message stringlengths 2 44.5k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 276 109k | cluster float64 23 23 | __index_level_0__ int64 552 217k |
|---|---|---|---|---|---|
Provide a correct Python 3 solution for this coding contest problem.
Seiji Hayashi had been a professor of the Nisshinkan Samurai School in the domain of Aizu for a long time in the 18th century. In order to reward him for his meritorious career in education, Katanobu Matsudaira, the lord of the domain of Aizu, had decided to grant him a rectangular estate within a large field in the Aizu Basin. Although the size (width and height) of the estate was strictly specified by the lord, he was allowed to choose any location for the estate in the field. Inside the field which had also a rectangular shape, many Japanese persimmon trees, whose fruit was one of the famous products of the Aizu region known as 'Mishirazu Persimmon', were planted. Since persimmon was Hayashi's favorite fruit, he wanted to have as many persimmon trees as possible in the estate given by the lord.
For example, in Figure 1, the entire field is a rectangular grid whose width and height are 10 and 8 respectively. Each asterisk (*) represents a place of a persimmon tree. If the specified width and height of the estate are 4 and 3 respectively, the area surrounded by the solid line contains the most persimmon trees. Similarly, if the estate's width is 6 and its height is 4, the area surrounded by the dashed line has the most, and if the estate's width and height are 3 and 4 respectively, the area surrounded by the dotted line contains the most persimmon trees. Note that the width and height cannot be swapped; the sizes 4 by 3 and 3 by 4 are different, as shown in Figure 1.
<image>
---
Figure 1: Examples of Rectangular Estates
Your task is to find the estate of a given size (width and height) that contains the largest number of persimmon trees.
Input
The input consists of multiple data sets. Each data set is given in the following format.
> N
> W` `H
> x1` `y1
> x2` `y2
> ...
> xN` `yN
> S` `T
>
N is the number of persimmon trees, which is a positive integer less than 500. W and H are the width and the height of the entire field respectively. You can assume that both W and H are positive integers whose values are less than 100. For each i (1 <= i <= N), xi and yi are coordinates of the i-th persimmon tree in the grid. Note that the origin of each coordinate is 1. You can assume that 1 <= xi <= W and 1 <= yi <= H, and no two trees have the same positions. But you should not assume that the persimmon trees are sorted in some order according to their positions. Lastly, S and T are positive integers of the width and height respectively of the estate given by the lord. You can also assume that 1 <= S <= W and 1 <= T <= H.
The end of the input is indicated by a line that solely contains a zero.
Output
For each data set, you are requested to print one line containing the maximum possible number of persimmon trees that can be included in an estate of the given size.
Example
Input
16
10 8
2 2
2 5
2 7
3 3
3 8
4 2
4 5
4 8
6 4
6 7
7 5
7 8
8 1
8 4
9 6
10 3
4 3
8
6 4
1 2
2 1
2 4
3 4
4 2
5 3
6 1
6 2
3 2
0
Output
4
3 | instruction | 0 | 18,493 | 23 | 36,986 |
"Correct Solution:
```
while(True):
N = int(input())
if N ==0 :break
W, H = map(int, input().split())
area = [ [ 0 for i in range(W)]for i in range(H)]
ans_max = 0
for i in range(N):
w, h = map(int, input().split())
area[h-1][w-1] = 1
S, T = map(int, input().split())
for i in range(H-T+1):
for j in range(W-S+1):
ans = 0
for k in range(i,i+T):
ans += sum(area[k][j:j+S])
if ans_max<ans:
ans_max = ans
print(ans_max)
``` | output | 1 | 18,493 | 23 | 36,987 |
Provide a correct Python 3 solution for this coding contest problem.
Seiji Hayashi had been a professor of the Nisshinkan Samurai School in the domain of Aizu for a long time in the 18th century. In order to reward him for his meritorious career in education, Katanobu Matsudaira, the lord of the domain of Aizu, had decided to grant him a rectangular estate within a large field in the Aizu Basin. Although the size (width and height) of the estate was strictly specified by the lord, he was allowed to choose any location for the estate in the field. Inside the field which had also a rectangular shape, many Japanese persimmon trees, whose fruit was one of the famous products of the Aizu region known as 'Mishirazu Persimmon', were planted. Since persimmon was Hayashi's favorite fruit, he wanted to have as many persimmon trees as possible in the estate given by the lord.
For example, in Figure 1, the entire field is a rectangular grid whose width and height are 10 and 8 respectively. Each asterisk (*) represents a place of a persimmon tree. If the specified width and height of the estate are 4 and 3 respectively, the area surrounded by the solid line contains the most persimmon trees. Similarly, if the estate's width is 6 and its height is 4, the area surrounded by the dashed line has the most, and if the estate's width and height are 3 and 4 respectively, the area surrounded by the dotted line contains the most persimmon trees. Note that the width and height cannot be swapped; the sizes 4 by 3 and 3 by 4 are different, as shown in Figure 1.
<image>
---
Figure 1: Examples of Rectangular Estates
Your task is to find the estate of a given size (width and height) that contains the largest number of persimmon trees.
Input
The input consists of multiple data sets. Each data set is given in the following format.
> N
> W` `H
> x1` `y1
> x2` `y2
> ...
> xN` `yN
> S` `T
>
N is the number of persimmon trees, which is a positive integer less than 500. W and H are the width and the height of the entire field respectively. You can assume that both W and H are positive integers whose values are less than 100. For each i (1 <= i <= N), xi and yi are coordinates of the i-th persimmon tree in the grid. Note that the origin of each coordinate is 1. You can assume that 1 <= xi <= W and 1 <= yi <= H, and no two trees have the same positions. But you should not assume that the persimmon trees are sorted in some order according to their positions. Lastly, S and T are positive integers of the width and height respectively of the estate given by the lord. You can also assume that 1 <= S <= W and 1 <= T <= H.
The end of the input is indicated by a line that solely contains a zero.
Output
For each data set, you are requested to print one line containing the maximum possible number of persimmon trees that can be included in an estate of the given size.
Example
Input
16
10 8
2 2
2 5
2 7
3 3
3 8
4 2
4 5
4 8
6 4
6 7
7 5
7 8
8 1
8 4
9 6
10 3
4 3
8
6 4
1 2
2 1
2 4
3 4
4 2
5 3
6 1
6 2
3 2
0
Output
4
3 | instruction | 0 | 18,494 | 23 | 36,988 |
"Correct Solution:
```
def get_data():
data = int(input())
return data
def get_data_list():
data_list = input().split()
for i, v in enumerate(data_list):
data_list[i] = int(v)
return data_list
class MakePositionEasy:
def __init__(self, data):
self.w = data[0]
self.h = data[1]
def get_all_persimmon_coordinate(persimmon_num):
all_persimmon_coordinate = [None] * persimmon_num
for i in range(persimmon_num):
all_persimmon_coordinate[i] = tuple(get_data_list())
all_persimmon_coordinate = set(all_persimmon_coordinate)
return all_persimmon_coordinate
def make_persimmon_map(area_size, all_persimmon_coordinate):
persimmon_map = [[0] * (area_size.w + 1) for i in range(area_size.h + 1)]
for w in range(1, area_size.w+1):
current_line_count = 0
for h in range(1, area_size.h+1):
if (w,h) in all_persimmon_coordinate:
current_line_count += 1
bef_num = persimmon_map[h][w-1]
persimmon_map[h][w] = (bef_num + current_line_count)
return persimmon_map
def get_max_num(persimmon_map,area_size, given_size):
max_num = 0
for w in range(given_size.w, area_size.w+1):
for h in range(given_size.h, area_size.h+1):
persimmon_num = calculate_persiommon_num(persimmon_map, given_size, w, h)
if persimmon_num > max_num:
max_num = persimmon_num
return max_num
def calculate_persiommon_num(persimmon_map , given_size, w, h):
persimmon_num = persimmon_map[h][w]
persimmon_num -= persimmon_map[h-given_size.h][w]
persimmon_num -= persimmon_map[h][w-given_size.w]
persimmon_num += persimmon_map[h-given_size.h][w-given_size.w]
return persimmon_num
if __name__ == "__main__":
while True:
persimmon_num = get_data()
if persimmon_num == 0:
break
area_size = MakePositionEasy(get_data_list())
all_persimmon_coordinate = get_all_persimmon_coordinate(persimmon_num)
given_size = MakePositionEasy(get_data_list())
persimmon_map = make_persimmon_map(area_size, all_persimmon_coordinate)
max_num = get_max_num(persimmon_map,area_size, given_size)
print(max_num)
``` | output | 1 | 18,494 | 23 | 36,989 |
Provide a correct Python 3 solution for this coding contest problem.
Seiji Hayashi had been a professor of the Nisshinkan Samurai School in the domain of Aizu for a long time in the 18th century. In order to reward him for his meritorious career in education, Katanobu Matsudaira, the lord of the domain of Aizu, had decided to grant him a rectangular estate within a large field in the Aizu Basin. Although the size (width and height) of the estate was strictly specified by the lord, he was allowed to choose any location for the estate in the field. Inside the field which had also a rectangular shape, many Japanese persimmon trees, whose fruit was one of the famous products of the Aizu region known as 'Mishirazu Persimmon', were planted. Since persimmon was Hayashi's favorite fruit, he wanted to have as many persimmon trees as possible in the estate given by the lord.
For example, in Figure 1, the entire field is a rectangular grid whose width and height are 10 and 8 respectively. Each asterisk (*) represents a place of a persimmon tree. If the specified width and height of the estate are 4 and 3 respectively, the area surrounded by the solid line contains the most persimmon trees. Similarly, if the estate's width is 6 and its height is 4, the area surrounded by the dashed line has the most, and if the estate's width and height are 3 and 4 respectively, the area surrounded by the dotted line contains the most persimmon trees. Note that the width and height cannot be swapped; the sizes 4 by 3 and 3 by 4 are different, as shown in Figure 1.
<image>
---
Figure 1: Examples of Rectangular Estates
Your task is to find the estate of a given size (width and height) that contains the largest number of persimmon trees.
Input
The input consists of multiple data sets. Each data set is given in the following format.
> N
> W` `H
> x1` `y1
> x2` `y2
> ...
> xN` `yN
> S` `T
>
N is the number of persimmon trees, which is a positive integer less than 500. W and H are the width and the height of the entire field respectively. You can assume that both W and H are positive integers whose values are less than 100. For each i (1 <= i <= N), xi and yi are coordinates of the i-th persimmon tree in the grid. Note that the origin of each coordinate is 1. You can assume that 1 <= xi <= W and 1 <= yi <= H, and no two trees have the same positions. But you should not assume that the persimmon trees are sorted in some order according to their positions. Lastly, S and T are positive integers of the width and height respectively of the estate given by the lord. You can also assume that 1 <= S <= W and 1 <= T <= H.
The end of the input is indicated by a line that solely contains a zero.
Output
For each data set, you are requested to print one line containing the maximum possible number of persimmon trees that can be included in an estate of the given size.
Example
Input
16
10 8
2 2
2 5
2 7
3 3
3 8
4 2
4 5
4 8
6 4
6 7
7 5
7 8
8 1
8 4
9 6
10 3
4 3
8
6 4
1 2
2 1
2 4
3 4
4 2
5 3
6 1
6 2
3 2
0
Output
4
3 | instruction | 0 | 18,495 | 23 | 36,990 |
"Correct Solution:
```
while True:
N = int(input())
if N==0:
break
W,H = map(int,input().split())
data = [[0 for i in range(W)] for j in range(H)]
for i in range(N):
x,y = map(int,input().split())
data[y-1][x-1] = 1
S,T = map(int,input().split())
ans = 0
for h in range(H-T+1):
for w in range(W-S+1):
#
a = 0
for x in range(T):
for y in range(S):
if data[h+x][w+y]==1:
a+=1
ans = max(ans,a)
print(ans)
``` | output | 1 | 18,495 | 23 | 36,991 |
Provide a correct Python 3 solution for this coding contest problem.
Seiji Hayashi had been a professor of the Nisshinkan Samurai School in the domain of Aizu for a long time in the 18th century. In order to reward him for his meritorious career in education, Katanobu Matsudaira, the lord of the domain of Aizu, had decided to grant him a rectangular estate within a large field in the Aizu Basin. Although the size (width and height) of the estate was strictly specified by the lord, he was allowed to choose any location for the estate in the field. Inside the field which had also a rectangular shape, many Japanese persimmon trees, whose fruit was one of the famous products of the Aizu region known as 'Mishirazu Persimmon', were planted. Since persimmon was Hayashi's favorite fruit, he wanted to have as many persimmon trees as possible in the estate given by the lord.
For example, in Figure 1, the entire field is a rectangular grid whose width and height are 10 and 8 respectively. Each asterisk (*) represents a place of a persimmon tree. If the specified width and height of the estate are 4 and 3 respectively, the area surrounded by the solid line contains the most persimmon trees. Similarly, if the estate's width is 6 and its height is 4, the area surrounded by the dashed line has the most, and if the estate's width and height are 3 and 4 respectively, the area surrounded by the dotted line contains the most persimmon trees. Note that the width and height cannot be swapped; the sizes 4 by 3 and 3 by 4 are different, as shown in Figure 1.
