing two (or more) balls of different colors (each set contains only balls of one color),
* there are no two sets such that the difference between their sizes is greater than 1.
Print the minimum possible number of sets.
Input
The first line contains one integer number n (1 ≤ n ≤ 500).
The second line contains n integer numbers a1, a2, ... , an (1 ≤ ai ≤ 109).
Output
Print one integer number — the minimum possible number of sets.
Examples
Input
3
4 7 8
Output
5
Input
2
2 7
Output
4
Note
In the first example the balls can be divided into sets like that: one set with 4 balls of the first color, two sets with 3 and 4 balls, respectively, of the second color, and two sets with 4 balls of the third color. | instruction | 0 | 9,567 | 7 | 19,134 |
Tags: greedy, math, number theory
Correct Solution:
```
#!/usr/bin/env python3
# solution after hint:(
from math import sqrt
n = int(input().strip())
ais = list(map(int, input().strip().split()))
ais.sort(reverse=True)
xmax = int(sqrt(ais[0]))
# {x, x + 1}
def check(x):
q = 0
for ai in ais:
qi = -((-ai) // (x + 1))
q += qi
if x * qi > ai:
return (False, None)
else:
return (True, q)
qbest = sum(ais) + 1
for x in range(1, xmax + 1): # {x, x + 1}
(valid, q) = check(x)
if valid:
qbest = min(qbest, q)
for k in range(1, xmax + 1): # a0 // k +/- 1
x = ais[0] // k
(valid, q) = check(x)
if valid:
qbest = min(qbest, q)
if ais[0] % k == 0:
x -= 1
(valid, q) = check(x)
if valid:
qbest = min(qbest, q)
print (qbest)
``` | output | 1 | 9,567 | 7 | 19,135 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n boxes with colored balls on the table. Colors are numbered from 1 to n. i-th box contains ai balls, all of which have color i. You have to write a program that will divide all balls into sets such that:
* each ball belongs to exactly one of the sets,
* there are no empty sets,
* there is no set containing two (or more) balls of different colors (each set contains only balls of one color),
* there are no two sets such that the difference between their sizes is greater than 1.
Print the minimum possible number of sets.
Input
The first line contains one integer number n (1 ≤ n ≤ 500).
The second line contains n integer numbers a1, a2, ... , an (1 ≤ ai ≤ 109).
Output
Print one integer number — the minimum possible number of sets.
Examples
Input
3
4 7 8
Output
5
Input
2
2 7
Output
4
Note
In the first example the balls can be divided into sets like that: one set with 4 balls of the first color, two sets with 3 and 4 balls, respectively, of the second color, and two sets with 4 balls of the third color. | instruction | 0 | 9,569 | 7 | 19,138 |
Tags: greedy, math, number theory
Correct Solution:
```
def judge(lists, x):
ans = 0
for i in lists:
t = i//x
d = i%x
if d == 0:
ans += t
elif t+d >= x-1:
ans += t+1
else:
return -1
return ans
while True:
try:
n = input()
balls = list(map(int, input().split()))
minn = min(balls)
for i in range(1, minn+1):
if judge(balls, minn//i + 1) >= 0:
ans = judge(balls, minn//i + 1)
break
elif judge(balls, minn//i) >= 0:
ans = judge(balls, minn//i)
break
print(ans)
except EOFError:
break
``` | output | 1 | 9,569 | 7 | 19,139 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n boxes with colored balls on the table. Colors are numbered from 1 to n. i-th box contains ai balls, all of which have color i. You have to write a program that will divide all balls into sets such that:
* each ball belongs to exactly one of the sets,
* there are no empty sets,
* there is no set containing two (or more) balls of different colors (each set contains only balls of one color),
* there are no two sets such that the difference between their sizes is greater than 1.
Print the minimum possible number of sets.
Input
The first line contains one integer number n (1 ≤ n ≤ 500).
The second line contains n integer numbers a1, a2, ... , an (1 ≤ ai ≤ 109).
Output
Print one integer number — the minimum possible number of sets.
Examples
Input
3
4 7 8
Output
5
Input
2
2 7
Output
4
Note
In the first example the balls can be divided into sets like that: one set with 4 balls of the first color, two sets with 3 and 4 balls, respectively, of the second color, and two sets with 4 balls of the third color. | instruction | 0 | 9,570 | 7 | 19,140 |
Tags: greedy, math, number theory
Correct Solution:
```
import time
import sys
from math import sqrt
n = int(input())
a = list(map(int, input().split()))
sq = int(sqrt(a[0]))+2
s = set()
for box in range(1, sq):
if a[0] % box == 0:
s.add(a[0] // box)
s.add(a[0] // box - 1)
else:
s.add(a[0] // box)
for balls in range(1, sq):
if a[0] % balls <= a[0] // balls:
s.add(balls)
for size in sorted(s, reverse=True):
ans = 0
for x in a:
if x / size >= x % size:
ans += (x+size) // (size+1)
else:
break
else:
print(ans)
exit()
``` | output | 1 | 9,570 | 7 | 19,141 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n boxes with colored balls on the table. Colors are numbered from 1 to n. i-th box contains ai balls, all of which have color i. You have to write a program that will divide all balls into sets such that:
* each ball belongs to exactly one of the sets,
* there are no empty sets,
* there is no set containing two (or more) balls of different colors (each set contains only balls of one color),
* there are no two sets such that the difference between their sizes is greater than 1.
