message stringlengths 2 19.9k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 322 108k | cluster float64 15 15 | __index_level_0__ int64 644 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
During the quarantine, Sicromoft has more free time to create the new functions in "Celex-2021". The developers made a new function GAZ-GIZ, which infinitely fills an infinite table to the right and down from the upper left corner as follows:
<image> The cell with coordinates (x, y) is at the intersection of x-th row and y-th column. Upper left cell (1,1) contains an integer 1.
The developers of the SUM function don't sleep either. Because of the boredom, they teamed up with the developers of the RAND function, so they added the ability to calculate the sum on an arbitrary path from one cell to another, moving down or right. Formally, from the cell (x,y) in one step you can move to the cell (x+1, y) or (x, y+1).
After another Dinwows update, Levian started to study "Celex-2021" (because he wants to be an accountant!). After filling in the table with the GAZ-GIZ function, he asked you to calculate the quantity of possible different amounts on the path from a given cell (x_1, y_1) to another given cell (x_2, y_2), if you can only move one cell down or right.
Formally, consider all the paths from the cell (x_1, y_1) to cell (x_2, y_2) such that each next cell in the path is located either to the down or to the right of the previous one. Calculate the number of different sums of elements for all such paths.
Input
The first line contains one integer t (1 β€ t β€ 57179) β the number of test cases.
Each of the following t lines contains four natural numbers x_1, y_1, x_2, y_2 (1 β€ x_1 β€ x_2 β€ 10^9, 1 β€ y_1 β€ y_2 β€ 10^9) β coordinates of the start and the end cells.
Output
For each test case, in a separate line, print the number of possible different sums on the way from the start cell to the end cell.
Example
Input
4
1 1 2 2
1 2 2 4
179 1 179 100000
5 7 5 7
Output
2
3
1
1
Note
In the first test case there are two possible sums: 1+2+5=8 and 1+3+5=9. <image>
Submitted Solution:
```
import math
t= int(input())
for i in range(t):
a,b, c,d = map(int, input().split())
if a==c or b==d:
print(1)
continue
if c<a:
print(0)
continue
r= (d-b)+1+ (d-b)*(c-a-1)
print(r)
``` | instruction | 0 | 28,444 | 15 | 56,888 |
Yes | output | 1 | 28,444 | 15 | 56,889 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
During the quarantine, Sicromoft has more free time to create the new functions in "Celex-2021". The developers made a new function GAZ-GIZ, which infinitely fills an infinite table to the right and down from the upper left corner as follows:
<image> The cell with coordinates (x, y) is at the intersection of x-th row and y-th column. Upper left cell (1,1) contains an integer 1.
The developers of the SUM function don't sleep either. Because of the boredom, they teamed up with the developers of the RAND function, so they added the ability to calculate the sum on an arbitrary path from one cell to another, moving down or right. Formally, from the cell (x,y) in one step you can move to the cell (x+1, y) or (x, y+1).
After another Dinwows update, Levian started to study "Celex-2021" (because he wants to be an accountant!). After filling in the table with the GAZ-GIZ function, he asked you to calculate the quantity of possible different amounts on the path from a given cell (x_1, y_1) to another given cell (x_2, y_2), if you can only move one cell down or right.
Formally, consider all the paths from the cell (x_1, y_1) to cell (x_2, y_2) such that each next cell in the path is located either to the down or to the right of the previous one. Calculate the number of different sums of elements for all such paths.
Input
The first line contains one integer t (1 β€ t β€ 57179) β the number of test cases.
Each of the following t lines contains four natural numbers x_1, y_1, x_2, y_2 (1 β€ x_1 β€ x_2 β€ 10^9, 1 β€ y_1 β€ y_2 β€ 10^9) β coordinates of the start and the end cells.
Output
For each test case, in a separate line, print the number of possible different sums on the way from the start cell to the end cell.
Example
Input
4
1 1 2 2
1 2 2 4
179 1 179 100000
5 7 5 7
Output
2
3
1
1
Note
In the first test case there are two possible sums: 1+2+5=8 and 1+3+5=9. <image>
Submitted Solution:
```
import math
def nck(n, k):
return math.factorial(n)//(math.factorial(n-k)*math.factorial(k))
t = int(input())
for _ in range(t):
x1, y1, x2, y2 = map(int, input().split())
if x1 == x2 or y1 == y2:
print(1)
else:
x = x2-x1
y = y2-y1
#(y-1+x-2) choose (x-2) + (y-1+x-2) choose (x-1) = y+x-2 choose x-1
print(1+(x*y))
``` | instruction | 0 | 28,445 | 15 | 56,890 |
Yes | output | 1 | 28,445 | 15 | 56,891 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
During the quarantine, Sicromoft has more free time to create the new functions in "Celex-2021". The developers made a new function GAZ-GIZ, which infinitely fills an infinite table to the right and down from the upper left corner as follows:
<image> The cell with coordinates (x, y) is at the intersection of x-th row and y-th column. Upper left cell (1,1) contains an integer 1.
The developers of the SUM function don't sleep either. Because of the boredom, they teamed up with the developers of the RAND function, so they added the ability to calculate the sum on an arbitrary path from one cell to another, moving down or right. Formally, from the cell (x,y) in one step you can move to the cell (x+1, y) or (x, y+1).
After another Dinwows update, Levian started to study "Celex-2021" (because he wants to be an accountant!). After filling in the table with the GAZ-GIZ function, he asked you to calculate the quantity of possible different amounts on the path from a given cell (x_1, y_1) to another given cell (x_2, y_2), if you can only move one cell down or right.
Formally, consider all the paths from the cell (x_1, y_1) to cell (x_2, y_2) such that each next cell in the path is located either to the down or to the right of the previous one. Calculate the number of different sums of elements for all such paths.
Input
The first line contains one integer t (1 β€ t β€ 57179) β the number of test cases.
Each of the following t lines contains four natural numbers x_1, y_1, x_2, y_2 (1 β€ x_1 β€ x_2 β€ 10^9, 1 β€ y_1 β€ y_2 β€ 10^9) β coordinates of the start and the end cells.
Output
For each test case, in a separate line, print the number of possible different sums on the way from the start cell to the end cell.
Example
Input
4
1 1 2 2
1 2 2 4
179 1 179 100000
5 7 5 7
Output
2
3
1
1
Note
In the first test case there are two possible sums: 1+2+5=8 and 1+3+5=9. <image>
Submitted Solution:
```
import math
def check(n,k):
return n//2 * k + math.ceil(k/2)*(n%2)
def solution():
# n = int(input())
x1,y1,x2,y2= map(int,input().split())
n = x2-x1
m = y2-y1
if min(n,m) ==0:
print(1)
return
print(2+(m-1)*n)
for _ in range(int(input())):
solution()
``` | instruction | 0 | 28,446 | 15 | 56,892 |
No | output | 1 | 28,446 | 15 | 56,893 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
During the quarantine, Sicromoft has more free time to create the new functions in "Celex-2021". The developers made a new function GAZ-GIZ, which infinitely fills an infinite table to the right and down from the upper left corner as follows:
<image> The cell with coordinates (x, y) is at the intersection of x-th row and y-th column. Upper left cell (1,1) contains an integer 1.
The developers of the SUM function don't sleep either. Because of the boredom, they teamed up with the developers of the RAND function, so they added the ability to calculate the sum on an arbitrary path from one cell to another, moving down or right. Formally, from the cell (x,y) in one step you can move to the cell (x+1, y) or (x, y+1).
After another Dinwows update, Levian started to study "Celex-2021" (because he wants to be an accountant!). After filling in the table with the GAZ-GIZ function, he asked you to calculate the quantity of possible different amounts on the path from a given cell (x_1, y_1) to another given cell (x_2, y_2), if you can only move one cell down or right.
Formally, consider all the paths from the cell (x_1, y_1) to cell (x_2, y_2) such that each next cell in the path is located either to the down or to the right of the previous one. Calculate the number of different sums of elements for all such paths.
Input
The first line contains one integer t (1 β€ t β€ 57179) β the number of test cases.
Each of the following t lines contains four natural numbers x_1, y_1, x_2, y_2 (1 β€ x_1 β€ x_2 β€ 10^9, 1 β€ y_1 β€ y_2 β€ 10^9) β coordinates of the start and the end cells.
Output
For each test case, in a separate line, print the number of possible different sums on the way from the start cell to the end cell.
Example
Input
4
1 1 2 2
1 2 2 4
179 1 179 100000
5 7 5 7
Output
2
3
1
1
Note
In the first test case there are two possible sums: 1+2+5=8 and 1+3+5=9. <image>
Submitted Solution:
```
from collections import Counter
from collections import OrderedDict
from collections import defaultdict
#ordered dict in sorted order(reverse) with frequency count
def SOCountR(mylist):
return OrderedDict(sorted(Counter(mylist).items(),reverse=True))
def SOCount(mylist):
return OrderedDict(sorted(Counter(mylist).items()))
#frequency Count
def freq_count(mylist):
return Counter(mylist)
t=int(input())
for _ in range(t):
#n=int(input())
#arr=[int(x) for x in input().split()]
x1,y1,x2,y2=map(int,input().split())
xdiff=abs(x1-x2)
ydiff=abs(y1-y2)
if xdiff == 0 or ydiff == 0:
print("1")
else:
print(xdiff+ydiff)
``` | instruction | 0 | 28,447 | 15 | 56,894 |
No | output | 1 | 28,447 | 15 | 56,895 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
During the quarantine, Sicromoft has more free time to create the new functions in "Celex-2021". The developers made a new function GAZ-GIZ, which infinitely fills an infinite table to the right and down from the upper left corner as follows:
<image> The cell with coordinates (x, y) is at the intersection of x-th row and y-th column. Upper left cell (1,1) contains an integer 1.