<image>
---
Figure 1: Examples of Rectangular Estates
Your task is to find the estate of a given size (width and height) that contains the largest number of persimmon trees.
Input
The input consists of multiple data sets. Each data set is given in the following format.
> N
> W` `H
> x1` `y1
> x2` `y2
> ...
> xN` `yN
> S` `T
>
N is the number of persimmon trees, which is a positive integer less than 500. W and H are the width and the height of the entire field respectively. You can assume that both W and H are positive integers whose values are less than 100. For each i (1 <= i <= N), xi and yi are coordinates of the i-th persimmon tree in the grid. Note that the origin of each coordinate is 1. You can assume that 1 <= xi <= W and 1 <= yi <= H, and no two trees have the same positions. But you should not assume that the persimmon trees are sorted in some order according to their positions. Lastly, S and T are positive integers of the width and height respectively of the estate given by the lord. You can also assume that 1 <= S <= W and 1 <= T <= H.
The end of the input is indicated by a line that solely contains a zero.
Output
For each data set, you are requested to print one line containing the maximum possible number of persimmon trees that can be included in an estate of the given size.
Example
Input
16
10 8
2 2
2 5
2 7
3 3
3 8
4 2
4 5
4 8
6 4
6 7
7 5
7 8
8 1
8 4
9 6
10 3
4 3
8
6 4
1 2
2 1
2 4
3 4
4 2
5 3
6 1
6 2
3 2
0
Output
4
3 | instruction | 0 | 18,496 | 23 | 36,992 |
"Correct Solution:
```
while True:
N = int(input())
if N == 0:
break
W, H = map(int, input().split())
field = [[0 for _ in range(W)] for _ in range(H)]
for _ in range(N):
x, y = map(int, input().split())
field[y - 1][x - 1] = 1
S, T = map(int, input().split())
c_list = []
for tate in range(H - T + 1):
for yoko in range(W - S + 1):
c = 0
for i in range(tate, T + tate):
for j in range(yoko, S + yoko):
if field[i][j] == 1:
# print(i + 1, j + 1)
c += 1
c_list.append(c)
# print(*field, sep="\n")
print(max(c_list))
``` | output | 1 | 18,496 | 23 | 36,993 |
Provide a correct Python 3 solution for this coding contest problem.
Seiji Hayashi had been a professor of the Nisshinkan Samurai School in the domain of Aizu for a long time in the 18th century. In order to reward him for his meritorious career in education, Katanobu Matsudaira, the lord of the domain of Aizu, had decided to grant him a rectangular estate within a large field in the Aizu Basin. Although the size (width and height) of the estate was strictly specified by the lord, he was allowed to choose any location for the estate in the field. Inside the field which had also a rectangular shape, many Japanese persimmon trees, whose fruit was one of the famous products of the Aizu region known as 'Mishirazu Persimmon', were planted. Since persimmon was Hayashi's favorite fruit, he wanted to have as many persimmon trees as possible in the estate given by the lord.
For example, in Figure 1, the entire field is a rectangular grid whose width and height are 10 and 8 respectively. Each asterisk (*) represents a place of a persimmon tree. If the specified width and height of the estate are 4 and 3 respectively, the area surrounded by the solid line contains the most persimmon trees. Similarly, if the estate's width is 6 and its height is 4, the area surrounded by the dashed line has the most, and if the estate's width and height are 3 and 4 respectively, the area surrounded by the dotted line contains the most persimmon trees. Note that the width and height cannot be swapped; the sizes 4 by 3 and 3 by 4 are different, as shown in Figure 1.
<image>
---
Figure 1: Examples of Rectangular Estates
Your task is to find the estate of a given size (width and height) that contains the largest number of persimmon trees.
Input
The input consists of multiple data sets. Each data set is given in the following format.
> N
> W` `H
> x1` `y1
> x2` `y2
> ...
> xN` `yN
> S` `T
>
N is the number of persimmon trees, which is a positive integer less than 500. W and H are the width and the height of the entire field respectively. You can assume that both W and H are positive integers whose values are less than 100. For each i (1 <= i <= N), xi and yi are coordinates of the i-th persimmon tree in the grid. Note that the origin of each coordinate is 1. You can assume that 1 <= xi <= W and 1 <= yi <= H, and no two trees have the same positions. But you should not assume that the persimmon trees are sorted in some order according to their positions. Lastly, S and T are positive integers of the width and height respectively of the estate given by the lord. You can also assume that 1 <= S <= W and 1 <= T <= H.
The end of the input is indicated by a line that solely contains a zero.
Output
For each data set, you are requested to print one line containing the maximum possible number of persimmon trees that can be included in an estate of the given size.
Example
Input
16
10 8
2 2
2 5
2 7
3 3
3 8
4 2
4 5
4 8
6 4
6 7
7 5
7 8
8 1
8 4
9 6
10 3
4 3
8
6 4
1 2
2 1
2 4
3 4
4 2
5 3
6 1
6 2
3 2
0
Output
4
3 | instruction | 0 | 18,497 | 23 | 36,994 |
"Correct Solution:
```
import sys
input = sys.stdin.readline
def main():
while True:
N = int(input())
if N == 0:
return
W, H = map(int, input().split())
board = [[0] * (W+1) for i in range(H+1)]
for i in range(N):
x, y = map(int, input().split())
board[y][x] = 1
for i in range(1, H+1):
for j in range(W+1):
board[i][j] += board[i-1][j]
for j in range(1, W+1):
for i in range(H+1):
board[i][j] += board[i][j-1]
S, T = map(int, input().split())
ans = 0
for y in range(1, H-T+2):
for x in range(1, W-S+2):
tx = x + S - 1
ty = y + T - 1
ans = max(ans, board[ty][tx] - board[ty][x-1] - board[y-1][tx] + board[y-1][x-1])
print(ans)
# for i in range(H):
# print(board[i])
if __name__ == "__main__":
main()
``` | output | 1 | 18,497 | 23 | 36,995 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Seiji Hayashi had been a professor of the Nisshinkan Samurai School in the domain of Aizu for a long time in the 18th century. In order to reward him for his meritorious career in education, Katanobu Matsudaira, the lord of the domain of Aizu, had decided to grant him a rectangular estate within a large field in the Aizu Basin. Although the size (width and height) of the estate was strictly specified by the lord, he was allowed to choose any location for the estate in the field. Inside the field which had also a rectangular shape, many Japanese persimmon trees, whose fruit was one of the famous products of the Aizu region known as 'Mishirazu Persimmon', were planted. Since persimmon was Hayashi's favorite fruit, he wanted to have as many persimmon trees as possible in the estate given by the lord.
For example, in Figure 1, the entire field is a rectangular grid whose width and height are 10 and 8 respectively. Each asterisk (*) represents a place of a persimmon tree. If the specified width and height of the estate are 4 and 3 respectively, the area surrounded by the solid line contains the most persimmon trees. Similarly, if the estate's width is 6 and its height is 4, the area surrounded by the dashed line has the most, and if the estate's width and height are 3 and 4 respectively, the area surrounded by the dotted line contains the most persimmon trees. Note that the width and height cannot be swapped; the sizes 4 by 3 and 3 by 4 are different, as shown in Figure 1.
<image>
---
Figure 1: Examples of Rectangular Estates
Your task is to find the estate of a given size (width and height) that contains the largest number of persimmon trees.
Input
The input consists of multiple data sets. Each data set is given in the following format.
> N
> W` `H
> x1` `y1
> x2` `y2
> ...
> xN` `yN
> S` `T
>
N is the number of persimmon trees, which is a positive integer less than 500. W and H are the width and the height of the entire field respectively. You can assume that both W and H are positive integers whose values are less than 100. For each i (1 <= i <= N), xi and yi are coordinates of the i-th persimmon tree in the grid. Note that the origin of each coordinate is 1. You can assume that 1 <= xi <= W and 1 <= yi <= H, and no two trees have the same positions. But you should not assume that the persimmon trees are sorted in some order according to their positions. Lastly, S and T are positive integers of the width and height respectively of the estate given by the lord. You can also assume that 1 <= S <= W and 1 <= T <= H.
The end of the input is indicated by a line that solely contains a zero.
Output
For each data set, you are requested to print one line containing the maximum possible number of persimmon trees that can be included in an estate of the given size.
Example
Input
16
10 8
2 2
2 5
2 7
3 3
3 8
4 2
4 5
4 8
6 4
6 7
7 5
7 8
8 1
8 4
9 6
10 3
4 3
8
6 4
1 2
2 1
2 4
3 4
4 2
5 3
6 1
6 2
3 2
0
Output
4
3
Submitted Solution:
```
#!/usr/bin/env python
# -*- coding: utf-8 -*-
while True:
N = int(input())
if N == 0:
break
W,H = map(int,input().split(" "))
Map = [[-1 for i in range(W)] for j in range(H)]
for i in range(N):
x,y = map(lambda x:int(x)-1,input().split(" "))
Map[y][x] = 1
S,T = map(int,input().split(" "))
count = 0
for i in range(0,H-T+1):
for j in range(0,W-S+1):
count = max(count,sum([1 if Map[k][l] == 1 else 0 for k in range(i,i+T) for l in range(j,j+S)]))
print(count)
``` | instruction | 0 | 18,498 | 23 | 36,996 |
Yes | output | 1 | 18,498 | 23 | 36,997 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Seiji Hayashi had been a professor of the Nisshinkan Samurai School in the domain of Aizu for a long time in the 18th century. In order to reward him for his meritorious career in education, Katanobu Matsudaira, the lord of the domain of Aizu, had decided to grant him a rectangular estate within a large field in the Aizu Basin. Although the size (width and height) of the estate was strictly specified by the lord, he was allowed to choose any location for the estate in the field. Inside the field which had also a rectangular shape, many Japanese persimmon trees, whose fruit was one of the famous products of the Aizu region known as 'Mishirazu Persimmon', were planted. Since persimmon was Hayashi's favorite fruit, he wanted to have as many persimmon trees as possible in the estate given by the lord.
For example, in Figure 1, the entire field is a rectangular grid whose width and height are 10 and 8 respectively. Each asterisk (*) represents a place of a persimmon tree. If the specified width and height of the estate are 4 and 3 respectively, the area surrounded by the solid line contains the most persimmon trees. Similarly, if the estate's width is 6 and its height is 4, the area surrounded by the dashed line has the most, and if the estate's width and height are 3 and 4 respectively, the area surrounded by the dotted line contains the most persimmon trees. Note that the width and height cannot be swapped; the sizes 4 by 3 and 3 by 4 are different, as shown in Figure 1.
<image>
---
Figure 1: Examples of Rectangular Estates
Your task is to find the estate of a given size (width and height) that contains the largest number of persimmon trees.
Input
The input consists of multiple data sets. Each data set is given in the following format.
> N
> W` `H
> x1` `y1
> x2` `y2
> ...
> xN` `yN
> S` `T
>
N is the number of persimmon trees, which is a positive integer less than 500. W and H are the width and the height of the entire field respectively. You can assume that both W and H are positive integers whose values are less than 100. For each i (1 <= i <= N), xi and yi are coordinates of the i-th persimmon tree in the grid. Note that the origin of each coordinate is 1. You can assume that 1 <= xi <= W and 1 <= yi <= H, and no two trees have the same positions. But you should not assume that the persimmon trees are sorted in some order according to their positions. Lastly, S and T are positive integers of the width and height respectively of the estate given by the lord. You can also assume that 1 <= S <= W and 1 <= T <= H.
The end of the input is indicated by a line that solely contains a zero.
Output
For each data set, you are requested to print one line containing the maximum possible number of persimmon trees that can be included in an estate of the given size.
Example
Input
16
10 8
2 2
2 5
2 7
3 3
3 8
4 2
4 5
4 8
6 4
6 7
7 5
7 8
8 1
8 4
9 6
10 3
4 3
8
6 4
1 2
2 1
2 4
3 4
4 2
5 3
6 1
6 2
3 2
0
Output
4
3
Submitted Solution:
```
while True:
N = int(input())
if N == 0: break
W, H = map(int, input().split())
board = [[0]*W for _ in range(H)]
for i in range(N):
x, y = map(lambda x:int(x)-1, input().split())
board[y][x] = 1
w, h = map(int, input().split())
ans = 0
rowSums = []
for row in board:
rowSum = [sum(row[:w])]
for i in range(W-w):
rowSum.append(rowSum[i]-row[i]+row[i+w])
rowSums.append(rowSum)
areaSums = [[0]*(W-w+1) for _ in range(H-h+1)]
for i in range(len(rowSums[0])):
areaSums[0][i] = sum(rowSums[j][i] for j in range(h))
for j in range(H-h):
areaSums[j+1][i] = areaSums[j][i] - rowSums[j][i] + rowSums[j+h][i]
print(max([max(row) for row in areaSums]))
``` | instruction | 0 | 18,499 | 23 | 36,998 |
Yes | output | 1 | 18,499 | 23 | 36,999 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Seiji Hayashi had been a professor of the Nisshinkan Samurai School in the domain of Aizu for a long time in the 18th century. In order to reward him for his meritorious career in education, Katanobu Matsudaira, the lord of the domain of Aizu, had decided to grant him a rectangular estate within a large field in the Aizu Basin. Although the size (width and height) of the estate was strictly specified by the lord, he was allowed to choose any location for the estate in the field. Inside the field which had also a rectangular shape, many Japanese persimmon trees, whose fruit was one of the famous products of the Aizu region known as 'Mishirazu Persimmon', were planted. Since persimmon was Hayashi's favorite fruit, he wanted to have as many persimmon trees as possible in the estate given by the lord.