Print the minimum possible number of sets.
Input
The first line contains one integer number n (1 ≤ n ≤ 500).
The second line contains n integer numbers a1, a2, ... , an (1 ≤ ai ≤ 109).
Output
Print one integer number — the minimum possible number of sets.
Examples
Input
3
4 7 8
Output
5
Input
2
2 7
Output
4
Note
In the first example the balls can be divided into sets like that: one set with 4 balls of the first color, two sets with 3 and 4 balls, respectively, of the second color, and two sets with 4 balls of the third color. | instruction | 0 | 9,571 | 7 | 19,142 |
Tags: greedy, math, number theory
Correct 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)
def modInverse(a, m):
m0 = m
y = 0
x = 1
if (m == 1):
return 0
while (a > 1):
q = a // m
t = m
m = a % m
a = t
t = y
y = x - q * y
x = t
if (x < 0):
x = x + m0
return x
#------------------------------------------------------code by AD18/apurva3455
n = int(input())
a = list(map(int, input().split()))
sq = int(math.sqrt(a[0]))+2
s = set()
for box in range(1, sq):
if a[0] % box == 0:
s.add(a[0] // box)
s.add(a[0] // box - 1)
else:
s.add(a[0] // box)
for balls in range(1, sq):
if a[0] % balls <= a[0] // balls:
s.add(balls)
for size in sorted(s, reverse=True):
ans = 0
for x in a:
if x / size >= x % size:
ans += (x+size) // (size+1)
else:
break
else:
print(ans)
exit()
``` | output | 1 | 9,571 | 7 | 19,143 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n boxes with colored balls on the table. Colors are numbered from 1 to n. i-th box contains ai balls, all of which have color i. You have to write a program that will divide all balls into sets such that:
* each ball belongs to exactly one of the sets,
* there are no empty sets,
* there is no set containing two (or more) balls of different colors (each set contains only balls of one color),
* there are no two sets such that the difference between their sizes is greater than 1.
Print the minimum possible number of sets.
Input
The first line contains one integer number n (1 ≤ n ≤ 500).
The second line contains n integer numbers a1, a2, ... , an (1 ≤ ai ≤ 109).
Output
Print one integer number — the minimum possible number of sets.
Examples
Input
3
4 7 8
Output
5
Input
2
2 7
Output
4
Note
In the first example the balls can be divided into sets like that: one set with 4 balls of the first color, two sets with 3 and 4 balls, respectively, of the second color, and two sets with 4 balls of the third color. | instruction | 0 | 9,572 | 7 | 19,144 |
Tags: greedy, math, number theory
Correct Solution:
```
from math import sqrt
def divide_into_p(p, a):
q = a // (p+1)
r = a % (p+1)
if r == 0:
return q
elif q + r >= p:
return q+1
else:
return None
def divisible_into_p(p, lst):
lgth = 0
for l in lst:
dp = divide_into_p(p, l)
if dp == None:
return None
else:
lgth += dp
return lgth
def index_lst(a,sqrta):
lst = []
for i in range(1, sqrta + 1):
p = a//i
lst.append(p)
if not a%p:
lst.append(p-1)
for i in range(sqrta+1,0,-1):
lst.append(i)
return lst
def check_all_p(balls):
a = balls[0]
sqrt_a = int(sqrt(a))
indices = index_lst(a,sqrt_a)
for i in indices:
dp = divisible_into_p(i, balls)
if dp != None:
return dp
input()
balls = sorted([int(x) for x in input().split()])
print(check_all_p(balls))
``` | output | 1 | 9,572 | 7 | 19,145 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n boxes with colored balls on the table. Colors are numbered from 1 to n. i-th box contains ai balls, all of which have color i. You have to write a program that will divide all balls into sets such that:
* each ball belongs to exactly one of the sets,
* there are no empty sets,
* there is no set containing two (or more) balls of different colors (each set contains only balls of one color),
* there are no two sets such that the difference between their sizes is greater than 1.
Print the minimum possible number of sets.
Input
The first line contains one integer number n (1 ≤ n ≤ 500).
The second line contains n integer numbers a1, a2, ... , an (1 ≤ ai ≤ 109).
Output
Print one integer number — the minimum possible number of sets.