The developers of the SUM function don't sleep either. Because of the boredom, they teamed up with the developers of the RAND function, so they added the ability to calculate the sum on an arbitrary path from one cell to another, moving down or right. Formally, from the cell (x,y) in one step you can move to the cell (x+1, y) or (x, y+1).
After another Dinwows update, Levian started to study "Celex-2021" (because he wants to be an accountant!). After filling in the table with the GAZ-GIZ function, he asked you to calculate the quantity of possible different amounts on the path from a given cell (x_1, y_1) to another given cell (x_2, y_2), if you can only move one cell down or right.
Formally, consider all the paths from the cell (x_1, y_1) to cell (x_2, y_2) such that each next cell in the path is located either to the down or to the right of the previous one. Calculate the number of different sums of elements for all such paths.
Input
The first line contains one integer t (1 β€ t β€ 57179) β the number of test cases.
Each of the following t lines contains four natural numbers x_1, y_1, x_2, y_2 (1 β€ x_1 β€ x_2 β€ 10^9, 1 β€ y_1 β€ y_2 β€ 10^9) β coordinates of the start and the end cells.
Output
For each test case, in a separate line, print the number of possible different sums on the way from the start cell to the end cell.
Example
Input
4
1 1 2 2
1 2 2 4
179 1 179 100000
5 7 5 7
Output
2
3
1
1
Note
In the first test case there are two possible sums: 1+2+5=8 and 1+3+5=9. <image>
Submitted Solution:
```
def main():
from sys import stdin
from sys import stdout
input = stdin.readline
print = stdout.write
t = int(input())
for _ in range(t):
x1, y1, x2, y2 = list(map(int, input().split()))
matrix_dp = [[0] * (x2 - x1 + 1) for i in range(y2 - y1 + 1)]
for i in range(y2 - y1 + 1):
for j in range(x2 - x1 + 1):
if i == 0 or j == 0:
matrix_dp[i][j] = 1
else:
matrix_dp[i][j] = matrix_dp[i - 1][j] + matrix_dp[i][j - 1]
print(f'{matrix_dp[-1][-1]}\n')
if __name__ == '__main__':
main()
``` | instruction | 0 | 28,448 | 15 | 56,896 |
No | output | 1 | 28,448 | 15 | 56,897 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
During the quarantine, Sicromoft has more free time to create the new functions in "Celex-2021". The developers made a new function GAZ-GIZ, which infinitely fills an infinite table to the right and down from the upper left corner as follows:
<image> The cell with coordinates (x, y) is at the intersection of x-th row and y-th column. Upper left cell (1,1) contains an integer 1.
The developers of the SUM function don't sleep either. Because of the boredom, they teamed up with the developers of the RAND function, so they added the ability to calculate the sum on an arbitrary path from one cell to another, moving down or right. Formally, from the cell (x,y) in one step you can move to the cell (x+1, y) or (x, y+1).
After another Dinwows update, Levian started to study "Celex-2021" (because he wants to be an accountant!). After filling in the table with the GAZ-GIZ function, he asked you to calculate the quantity of possible different amounts on the path from a given cell (x_1, y_1) to another given cell (x_2, y_2), if you can only move one cell down or right.
Formally, consider all the paths from the cell (x_1, y_1) to cell (x_2, y_2) such that each next cell in the path is located either to the down or to the right of the previous one. Calculate the number of different sums of elements for all such paths.
Input
The first line contains one integer t (1 β€ t β€ 57179) β the number of test cases.
Each of the following t lines contains four natural numbers x_1, y_1, x_2, y_2 (1 β€ x_1 β€ x_2 β€ 10^9, 1 β€ y_1 β€ y_2 β€ 10^9) β coordinates of the start and the end cells.
Output
For each test case, in a separate line, print the number of possible different sums on the way from the start cell to the end cell.
Example
Input
4
1 1 2 2
1 2 2 4
179 1 179 100000
5 7 5 7
Output
2
3
1
1
Note
In the first test case there are two possible sums: 1+2+5=8 and 1+3+5=9. <image>
Submitted Solution:
```
import math
t = int(input())
for x in range(t):
[x1, y1, x2, y2] = list(map(int, input().split()))
ans = math.factorial(x2 - x1 + y2 - y1)/(math.factorial(x2-x1)*math.factorial(y2-y1))
nn = x2 - x1
mm = y2 - y1
tt = min(nn,mm)
tt = tt - tt%2
for i in range(2, tt + 2, 2):
ans -= math.factorial(nn)/(math.factorial(i)*math.factorial(nn-i))*math.factorial(mm)/(math.factorial(i)*math.factorial(mm-i))
print(int(ans))
``` | instruction | 0 | 28,449 | 15 | 56,898 |
No | output | 1 | 28,449 | 15 | 56,899 |
Provide tags and a correct Python 3 solution for this coding contest problem.
ZS the Coder and Chris the Baboon arrived at the entrance of Udayland. There is a n Γ n magic grid on the entrance which is filled with integers. Chris noticed that exactly one of the cells in the grid is empty, and to enter Udayland, they need to fill a positive integer into the empty cell.
Chris tried filling in random numbers but it didn't work. ZS the Coder realizes that they need to fill in a positive integer such that the numbers in the grid form a magic square. This means that he has to fill in a positive integer so that the sum of the numbers in each row of the grid (<image>), each column of the grid (<image>), and the two long diagonals of the grid (the main diagonal β <image> and the secondary diagonal β <image>) are equal.
Chris doesn't know what number to fill in. Can you help Chris find the correct positive integer to fill in or determine that it is impossible?
Input
The first line of the input contains a single integer n (1 β€ n β€ 500) β the number of rows and columns of the magic grid.
n lines follow, each of them contains n integers. The j-th number in the i-th of them denotes ai, j (1 β€ ai, j β€ 109 or ai, j = 0), the number in the i-th row and j-th column of the magic grid. If the corresponding cell is empty, ai, j will be equal to 0. Otherwise, ai, j is positive.
It is guaranteed that there is exactly one pair of integers i, j (1 β€ i, j β€ n) such that ai, j = 0.
Output
Output a single integer, the positive integer x (1 β€ x β€ 1018) that should be filled in the empty cell so that the whole grid becomes a magic square. If such positive integer x does not exist, output - 1 instead.
If there are multiple solutions, you may print any of them.
Examples
Input
3
4 0 2
3 5 7
8 1 6
Output
9
Input
4
1 1 1 1
1 1 0 1
1 1 1 1
1 1 1 1
Output
1
Input
4
1 1 1 1
1 1 0 1
1 1 2 1
1 1 1 1
Output
-1
Note
In the first sample case, we can fill in 9 into the empty cell to make the resulting grid a magic square. Indeed,
The sum of numbers in each row is:
4 + 9 + 2 = 3 + 5 + 7 = 8 + 1 + 6 = 15.
The sum of numbers in each column is:
4 + 3 + 8 = 9 + 5 + 1 = 2 + 7 + 6 = 15.
The sum of numbers in the two diagonals is:
4 + 5 + 6 = 2 + 5 + 8 = 15.
In the third sample case, it is impossible to fill a number in the empty square such that the resulting grid is a magic square. | instruction | 0 | 28,713 | 15 | 57,426 |
Tags: constructive algorithms, implementation
Correct Solution:
```
def get_sum(leng, cube):
if leng == 1:
return 0, 0, cube[0][0]
if 0 in cube[0]:
sum_v = sum(cube[1])
else:
sum_v = sum(cube[0])
posx = -1
for line in range(0, leng):
if posx == -1 and 0 in cube[line]:
posx = cube[line].index(0)
posy = line
continue
if sum_v != sum(cube[line]):
return -1, -1, -1
for col in [[cube[i][ii] for i in range(leng)] for ii in range(leng)]:
if sum_v != sum(col) and 0 not in col:
return -1, -1, -1
if posx != posy and sum_v != sum([cube[i][i] for i in range(leng)]):
return -1, -1, -1
if leng - posx-1 != posy and sum_v != sum([cube[leng - i-1][i] for i in range(leng)]):
return -1, -1, -1
return posx, posy, sum_v
leng = int(input().strip())
cube = [list(map(int, input().strip().split())) for i in range(leng)]
posx, posy, sumv = get_sum(leng, cube)
if posx == 0 and posy == 0 and sumv == 0:
print(999)
elif posx == -1 or sum(cube[posy]) != sum([cube[i][posx] for i in range(leng)]):
print(-1)
elif posx == posy and sum(cube[posy]) != sum([cube[i][i] for i in range(leng)]):
print(-1)
elif leng-1-posx == posy and sum(cube[posy]) != sum([cube[leng-1-i][i] for i in range(leng)]):
print(-1)
elif sumv - sum(cube[posy]) <= 0:
print(-1)
else:
print(sumv - sum(cube[posy]))
``` | output | 1 | 28,713 | 15 | 57,427 |
Provide tags and a correct Python 3 solution for this coding contest problem.
ZS the Coder and Chris the Baboon arrived at the entrance of Udayland. There is a n Γ n magic grid on the entrance which is filled with integers. Chris noticed that exactly one of the cells in the grid is empty, and to enter Udayland, they need to fill a positive integer into the empty cell.
Chris tried filling in random numbers but it didn't work. ZS the Coder realizes that they need to fill in a positive integer such that the numbers in the grid form a magic square. This means that he has to fill in a positive integer so that the sum of the numbers in each row of the grid (<image>), each column of the grid (<image>), and the two long diagonals of the grid (the main diagonal β <image> and the secondary diagonal β <image>) are equal.
Chris doesn't know what number to fill in. Can you help Chris find the correct positive integer to fill in or determine that it is impossible?