For example, in Figure 1, the entire field is a rectangular grid whose width and height are 10 and 8 respectively. Each asterisk (*) represents a place of a persimmon tree. If the specified width and height of the estate are 4 and 3 respectively, the area surrounded by the solid line contains the most persimmon trees. Similarly, if the estate's width is 6 and its height is 4, the area surrounded by the dashed line has the most, and if the estate's width and height are 3 and 4 respectively, the area surrounded by the dotted line contains the most persimmon trees. Note that the width and height cannot be swapped; the sizes 4 by 3 and 3 by 4 are different, as shown in Figure 1.
<image>
---
Figure 1: Examples of Rectangular Estates
Your task is to find the estate of a given size (width and height) that contains the largest number of persimmon trees.
Input
The input consists of multiple data sets. Each data set is given in the following format.
> N
> W` `H
> x1` `y1
> x2` `y2
> ...
> xN` `yN
> S` `T
>
N is the number of persimmon trees, which is a positive integer less than 500. W and H are the width and the height of the entire field respectively. You can assume that both W and H are positive integers whose values are less than 100. For each i (1 <= i <= N), xi and yi are coordinates of the i-th persimmon tree in the grid. Note that the origin of each coordinate is 1. You can assume that 1 <= xi <= W and 1 <= yi <= H, and no two trees have the same positions. But you should not assume that the persimmon trees are sorted in some order according to their positions. Lastly, S and T are positive integers of the width and height respectively of the estate given by the lord. You can also assume that 1 <= S <= W and 1 <= T <= H.
The end of the input is indicated by a line that solely contains a zero.
Output
For each data set, you are requested to print one line containing the maximum possible number of persimmon trees that can be included in an estate of the given size.
Example
Input
16
10 8
2 2
2 5
2 7
3 3
3 8
4 2
4 5
4 8
6 4
6 7
7 5
7 8
8 1
8 4
9 6
10 3
4 3
8
6 4
1 2
2 1
2 4
3 4
4 2
5 3
6 1
6 2
3 2
0
Output
4
3
Submitted Solution:
```
while True:
n = int(input())
if n == 0:
break
w, h = map(int, input().split())
field = [[False for y in range(h)] for x in range(w)]
for i in range(n):
x, y = map(int, input().split())
field[x - 1][y - 1] = True
s, t = map(int, input().split())
ans = 0
for x0 in range(w - s + 1):
for y0 in range(h - t + 1):
cnt = 0
for dx in range(s):
for dy in range(t):
if field[x0 + dx][y0 + dy]:
cnt += 1
ans = max(ans, cnt)
print(ans)
``` | instruction | 0 | 18,500 | 23 | 37,000 |
Yes | output | 1 | 18,500 | 23 | 37,001 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Seiji Hayashi had been a professor of the Nisshinkan Samurai School in the domain of Aizu for a long time in the 18th century. In order to reward him for his meritorious career in education, Katanobu Matsudaira, the lord of the domain of Aizu, had decided to grant him a rectangular estate within a large field in the Aizu Basin. Although the size (width and height) of the estate was strictly specified by the lord, he was allowed to choose any location for the estate in the field. Inside the field which had also a rectangular shape, many Japanese persimmon trees, whose fruit was one of the famous products of the Aizu region known as 'Mishirazu Persimmon', were planted. Since persimmon was Hayashi's favorite fruit, he wanted to have as many persimmon trees as possible in the estate given by the lord.
For example, in Figure 1, the entire field is a rectangular grid whose width and height are 10 and 8 respectively. Each asterisk (*) represents a place of a persimmon tree. If the specified width and height of the estate are 4 and 3 respectively, the area surrounded by the solid line contains the most persimmon trees. Similarly, if the estate's width is 6 and its height is 4, the area surrounded by the dashed line has the most, and if the estate's width and height are 3 and 4 respectively, the area surrounded by the dotted line contains the most persimmon trees. Note that the width and height cannot be swapped; the sizes 4 by 3 and 3 by 4 are different, as shown in Figure 1.
<image>
---
Figure 1: Examples of Rectangular Estates
Your task is to find the estate of a given size (width and height) that contains the largest number of persimmon trees.
Input
The input consists of multiple data sets. Each data set is given in the following format.
> N
> W` `H
> x1` `y1
> x2` `y2
> ...
> xN` `yN
> S` `T
>
N is the number of persimmon trees, which is a positive integer less than 500. W and H are the width and the height of the entire field respectively. You can assume that both W and H are positive integers whose values are less than 100. For each i (1 <= i <= N), xi and yi are coordinates of the i-th persimmon tree in the grid. Note that the origin of each coordinate is 1. You can assume that 1 <= xi <= W and 1 <= yi <= H, and no two trees have the same positions. But you should not assume that the persimmon trees are sorted in some order according to their positions. Lastly, S and T are positive integers of the width and height respectively of the estate given by the lord. You can also assume that 1 <= S <= W and 1 <= T <= H.
The end of the input is indicated by a line that solely contains a zero.
Output
For each data set, you are requested to print one line containing the maximum possible number of persimmon trees that can be included in an estate of the given size.
Example
Input
16
10 8
2 2
2 5
2 7
3 3
3 8
4 2
4 5
4 8
6 4
6 7
7 5
7 8
8 1
8 4
9 6
10 3
4 3
8
6 4
1 2
2 1
2 4
3 4
4 2
5 3
6 1
6 2
3 2
0
Output
4
3
Submitted Solution:
```
while True:
n = int(input())
if n == 0:
break
w, h = map(int, input().split())
count = 0
table = [[False]*h for i in range(w)]
for i in range(n):
x, y = map(int, input().split())
table[x-1][y-1] = True
s, t = map(int, input().split())
for i in range(w-s+1):
for j in range(h-t+1):
rc = ([row[j:j+t] for row in table[i:i+s]])
count = max(count, (sum(rc, []).count(True)))
print(count)
``` | instruction | 0 | 18,501 | 23 | 37,002 |
Yes | output | 1 | 18,501 | 23 | 37,003 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Seiji Hayashi had been a professor of the Nisshinkan Samurai School in the domain of Aizu for a long time in the 18th century. In order to reward him for his meritorious career in education, Katanobu Matsudaira, the lord of the domain of Aizu, had decided to grant him a rectangular estate within a large field in the Aizu Basin. Although the size (width and height) of the estate was strictly specified by the lord, he was allowed to choose any location for the estate in the field. Inside the field which had also a rectangular shape, many Japanese persimmon trees, whose fruit was one of the famous products of the Aizu region known as 'Mishirazu Persimmon', were planted. Since persimmon was Hayashi's favorite fruit, he wanted to have as many persimmon trees as possible in the estate given by the lord.
For example, in Figure 1, the entire field is a rectangular grid whose width and height are 10 and 8 respectively. Each asterisk (*) represents a place of a persimmon tree. If the specified width and height of the estate are 4 and 3 respectively, the area surrounded by the solid line contains the most persimmon trees. Similarly, if the estate's width is 6 and its height is 4, the area surrounded by the dashed line has the most, and if the estate's width and height are 3 and 4 respectively, the area surrounded by the dotted line contains the most persimmon trees. Note that the width and height cannot be swapped; the sizes 4 by 3 and 3 by 4 are different, as shown in Figure 1.
<image>
---
Figure 1: Examples of Rectangular Estates
Your task is to find the estate of a given size (width and height) that contains the largest number of persimmon trees.
Input
The input consists of multiple data sets. Each data set is given in the following format.
> N
> W` `H
> x1` `y1
> x2` `y2
> ...
> xN` `yN
> S` `T
>
N is the number of persimmon trees, which is a positive integer less than 500. W and H are the width and the height of the entire field respectively. You can assume that both W and H are positive integers whose values are less than 100. For each i (1 <= i <= N), xi and yi are coordinates of the i-th persimmon tree in the grid. Note that the origin of each coordinate is 1. You can assume that 1 <= xi <= W and 1 <= yi <= H, and no two trees have the same positions. But you should not assume that the persimmon trees are sorted in some order according to their positions. Lastly, S and T are positive integers of the width and height respectively of the estate given by the lord. You can also assume that 1 <= S <= W and 1 <= T <= H.
The end of the input is indicated by a line that solely contains a zero.
Output
For each data set, you are requested to print one line containing the maximum possible number of persimmon trees that can be included in an estate of the given size.
Example
Input
16
10 8
2 2
2 5
2 7
3 3
3 8
4 2
4 5
4 8
6 4
6 7
7 5
7 8
8 1
8 4
9 6
10 3
4 3
8
6 4
1 2
2 1
2 4
3 4
4 2
5 3
6 1
6 2
3 2
0
Output
4
3
Submitted Solution:
```
trial = int(input())
while True:
area = [int(n) for n in input().split(" ")]
trees = []
for t in range(trial):
trees.append(int(n) for n in input().split(" "))
cnt,answer = 0,0
trees = sorted(trees,key=lambda x:(x[0],x[1]))
building = [int(n) for n in input().split(" ")]
for x in range(1,1+area[0]-building[0]):
for y in range(1,1+area[1]-building[1]):
cnt = 0
for tree in trees:
if x <= tree[0] <= x+building[0] and y <= tree[1] <= y+building[1]:
cnt += 1
elif x+building[0] < tree[0] and y+building[1] < tree[1]:
if answer < cnt:
answer = cnt
break
else:
if answer < cnt:
answer = cnt
else:
print(answer)
trial = int(input())
if trial == 0:
break
``` | instruction | 0 | 18,502 | 23 | 37,004 |
No | output | 1 | 18,502 | 23 | 37,005 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Seiji Hayashi had been a professor of the Nisshinkan Samurai School in the domain of Aizu for a long time in the 18th century. In order to reward him for his meritorious career in education, Katanobu Matsudaira, the lord of the domain of Aizu, had decided to grant him a rectangular estate within a large field in the Aizu Basin. Although the size (width and height) of the estate was strictly specified by the lord, he was allowed to choose any location for the estate in the field. Inside the field which had also a rectangular shape, many Japanese persimmon trees, whose fruit was one of the famous products of the Aizu region known as 'Mishirazu Persimmon', were planted. Since persimmon was Hayashi's favorite fruit, he wanted to have as many persimmon trees as possible in the estate given by the lord.
For example, in Figure 1, the entire field is a rectangular grid whose width and height are 10 and 8 respectively. Each asterisk (*) represents a place of a persimmon tree. If the specified width and height of the estate are 4 and 3 respectively, the area surrounded by the solid line contains the most persimmon trees. Similarly, if the estate's width is 6 and its height is 4, the area surrounded by the dashed line has the most, and if the estate's width and height are 3 and 4 respectively, the area surrounded by the dotted line contains the most persimmon trees. Note that the width and height cannot be swapped; the sizes 4 by 3 and 3 by 4 are different, as shown in Figure 1.
<image>
---
Figure 1: Examples of Rectangular Estates
Your task is to find the estate of a given size (width and height) that contains the largest number of persimmon trees.
Input
The input consists of multiple data sets. Each data set is given in the following format.
> N
> W` `H
> x1` `y1
> x2` `y2
> ...
> xN` `yN
> S` `T
>
N is the number of persimmon trees, which is a positive integer less than 500. W and H are the width and the height of the entire field respectively. You can assume that both W and H are positive integers whose values are less than 100. For each i (1 <= i <= N), xi and yi are coordinates of the i-th persimmon tree in the grid. Note that the origin of each coordinate is 1. You can assume that 1 <= xi <= W and 1 <= yi <= H, and no two trees have the same positions. But you should not assume that the persimmon trees are sorted in some order according to their positions. Lastly, S and T are positive integers of the width and height respectively of the estate given by the lord. You can also assume that 1 <= S <= W and 1 <= T <= H.
The end of the input is indicated by a line that solely contains a zero.
Output
For each data set, you are requested to print one line containing the maximum possible number of persimmon trees that can be included in an estate of the given size.
Example
Input
16
10 8
2 2
2 5
2 7
3 3
3 8
4 2
4 5
4 8
6 4
6 7
7 5
7 8
8 1
8 4
9 6
10 3
4 3
8
6 4
1 2
2 1
2 4
3 4
4 2
5 3
6 1
6 2
3 2
0
Output
4
3
Submitted Solution:
```
while True:
N = int(input())
if N == 0: break
W,H = map(int,input().split())
fld = [[0 for w in range(W)] for h in range(H)]
for i in range(N):
x,y = map(lambda s:int(s)-1,raw_input().split())
fld[y][x] = 1
cums = [[0 for w in range(W+1)] for h in range(H+1)]
for y in range(H):
for x in range(W):
cums[y+1][x+1] = fld[y][x] + cums[y][x+1] + cums[y+1][x] - cums[y][x]
S,T = map(int,input().split())
ans = 0
for y in range(H-T+1):
for x in range(W-S+1):
ans = max(ans, cums[y+T][x+S] - cums[y+T][x] - cums[y][x+S] + cums[y][x])
print(ans)
``` | instruction | 0 | 18,503 | 23 | 37,006 |
No | output | 1 | 18,503 | 23 | 37,007 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Seiji Hayashi had been a professor of the Nisshinkan Samurai School in the domain of Aizu for a long time in the 18th century. In order to reward him for his meritorious career in education, Katanobu Matsudaira, the lord of the domain of Aizu, had decided to grant him a rectangular estate within a large field in the Aizu Basin. Although the size (width and height) of the estate was strictly specified by the lord, he was allowed to choose any location for the estate in the field. Inside the field which had also a rectangular shape, many Japanese persimmon trees, whose fruit was one of the famous products of the Aizu region known as 'Mishirazu Persimmon', were planted. Since persimmon was Hayashi's favorite fruit, he wanted to have as many persimmon trees as possible in the estate given by the lord.