Examples
Input
3
4 7 8
Output
5
Input
2
2 7
Output
4
Note
In the first example the balls can be divided into sets like that: one set with 4 balls of the first color, two sets with 3 and 4 balls, respectively, of the second color, and two sets with 4 balls of the third color.
Submitted Solution:
```
import math
def check(n,a,m):
t=[]
for j in range(n):
t.append(a[j]%m)
t=set(t)
t=list(t)
t.sort()
if(len(t)==2):
if((t[0]==0 and t[1]==1 )or (t[0]==0 and t[1]==m-1)):
return 1
return 0
n=int(input())
a=list(map(int,input().split()))
m=min(a)
d=[]
j=1
val=2
while(j*j<=m):
if(m%j==0):
d.append(j)
d.append(j+1)
d.append(j-1)
if(m//j!=j):
d.append(m//j)
d.append((m//j)+1)
d.append((m//j)-1)
j+=1
d=set(d)
d=list(d)
d.sort()
#print(d)
if(d[0]==0):
d.pop(0)
#print(d)
for j in range(len(d)-1,-1,-1):
#print(check(n,a,d[j]))
if(check(n,a,d[j])==1):
val=d[j]
break
#print(val)
count=0
for j in range(n):
if(a[j]%val==0 or a[j]%val==1):
count+=a[j]//val
elif(a[j]%val==val-1):
count+=math.ceil(a[j]/val)
print(count)
``` | instruction | 0 | 9,573 | 7 | 19,146 |
No | output | 1 | 9,573 | 7 | 19,147 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n boxes with colored balls on the table. Colors are numbered from 1 to n. i-th box contains ai balls, all of which have color i. You have to write a program that will divide all balls into sets such that:
* each ball belongs to exactly one of the sets,
* there are no empty sets,
* there is no set containing two (or more) balls of different colors (each set contains only balls of one color),
* there are no two sets such that the difference between their sizes is greater than 1.
Print the minimum possible number of sets.
Input
The first line contains one integer number n (1 ≤ n ≤ 500).
The second line contains n integer numbers a1, a2, ... , an (1 ≤ ai ≤ 109).
Output
Print one integer number — the minimum possible number of sets.
Examples
Input
3
4 7 8
Output
5
Input
2
2 7
Output
4
Note
In the first example the balls can be divided into sets like that: one set with 4 balls of the first color, two sets with 3 and 4 balls, respectively, of the second color, and two sets with 4 balls of the third color.
Submitted Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time
sys.setrecursionlimit(10**7)
inf = 10**20
mod = 10**9 + 7
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def main():
n = I()
a = LI()
def f(t):
for c in a:
if c < t:
return False
if c % (t+1) == 0:
continue
d = c//t
e = c - t*d
if d<e:
return False
return True
ma = min(a)
t = 1
for i in range(1,ma+1):
if f(ma//i+1):
t = ma//i+1
break
if f(ma//i):
t = ma//i
break
r = 0
for c in a:
if c % (t+1) == 0:
r += c//(t+1)
continue
r += c//t
return r
print(main())
``` | instruction | 0 | 9,574 | 7 | 19,148 |
No | output | 1 | 9,574 | 7 | 19,149 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n boxes with colored balls on the table. Colors are numbered from 1 to n. i-th box contains ai balls, all of which have color i. You have to write a program that will divide all balls into sets such that:
* each ball belongs to exactly one of the sets,
* there are no empty sets,
* there is no set containing two (or more) balls of different colors (each set contains only balls of one color),
* there are no two sets such that the difference between their sizes is greater than 1.
Print the minimum possible number of sets.
Input
The first line contains one integer number n (1 ≤ n ≤ 500).
The second line contains n integer numbers a1, a2, ... , an (1 ≤ ai ≤ 109).
Output
Print one integer number — the minimum possible number of sets.
Examples
Input
3
4 7 8
Output
5
Input
2
2 7
Output
4
Note
In the first example the balls can be divided into sets like that: one set with 4 balls of the first color, two sets with 3 and 4 balls, respectively, of the second color, and two sets with 4 balls of the third color.
Submitted Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time
sys.setrecursionlimit(10**7)
inf = 10**20
mod = 10**9 + 7
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def main():
n = I()
a = LI()
def f(t):
for c in a:
if c == t-1:
continue
d = c//t
e = c - t*d
if d<e:
return False
return True
ma = min(a)
t = 1
if f(ma) :
t = ma
else:
for i in range(2,ma+1):
if ma%i == 0 and f(ma//i+1):
t = ma//i+1
break
if f(ma//i):
t = ma//i
break
r = 0
for c in a:
if c % (t+1) == 0:
r += c//(t+1)
continue
r += c//t
return r
print(main())
``` | instruction | 0 | 9,575 | 7 | 19,150 |
No | output | 1 | 9,575 | 7 | 19,151 |
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