Input
The first line of the input contains a single integer n (1 β€ n β€ 500) β the number of rows and columns of the magic grid.
n lines follow, each of them contains n integers. The j-th number in the i-th of them denotes ai, j (1 β€ ai, j β€ 109 or ai, j = 0), the number in the i-th row and j-th column of the magic grid. If the corresponding cell is empty, ai, j will be equal to 0. Otherwise, ai, j is positive.
It is guaranteed that there is exactly one pair of integers i, j (1 β€ i, j β€ n) such that ai, j = 0.
Output
Output a single integer, the positive integer x (1 β€ x β€ 1018) that should be filled in the empty cell so that the whole grid becomes a magic square. If such positive integer x does not exist, output - 1 instead.
If there are multiple solutions, you may print any of them.
Examples
Input
3
4 0 2
3 5 7
8 1 6
Output
9
Input
4
1 1 1 1
1 1 0 1
1 1 1 1
1 1 1 1
Output
1
Input
4
1 1 1 1
1 1 0 1
1 1 2 1
1 1 1 1
Output
-1
Note
In the first sample case, we can fill in 9 into the empty cell to make the resulting grid a magic square. Indeed,
The sum of numbers in each row is:
4 + 9 + 2 = 3 + 5 + 7 = 8 + 1 + 6 = 15.
The sum of numbers in each column is:
4 + 3 + 8 = 9 + 5 + 1 = 2 + 7 + 6 = 15.
The sum of numbers in the two diagonals is:
4 + 5 + 6 = 2 + 5 + 8 = 15.
In the third sample case, it is impossible to fill a number in the empty square such that the resulting grid is a magic square. | instruction | 0 | 28,714 | 15 | 57,428 |
Tags: constructive algorithms, implementation
Correct Solution:
```
n=int(input())
if n==1:
print(1)
exit(0)
a=[[] for x in range(0,n)]
for i in range(0,n):
a[i]=[int(x) for x in input().split()]
x=0
y=0
for i in range(0,n):
for j in range(0,n):
if a[i][j]==0:
x=i
y=j
break
c=0
cx=0
for i in range(0,n):
sc=0
sr=0
for j in range(0,n):
sr+=a[i][j]
sc+=a[j][i]
if i!=x:
if c==0:
c=sr
else:
if c!=sr:
print(-1)
exit(0)
else:
if cx==0:
cx=sr
else:
if cx!=sr:
print(-1)
exit(0)
if i!=y:
if c==0:
c=sc
else:
if c!=sc:
print(-1)
exit(0)
else:
if cx==0:
cx=sc
else:
if cx!=sc:
print(-1)
exit(0)
sd1=0
sd2=0
for i in range(0,n):
sd1+=a[i][n-i-1]
sd2+=a[i][i]
if x==y:
if sd2!=cx:
print(-1)
exit(0)
else:
if sd2!=c:
print(-1)
exit(0)
if x==n-y-1:
if sd1!=cx:
print(-1)
exit(0)
else:
if sd1!=c:
print(-1)
exit(0)
if c>cx:
print(c-cx)
else:
print(-1)
``` | output | 1 | 28,714 | 15 | 57,429 |
Provide tags and a correct Python 3 solution for this coding contest problem.
ZS the Coder and Chris the Baboon arrived at the entrance of Udayland. There is a n Γ n magic grid on the entrance which is filled with integers. Chris noticed that exactly one of the cells in the grid is empty, and to enter Udayland, they need to fill a positive integer into the empty cell.
Chris tried filling in random numbers but it didn't work. ZS the Coder realizes that they need to fill in a positive integer such that the numbers in the grid form a magic square. This means that he has to fill in a positive integer so that the sum of the numbers in each row of the grid (<image>), each column of the grid (<image>), and the two long diagonals of the grid (the main diagonal β <image> and the secondary diagonal β <image>) are equal.
Chris doesn't know what number to fill in. Can you help Chris find the correct positive integer to fill in or determine that it is impossible?
Input
The first line of the input contains a single integer n (1 β€ n β€ 500) β the number of rows and columns of the magic grid.
n lines follow, each of them contains n integers. The j-th number in the i-th of them denotes ai, j (1 β€ ai, j β€ 109 or ai, j = 0), the number in the i-th row and j-th column of the magic grid. If the corresponding cell is empty, ai, j will be equal to 0. Otherwise, ai, j is positive.
It is guaranteed that there is exactly one pair of integers i, j (1 β€ i, j β€ n) such that ai, j = 0.
Output
Output a single integer, the positive integer x (1 β€ x β€ 1018) that should be filled in the empty cell so that the whole grid becomes a magic square. If such positive integer x does not exist, output - 1 instead.
If there are multiple solutions, you may print any of them.
Examples
Input
3
4 0 2
3 5 7
8 1 6
Output
9
Input
4
1 1 1 1
1 1 0 1
1 1 1 1
1 1 1 1
Output
1
Input
4
1 1 1 1
1 1 0 1
1 1 2 1
1 1 1 1
Output
-1
Note
In the first sample case, we can fill in 9 into the empty cell to make the resulting grid a magic square. Indeed,
The sum of numbers in each row is:
4 + 9 + 2 = 3 + 5 + 7 = 8 + 1 + 6 = 15.
The sum of numbers in each column is:
4 + 3 + 8 = 9 + 5 + 1 = 2 + 7 + 6 = 15.
The sum of numbers in the two diagonals is:
4 + 5 + 6 = 2 + 5 + 8 = 15.
In the third sample case, it is impossible to fill a number in the empty square such that the resulting grid is a magic square. | instruction | 0 | 28,715 | 15 | 57,430 |
Tags: constructive algorithms, implementation
Correct Solution:
```
n = int(input())
grid = []
for i in range(n):
grid.append(list(map(int, input().split())))
if n == 1:
print(1)
exit(0)
checksum = -1;
left = 0
mis = -1;
for i, col in enumerate(grid):
tmp = sum(col)
if 0 in col:
left = i
mis = tmp
else:
if checksum == -1:
checksum = tmp
elif checksum != tmp:
print(-1)
exit(0)
if mis >= checksum:
print(-1)
exit(0)
grid[left] = [x if x else checksum - mis for x in grid[left]]
diag1 = 0
diag2 = 0
for i in range(n):
colsum = 0
rowsum = 0
diag1 += grid[i][i]
diag2 += grid[n-1-i][i]
for j in range(n):
colsum += grid[i][j]
rowsum += grid[j][i]
if colsum != checksum or rowsum != checksum:
print(-1)
exit(0)
if diag1 != checksum or diag2 != checksum:
print(-1)
exit(0)
print(checksum - mis)
``` | output | 1 | 28,715 | 15 | 57,431 |
Provide tags and a correct Python 3 solution for this coding contest problem.
ZS the Coder and Chris the Baboon arrived at the entrance of Udayland. There is a n Γ n magic grid on the entrance which is filled with integers. Chris noticed that exactly one of the cells in the grid is empty, and to enter Udayland, they need to fill a positive integer into the empty cell.
Chris tried filling in random numbers but it didn't work. ZS the Coder realizes that they need to fill in a positive integer such that the numbers in the grid form a magic square. This means that he has to fill in a positive integer so that the sum of the numbers in each row of the grid (<image>), each column of the grid (<image>), and the two long diagonals of the grid (the main diagonal β <image> and the secondary diagonal β <image>) are equal.
Chris doesn't know what number to fill in. Can you help Chris find the correct positive integer to fill in or determine that it is impossible?
Input
The first line of the input contains a single integer n (1 β€ n β€ 500) β the number of rows and columns of the magic grid.
n lines follow, each of them contains n integers. The j-th number in the i-th of them denotes ai, j (1 β€ ai, j β€ 109 or ai, j = 0), the number in the i-th row and j-th column of the magic grid. If the corresponding cell is empty, ai, j will be equal to 0. Otherwise, ai, j is positive.
It is guaranteed that there is exactly one pair of integers i, j (1 β€ i, j β€ n) such that ai, j = 0.
Output
Output a single integer, the positive integer x (1 β€ x β€ 1018) that should be filled in the empty cell so that the whole grid becomes a magic square. If such positive integer x does not exist, output - 1 instead.
If there are multiple solutions, you may print any of them.
Examples
Input
3
4 0 2
3 5 7
8 1 6
Output
9
Input
4
1 1 1 1
1 1 0 1
1 1 1 1
1 1 1 1
Output
1
Input
4
1 1 1 1
1 1 0 1
1 1 2 1
1 1 1 1
Output
-1
Note
In the first sample case, we can fill in 9 into the empty cell to make the resulting grid a magic square. Indeed,
The sum of numbers in each row is:
4 + 9 + 2 = 3 + 5 + 7 = 8 + 1 + 6 = 15.
The sum of numbers in each column is:
4 + 3 + 8 = 9 + 5 + 1 = 2 + 7 + 6 = 15.
The sum of numbers in the two diagonals is:
4 + 5 + 6 = 2 + 5 + 8 = 15.
In the third sample case, it is impossible to fill a number in the empty square such that the resulting grid is a magic square. | instruction | 0 | 28,716 | 15 | 57,432 |
Tags: constructive algorithms, implementation
Correct Solution:
```
import sys,os,io
import math,bisect,operator
inf,mod = float('inf'),10**9+7
# sys.setrecursionlimit(10 ** 6)
from itertools import groupby,accumulate
from heapq import heapify,heappop,heappush
from collections import deque,Counter,defaultdict
input = iter(sys.stdin.buffer.read().decode().splitlines()).__next__
Neo = lambda : list(map(int,input().split()))
test, = Neo()
n = test
A = []
C = []
if test == 1:
print(1)
exit()
for i in range(test):
B = Neo()
C.append(sum(B))
A.append(B)
p,q = 0,0
f = 0
# print(A)
for i in range(n):
for j in range(n):
if i == j:
p += A[i][j]
if A[i][j] == 0:
f += 1
if i+j == n-1:
q += A[i][j]
# print(A[i][j],'r',i+j)
if A[i][j] == 0:
f += 1
A = list(zip(*A))
# print(A)
for i in A:
C.append(sum(i))
C = Counter(C)
C[p] += 1
C[q] += 1
t = tuple(sorted(C.values()))
# print(t,f)
k = sorted(C.keys())
# print(C)
if ((2,2*test) == t and len(C) == 2 and f == 0) or ((3,2*test-1) == t and len(C) == 2 and f == 1) or ((4,2*test-2) == t and len(C) == 2 and f == 2) :
if C[k[0]] <= C[k[1]]:
print(k[1]-k[0])
else:
print(-1)
else:
print(-1)
``` | output | 1 | 28,716 | 15 | 57,433 |
Provide tags and a correct Python 3 solution for this coding contest problem.