For example, in Figure 1, the entire field is a rectangular grid whose width and height are 10 and 8 respectively. Each asterisk (*) represents a place of a persimmon tree. If the specified width and height of the estate are 4 and 3 respectively, the area surrounded by the solid line contains the most persimmon trees. Similarly, if the estate's width is 6 and its height is 4, the area surrounded by the dashed line has the most, and if the estate's width and height are 3 and 4 respectively, the area surrounded by the dotted line contains the most persimmon trees. Note that the width and height cannot be swapped; the sizes 4 by 3 and 3 by 4 are different, as shown in Figure 1.
<image>
---
Figure 1: Examples of Rectangular Estates
Your task is to find the estate of a given size (width and height) that contains the largest number of persimmon trees.
Input
The input consists of multiple data sets. Each data set is given in the following format.
> N
> W` `H
> x1` `y1
> x2` `y2
> ...
> xN` `yN
> S` `T
>
N is the number of persimmon trees, which is a positive integer less than 500. W and H are the width and the height of the entire field respectively. You can assume that both W and H are positive integers whose values are less than 100. For each i (1 <= i <= N), xi and yi are coordinates of the i-th persimmon tree in the grid. Note that the origin of each coordinate is 1. You can assume that 1 <= xi <= W and 1 <= yi <= H, and no two trees have the same positions. But you should not assume that the persimmon trees are sorted in some order according to their positions. Lastly, S and T are positive integers of the width and height respectively of the estate given by the lord. You can also assume that 1 <= S <= W and 1 <= T <= H.
The end of the input is indicated by a line that solely contains a zero.
Output
For each data set, you are requested to print one line containing the maximum possible number of persimmon trees that can be included in an estate of the given size.
Example
Input
16
10 8
2 2
2 5
2 7
3 3
3 8
4 2
4 5
4 8
6 4
6 7
7 5
7 8
8 1
8 4
9 6
10 3
4 3
8
6 4
1 2
2 1
2 4
3 4
4 2
5 3
6 1
6 2
3 2
0
Output
4
3
Submitted Solution:
```
while True:
n = int(input())
if n == 0: break
w, h = map(int, input().split())
trees = []
for i in range(n):
trees.append(tuple(map(int, input().split())))
s, t = map(int, input().split())
max_num = 0
# loops in upper left coordinate
for i in range(w - s + 1):
for j in range(h - t + 1):
num = 0
for k in range(s):
for l in range(t):
if (i+k+1, j+l+1) in trees:
num += 1
if num > max_num:
max_num = num
print(max_num)
``` | instruction | 0 | 18,504 | 23 | 37,008 |
No | output | 1 | 18,504 | 23 | 37,009 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Seiji Hayashi had been a professor of the Nisshinkan Samurai School in the domain of Aizu for a long time in the 18th century. In order to reward him for his meritorious career in education, Katanobu Matsudaira, the lord of the domain of Aizu, had decided to grant him a rectangular estate within a large field in the Aizu Basin. Although the size (width and height) of the estate was strictly specified by the lord, he was allowed to choose any location for the estate in the field. Inside the field which had also a rectangular shape, many Japanese persimmon trees, whose fruit was one of the famous products of the Aizu region known as 'Mishirazu Persimmon', were planted. Since persimmon was Hayashi's favorite fruit, he wanted to have as many persimmon trees as possible in the estate given by the lord.
For example, in Figure 1, the entire field is a rectangular grid whose width and height are 10 and 8 respectively. Each asterisk (*) represents a place of a persimmon tree. If the specified width and height of the estate are 4 and 3 respectively, the area surrounded by the solid line contains the most persimmon trees. Similarly, if the estate's width is 6 and its height is 4, the area surrounded by the dashed line has the most, and if the estate's width and height are 3 and 4 respectively, the area surrounded by the dotted line contains the most persimmon trees. Note that the width and height cannot be swapped; the sizes 4 by 3 and 3 by 4 are different, as shown in Figure 1.
<image>
---
Figure 1: Examples of Rectangular Estates
Your task is to find the estate of a given size (width and height) that contains the largest number of persimmon trees.
Input
The input consists of multiple data sets. Each data set is given in the following format.
> N
> W` `H
> x1` `y1
> x2` `y2
> ...
> xN` `yN
> S` `T
>
N is the number of persimmon trees, which is a positive integer less than 500. W and H are the width and the height of the entire field respectively. You can assume that both W and H are positive integers whose values are less than 100. For each i (1 <= i <= N), xi and yi are coordinates of the i-th persimmon tree in the grid. Note that the origin of each coordinate is 1. You can assume that 1 <= xi <= W and 1 <= yi <= H, and no two trees have the same positions. But you should not assume that the persimmon trees are sorted in some order according to their positions. Lastly, S and T are positive integers of the width and height respectively of the estate given by the lord. You can also assume that 1 <= S <= W and 1 <= T <= H.
The end of the input is indicated by a line that solely contains a zero.
Output
For each data set, you are requested to print one line containing the maximum possible number of persimmon trees that can be included in an estate of the given size.
Example
Input
16
10 8
2 2
2 5
2 7
3 3
3 8
4 2
4 5
4 8
6 4
6 7
7 5
7 8
8 1
8 4
9 6
10 3
4 3
8
6 4
1 2
2 1
2 4
3 4
4 2
5 3
6 1
6 2
3 2
0
Output
4
3
Submitted Solution:
```
trial = int(input())
while True:
area = [int(n) for n in input().split(" ")]
trees = []
for t in range(trial):
trees.append([int(n) for n in input().split(" ")])
cnt,answer = 0,0
trees = sorted(trees,key=lambda x:(x[0],x[1]))
building = [int(n) for n in input().split(" ")]
for x in range(1,area[0]-building[0]+2):
for y in range(1,area[1]-building[1]+2):
cnt = 0
for tree in trees:
if x <= tree[0] < x+building[0] and y <= tree[1] < y+building[1]:
#print(tree,x,y)
cnt += 1
elif x+building[0] < tree[0] and y+building[1] < tree[1]:
if answer < cnt:
answer = cnt
break
else:
print("")
if answer < cnt:
answer = cnt
else:
print(answer)
trial = int(input())
if trial == 0:
break
``` | instruction | 0 | 18,505 | 23 | 37,010 |
No | output | 1 | 18,505 | 23 | 37,011 |
Provide a correct Python 3 solution for this coding contest problem.
problem
Given the lengths of $ 2 $ sides that are not the hypotenuse of a right triangle, $ A $ and $ B $. The side of length $ A $ overlaps the $ x $ axis, and the side of length $ B $ overlaps the $ y $ axis.
Do the following:
1. Rotate the triangle around the $ x $ axis.
2. Rotate the shape created by performing the operation $ 1 $ around the $ y $ axis.
Perform the operation $ 2 $ to find the volume of the created figure.
output
Output the volume of the figure. Also, output a line break at the end. Absolute or relative errors less than $ 0.000001 $ are allowed.
Example
Input
1 2
Output
33.510322 | instruction | 0 | 18,527 | 23 | 37,054 |
"Correct Solution:
```
import math
a, b = map(int, input().split())
print(math.pi * float(b) ** 3 * float(4/3))
``` | output | 1 | 18,527 | 23 | 37,055 |
Provide a correct Python 3 solution for this coding contest problem.
problem
Given the lengths of $ 2 $ sides that are not the hypotenuse of a right triangle, $ A $ and $ B $. The side of length $ A $ overlaps the $ x $ axis, and the side of length $ B $ overlaps the $ y $ axis.
Do the following:
1. Rotate the triangle around the $ x $ axis.
2. Rotate the shape created by performing the operation $ 1 $ around the $ y $ axis.
Perform the operation $ 2 $ to find the volume of the created figure.
output
Output the volume of the figure. Also, output a line break at the end. Absolute or relative errors less than $ 0.000001 $ are allowed.
Example
Input
1 2
Output
33.510322 | instruction | 0 | 18,528 | 23 | 37,056 |
"Correct Solution:
```
import math
a, b = map(int, input().split())
if a > b:
print(2 / 3 * math.pi * a**2 * b)
else:
print(4 / 3 *math.pi * b**3)
``` | output | 1 | 18,528 | 23 | 37,057 |
Provide a correct Python 3 solution for this coding contest problem.
problem
Given the lengths of $ 2 $ sides that are not the hypotenuse of a right triangle, $ A $ and $ B $. The side of length $ A $ overlaps the $ x $ axis, and the side of length $ B $ overlaps the $ y $ axis.
Do the following:
1. Rotate the triangle around the $ x $ axis.
2. Rotate the shape created by performing the operation $ 1 $ around the $ y $ axis.
Perform the operation $ 2 $ to find the volume of the created figure.
output
Output the volume of the figure. Also, output a line break at the end. Absolute or relative errors less than $ 0.000001 $ are allowed.
Example
Input
1 2
Output
33.510322 | instruction | 0 | 18,529 | 23 | 37,058 |
"Correct Solution:
```
import math
a,b=map(int,input().split())
print(4/3*b**3*math.pi)
``` | output | 1 | 18,529 | 23 | 37,059 |
Provide a correct Python 3 solution for this coding contest problem.
problem
Given the lengths of $ 2 $ sides that are not the hypotenuse of a right triangle, $ A $ and $ B $. The side of length $ A $ overlaps the $ x $ axis, and the side of length $ B $ overlaps the $ y $ axis.
Do the following:
1. Rotate the triangle around the $ x $ axis.
2. Rotate the shape created by performing the operation $ 1 $ around the $ y $ axis.
Perform the operation $ 2 $ to find the volume of the created figure.
output
Output the volume of the figure. Also, output a line break at the end. Absolute or relative errors less than $ 0.000001 $ are allowed.
Example
Input
1 2
Output
33.510322 | instruction | 0 | 18,530 | 23 | 37,060 |
"Correct Solution:
```
import math
a,b = map(int,input().split())
print(b**3 * math.pi * 4/3)
``` | output | 1 | 18,530 | 23 | 37,061 |
Provide a correct Python 3 solution for this coding contest problem.
problem
Given the lengths of $ 2 $ sides that are not the hypotenuse of a right triangle, $ A $ and $ B $. The side of length $ A $ overlaps the $ x $ axis, and the side of length $ B $ overlaps the $ y $ axis.
Do the following:
1. Rotate the triangle around the $ x $ axis.
2. Rotate the shape created by performing the operation $ 1 $ around the $ y $ axis.
Perform the operation $ 2 $ to find the volume of the created figure.
output
Output the volume of the figure. Also, output a line break at the end. Absolute or relative errors less than $ 0.000001 $ are allowed.
Example
Input
1 2
Output
33.510322 | instruction | 0 | 18,531 | 23 | 37,062 |
"Correct Solution:
```
import math
a,b=map(int,input().split())
print(b**3*math.pi*(4/3))
``` | output | 1 | 18,531 | 23 | 37,063 |
Provide a correct Python 3 solution for this coding contest problem.
problem
Given the lengths of $ 2 $ sides that are not the hypotenuse of a right triangle, $ A $ and $ B $. The side of length $ A $ overlaps the $ x $ axis, and the side of length $ B $ overlaps the $ y $ axis.
Do the following:
1. Rotate the triangle around the $ x $ axis.
2. Rotate the shape created by performing the operation $ 1 $ around the $ y $ axis.
Perform the operation $ 2 $ to find the volume of the created figure.
output
Output the volume of the figure. Also, output a line break at the end. Absolute or relative errors less than $ 0.000001 $ are allowed.
Example
Input
1 2
Output
33.510322 | instruction | 0 | 18,532 | 23 | 37,064 |
"Correct Solution:
```
from math import pi
A, B = map(float, input().split())
print('{:.10f}'.format(4/3*pi*B**3))
``` | output | 1 | 18,532 | 23 | 37,065 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers n and m (m < n). Consider a convex regular polygon of n vertices. Recall that a regular polygon is a polygon that is equiangular (all angles are equal in measure) and equilateral (all sides have the same length).
<image> Examples of convex regular polygons
Your task is to say if it is possible to build another convex regular polygon with m vertices such that its center coincides with the center of the initial polygon and each of its vertices is some vertex of the initial polygon.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The next t lines describe test cases. Each test case is given as two space-separated integers n and m (3 ≤ m < n ≤ 100) — the number of vertices in the initial polygon and the number of vertices in the polygon you want to build.
Output
For each test case, print the answer — "YES" (without quotes), if it is possible to build another convex regular polygon with m vertices such that its center coincides with the center of the initial polygon and each of its vertices is some vertex of the initial polygon and "NO" otherwise.