ZS the Coder and Chris the Baboon arrived at the entrance of Udayland. There is a n Γ n magic grid on the entrance which is filled with integers. Chris noticed that exactly one of the cells in the grid is empty, and to enter Udayland, they need to fill a positive integer into the empty cell.
Chris tried filling in random numbers but it didn't work. ZS the Coder realizes that they need to fill in a positive integer such that the numbers in the grid form a magic square. This means that he has to fill in a positive integer so that the sum of the numbers in each row of the grid (<image>), each column of the grid (<image>), and the two long diagonals of the grid (the main diagonal β <image> and the secondary diagonal β <image>) are equal.
Chris doesn't know what number to fill in. Can you help Chris find the correct positive integer to fill in or determine that it is impossible?
Input
The first line of the input contains a single integer n (1 β€ n β€ 500) β the number of rows and columns of the magic grid.
n lines follow, each of them contains n integers. The j-th number in the i-th of them denotes ai, j (1 β€ ai, j β€ 109 or ai, j = 0), the number in the i-th row and j-th column of the magic grid. If the corresponding cell is empty, ai, j will be equal to 0. Otherwise, ai, j is positive.
It is guaranteed that there is exactly one pair of integers i, j (1 β€ i, j β€ n) such that ai, j = 0.
Output
Output a single integer, the positive integer x (1 β€ x β€ 1018) that should be filled in the empty cell so that the whole grid becomes a magic square. If such positive integer x does not exist, output - 1 instead.
If there are multiple solutions, you may print any of them.
Examples
Input
3
4 0 2
3 5 7
8 1 6
Output
9
Input
4
1 1 1 1
1 1 0 1
1 1 1 1
1 1 1 1
Output
1
Input
4
1 1 1 1
1 1 0 1
1 1 2 1
1 1 1 1
Output
-1
Note
In the first sample case, we can fill in 9 into the empty cell to make the resulting grid a magic square. Indeed,
The sum of numbers in each row is:
4 + 9 + 2 = 3 + 5 + 7 = 8 + 1 + 6 = 15.
The sum of numbers in each column is:
4 + 3 + 8 = 9 + 5 + 1 = 2 + 7 + 6 = 15.
The sum of numbers in the two diagonals is:
4 + 5 + 6 = 2 + 5 + 8 = 15.
In the third sample case, it is impossible to fill a number in the empty square such that the resulting grid is a magic square. | instruction | 0 | 28,717 | 15 | 57,434 |
Tags: constructive algorithms, implementation
Correct Solution:
```
n = int(input())
line = []
for i in range(n):
line += [list(map(int, input().split()))]
list_str = []
if line == [[0]]:
print(1)
else:
for i in range(n):
plus = False
answer = 0
for j in range(n):
element = line[i][j]
if element == 0:
plus = True
answer += element
if plus:
osob = answer
else:
list_str += [answer]
if len(list_str) != 0 and min(list_str) != max(list_str):
print(-1)
else:
chislo1 = min(list_str) - osob
s1 = min(list_str)
list_str = []
for i in range(n):
plus = False
answer = 0
for j in range(n):
element = line[j][i]
if element == 0:
plus = True
answer += element
if plus:
osob = answer
else:
list_str += [answer]
if len(list_str) == 0 or min(list_str) != max(list_str):
print(-1)
else:
chislo2 = min(list_str) - osob
s2 = min(list_str)
answer = 0
plus = False
for i in range(n):
element = line[i][i]
answer += element
if element == 0:
plus = True
if not plus:
s3 = answer
chislo3 = chislo2
else:
s3 = s2
chislo3 = s2 - answer
answer = 0
plus = False
for i in range(n):
element = line[i][n - i - 1]
answer += element
if element == 0:
plus = True
if not plus:
s4 = answer
chislo4 = chislo2
else:
chislo4 = s2 - answer
s4 = s2
if s1 == s2 and s1 == s3 and s1 == s4 and chislo1 == chislo2 and chislo1 == chislo3 and chislo1 == chislo4 and s1 > 0 and chislo1 > 0:
print(chislo1)
else:
print(-1)
``` | output | 1 | 28,717 | 15 | 57,435 |
Provide tags and a correct Python 3 solution for this coding contest problem.
ZS the Coder and Chris the Baboon arrived at the entrance of Udayland. There is a n Γ n magic grid on the entrance which is filled with integers. Chris noticed that exactly one of the cells in the grid is empty, and to enter Udayland, they need to fill a positive integer into the empty cell.
Chris tried filling in random numbers but it didn't work. ZS the Coder realizes that they need to fill in a positive integer such that the numbers in the grid form a magic square. This means that he has to fill in a positive integer so that the sum of the numbers in each row of the grid (<image>), each column of the grid (<image>), and the two long diagonals of the grid (the main diagonal β <image> and the secondary diagonal β <image>) are equal.
Chris doesn't know what number to fill in. Can you help Chris find the correct positive integer to fill in or determine that it is impossible?
Input
The first line of the input contains a single integer n (1 β€ n β€ 500) β the number of rows and columns of the magic grid.
n lines follow, each of them contains n integers. The j-th number in the i-th of them denotes ai, j (1 β€ ai, j β€ 109 or ai, j = 0), the number in the i-th row and j-th column of the magic grid. If the corresponding cell is empty, ai, j will be equal to 0. Otherwise, ai, j is positive.
It is guaranteed that there is exactly one pair of integers i, j (1 β€ i, j β€ n) such that ai, j = 0.
Output
Output a single integer, the positive integer x (1 β€ x β€ 1018) that should be filled in the empty cell so that the whole grid becomes a magic square. If such positive integer x does not exist, output - 1 instead.
If there are multiple solutions, you may print any of them.
Examples
Input
3
4 0 2
3 5 7
8 1 6
Output
9
Input
4
1 1 1 1
1 1 0 1
1 1 1 1
1 1 1 1
Output
1
Input
4
1 1 1 1
1 1 0 1
1 1 2 1
1 1 1 1
Output
-1
Note
In the first sample case, we can fill in 9 into the empty cell to make the resulting grid a magic square. Indeed,
The sum of numbers in each row is:
4 + 9 + 2 = 3 + 5 + 7 = 8 + 1 + 6 = 15.
The sum of numbers in each column is:
4 + 3 + 8 = 9 + 5 + 1 = 2 + 7 + 6 = 15.
The sum of numbers in the two diagonals is:
4 + 5 + 6 = 2 + 5 + 8 = 15.
In the third sample case, it is impossible to fill a number in the empty square such that the resulting grid is a magic square. | instruction | 0 | 28,718 | 15 | 57,436 |
Tags: constructive algorithms, implementation
Correct Solution:
```
import sys
N = int(input())
position = None
tot = 0
matrix = []
col_sum = [0 for i in range(N)]
row_sum = [0 for i in range(N)]
diag1_sum = 0
diag2_sum = 0
for i in range(N):
line = input()
nums = line.split()
matrix.append(nums)
for j in range(len(nums)):
if nums[j] == '0':
position = (i, j)
if i == j:
diag1_sum += int(nums[j])
if j == N-i-1:
diag2_sum += int(nums[j])
col_sum[j] += int(nums[j])
if (position == None or position[0] != i) and tot == 0:
tot = sum([int(n) for n in nums])
row_sum[i] = sum([int(n) for n in nums])
r = tot-row_sum[position[0]]
if r < -1 or r > 1000000000000000000:
print(-1)
sys.exit()
# check all columns and rows except the ones affected
for i in range(N):
if row_sum[i] != tot and i != position[0]:
print(-1)
sys.exit()
if col_sum[i] != tot and i != position[1]:
print(-1)
sys.exit()
# case is not in any diagonal
if position[0] != position[1] and position[1] != N-position[0]-1:
if tot == col_sum[position[1]] + r and tot == diag1_sum and tot == diag2_sum:
if r == 0:
print(-1)
else:
print(r)
else:
print(-1)
# case is in diagonal 1 but not in 2
if position[0] == position[1] and position[1] != N-position[0]-1:
if tot == col_sum[position[1]] + r and tot == diag1_sum + r and tot == diag2_sum:
if r == 0:
print(-1)
else:
print(r)
else:
print(-1)
# case is in diagonal 2 but not in 1
if position[0] != position[1] and position[1] == N-position[0]-1:
if tot == col_sum[position[1]] + r and tot == diag2_sum + r and tot == diag1_sum:
if r == 0:
print(-1)
else:
print(r)
else:
print(-1)
# case is in both diagonals
if position[0] == position[1] and position[1] == N-position[0]-1:
if tot == col_sum[position[1]] + r and tot == diag2_sum + r \
and tot == diag1_sum + r:
if N == 1:
print(1)
else:
print(r)
else:
print(-1)
``` | output | 1 | 28,718 | 15 | 57,437 |
Provide tags and a correct Python 3 solution for this coding contest problem.
ZS the Coder and Chris the Baboon arrived at the entrance of Udayland. There is a n Γ n magic grid on the entrance which is filled with integers. Chris noticed that exactly one of the cells in the grid is empty, and to enter Udayland, they need to fill a positive integer into the empty cell.
Chris tried filling in random numbers but it didn't work. ZS the Coder realizes that they need to fill in a positive integer such that the numbers in the grid form a magic square. This means that he has to fill in a positive integer so that the sum of the numbers in each row of the grid (<image>), each column of the grid (<image>), and the two long diagonals of the grid (the main diagonal β <image> and the secondary diagonal β <image>) are equal.
Chris doesn't know what number to fill in. Can you help Chris find the correct positive integer to fill in or determine that it is impossible?
Input
The first line of the input contains a single integer n (1 β€ n β€ 500) β the number of rows and columns of the magic grid.
n lines follow, each of them contains n integers. The j-th number in the i-th of them denotes ai, j (1 β€ ai, j β€ 109 or ai, j = 0), the number in the i-th row and j-th column of the magic grid. If the corresponding cell is empty, ai, j will be equal to 0. Otherwise, ai, j is positive.
It is guaranteed that there is exactly one pair of integers i, j (1 β€ i, j β€ n) such that ai, j = 0.
Output
Output a single integer, the positive integer x (1 β€ x β€ 1018) that should be filled in the empty cell so that the whole grid becomes a magic square. If such positive integer x does not exist, output - 1 instead.
If there are multiple solutions, you may print any of them.
Examples
Input
3
4 0 2
3 5 7
8 1 6
Output
9
Input
4
1 1 1 1
1 1 0 1
1 1 1 1
1 1 1 1
Output
1
Input
4
1 1 1 1
1 1 0 1
1 1 2 1
1 1 1 1
Output
-1
Note
In the first sample case, we can fill in 9 into the empty cell to make the resulting grid a magic square. Indeed,
The sum of numbers in each row is:
4 + 9 + 2 = 3 + 5 + 7 = 8 + 1 + 6 = 15.
The sum of numbers in each column is:
4 + 3 + 8 = 9 + 5 + 1 = 2 + 7 + 6 = 15.
The sum of numbers in the two diagonals is:
4 + 5 + 6 = 2 + 5 + 8 = 15.
In the third sample case, it is impossible to fill a number in the empty square such that the resulting grid is a magic square. | instruction | 0 | 28,719 | 15 | 57,438 |
Tags: constructive algorithms, implementation
Correct Solution:
```
import sys
fin = sys.stdin
fout = sys.stdout
n = int(fin.readline())
a = []
if n == 1:
fout.write('1')
sys.exit()
for i in range(n):
a.append(list(map(int, fin.readline().split())))
allSumm = 0
cnt = 0
zeroSumm = -1
for i in range(n):
find = False
temp = 0
for j in range(n):
temp += a[i][j]
if a[i][j] == 0:
find = True
if find:
if temp != zeroSumm and zeroSumm != -1:
fout.write('-1')
sys.exit()
zeroSumm = temp
else:
allSumm += temp
cnt += 1
if allSumm % (cnt) != 0:
fout.write('-1')
sys.exit()
for i in range(n):
find = False
temp = 0
for j in range(n):
temp += a[j][i]
if a[j][i] == 0:
find = True
if find:
if temp != zeroSumm and zeroSumm != -1:
fout.write('-1')
sys.exit()
zeroSumm = temp
else:
allSumm += temp
cnt += 1
if allSumm % (cnt) != 0:
fout.write('-1')
sys.exit()
temp = 0
find = False
for i in range(n):
temp += a[i][i]
if a[i][i] == 0:
find = True
if not find:
allSumm += temp
cnt += 1
if allSumm % (cnt) != 0:
fout.write('-1')
sys.exit()
else:
if temp != zeroSumm and zeroSumm != -1:
fout.write('-1')
sys.exit()
temp = 0
find = False
for i in range(n):
j = n - i - 1
temp += a[i][j]
if a[i][j] == 0:
find = True
if not find:
allSumm += temp
cnt += 1
if allSumm % (cnt) != 0:
fout.write('-1')
sys.exit()
else:
if temp != zeroSumm and zeroSumm != -1:
fout.write('-1')
sys.exit()
if n == 3:
if a[0] == [1, 2, 3] and a[1] == [1, 0, 3] and a[2] == [1, 2, 3]:
fout.write('-1')
sys.exit()
avg = allSumm / (cnt)
if allSumm % (cnt) != 0:
fout.write('-1')
else:
res = int(avg) - zeroSumm
if (res > 0) and (res <= 10 ** 18):
fout.write(str(res))
else:
fout.write('-1')
``` | output | 1 | 28,719 | 15 | 57,439 |
Provide tags and a correct Python 3 solution for this coding contest problem.
ZS the Coder and Chris the Baboon arrived at the entrance of Udayland. There is a n Γ n magic grid on the entrance which is filled with integers. Chris noticed that exactly one of the cells in the grid is empty, and to enter Udayland, they need to fill a positive integer into the empty cell.
Chris tried filling in random numbers but it didn't work. ZS the Coder realizes that they need to fill in a positive integer such that the numbers in the grid form a magic square. This means that he has to fill in a positive integer so that the sum of the numbers in each row of the grid (<image>), each column of the grid (<image>), and the two long diagonals of the grid (the main diagonal β <image> and the secondary diagonal β <image>) are equal.
Chris doesn't know what number to fill in. Can you help Chris find the correct positive integer to fill in or determine that it is impossible?
Input
The first line of the input contains a single integer n (1 β€ n β€ 500) β the number of rows and columns of the magic grid.
n lines follow, each of them contains n integers. The j-th number in the i-th of them denotes ai, j (1 β€ ai, j β€ 109 or ai, j = 0), the number in the i-th row and j-th column of the magic grid. If the corresponding cell is empty, ai, j will be equal to 0. Otherwise, ai, j is positive.
It is guaranteed that there is exactly one pair of integers i, j (1 β€ i, j β€ n) such that ai, j = 0.
Output
Output a single integer, the positive integer x (1 β€ x β€ 1018) that should be filled in the empty cell so that the whole grid becomes a magic square. If such positive integer x does not exist, output - 1 instead.
If there are multiple solutions, you may print any of them.
Examples
Input
3
4 0 2
3 5 7
8 1 6
Output
9
Input
4
1 1 1 1
1 1 0 1
1 1 1 1
1 1 1 1
Output
1
Input
4
1 1 1 1
1 1 0 1
1 1 2 1
1 1 1 1
Output
-1
Note
In the first sample case, we can fill in 9 into the empty cell to make the resulting grid a magic square. Indeed,
The sum of numbers in each row is:
4 + 9 + 2 = 3 + 5 + 7 = 8 + 1 + 6 = 15.
The sum of numbers in each column is:
4 + 3 + 8 = 9 + 5 + 1 = 2 + 7 + 6 = 15.
The sum of numbers in the two diagonals is:
4 + 5 + 6 = 2 + 5 + 8 = 15.
In the third sample case, it is impossible to fill a number in the empty square such that the resulting grid is a magic square. | instruction | 0 | 28,720 | 15 | 57,440 |
Tags: constructive algorithms, implementation
Correct Solution:
```
import math
import itertools
import collections
def getdict(n):
d = {}
if type(n) is list or type(n) is str:
for i in n:
if i in d:
d[i] += 1
else:
d[i] = 1
else:
for i in range(n):
t = ii()
if t in d:
d[t] += 1
else:
d[t] = 1
return d
def cdiv(n, k): return n // k + (n % k != 0)
def ii(): return int(input())
def mi(): return map(int, input().split())
def li(): return list(map(int, input().split()))
def lcm(a, b): return abs(a*b) // math.gcd(a, b)
def wr(arr): return ' '.join(map(str, arr))
def revn(n): return int(str(n)[::-1])
def prime(n):
if n == 2:
return True
if n % 2 == 0 or n <= 1:
return False
sqr = int(math.sqrt(n)) + 1
for d in range(3, sqr, 2):
if n % d == 0:
return False
return True
n = ii()
if n == 1:
print(1)
exit()
ms = []
for i in range(n):
ms.append(li())
sumcolms = [0] * n
diags = [0] * 2
for j in range(n):
for i in range(n):
sumcolms[j] += ms[i][j]
if i == j:
diags[0] += ms[i][j]
if i + j == n - 1:
diags[1] += ms[i][j]
if ms[i][j] == 0:
pos = (i, j)
setrowms = list(set(map(sum, ms)))
setcolms = list(set(sumcolms))
if len(setrowms) != 2 or len(setcolms) != 2 or diags[0] not in setcolms or diags[1] not in setcolms or diags[0] not in setrowms or diags[1] not in setrowms:
print(-1)
elif max(setrowms) - min(setrowms) != max(setcolms) - min(setcolms) or max(setrowms) - min(setrowms) < 1 or max(setrowms) - min(setrowms) < 1:
print(-1)
else:
ms[pos[0]][pos[1]] = abs(setrowms[0] - setrowms[1])
sumcolms = [0] * n
diags = [0] * 2
for j in range(n):
for i in range(n):
sumcolms[j] += ms[i][j]
if i == j:
diags[0] += ms[i][j]
if i + j == n - 1:
diags[1] += ms[i][j]
if ms[i][j] == 0:
pos = (i, j)
setrowms = list(set(map(sum, ms)))
setcolms = list(set(sumcolms))
r = []
r.extend(setcolms)
r.extend(setrowms)
r.extend(diags)
if len(set(r)) == 1:
print(ms[pos[0]][pos[1]])
else:
print(-1)
``` | output | 1 | 28,720 | 15 | 57,441 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
ZS the Coder and Chris the Baboon arrived at the entrance of Udayland. There is a n Γ n magic grid on the entrance which is filled with integers. Chris noticed that exactly one of the cells in the grid is empty, and to enter Udayland, they need to fill a positive integer into the empty cell.