Example
Input
2
6 3
7 3
Output
YES
NO
Note
<image> The first test case of the example
It can be shown that the answer for the second test case of the example is "NO". | instruction | 0 | 18,651 | 23 | 37,302 |
Tags: geometry, greedy, math, number theory
Correct Solution:
```
t = int(input())
for _ in range(t):
l = list(map(int,input().split()))
if (l[0]-l[1])%l[1]==0:
print("YES")
else:
print("NO")
``` | output | 1 | 18,651 | 23 | 37,303 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers n and m (m < n). Consider a convex regular polygon of n vertices. Recall that a regular polygon is a polygon that is equiangular (all angles are equal in measure) and equilateral (all sides have the same length).
<image> Examples of convex regular polygons
Your task is to say if it is possible to build another convex regular polygon with m vertices such that its center coincides with the center of the initial polygon and each of its vertices is some vertex of the initial polygon.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The next t lines describe test cases. Each test case is given as two space-separated integers n and m (3 ≤ m < n ≤ 100) — the number of vertices in the initial polygon and the number of vertices in the polygon you want to build.
Output
For each test case, print the answer — "YES" (without quotes), if it is possible to build another convex regular polygon with m vertices such that its center coincides with the center of the initial polygon and each of its vertices is some vertex of the initial polygon and "NO" otherwise.
Example
Input
2
6 3
7 3
Output
YES
NO
Note
<image> The first test case of the example
It can be shown that the answer for the second test case of the example is "NO". | instruction | 0 | 18,652 | 23 | 37,304 |
Tags: geometry, greedy, math, number theory
Correct Solution:
```
ans = []
for t in range(int(input())):
n, m = map(int, input().split())
if n % m: ans += ['NO']
else: ans += ['YES']
print('\n'.join(ans))
``` | output | 1 | 18,652 | 23 | 37,305 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers n and m (m < n). Consider a convex regular polygon of n vertices. Recall that a regular polygon is a polygon that is equiangular (all angles are equal in measure) and equilateral (all sides have the same length).
<image> Examples of convex regular polygons
Your task is to say if it is possible to build another convex regular polygon with m vertices such that its center coincides with the center of the initial polygon and each of its vertices is some vertex of the initial polygon.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The next t lines describe test cases. Each test case is given as two space-separated integers n and m (3 ≤ m < n ≤ 100) — the number of vertices in the initial polygon and the number of vertices in the polygon you want to build.
Output
For each test case, print the answer — "YES" (without quotes), if it is possible to build another convex regular polygon with m vertices such that its center coincides with the center of the initial polygon and each of its vertices is some vertex of the initial polygon and "NO" otherwise.
Example
Input
2
6 3
7 3
Output
YES
NO
Note
<image> The first test case of the example
It can be shown that the answer for the second test case of the example is "NO". | instruction | 0 | 18,653 | 23 | 37,306 |
Tags: geometry, greedy, math, number theory
Correct Solution:
```
for i in range(int(input())):
n, m = list(map(int, input().split()))
if (n % m == 0):
print("YES")
else:
print("NO")
``` | output | 1 | 18,653 | 23 | 37,307 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers n and m (m < n). Consider a convex regular polygon of n vertices. Recall that a regular polygon is a polygon that is equiangular (all angles are equal in measure) and equilateral (all sides have the same length).
<image> Examples of convex regular polygons
Your task is to say if it is possible to build another convex regular polygon with m vertices such that its center coincides with the center of the initial polygon and each of its vertices is some vertex of the initial polygon.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The next t lines describe test cases. Each test case is given as two space-separated integers n and m (3 ≤ m < n ≤ 100) — the number of vertices in the initial polygon and the number of vertices in the polygon you want to build.
Output
For each test case, print the answer — "YES" (without quotes), if it is possible to build another convex regular polygon with m vertices such that its center coincides with the center of the initial polygon and each of its vertices is some vertex of the initial polygon and "NO" otherwise.
Example
Input
2
6 3
7 3
Output
YES
NO
Note
<image> The first test case of the example
It can be shown that the answer for the second test case of the example is "NO". | instruction | 0 | 18,654 | 23 | 37,308 |
Tags: geometry, greedy, math, number theory
Correct Solution:
```
for t in range(int(input())):
l,m=map(int,input().split())
if l%m==0 and m>=3:
print("YES")
else:
print("NO")
``` | output | 1 | 18,654 | 23 | 37,309 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers n and m (m < n). Consider a convex regular polygon of n vertices. Recall that a regular polygon is a polygon that is equiangular (all angles are equal in measure) and equilateral (all sides have the same length).
<image> Examples of convex regular polygons
Your task is to say if it is possible to build another convex regular polygon with m vertices such that its center coincides with the center of the initial polygon and each of its vertices is some vertex of the initial polygon.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The next t lines describe test cases. Each test case is given as two space-separated integers n and m (3 ≤ m < n ≤ 100) — the number of vertices in the initial polygon and the number of vertices in the polygon you want to build.
Output
For each test case, print the answer — "YES" (without quotes), if it is possible to build another convex regular polygon with m vertices such that its center coincides with the center of the initial polygon and each of its vertices is some vertex of the initial polygon and "NO" otherwise.
Example
Input
2
6 3
7 3
Output
YES
NO
Note
<image> The first test case of the example
It can be shown that the answer for the second test case of the example is "NO". | instruction | 0 | 18,655 | 23 | 37,310 |
Tags: geometry, greedy, math, number theory
Correct Solution:
```
for _ in range(int(input())):
first, sec = list(map(int, input().split()))
if first%sec == 0:
print("YES")
else:
print("NO")
``` | output | 1 | 18,655 | 23 | 37,311 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers n and m (m < n). Consider a convex regular polygon of n vertices. Recall that a regular polygon is a polygon that is equiangular (all angles are equal in measure) and equilateral (all sides have the same length).
<image> Examples of convex regular polygons
Your task is to say if it is possible to build another convex regular polygon with m vertices such that its center coincides with the center of the initial polygon and each of its vertices is some vertex of the initial polygon.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The next t lines describe test cases. Each test case is given as two space-separated integers n and m (3 ≤ m < n ≤ 100) — the number of vertices in the initial polygon and the number of vertices in the polygon you want to build.
Output
For each test case, print the answer — "YES" (without quotes), if it is possible to build another convex regular polygon with m vertices such that its center coincides with the center of the initial polygon and each of its vertices is some vertex of the initial polygon and "NO" otherwise.
Example
Input
2
6 3
7 3
Output
YES
NO
Note
<image> The first test case of the example
It can be shown that the answer for the second test case of the example is "NO". | instruction | 0 | 18,656 | 23 | 37,312 |
Tags: geometry, greedy, math, number theory
Correct Solution:
```
import os
import heapq
import sys
import math
from collections import defaultdict
from io import BytesIO, IOBase
def main():
for _ in range(int(input())):
n,m=map(int,input().split())
if n%m==0:
print("YES")
else:
print("NO")
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")
if __name__ == "__main__":
main()
``` | output | 1 | 18,656 | 23 | 37,313 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers n and m (m < n). Consider a convex regular polygon of n vertices. Recall that a regular polygon is a polygon that is equiangular (all angles are equal in measure) and equilateral (all sides have the same length).
<image> Examples of convex regular polygons
Your task is to say if it is possible to build another convex regular polygon with m vertices such that its center coincides with the center of the initial polygon and each of its vertices is some vertex of the initial polygon.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The next t lines describe test cases. Each test case is given as two space-separated integers n and m (3 ≤ m < n ≤ 100) — the number of vertices in the initial polygon and the number of vertices in the polygon you want to build.
Output
For each test case, print the answer — "YES" (without quotes), if it is possible to build another convex regular polygon with m vertices such that its center coincides with the center of the initial polygon and each of its vertices is some vertex of the initial polygon and "NO" otherwise.
Example
Input
2
6 3
7 3
Output
YES
NO
Note
<image> The first test case of the example
It can be shown that the answer for the second test case of the example is "NO". | instruction | 0 | 18,657 | 23 | 37,314 |
Tags: geometry, greedy, math, number theory
Correct Solution:
```
c = int(input())
for i in range(c):
n, m = map(int, input().split())
if n%m == 0:
print("YES")
else:
print("NO")
``` | output | 1 | 18,657 | 23 | 37,315 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers n and m (m < n). Consider a convex regular polygon of n vertices. Recall that a regular polygon is a polygon that is equiangular (all angles are equal in measure) and equilateral (all sides have the same length).
<image> Examples of convex regular polygons
Your task is to say if it is possible to build another convex regular polygon with m vertices such that its center coincides with the center of the initial polygon and each of its vertices is some vertex of the initial polygon.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The next t lines describe test cases. Each test case is given as two space-separated integers n and m (3 ≤ m < n ≤ 100) — the number of vertices in the initial polygon and the number of vertices in the polygon you want to build.
Output
For each test case, print the answer — "YES" (without quotes), if it is possible to build another convex regular polygon with m vertices such that its center coincides with the center of the initial polygon and each of its vertices is some vertex of the initial polygon and "NO" otherwise.
Example
Input
2
6 3
7 3
Output
YES
NO
Note
<image> The first test case of the example
It can be shown that the answer for the second test case of the example is "NO". | instruction | 0 | 18,658 | 23 | 37,316 |
Tags: geometry, greedy, math, number theory
Correct Solution:
```
import sys
n = int(sys.stdin.readline())
for i in range(n):
a, b = map(int, sys.stdin.readline().split())
if a%b==0:
print("YES")
else:
print("NO")
``` | output | 1 | 18,658 | 23 | 37,317 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integers n and m (m < n). Consider a convex regular polygon of n vertices. Recall that a regular polygon is a polygon that is equiangular (all angles are equal in measure) and equilateral (all sides have the same length).
<image> Examples of convex regular polygons
Your task is to say if it is possible to build another convex regular polygon with m vertices such that its center coincides with the center of the initial polygon and each of its vertices is some vertex of the initial polygon.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The next t lines describe test cases. Each test case is given as two space-separated integers n and m (3 ≤ m < n ≤ 100) — the number of vertices in the initial polygon and the number of vertices in the polygon you want to build.
Output
For each test case, print the answer — "YES" (without quotes), if it is possible to build another convex regular polygon with m vertices such that its center coincides with the center of the initial polygon and each of its vertices is some vertex of the initial polygon and "NO" otherwise.
Example
Input
2
6 3
7 3
Output
YES
NO
Note
<image> The first test case of the example
It can be shown that the answer for the second test case of the example is "NO".
Submitted Solution:
```
for t in range(int(input())):
n, m = [int(x) for x in input().split()]
if n % m:
print("NO")
else:
print("YES")
``` | instruction | 0 | 18,659 | 23 | 37,318 |
Yes | output | 1 | 18,659 | 23 | 37,319 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integers n and m (m < n). Consider a convex regular polygon of n vertices. Recall that a regular polygon is a polygon that is equiangular (all angles are equal in measure) and equilateral (all sides have the same length).
<image> Examples of convex regular polygons
Your task is to say if it is possible to build another convex regular polygon with m vertices such that its center coincides with the center of the initial polygon and each of its vertices is some vertex of the initial polygon.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The next t lines describe test cases. Each test case is given as two space-separated integers n and m (3 ≤ m < n ≤ 100) — the number of vertices in the initial polygon and the number of vertices in the polygon you want to build.
Output
For each test case, print the answer — "YES" (without quotes), if it is possible to build another convex regular polygon with m vertices such that its center coincides with the center of the initial polygon and each of its vertices is some vertex of the initial polygon and "NO" otherwise.
Example
Input
2
6 3
7 3
Output
YES
NO
Note
<image> The first test case of the example
It can be shown that the answer for the second test case of the example is "NO".
Submitted Solution:
```
from sys import stdin, stdout
import math
def main():
t=int(input())
while t:
n,m=map(int,input().split())
if n%m==0:
print("YES")
else:
print("NO")
t-=1
if __name__ == "__main__":
main()
``` | instruction | 0 | 18,660 | 23 | 37,320 |
Yes | output | 1 | 18,660 | 23 | 37,321 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integers n and m (m < n). Consider a convex regular polygon of n vertices. Recall that a regular polygon is a polygon that is equiangular (all angles are equal in measure) and equilateral (all sides have the same length).
<image> Examples of convex regular polygons
Your task is to say if it is possible to build another convex regular polygon with m vertices such that its center coincides with the center of the initial polygon and each of its vertices is some vertex of the initial polygon.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The next t lines describe test cases. Each test case is given as two space-separated integers n and m (3 ≤ m < n ≤ 100) — the number of vertices in the initial polygon and the number of vertices in the polygon you want to build.
Output
For each test case, print the answer — "YES" (without quotes), if it is possible to build another convex regular polygon with m vertices such that its center coincides with the center of the initial polygon and each of its vertices is some vertex of the initial polygon and "NO" otherwise.
Example
Input
2
6 3
7 3
Output
YES
NO
Note
<image> The first test case of the example
It can be shown that the answer for the second test case of the example is "NO".