Chris tried filling in random numbers but it didn't work. ZS the Coder realizes that they need to fill in a positive integer such that the numbers in the grid form a magic square. This means that he has to fill in a positive integer so that the sum of the numbers in each row of the grid (<image>), each column of the grid (<image>), and the two long diagonals of the grid (the main diagonal β <image> and the secondary diagonal β <image>) are equal.
Chris doesn't know what number to fill in. Can you help Chris find the correct positive integer to fill in or determine that it is impossible?
Input
The first line of the input contains a single integer n (1 β€ n β€ 500) β the number of rows and columns of the magic grid.
n lines follow, each of them contains n integers. The j-th number in the i-th of them denotes ai, j (1 β€ ai, j β€ 109 or ai, j = 0), the number in the i-th row and j-th column of the magic grid. If the corresponding cell is empty, ai, j will be equal to 0. Otherwise, ai, j is positive.
It is guaranteed that there is exactly one pair of integers i, j (1 β€ i, j β€ n) such that ai, j = 0.
Output
Output a single integer, the positive integer x (1 β€ x β€ 1018) that should be filled in the empty cell so that the whole grid becomes a magic square. If such positive integer x does not exist, output - 1 instead.
If there are multiple solutions, you may print any of them.
Examples
Input
3
4 0 2
3 5 7
8 1 6
Output
9
Input
4
1 1 1 1
1 1 0 1
1 1 1 1
1 1 1 1
Output
1
Input
4
1 1 1 1
1 1 0 1
1 1 2 1
1 1 1 1
Output
-1
Note
In the first sample case, we can fill in 9 into the empty cell to make the resulting grid a magic square. Indeed,
The sum of numbers in each row is:
4 + 9 + 2 = 3 + 5 + 7 = 8 + 1 + 6 = 15.
The sum of numbers in each column is:
4 + 3 + 8 = 9 + 5 + 1 = 2 + 7 + 6 = 15.
The sum of numbers in the two diagonals is:
4 + 5 + 6 = 2 + 5 + 8 = 15.
In the third sample case, it is impossible to fill a number in the empty square such that the resulting grid is a magic square.
Submitted Solution:
```
n = int(input())
matrix = []
for i in range(n):
matrix.append([int(i) for i in input().split()])
zero_col = zero_row = 0
if n == 1:
print(1)
else:
for i in range(n):
for j in range(n):
if matrix[i][j] == 0:
zero_col = j
zero_row = i
cur = (zero_row + 1) % n
ans = 0
for i in range(n):
ans += matrix[cur][i]
a1 = 0
for i in range(n):
a1 += matrix[zero_row][i]
rem = ans - a1
matrix[zero_row][zero_col] = rem
flag = 0
# row check
for i in range(n):
s1 = 0
for j in range(n):
s1 += matrix[i][j]
if s1 != ans:
flag = 1
break
# col check
for i in range(n):
s1 = 0
for j in range(n):
s1 += matrix[j][i]
if s1 != ans:
flag = 1
break
# diagonal check
s1 = 0
for i in range(n):
s1 += matrix[i][i]
if s1 != ans:
flag = 1
s1 = 0
for i in range(n):
s1 += matrix[i][n-1-i]
if s1 != ans:
flag = 1
if flag == 1:
print(-1)
else:
if rem <= 0:
print(-1)
else:
print(rem)
``` | instruction | 0 | 28,721 | 15 | 57,442 |
Yes | output | 1 | 28,721 | 15 | 57,443 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
ZS the Coder and Chris the Baboon arrived at the entrance of Udayland. There is a n Γ n magic grid on the entrance which is filled with integers. Chris noticed that exactly one of the cells in the grid is empty, and to enter Udayland, they need to fill a positive integer into the empty cell.
Chris tried filling in random numbers but it didn't work. ZS the Coder realizes that they need to fill in a positive integer such that the numbers in the grid form a magic square. This means that he has to fill in a positive integer so that the sum of the numbers in each row of the grid (<image>), each column of the grid (<image>), and the two long diagonals of the grid (the main diagonal β <image> and the secondary diagonal β <image>) are equal.
Chris doesn't know what number to fill in. Can you help Chris find the correct positive integer to fill in or determine that it is impossible?
Input
The first line of the input contains a single integer n (1 β€ n β€ 500) β the number of rows and columns of the magic grid.
n lines follow, each of them contains n integers. The j-th number in the i-th of them denotes ai, j (1 β€ ai, j β€ 109 or ai, j = 0), the number in the i-th row and j-th column of the magic grid. If the corresponding cell is empty, ai, j will be equal to 0. Otherwise, ai, j is positive.
It is guaranteed that there is exactly one pair of integers i, j (1 β€ i, j β€ n) such that ai, j = 0.
Output
Output a single integer, the positive integer x (1 β€ x β€ 1018) that should be filled in the empty cell so that the whole grid becomes a magic square. If such positive integer x does not exist, output - 1 instead.
If there are multiple solutions, you may print any of them.
Examples
Input
3
4 0 2
3 5 7
8 1 6
Output
9
Input
4
1 1 1 1
1 1 0 1
1 1 1 1
1 1 1 1
Output
1
Input
4
1 1 1 1
1 1 0 1
1 1 2 1
1 1 1 1
Output
-1
Note
In the first sample case, we can fill in 9 into the empty cell to make the resulting grid a magic square. Indeed,
The sum of numbers in each row is:
4 + 9 + 2 = 3 + 5 + 7 = 8 + 1 + 6 = 15.
The sum of numbers in each column is:
4 + 3 + 8 = 9 + 5 + 1 = 2 + 7 + 6 = 15.
The sum of numbers in the two diagonals is:
4 + 5 + 6 = 2 + 5 + 8 = 15.
In the third sample case, it is impossible to fill a number in the empty square such that the resulting grid is a magic square.
Submitted Solution:
```
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
def func():
N = int(input())
cells = [0] * N
if N == 1:
return 1
mx = 0
for n in range(N):
cells[n] = list(map(int,input().split()))
mx = max(mx, sum(cells[n]))
ans = None
for j in range(N):
for i in range(N):
if cells[j][i] == 0:
ans = mx - sum(cells[j])
cells[j][i] = ans
if ans <= 0:
return -1
# validation
for j in range(N):
if sum(cells[j]) != mx:
return -1
for i in range(N):
if mx != sum([cells[j][i] for j in range(N)]):
return -1
if mx != sum([cells[j][j] for j in range(N)]):
return -1
if mx != sum([cells[j][N-1-j] for j in range(N)]):
return -1
return ans
print(func())
``` | instruction | 0 | 28,722 | 15 | 57,444 |
Yes | output | 1 | 28,722 | 15 | 57,445 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
ZS the Coder and Chris the Baboon arrived at the entrance of Udayland. There is a n Γ n magic grid on the entrance which is filled with integers. Chris noticed that exactly one of the cells in the grid is empty, and to enter Udayland, they need to fill a positive integer into the empty cell.
Chris tried filling in random numbers but it didn't work. ZS the Coder realizes that they need to fill in a positive integer such that the numbers in the grid form a magic square. This means that he has to fill in a positive integer so that the sum of the numbers in each row of the grid (<image>), each column of the grid (<image>), and the two long diagonals of the grid (the main diagonal β <image> and the secondary diagonal β <image>) are equal.
Chris doesn't know what number to fill in. Can you help Chris find the correct positive integer to fill in or determine that it is impossible?
Input
The first line of the input contains a single integer n (1 β€ n β€ 500) β the number of rows and columns of the magic grid.
n lines follow, each of them contains n integers. The j-th number in the i-th of them denotes ai, j (1 β€ ai, j β€ 109 or ai, j = 0), the number in the i-th row and j-th column of the magic grid. If the corresponding cell is empty, ai, j will be equal to 0. Otherwise, ai, j is positive.
It is guaranteed that there is exactly one pair of integers i, j (1 β€ i, j β€ n) such that ai, j = 0.
Output
Output a single integer, the positive integer x (1 β€ x β€ 1018) that should be filled in the empty cell so that the whole grid becomes a magic square. If such positive integer x does not exist, output - 1 instead.
If there are multiple solutions, you may print any of them.
Examples
Input
3
4 0 2
3 5 7
8 1 6
Output
9
Input
4
1 1 1 1
1 1 0 1
1 1 1 1
1 1 1 1
Output
1
Input
4
1 1 1 1
1 1 0 1
1 1 2 1
1 1 1 1
Output
-1
Note
In the first sample case, we can fill in 9 into the empty cell to make the resulting grid a magic square. Indeed,
The sum of numbers in each row is:
4 + 9 + 2 = 3 + 5 + 7 = 8 + 1 + 6 = 15.
The sum of numbers in each column is:
4 + 3 + 8 = 9 + 5 + 1 = 2 + 7 + 6 = 15.
The sum of numbers in the two diagonals is:
4 + 5 + 6 = 2 + 5 + 8 = 15.
In the third sample case, it is impossible to fill a number in the empty square such that the resulting grid is a magic square.
Submitted Solution:
```
n = int(input())
a = []
s = 0
zero_i = None
zero_j = None
zero_in_1 = True
for i in range(n):
a.append(list(map(int, input().split())))
for j in range(n):
if a[i][j] == 0:
zero_i = i
zero_j = j
if n == 1:
print(1)
exit(0)
zero_s = sum(a[zero_i])
for i in range(len(a)):
if i != zero_i:
s = sum(a[i])
break
zero_value = s - zero_s
if zero_value <= 0:
print(-1)
exit(0)
a[zero_i][zero_j] = zero_value
for i in range(n):
if sum(a[i]) != s:
print(-1)
exit(0)
for j in range(n):
column_s = 0
for i in range(n):
column_s += a[i][j]
if column_s != s:
print(-1)
exit(0)
d1_s = 0
for i in range(n):
d1_s += a[i][i]
if d1_s != s:
print(-1)
exit(0)
d2_s = 0
for i in range(n):
d2_s += a[i][n - i - 1]
if d2_s != s:
print(-1)
exit(0)
print(zero_value)
``` | instruction | 0 | 28,723 | 15 | 57,446 |
Yes | output | 1 | 28,723 | 15 | 57,447 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
ZS the Coder and Chris the Baboon arrived at the entrance of Udayland. There is a n Γ n magic grid on the entrance which is filled with integers. Chris noticed that exactly one of the cells in the grid is empty, and to enter Udayland, they need to fill a positive integer into the empty cell.