Submitted Solution:
```
T = (int)(input())
while T>0:
n,m = [int(x) for x in input().split()]
if(n%m==0):
print("YES")
else:
print("NO")
T-=1
``` | instruction | 0 | 18,661 | 23 | 37,322 |
Yes | output | 1 | 18,661 | 23 | 37,323 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integers n and m (m < n). Consider a convex regular polygon of n vertices. Recall that a regular polygon is a polygon that is equiangular (all angles are equal in measure) and equilateral (all sides have the same length).
<image> Examples of convex regular polygons
Your task is to say if it is possible to build another convex regular polygon with m vertices such that its center coincides with the center of the initial polygon and each of its vertices is some vertex of the initial polygon.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The next t lines describe test cases. Each test case is given as two space-separated integers n and m (3 ≤ m < n ≤ 100) — the number of vertices in the initial polygon and the number of vertices in the polygon you want to build.
Output
For each test case, print the answer — "YES" (without quotes), if it is possible to build another convex regular polygon with m vertices such that its center coincides with the center of the initial polygon and each of its vertices is some vertex of the initial polygon and "NO" otherwise.
Example
Input
2
6 3
7 3
Output
YES
NO
Note
<image> The first test case of the example
It can be shown that the answer for the second test case of the example is "NO".
Submitted Solution:
```
for _ in range(int(input())):
n,m=map(int,input().split())
#a=[int(i) for i in input().split()]
if n%m==0:
print('YES')
else:
print('NO')
``` | instruction | 0 | 18,662 | 23 | 37,324 |
Yes | output | 1 | 18,662 | 23 | 37,325 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integers n and m (m < n). Consider a convex regular polygon of n vertices. Recall that a regular polygon is a polygon that is equiangular (all angles are equal in measure) and equilateral (all sides have the same length).
<image> Examples of convex regular polygons
Your task is to say if it is possible to build another convex regular polygon with m vertices such that its center coincides with the center of the initial polygon and each of its vertices is some vertex of the initial polygon.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The next t lines describe test cases. Each test case is given as two space-separated integers n and m (3 ≤ m < n ≤ 100) — the number of vertices in the initial polygon and the number of vertices in the polygon you want to build.
Output
For each test case, print the answer — "YES" (without quotes), if it is possible to build another convex regular polygon with m vertices such that its center coincides with the center of the initial polygon and each of its vertices is some vertex of the initial polygon and "NO" otherwise.
Example
Input
2
6 3
7 3
Output
YES
NO
Note
<image> The first test case of the example
It can be shown that the answer for the second test case of the example is "NO".
Submitted Solution:
```
for i in range(int(input())):
n,m=map(int,input().split())
if n%2==0 and m==n//2:
print("YES")
else:
print("NO")
``` | instruction | 0 | 18,663 | 23 | 37,326 |
No | output | 1 | 18,663 | 23 | 37,327 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integers n and m (m < n). Consider a convex regular polygon of n vertices. Recall that a regular polygon is a polygon that is equiangular (all angles are equal in measure) and equilateral (all sides have the same length).
<image> Examples of convex regular polygons
Your task is to say if it is possible to build another convex regular polygon with m vertices such that its center coincides with the center of the initial polygon and each of its vertices is some vertex of the initial polygon.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The next t lines describe test cases. Each test case is given as two space-separated integers n and m (3 ≤ m < n ≤ 100) — the number of vertices in the initial polygon and the number of vertices in the polygon you want to build.
Output
For each test case, print the answer — "YES" (without quotes), if it is possible to build another convex regular polygon with m vertices such that its center coincides with the center of the initial polygon and each of its vertices is some vertex of the initial polygon and "NO" otherwise.
Example
Input
2
6 3
7 3
Output
YES
NO
Note
<image> The first test case of the example
It can be shown that the answer for the second test case of the example is "NO".
Submitted Solution:
```
t=int(input())
for i in range(t):
n,m=map(int,input().split())
q=(n-2)/n
r=(m-2)/m
if m>2:
k=q%r
if n%m==0 and k==0:
print("YES")
else:
print("NO")
``` | instruction | 0 | 18,664 | 23 | 37,328 |
No | output | 1 | 18,664 | 23 | 37,329 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integers n and m (m < n). Consider a convex regular polygon of n vertices. Recall that a regular polygon is a polygon that is equiangular (all angles are equal in measure) and equilateral (all sides have the same length).
<image> Examples of convex regular polygons
Your task is to say if it is possible to build another convex regular polygon with m vertices such that its center coincides with the center of the initial polygon and each of its vertices is some vertex of the initial polygon.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The next t lines describe test cases. Each test case is given as two space-separated integers n and m (3 ≤ m < n ≤ 100) — the number of vertices in the initial polygon and the number of vertices in the polygon you want to build.
Output
For each test case, print the answer — "YES" (without quotes), if it is possible to build another convex regular polygon with m vertices such that its center coincides with the center of the initial polygon and each of its vertices is some vertex of the initial polygon and "NO" otherwise.
Example
Input
2
6 3
7 3
Output
YES
NO
Note
<image> The first test case of the example
It can be shown that the answer for the second test case of the example is "NO".
Submitted Solution:
```
for _ in range(int(input())):
n,m = map(int, input().split())
if n % m == 0 and n//m == 2:
print("YES")
else:
print("NO")
``` | instruction | 0 | 18,665 | 23 | 37,330 |
No | output | 1 | 18,665 | 23 | 37,331 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integers n and m (m < n). Consider a convex regular polygon of n vertices. Recall that a regular polygon is a polygon that is equiangular (all angles are equal in measure) and equilateral (all sides have the same length).
<image> Examples of convex regular polygons
Your task is to say if it is possible to build another convex regular polygon with m vertices such that its center coincides with the center of the initial polygon and each of its vertices is some vertex of the initial polygon.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases.
The next t lines describe test cases. Each test case is given as two space-separated integers n and m (3 ≤ m < n ≤ 100) — the number of vertices in the initial polygon and the number of vertices in the polygon you want to build.
Output
For each test case, print the answer — "YES" (without quotes), if it is possible to build another convex regular polygon with m vertices such that its center coincides with the center of the initial polygon and each of its vertices is some vertex of the initial polygon and "NO" otherwise.
Example
Input
2
6 3
7 3
Output
YES
NO
Note
<image> The first test case of the example
It can be shown that the answer for the second test case of the example is "NO".
Submitted Solution:
```
k=int(input())
for i in range(k):
arr=list(map(int,input().split()))
m=arr[0]
n=arr[1]
# print(m)
# print(n)
x=((m-2)*180/m)
print(x)
y=360/n
print(y)
if( x==y):
print("YES")
else:
print("NO")
``` | instruction | 0 | 18,666 | 23 | 37,332 |
No | output | 1 | 18,666 | 23 | 37,333 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A line on the plane is described by an equation Ax + By + C = 0. You are to find any point on this line, whose coordinates are integer numbers from - 5·1018 to 5·1018 inclusive, or to find out that such points do not exist.
Input
The first line contains three integers A, B and C ( - 2·109 ≤ A, B, C ≤ 2·109) — corresponding coefficients of the line equation. It is guaranteed that A2 + B2 > 0.
Output
If the required point exists, output its coordinates, otherwise output -1.
Examples
Input
2 5 3
Output
6 -3
Submitted Solution:
```
def gcd(a,b,x,y):
if a==0:
return b
g=gcd(b%a,a,x,y)
x[0],y[0]=y[0]-(b//a)*x[0],x[0]
#print(x,y)
return g
a,b,c=list(map(int,input().split()))
x,y=[0],[1]
g=gcd(a,b,x,y)
if c%g!=0:
print(-1)
else:
c=c//g
print(x[0]*(-c),y[0]*(-c))
``` | instruction | 0 | 19,035 | 23 | 38,070 |
Yes | output | 1 | 19,035 | 23 | 38,071 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A line on the plane is described by an equation Ax + By + C = 0. You are to find any point on this line, whose coordinates are integer numbers from - 5·1018 to 5·1018 inclusive, or to find out that such points do not exist.
Input
The first line contains three integers A, B and C ( - 2·109 ≤ A, B, C ≤ 2·109) — corresponding coefficients of the line equation. It is guaranteed that A2 + B2 > 0.
Output
If the required point exists, output its coordinates, otherwise output -1.
Examples
Input
2 5 3
Output
6 -3
Submitted Solution:
```
def gcd(x,y):
if x % y == 0:
return y
else:
return gcd(y,x%y)
def bezout(x,y):
if x % y == 0:
return (0,1)
else:
t,s = bezout (y,x%y)
return (s,t-(x//y)*s)
a,b,c = map(int,input().split())
if a == 0:
if c % b == 0:
print("%d %d" % (0,-c//b))
else:
print(-1)
exit()
if b == 0:
if c % a == 0:
print("%d %d" % (-c//a,0))
else:
print(-1)
exit()
if c % gcd(a,b) != 0:
print(-1)
else:
x,y = bezout(a,b)
m = -c//gcd(a,b)
print("%d %d" % (m*x,m*y))
``` | instruction | 0 | 19,036 | 23 | 38,072 |
Yes | output | 1 | 19,036 | 23 | 38,073 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A line on the plane is described by an equation Ax + By + C = 0. You are to find any point on this line, whose coordinates are integer numbers from - 5·1018 to 5·1018 inclusive, or to find out that such points do not exist.
Input
The first line contains three integers A, B and C ( - 2·109 ≤ A, B, C ≤ 2·109) — corresponding coefficients of the line equation. It is guaranteed that A2 + B2 > 0.
Output
If the required point exists, output its coordinates, otherwise output -1.
Examples
Input
2 5 3
Output
6 -3
Submitted Solution:
```
def gcd(a, b):
if a == 0:
return 0, 1, b;
x, y, g = gcd(b % a, a)
return y - (b // a) * x, x, g
a, b, c = map(int, input().split())
c = -c
x, y, g = gcd(a, b)
if c % g != 0:
print("-1")
else:
x *= c // g
y *= c // g
print(x,y)
``` | instruction | 0 | 19,037 | 23 | 38,074 |
Yes | output | 1 | 19,037 | 23 | 38,075 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A line on the plane is described by an equation Ax + By + C = 0. You are to find any point on this line, whose coordinates are integer numbers from - 5·1018 to 5·1018 inclusive, or to find out that such points do not exist.
Input
The first line contains three integers A, B and C ( - 2·109 ≤ A, B, C ≤ 2·109) — corresponding coefficients of the line equation. It is guaranteed that A2 + B2 > 0.
Output
If the required point exists, output its coordinates, otherwise output -1.
Examples
Input
2 5 3
Output
6 -3
Submitted Solution:
```
import math
def ii(): return int(input())
def ill(): return input().split(' ')
def ili(): return [int(i) for i in input().split(' ')]
def ilis(): return (int(i) for i in input().split(' '))
def gcd(a, b): return a if b == 0 else gcd(b, a % b)
def extgcd(a, b, x=0, y=0):
if b == 0:
return (a, 1, 0)
d, m, n = extgcd(b, a % b, x, y)
return (d, n, m-(a//b)*n)
def main():
a, b, c = ilis()
d, x, y = extgcd(abs(a), abs(b))
if abs(c) % d == 0:
x *= (-c) // d
y *= (-c) // d
if a < 0:
x = -x
if b < 0:
y = -y
print(x, y)
return
else:
print(-1)
return
if __name__ == "__main__":
main()
``` | instruction | 0 | 19,038 | 23 | 38,076 |
Yes | output | 1 | 19,038 | 23 | 38,077 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A line on the plane is described by an equation Ax + By + C = 0. You are to find any point on this line, whose coordinates are integer numbers from - 5·1018 to 5·1018 inclusive, or to find out that such points do not exist.
Input
The first line contains three integers A, B and C ( - 2·109 ≤ A, B, C ≤ 2·109) — corresponding coefficients of the line equation. It is guaranteed that A2 + B2 > 0.
Output
If the required point exists, output its coordinates, otherwise output -1.