Chris tried filling in random numbers but it didn't work. ZS the Coder realizes that they need to fill in a positive integer such that the numbers in the grid form a magic square. This means that he has to fill in a positive integer so that the sum of the numbers in each row of the grid (<image>), each column of the grid (<image>), and the two long diagonals of the grid (the main diagonal β <image> and the secondary diagonal β <image>) are equal.
Chris doesn't know what number to fill in. Can you help Chris find the correct positive integer to fill in or determine that it is impossible?
Input
The first line of the input contains a single integer n (1 β€ n β€ 500) β the number of rows and columns of the magic grid.
n lines follow, each of them contains n integers. The j-th number in the i-th of them denotes ai, j (1 β€ ai, j β€ 109 or ai, j = 0), the number in the i-th row and j-th column of the magic grid. If the corresponding cell is empty, ai, j will be equal to 0. Otherwise, ai, j is positive.
It is guaranteed that there is exactly one pair of integers i, j (1 β€ i, j β€ n) such that ai, j = 0.
Output
Output a single integer, the positive integer x (1 β€ x β€ 1018) that should be filled in the empty cell so that the whole grid becomes a magic square. If such positive integer x does not exist, output - 1 instead.
If there are multiple solutions, you may print any of them.
Examples
Input
3
4 0 2
3 5 7
8 1 6
Output
9
Input
4
1 1 1 1
1 1 0 1
1 1 1 1
1 1 1 1
Output
1
Input
4
1 1 1 1
1 1 0 1
1 1 2 1
1 1 1 1
Output
-1
Note
In the first sample case, we can fill in 9 into the empty cell to make the resulting grid a magic square. Indeed,
The sum of numbers in each row is:
4 + 9 + 2 = 3 + 5 + 7 = 8 + 1 + 6 = 15.
The sum of numbers in each column is:
4 + 3 + 8 = 9 + 5 + 1 = 2 + 7 + 6 = 15.
The sum of numbers in the two diagonals is:
4 + 5 + 6 = 2 + 5 + 8 = 15.
In the third sample case, it is impossible to fill a number in the empty square such that the resulting grid is a magic square.
Submitted Solution:
```
n = int(input())
a = 0
r,c = 0,-1
dl = []
for i in range(n):
tl = list(map(int, input().split()))
if c<0 and tl.count(0):
r = i
c = tl.index(0)
dl+=[tl]
s = sum(dl[(r+1)%n])
a = s-sum(dl[r])
dl[r][c] = a
if n>1 and a<1:
print(-1)
exit()
# print(dl)
for i in range(n):
if s!=sum(dl[i]):
print(-1)
exit()
dl = list(map(list,zip(*dl)))
for i in range(n):
if s!=sum(dl[i]):
print(-1)
exit()
s1 = s2 = 0
for i in range(n):
s1+=dl[i][i]
s2+=dl[i][n-i-1]
if not(s1==s2==s):
print(-1)
exit()
if n==1:
a=1
print(a)
``` | instruction | 0 | 28,724 | 15 | 57,448 |
Yes | output | 1 | 28,724 | 15 | 57,449 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
ZS the Coder and Chris the Baboon arrived at the entrance of Udayland. There is a n Γ n magic grid on the entrance which is filled with integers. Chris noticed that exactly one of the cells in the grid is empty, and to enter Udayland, they need to fill a positive integer into the empty cell.
Chris tried filling in random numbers but it didn't work. ZS the Coder realizes that they need to fill in a positive integer such that the numbers in the grid form a magic square. This means that he has to fill in a positive integer so that the sum of the numbers in each row of the grid (<image>), each column of the grid (<image>), and the two long diagonals of the grid (the main diagonal β <image> and the secondary diagonal β <image>) are equal.
Chris doesn't know what number to fill in. Can you help Chris find the correct positive integer to fill in or determine that it is impossible?
Input
The first line of the input contains a single integer n (1 β€ n β€ 500) β the number of rows and columns of the magic grid.
n lines follow, each of them contains n integers. The j-th number in the i-th of them denotes ai, j (1 β€ ai, j β€ 109 or ai, j = 0), the number in the i-th row and j-th column of the magic grid. If the corresponding cell is empty, ai, j will be equal to 0. Otherwise, ai, j is positive.
It is guaranteed that there is exactly one pair of integers i, j (1 β€ i, j β€ n) such that ai, j = 0.
Output
Output a single integer, the positive integer x (1 β€ x β€ 1018) that should be filled in the empty cell so that the whole grid becomes a magic square. If such positive integer x does not exist, output - 1 instead.
If there are multiple solutions, you may print any of them.
Examples
Input
3
4 0 2
3 5 7
8 1 6
Output
9
Input
4
1 1 1 1
1 1 0 1
1 1 1 1
1 1 1 1
Output
1
Input
4
1 1 1 1
1 1 0 1
1 1 2 1
1 1 1 1
Output
-1
Note
In the first sample case, we can fill in 9 into the empty cell to make the resulting grid a magic square. Indeed,
The sum of numbers in each row is:
4 + 9 + 2 = 3 + 5 + 7 = 8 + 1 + 6 = 15.
The sum of numbers in each column is:
4 + 3 + 8 = 9 + 5 + 1 = 2 + 7 + 6 = 15.
The sum of numbers in the two diagonals is:
4 + 5 + 6 = 2 + 5 + 8 = 15.
In the third sample case, it is impossible to fill a number in the empty square such that the resulting grid is a magic square.
Submitted Solution:
```
# Author : nitish420 --------------------------------------------------------------------
import os
import sys
from io import BytesIO, IOBase
mod=10**9+7
# sys.setrecursionlimit(10**6)
def main():
n=int(input())
arr=[]
for i in range(n):
arr.append(list(map(int,input().split())))
if n==1:
print(1)
exit(0)
x,y=0,0
flag=0
for i in range(n):
for j in range(n):
if arr[i][j]==0:
x=i
y=j
flag=1
break
if flag:
break
sumd1=0
sumd2=0
rows=0
rowi=0
columns=[0]*n
ans=0
for i in range(n):
rowtemp=0
for j in range(n):
rowtemp+=arr[i][j]
if i==j:
sumd1+=arr[i][j]
elif i==n-j-1:
sumd2+=arr[i][j]
columns[j]+=arr[i][j]
if rows!=rowtemp and i==x:
rowi=rowtemp
elif rows!=rowtemp and rows==0:
rows=rowtemp
elif rows==rowtemp:
pass
else:
ans=-1
break
if ans!=-1:
# print(rows-rowi)
arr[x][y]=(rows-rowi)
if len(set(columns))==2:
print(arr[x][y])
else:
print(-1)
else:
print(ans)
#----------------------------------------------------------------------------------------
def nouse0():
# This is to save my code from plag due to use of FAST IO template in it.
a=420
b=420
print(f'i am nitish{(a+b)//2}')
def nouse1():
# This is to save my code from plag due to use of FAST IO template in it.
a=420
b=420
print(f'i am nitish{(a+b)//2}')
def nouse2():
# This is to save my code from plag due to use of FAST IO template in it.
a=420
b=420
print(f'i am nitish{(a+b)//2}')
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = 'x' in file.mode or 'r' not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b'\n') + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode('ascii'))
self.read = lambda: self.buffer.read().decode('ascii')
self.readline = lambda: self.buffer.readline().decode('ascii')
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip('\r\n')
def nouse3():
# This is to save my code from plag due to use of FAST IO template in it.
a=420
b=420
print(f'i am nitish{(a+b)//2}')
def nouse4():
# This is to save my code from plag due to use of FAST IO template in it.
a=420
b=420
print(f'i am nitish{(a+b)//2}')
def nouse5():
# This is to save my code from plag due to use of FAST IO template in it.
a=420
b=420
print(f'i am nitish{(a+b)//2}')
# endregion
if __name__ == '__main__':
main()
``` | instruction | 0 | 28,725 | 15 | 57,450 |
No | output | 1 | 28,725 | 15 | 57,451 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
ZS the Coder and Chris the Baboon arrived at the entrance of Udayland. There is a n Γ n magic grid on the entrance which is filled with integers. Chris noticed that exactly one of the cells in the grid is empty, and to enter Udayland, they need to fill a positive integer into the empty cell.
Chris tried filling in random numbers but it didn't work. ZS the Coder realizes that they need to fill in a positive integer such that the numbers in the grid form a magic square. This means that he has to fill in a positive integer so that the sum of the numbers in each row of the grid (<image>), each column of the grid (<image>), and the two long diagonals of the grid (the main diagonal β <image> and the secondary diagonal β <image>) are equal.
Chris doesn't know what number to fill in. Can you help Chris find the correct positive integer to fill in or determine that it is impossible?
Input
The first line of the input contains a single integer n (1 β€ n β€ 500) β the number of rows and columns of the magic grid.
n lines follow, each of them contains n integers. The j-th number in the i-th of them denotes ai, j (1 β€ ai, j β€ 109 or ai, j = 0), the number in the i-th row and j-th column of the magic grid. If the corresponding cell is empty, ai, j will be equal to 0. Otherwise, ai, j is positive.