Examples
Input
2 5 3
Output
6 -3
Submitted Solution:
```
from math import *
a,b,c=list(map(int,input().split()))
if b==0:
if c%a==0:
print(round(-c/a),0)
else:
print(-1)
elif a==0:
if c%b==0:
print(0,round(-c/b))
else:
print(-1)
else:
if c%gcd(a,b)!=0:
print(-1)
else:
if abs(a)==abs(b):
x=0
y=(-c-a*x)/b
y=round(y)
print(x,y)
else:
if abs(b)>abs(a):
z=[abs(b),abs(a)]
c1=abs(c)%abs(b)
if (a>=0 and b<0) or (a<0 and b>=0):
if (c>=0 and a<0) or (c<0 and a>=0):
v=[c1,-(c1%abs(a))]
h=2
else:
v=[-c1,(c1%abs(a))]
h=3
else:
if (c>=0 and a>=0) or (c<0 and a<0):
v=[-c1,-(c1%abs(a))]
h=-1
else:
v=[c1,(c1%abs(a))]
h=1
i=2
while z[-2]%z[-1]!=1:
z.append(z[-2]%z[-1])
i+=1
c1%=abs(z[-1])
if h==2:
v.append(c1*(-1)**(i+1))
elif h==3:
v.append(c1*(-1)**i)
else:
v.append(h*c1)
if h==-1 or h==1:
for j in range(i):
z[j]=-z[j]
x=v[-1]
for j in range(i-2,-1,-1):
x=(x*z[j]+v[j])//abs(z[j+1])
y=(-c-a*x)//b
print(x,y)
else:
w=a
a=b
b=w
z=[abs(b),abs(a)]
c1=abs(c)%abs(b)
if (a>=0 and b<0) or (a<0 and b>=0):
if (c>=0 and a<0) or (c<0 and a>=0):
v=[c1,-(c1%abs(a))]
h=2
else:
v=[-c1,(c1%abs(a))]
h=3
else:
if (c>=0 and a>=0) or (c<0 and a<0):
v=[-c1,-(c1%abs(a))]
h=-1
else:
v=[c1,(c1%abs(a))]
h=1
i=2
while z[-2]%z[-1]!=1:
z.append(z[-2]%z[-1])
i+=1
c1%=abs(z[-1])
if h==2:
v.append(c1*(-1)**(i+1))
elif h==3:
v.append(c1*(-1)**i)
else:
v.append(h*c1)
if h==-1 or h==1:
for j in range(i):
z[j]=-z[j]
x=v[-1]
for j in range(i-2,-1,-1):
x=(x*z[j]+v[j])//abs(z[j+1])
y=(-c-a*x)//b
print(y,x)
``` | instruction | 0 | 19,039 | 23 | 38,078 |
No | output | 1 | 19,039 | 23 | 38,079 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A line on the plane is described by an equation Ax + By + C = 0. You are to find any point on this line, whose coordinates are integer numbers from - 5·1018 to 5·1018 inclusive, or to find out that such points do not exist.
Input
The first line contains three integers A, B and C ( - 2·109 ≤ A, B, C ≤ 2·109) — corresponding coefficients of the line equation. It is guaranteed that A2 + B2 > 0.
Output
If the required point exists, output its coordinates, otherwise output -1.
Examples
Input
2 5 3
Output
6 -3
Submitted Solution:
```
A=[int(i)for i in input().split(" ")]
a=A[0]
b=A[1]
c=A[2]
x=6
if((a*a+b*b)>0):
y=(-c-(a*x))/b
print(x, y)
else:
print(-1)
``` | instruction | 0 | 19,040 | 23 | 38,080 |
No | output | 1 | 19,040 | 23 | 38,081 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A line on the plane is described by an equation Ax + By + C = 0. You are to find any point on this line, whose coordinates are integer numbers from - 5·1018 to 5·1018 inclusive, or to find out that such points do not exist.
Input
The first line contains three integers A, B and C ( - 2·109 ≤ A, B, C ≤ 2·109) — corresponding coefficients of the line equation. It is guaranteed that A2 + B2 > 0.
Output
If the required point exists, output its coordinates, otherwise output -1.
Examples
Input
2 5 3
Output
6 -3
Submitted Solution:
```
#Line
A,B,C=map(int,input().split())
c=0
for i in (((-5)*10**18),((5)*10**18)+1):
y=(((-1)*C)+((-1)*A*i))/B
if (y%1==0):
print(str(i)+' '+str(y))
c=1
break
if(c==0):
print(-1)
``` | instruction | 0 | 19,041 | 23 | 38,082 |
No | output | 1 | 19,041 | 23 | 38,083 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A line on the plane is described by an equation Ax + By + C = 0. You are to find any point on this line, whose coordinates are integer numbers from - 5·1018 to 5·1018 inclusive, or to find out that such points do not exist.
Input
The first line contains three integers A, B and C ( - 2·109 ≤ A, B, C ≤ 2·109) — corresponding coefficients of the line equation. It is guaranteed that A2 + B2 > 0.
Output
If the required point exists, output its coordinates, otherwise output -1.
Examples
Input
2 5 3
Output
6 -3
Submitted Solution:
```
def find_point(a, b, c):
stop = False
for x in range(-2*10^9, 2*10^9):
if stop:
break
for y in range(-2*10^9, 2*10^9):
if stop:
break
if (a*x + b*y + c) == 0:
print("{} {}".format(x, y))
stop = True
if not stop:
print(-1)
new_input = input()
new_input = new_input.split(' ')
find_point(int(new_input[0]), int(new_input[1]), int(new_input[2]))
``` | instruction | 0 | 19,042 | 23 | 38,084 |
No | output | 1 | 19,042 | 23 | 38,085 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's denote a m-free matrix as a binary (that is, consisting of only 1's and 0's) matrix such that every square submatrix of size m × m of this matrix contains at least one zero.
Consider the following problem:
You are given two integers n and m. You have to construct an m-free square matrix of size n × n such that the number of 1's in this matrix is maximum possible. Print the maximum possible number of 1's in such matrix.
You don't have to solve this problem. Instead, you have to construct a few tests for it.
You will be given t numbers x1, x2, ..., xt. For every <image>, find two integers ni and mi (ni ≥ mi) such that the answer for the aforementioned problem is exactly xi if we set n = ni and m = mi.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of tests you have to construct.
Then t lines follow, i-th line containing one integer xi (0 ≤ xi ≤ 109).
Note that in hacks you have to set t = 1.
Output
For each test you have to construct, output two positive numbers ni and mi (1 ≤ mi ≤ ni ≤ 109) such that the maximum number of 1's in a mi-free ni × ni matrix is exactly xi. If there are multiple solutions, you may output any of them; and if this is impossible to construct a test, output a single integer - 1.
Example
Input
3
21
0
1
Output
5 2
1 1
-1 | instruction | 0 | 19,093 | 23 | 38,186 |
Tags: binary search, brute force, constructive algorithms
Correct Solution:
```
import sys
import math
lines = sys.stdin.read().splitlines()
lincnt = -1
def input():
global lincnt
lincnt += 1
return lines[lincnt]
def solve():
x = int(input())
for n in range(max(1,int(math.sqrt(x) - 5)),int(10**4.7)):
if n ** 2 <= x:
continue
m = math.sqrt(n** 2 / (n** 2 - x))
for k in range(int(m) - 3, int(m) + 3):
if k < 1:
continue
if n ** 2 - (n // k) ** 2 == x:
print(n,k)
return
print(-1)
for _ in range(int(input())):
solve()
``` | output | 1 | 19,093 | 23 | 38,187 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's denote a m-free matrix as a binary (that is, consisting of only 1's and 0's) matrix such that every square submatrix of size m × m of this matrix contains at least one zero.
Consider the following problem:
You are given two integers n and m. You have to construct an m-free square matrix of size n × n such that the number of 1's in this matrix is maximum possible. Print the maximum possible number of 1's in such matrix.
You don't have to solve this problem. Instead, you have to construct a few tests for it.
You will be given t numbers x1, x2, ..., xt. For every <image>, find two integers ni and mi (ni ≥ mi) such that the answer for the aforementioned problem is exactly xi if we set n = ni and m = mi.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of tests you have to construct.
Then t lines follow, i-th line containing one integer xi (0 ≤ xi ≤ 109).
Note that in hacks you have to set t = 1.
Output
For each test you have to construct, output two positive numbers ni and mi (1 ≤ mi ≤ ni ≤ 109) such that the maximum number of 1's in a mi-free ni × ni matrix is exactly xi. If there are multiple solutions, you may output any of them; and if this is impossible to construct a test, output a single integer - 1.
Example
Input
3
21
0
1
Output
5 2
1 1
-1
Submitted Solution:
```
# ---------------------------iye ha aam zindegi---------------------------------------------
import math
import random
import heapq, bisect
import sys
from collections import deque, defaultdict
from fractions import Fraction
import sys
import threading
from collections import defaultdict
threading.stack_size(10**8)
mod = 10 ** 9 + 7
mod1 = 998244353
# ------------------------------warmup----------------------------
import os
import sys
from io import BytesIO, IOBase
sys.setrecursionlimit(300000)
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")
# -------------------game starts now----------------------------------------------------import math
class TreeNode:
def __init__(self, k, v):
self.key = k
self.value = v
self.left = None
self.right = None
self.parent = None
self.height = 1
self.num_left = 1
self.num_total = 1
class AvlTree:
def __init__(self):
self._tree = None
def add(self, k, v):
if not self._tree:
self._tree = TreeNode(k, v)
return
node = self._add(k, v)
if node:
self._rebalance(node)
def _add(self, k, v):
node = self._tree
while node:
if k < node.key:
if node.left:
node = node.left
else:
node.left = TreeNode(k, v)
node.left.parent = node
return node.left
elif node.key < k:
if node.right:
node = node.right
else:
node.right = TreeNode(k, v)
node.right.parent = node
return node.right
else:
node.value = v
return
@staticmethod
def get_height(x):
return x.height if x else 0
@staticmethod
def get_num_total(x):
return x.num_total if x else 0
def _rebalance(self, node):
n = node
while n:
lh = self.get_height(n.left)
rh = self.get_height(n.right)
n.height = max(lh, rh) + 1
balance_factor = lh - rh
n.num_total = 1 + self.get_num_total(n.left) + self.get_num_total(n.right)
n.num_left = 1 + self.get_num_total(n.left)
if balance_factor > 1:
if self.get_height(n.left.left) < self.get_height(n.left.right):
self._rotate_left(n.left)
self._rotate_right(n)
elif balance_factor < -1:
if self.get_height(n.right.right) < self.get_height(n.right.left):
self._rotate_right(n.right)
self._rotate_left(n)
else:
n = n.parent
def _remove_one(self, node):
"""
Side effect!!! Changes node. Node should have exactly one child
"""
replacement = node.left or node.right
if node.parent:
if AvlTree._is_left(node):
node.parent.left = replacement
else:
node.parent.right = replacement
replacement.parent = node.parent
node.parent = None
else:
self._tree = replacement
replacement.parent = None
node.left = None
node.right = None
node.parent = None
self._rebalance(replacement)
def _remove_leaf(self, node):
if node.parent:
if AvlTree._is_left(node):
node.parent.left = None
else:
node.parent.right = None
self._rebalance(node.parent)
else:
self._tree = None
node.parent = None
node.left = None
node.right = None
def remove(self, k):
node = self._get_node(k)
if not node:
return
if AvlTree._is_leaf(node):
self._remove_leaf(node)
return
if node.left and node.right:
nxt = AvlTree._get_next(node)
node.key = nxt.key
node.value = nxt.value
if self._is_leaf(nxt):
self._remove_leaf(nxt)
else:
self._remove_one(nxt)
self._rebalance(node)
else:
self._remove_one(node)
def get(self, k):
node = self._get_node(k)
return node.value if node else -1
def _get_node(self, k):
if not self._tree:
return None
node = self._tree
while node:
if k < node.key:
node = node.left
elif node.key < k:
node = node.right
else:
return node
return None
def get_at(self, pos):
x = pos + 1
node = self._tree
while node:
if x < node.num_left:
node = node.left
elif node.num_left < x:
x -= node.num_left
node = node.right
else:
return (node.key, node.value)
raise IndexError("Out of ranges")
@staticmethod
def _is_left(node):
return node.parent.left and node.parent.left == node
@staticmethod
def _is_leaf(node):
return node.left is None and node.right is None
def _rotate_right(self, node):
if not node.parent:
self._tree = node.left
node.left.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.left
node.left.parent = node.parent
else:
node.parent.right = node.left
node.left.parent = node.parent
bk = node.left.right
node.left.right = node
node.parent = node.left
node.left = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
def _rotate_left(self, node):
if not node.parent:
self._tree = node.right
node.right.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.right
node.right.parent = node.parent
else:
node.parent.right = node.right
node.right.parent = node.parent
bk = node.right.left
node.right.left = node
node.parent = node.right
node.right = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
@staticmethod
def _get_next(node):
if not node.right:
return node.parent
n = node.right
while n.left:
n = n.left
return n
# -----------------------------------------------binary seacrh tree---------------------------------------
class SegmentTree1:
def __init__(self, data, default=2**51, func=lambda a, b: a & b):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------game starts now----------------------------------------------------import math
class SegmentTree:
def __init__(self, data, default=0, func=lambda a, b: a + b):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------------------iye ha chutiya zindegi-------------------------------------
class Factorial:
def __init__(self, MOD):
self.MOD = MOD
self.factorials = [1, 1]
self.invModulos = [0, 1]
self.invFactorial_ = [1, 1]
def calc(self, n):
if n <= -1:
print("Invalid argument to calculate n!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.factorials):
return self.factorials[n]
nextArr = [0] * (n + 1 - len(self.factorials))
initialI = len(self.factorials)
prev = self.factorials[-1]
m = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = prev * i % m
self.factorials += nextArr
return self.factorials[n]
def inv(self, n):
if n <= -1:
print("Invalid argument to calculate n^(-1)")
print("n must be non-negative value. But the argument was " + str(n))
exit()
p = self.MOD
pi = n % p
if pi < len(self.invModulos):
return self.invModulos[pi]
nextArr = [0] * (n + 1 - len(self.invModulos))
initialI = len(self.invModulos)
for i in range(initialI, min(p, n + 1)):
next = -self.invModulos[p % i] * (p // i) % p
self.invModulos.append(next)
return self.invModulos[pi]
def invFactorial(self, n):
if n <= -1:
print("Invalid argument to calculate (n^(-1))!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.invFactorial_):
return self.invFactorial_[n]
self.inv(n) # To make sure already calculated n^-1
nextArr = [0] * (n + 1 - len(self.invFactorial_))
initialI = len(self.invFactorial_)
prev = self.invFactorial_[-1]
p = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = (prev * self.invModulos[i % p]) % p
self.invFactorial_ += nextArr
return self.invFactorial_[n]
class Combination:
def __init__(self, MOD):
self.MOD = MOD
self.factorial = Factorial(MOD)
def ncr(self, n, k):
if k < 0 or n < k:
return 0
k = min(k, n - k)
f = self.factorial
return f.calc(n) * f.invFactorial(max(n - k, k)) * f.invFactorial(min(k, n - k)) % self.MOD
# --------------------------------------iye ha combinations ka zindegi---------------------------------
def powm(a, n, m):
if a == 1 or n == 0:
return 1
if n % 2 == 0:
s = powm(a, n // 2, m)
return s * s % m
else:
return a * powm(a, n - 1, m) % m
# --------------------------------------iye ha power ka zindegi---------------------------------
def sort_list(list1, list2):
zipped_pairs = zip(list2, list1)
z = [x for _, x in sorted(zipped_pairs)]
return z
# --------------------------------------------------product----------------------------------------
def product(l):
por = 1
for i in range(len(l)):
por *= l[i]
return por
# --------------------------------------------------binary----------------------------------------
def binarySearchCount(arr, n, key):
left = 0
right = n - 1
count = 0
while (left <= right):
mid = int((right + left) / 2)
# Check if middle element is
# less than or equal to key
if (arr[mid] < key):
count = mid + 1
left = mid + 1
# If key is smaller, ignore right half
else:
right = mid - 1
return count
# --------------------------------------------------binary----------------------------------------
def countdig(n):
c = 0
while (n > 0):
n //= 10
c += 1
return c
def binary(x, length):
y = bin(x)[2:]
return y if len(y) >= length else "0" * (length - len(y)) + y
def countGreater(arr, n, k):
l = 0
r = n - 1
# Stores the index of the left most element
# from the array which is greater than k
leftGreater = n
# Finds number of elements greater than k
while (l <= r):
m = int(l + (r - l) / 2)
if (arr[m] >= k):
leftGreater = m
r = m - 1
# If mid element is less than
# or equal to k update l
else:
l = m + 1
# Return the count of elements
# greater than k
return (n - leftGreater)
# --------------------------------------------------binary------------------------------------
for ik in range(int(input())):
n=int(input())
f=0
for i in range(int(math.sqrt(n)),40000):
w=i*i-n
if w<0:
continue
sq=int(math.sqrt(w))
if sq*sq==w:
tr=sq
j=-1
st=1
end=i
while(st<=end):
mid=(st+end)//2
if i//mid==tr:
j=mid
break
else:
if (i//mid)<tr:
end=mid-1
else:
st=mid+1
if j!=-1:
print(i,j)
f=1
break
if f==0:
print(-1)
``` | instruction | 0 | 19,099 | 23 | 38,198 |
Yes | output | 1 | 19,099 | 23 | 38,199 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's denote a m-free matrix as a binary (that is, consisting of only 1's and 0's) matrix such that every square submatrix of size m × m of this matrix contains at least one zero.
Consider the following problem:
You are given two integers n and m. You have to construct an m-free square matrix of size n × n such that the number of 1's in this matrix is maximum possible. Print the maximum possible number of 1's in such matrix.
You don't have to solve this problem. Instead, you have to construct a few tests for it.
You will be given t numbers x1, x2, ..., xt. For every <image>, find two integers ni and mi (ni ≥ mi) such that the answer for the aforementioned problem is exactly xi if we set n = ni and m = mi.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of tests you have to construct.
Then t lines follow, i-th line containing one integer xi (0 ≤ xi ≤ 109).
Note that in hacks you have to set t = 1.
Output
For each test you have to construct, output two positive numbers ni and mi (1 ≤ mi ≤ ni ≤ 109) such that the maximum number of 1's in a mi-free ni × ni matrix is exactly xi. If there are multiple solutions, you may output any of them; and if this is impossible to construct a test, output a single integer - 1.
Example
Input
3
21
0
1
Output
5 2
1 1
-1
Submitted Solution:
```
import time
import itertools
import math
x = 10**9+2
def find_factor(x):
k = int(math.ceil(math.sqrt(x)))
if x//k == k:
k -=1
for i in range(k,2,-1):
if not x%i:
yield i
yield 1
def find_factor_div2(x):
k = int((math.ceil(math.sqrt(x))//2)*2)
if x//k == k:
k -=2
for i in range(k, 1, -2):
if not x%i and not (x//i)%2:
yield i
return 0
def _sol_kMn_kPn_for_x(x):
if x%4 == 0:
f = find_factor_div2(x)
if f:
for k in f:
yield k, x/k
return 0
if x%2 == 0:
return 0
f = find_factor(x)
if f:
for k in f:
yield k, x/k
return 0
def sol_k_n_for_x(x):
if x < 4:
if x == 3:
yield (1, 2)
elif x == 0:
yield (1, 1)
else:
return 0
return
k_n = _sol_kMn_kPn_for_x(x)
if not k_n:
return 0
for _k, _n in k_n:
n, k = ((_n - _k)//2, (_k + _n)//2)
yield n,k
def sol_m_n(x):
if x == 0:
return (1, 1)
kn = sol_k_n_for_x(x)
if not kn:
return -1, -1
for k, n in kn:
l = math.ceil((n+1)/(k+1))
r = math.floor((n)/(k))
#if l == (n+1)/(k+1):
# l +=1
if l <= r:
return r, n
return -1,-1
if __name__ == "__main__":
n = int(input())
for i in range(n):
q = int(input())
m, n = sol_m_n(q)
if m == -1:
print(-1)
else:
print(int(n), int(m))
# 5, 8
``` | instruction | 0 | 19,100 | 23 | 38,200 |
Yes | output | 1 | 19,100 | 23 | 38,201 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's denote a m-free matrix as a binary (that is, consisting of only 1's and 0's) matrix such that every square submatrix of size m × m of this matrix contains at least one zero.
Consider the following problem:
You are given two integers n and m. You have to construct an m-free square matrix of size n × n such that the number of 1's in this matrix is maximum possible. Print the maximum possible number of 1's in such matrix.
You don't have to solve this problem. Instead, you have to construct a few tests for it.
You will be given t numbers x1, x2, ..., xt. For every <image>, find two integers ni and mi (ni ≥ mi) such that the answer for the aforementioned problem is exactly xi if we set n = ni and m = mi.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of tests you have to construct.
Then t lines follow, i-th line containing one integer xi (0 ≤ xi ≤ 109).
Note that in hacks you have to set t = 1.
Output
For each test you have to construct, output two positive numbers ni and mi (1 ≤ mi ≤ ni ≤ 109) such that the maximum number of 1's in a mi-free ni × ni matrix is exactly xi. If there are multiple solutions, you may output any of them; and if this is impossible to construct a test, output a single integer - 1.
Example
Input
3
21
0
1
Output
5 2
1 1
-1
Submitted Solution:
```
#Bhargey Mehta (Junior)
#DA-IICT, Gandhinagar
import sys, math, queue, bisect
#sys.stdin = open("input.txt", "r")
MOD = 10**9+7
sys.setrecursionlimit(1000000)
N = 50000
for _ in range(int(input())):
x = int(input())
found = None
for i in range(1, N):
if i*i-x <= 0: continue
low, high = 1, N-1
while low <= high:
mid = (low+high)>>1
lhs = i*i - (i//mid)*(i//mid)
if lhs == x:
found = (i, mid)
break
elif lhs < x:
low = mid+1
else:
high = mid-1
if found is not None:
break
if found is None: print(-1)
else: print(*found)
``` | instruction | 0 | 19,102 | 23 | 38,204 |
Yes | output | 1 | 19,102 | 23 | 38,205 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's denote a m-free matrix as a binary (that is, consisting of only 1's and 0's) matrix such that every square submatrix of size m × m of this matrix contains at least one zero.
Consider the following problem:
You are given two integers n and m. You have to construct an m-free square matrix of size n × n such that the number of 1's in this matrix is maximum possible. Print the maximum possible number of 1's in such matrix.
You don't have to solve this problem. Instead, you have to construct a few tests for it.
You will be given t numbers x1, x2, ..., xt. For every <image>, find two integers ni and mi (ni ≥ mi) such that the answer for the aforementioned problem is exactly xi if we set n = ni and m = mi.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of tests you have to construct.
Then t lines follow, i-th line containing one integer xi (0 ≤ xi ≤ 109).
Note that in hacks you have to set t = 1.
Output
For each test you have to construct, output two positive numbers ni and mi (1 ≤ mi ≤ ni ≤ 109) such that the maximum number of 1's in a mi-free ni × ni matrix is exactly xi. If there are multiple solutions, you may output any of them; and if this is impossible to construct a test, output a single integer - 1.
Example
Input
3
21
0
1
Output
5 2
1 1
-1
Submitted Solution:
```
t = int(input())
def sqr(x):
if x < 0:
return -1
an = x ** 0.5
for i in range(-2, 3):
if an + i >= 0 and (an + i) ** 2 == x:
return int(an + i)
return -1
while t:
t -= 1
x = int(input())
n = 1
t = 0
while n == 1 or n ** 2 - (n // 2) ** 2 <= x:
s = sqr(n ** 2 - x)
if s <= 0:
n += 1
continue
m = n // s
if n ** 2 - (n // m) ** 2 == x:
print(n, m)
t = 1
break
n += 1
if t == 0:
print(0)
``` | instruction | 0 | 19,103 | 23 | 38,206 |
No | output | 1 | 19,103 | 23 | 38,207 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's denote a m-free matrix as a binary (that is, consisting of only 1's and 0's) matrix such that every square submatrix of size m × m of this matrix contains at least one zero.
Consider the following problem:
You are given two integers n and m. You have to construct an m-free square matrix of size n × n such that the number of 1's in this matrix is maximum possible. Print the maximum possible number of 1's in such matrix.
You don't have to solve this problem. Instead, you have to construct a few tests for it.
You will be given t numbers x1, x2, ..., xt. For every <image>, find two integers ni and mi (ni ≥ mi) such that the answer for the aforementioned problem is exactly xi if we set n = ni and m = mi.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of tests you have to construct.
Then t lines follow, i-th line containing one integer xi (0 ≤ xi ≤ 109).
Note that in hacks you have to set t = 1.
Output
For each test you have to construct, output two positive numbers ni and mi (1 ≤ mi ≤ ni ≤ 109) such that the maximum number of 1's in a mi-free ni × ni matrix is exactly xi. If there are multiple solutions, you may output any of them; and if this is impossible to construct a test, output a single integer - 1.
Example
Input
3
21
0
1
Output
5 2
1 1
-1
Submitted Solution:
```
import math
li = [i*i for i in range(1, 31622)]
for _ in range(int(input().strip())):
x = int(input().strip())
if x==0:
print(1, 1)
continue
for i in range(31621):
if li[i] > x:
break
else:
print(-1)
continue
item = int(math.sqrt(li[i]))
t = item//2
if x==li[i]-1:
print(item, (item//2)+1)
continue
for j in range(t, 1, -1):
tmp = li[i]-(j*j)
if tmp==x:
print(item, j)
break
else:
print(-1)
``` | instruction | 0 | 19,104 | 23 | 38,208 |
No | output | 1 | 19,104 | 23 | 38,209 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's denote a m-free matrix as a binary (that is, consisting of only 1's and 0's) matrix such that every square submatrix of size m × m of this matrix contains at least one zero.
Consider the following problem:
You are given two integers n and m. You have to construct an m-free square matrix of size n × n such that the number of 1's in this matrix is maximum possible. Print the maximum possible number of 1's in such matrix.
You don't have to solve this problem. Instead, you have to construct a few tests for it.
You will be given t numbers x1, x2, ..., xt. For every <image>, find two integers ni and mi (ni ≥ mi) such that the answer for the aforementioned problem is exactly xi if we set n = ni and m = mi.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of tests you have to construct.
Then t lines follow, i-th line containing one integer xi (0 ≤ xi ≤ 109).
Note that in hacks you have to set t = 1.
Output
For each test you have to construct, output two positive numbers ni and mi (1 ≤ mi ≤ ni ≤ 109) such that the maximum number of 1's in a mi-free ni × ni matrix is exactly xi. If there are multiple solutions, you may output any of them; and if this is impossible to construct a test, output a single integer - 1.
Example
Input
3
21
0
1
Output
5 2
1 1
-1
Submitted Solution:
```
import math
def nextsquare(x):
y = int(math.sqrt(x))
return ((y+1)*(y+1))
a = input()
for i in range(int(a)):
k = int(input())
j = nextsquare(k)
answer1 = -1
answer2 = 0
while (j <= (4 * k / 3) + 1):
l = j - k
if not (l == 0):
if (math.sqrt(l) * math.sqrt(l) == l):
n = int(math.sqrt(j))
m = int(math.sqrt(l))
m = n // m
answer1 = n
answer2 = m
j =nextsquare(j)
if answer1 == -1 :
print(answer1)
else:
print(answer1,answer2)
``` | instruction | 0 | 19,105 | 23 | 38,210 |
No | output | 1 | 19,105 | 23 | 38,211 |
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