It is guaranteed that there is exactly one pair of integers i, j (1 β€ i, j β€ n) such that ai, j = 0.
Output
Output a single integer, the positive integer x (1 β€ x β€ 1018) that should be filled in the empty cell so that the whole grid becomes a magic square. If such positive integer x does not exist, output - 1 instead.
If there are multiple solutions, you may print any of them.
Examples
Input
3
4 0 2
3 5 7
8 1 6
Output
9
Input
4
1 1 1 1
1 1 0 1
1 1 1 1
1 1 1 1
Output
1
Input
4
1 1 1 1
1 1 0 1
1 1 2 1
1 1 1 1
Output
-1
Note
In the first sample case, we can fill in 9 into the empty cell to make the resulting grid a magic square. Indeed,
The sum of numbers in each row is:
4 + 9 + 2 = 3 + 5 + 7 = 8 + 1 + 6 = 15.
The sum of numbers in each column is:
4 + 3 + 8 = 9 + 5 + 1 = 2 + 7 + 6 = 15.
The sum of numbers in the two diagonals is:
4 + 5 + 6 = 2 + 5 + 8 = 15.
In the third sample case, it is impossible to fill a number in the empty square such that the resulting grid is a magic square.
Submitted Solution:
```
##n = int(input())
##a = list(map(int, input().split()))
##print(" ".join(map(str, res)))
n = int(input())
if n == 1:
print('1')
exit(0)
a = []
for i in range(n):
tmp = list(map(int, input().split()))
a.append(tmp)
r = -1
c = -1
for i in range(n):
for j in range(n):
if a[i][j] == 0:
r = i
c = j
r2 = (r+1)%n
tot = 0
for j in range(n):
tot += a[r2][j]
cur = 0
for j in range(n):
cur += a[r][j]
x = tot-cur
a[r][c] = x
res = True
for i in range(n):
tmp = 0
tmp2 = 0
for j in range(n):
tmp += a[i][j]
tmp2 += a[j][i]
if tmp != tot or tmp2 != tot:
res = False
break
tmp = 0
tmp2 = 0
for i in range(n):
tmp += a[i][i]
tmp2 += a[i][n-1-i]
if tmp != tot or tmp2 != tot:
res = False
if res == True:
print(x)
else:
print('-1')
``` | instruction | 0 | 28,726 | 15 | 57,452 |
No | output | 1 | 28,726 | 15 | 57,453 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
ZS the Coder and Chris the Baboon arrived at the entrance of Udayland. There is a n Γ n magic grid on the entrance which is filled with integers. Chris noticed that exactly one of the cells in the grid is empty, and to enter Udayland, they need to fill a positive integer into the empty cell.
Chris tried filling in random numbers but it didn't work. ZS the Coder realizes that they need to fill in a positive integer such that the numbers in the grid form a magic square. This means that he has to fill in a positive integer so that the sum of the numbers in each row of the grid (<image>), each column of the grid (<image>), and the two long diagonals of the grid (the main diagonal β <image> and the secondary diagonal β <image>) are equal.
Chris doesn't know what number to fill in. Can you help Chris find the correct positive integer to fill in or determine that it is impossible?
Input
The first line of the input contains a single integer n (1 β€ n β€ 500) β the number of rows and columns of the magic grid.
n lines follow, each of them contains n integers. The j-th number in the i-th of them denotes ai, j (1 β€ ai, j β€ 109 or ai, j = 0), the number in the i-th row and j-th column of the magic grid. If the corresponding cell is empty, ai, j will be equal to 0. Otherwise, ai, j is positive.
It is guaranteed that there is exactly one pair of integers i, j (1 β€ i, j β€ n) such that ai, j = 0.
Output
Output a single integer, the positive integer x (1 β€ x β€ 1018) that should be filled in the empty cell so that the whole grid becomes a magic square. If such positive integer x does not exist, output - 1 instead.
If there are multiple solutions, you may print any of them.
Examples
Input
3
4 0 2
3 5 7
8 1 6
Output
9
Input
4
1 1 1 1
1 1 0 1
1 1 1 1
1 1 1 1
Output
1
Input
4
1 1 1 1
1 1 0 1
1 1 2 1
1 1 1 1
Output
-1
Note
In the first sample case, we can fill in 9 into the empty cell to make the resulting grid a magic square. Indeed,
The sum of numbers in each row is:
4 + 9 + 2 = 3 + 5 + 7 = 8 + 1 + 6 = 15.
The sum of numbers in each column is:
4 + 3 + 8 = 9 + 5 + 1 = 2 + 7 + 6 = 15.
The sum of numbers in the two diagonals is:
4 + 5 + 6 = 2 + 5 + 8 = 15.
In the third sample case, it is impossible to fill a number in the empty square such that the resulting grid is a magic square.
Submitted Solution:
```
l = []
n = int(input())
st1 = 0
for i in range(n):
l1 = (list(map(int,input().split())))
l.append(l1)
if 0 in l1:
st1 = sum(l1)
if n==1:
print(1)
else:
l1,l2 = [], []
for i in range(n):
s1,s2 = 0, 0
for j in range(n):
s1+=l[i][j]
s2+=l[j][i]
l1.append(s1)
l2.append(s2)
l1,l2 = set(l1),set(l2)
l3 = 0
for i in range(n):
l3 += l[i][i]
l4 = 0
for i in range(n):
l4 += l[i][n-i-1]
sm = 0
n = 0
if len(l1)==2 and len(l2)==2:
sm = max(l1)
if sm>st1:
for i in l1:
if i!=sm:
n = abs(sm-i)
if ((l3==sm) or ((l3+n)==sm)) and ((l4==sm) or ((l4+n)==sm)) and ((st1+n)==sm):
print(n)
else:
print(-1)
else:
print(-1)
else:
print(-1)
``` | instruction | 0 | 28,727 | 15 | 57,454 |
No | output | 1 | 28,727 | 15 | 57,455 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
ZS the Coder and Chris the Baboon arrived at the entrance of Udayland. There is a n Γ n magic grid on the entrance which is filled with integers. Chris noticed that exactly one of the cells in the grid is empty, and to enter Udayland, they need to fill a positive integer into the empty cell.
Chris tried filling in random numbers but it didn't work. ZS the Coder realizes that they need to fill in a positive integer such that the numbers in the grid form a magic square. This means that he has to fill in a positive integer so that the sum of the numbers in each row of the grid (<image>), each column of the grid (<image>), and the two long diagonals of the grid (the main diagonal β <image> and the secondary diagonal β <image>) are equal.
Chris doesn't know what number to fill in. Can you help Chris find the correct positive integer to fill in or determine that it is impossible?
Input
The first line of the input contains a single integer n (1 β€ n β€ 500) β the number of rows and columns of the magic grid.
n lines follow, each of them contains n integers. The j-th number in the i-th of them denotes ai, j (1 β€ ai, j β€ 109 or ai, j = 0), the number in the i-th row and j-th column of the magic grid. If the corresponding cell is empty, ai, j will be equal to 0. Otherwise, ai, j is positive.
It is guaranteed that there is exactly one pair of integers i, j (1 β€ i, j β€ n) such that ai, j = 0.
Output
Output a single integer, the positive integer x (1 β€ x β€ 1018) that should be filled in the empty cell so that the whole grid becomes a magic square. If such positive integer x does not exist, output - 1 instead.
If there are multiple solutions, you may print any of them.
Examples
Input
3
4 0 2
3 5 7
8 1 6
Output
9
Input
4
1 1 1 1
1 1 0 1
1 1 1 1
1 1 1 1
Output
1
Input
4
1 1 1 1
1 1 0 1
1 1 2 1
1 1 1 1
Output
-1
Note
In the first sample case, we can fill in 9 into the empty cell to make the resulting grid a magic square. Indeed,
The sum of numbers in each row is:
4 + 9 + 2 = 3 + 5 + 7 = 8 + 1 + 6 = 15.
The sum of numbers in each column is:
4 + 3 + 8 = 9 + 5 + 1 = 2 + 7 + 6 = 15.
The sum of numbers in the two diagonals is:
4 + 5 + 6 = 2 + 5 + 8 = 15.
In the third sample case, it is impossible to fill a number in the empty square such that the resulting grid is a magic square.
Submitted Solution:
```
'''input
3
1 2 2
2 2 1
0 1 2
'''
n = int(input())
s, t, y = [], 0, []
for _ in range(n):
i = list(map(int, input().split()))
if 0 not in i:
s.append(sum(i))
else:
t = i
y.append(i)
if len(set(s)) == 1:
x = s[0] - sum(t)
z = t.index(0)
if sum([y[j][z] for j in range(n)]) + x == s[0] and x != 0:
if sum([y[k][k] for k in range(n)]) == s[0]:
if sum([y[l][n-l-1] for l in range(n)]) == s[0]:
print(x)
quit()
print(-1)
elif len(set(s)) == 0:
print(100)
else:
print(-1)
``` | instruction | 0 | 28,728 | 15 | 57,456 |
No | output | 1 | 28,728 | 15 | 57,457 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Arkady is playing Battleship. The rules of this game aren't really important.
There is a field of n Γ n cells. There should be exactly one k-decker on the field, i. e. a ship that is k cells long oriented either horizontally or vertically. However, Arkady doesn't know where it is located. For each cell Arkady knows if it is definitely empty or can contain a part of the ship.
Consider all possible locations of the ship. Find such a cell that belongs to the maximum possible number of different locations of the ship.
Input
The first line contains two integers n and k (1 β€ k β€ n β€ 100) β the size of the field and the size of the ship.
The next n lines contain the field. Each line contains n characters, each of which is either '#' (denotes a definitely empty cell) or '.' (denotes a cell that can belong to the ship).
Output
Output two integers β the row and the column of a cell that belongs to the maximum possible number of different locations of the ship.
If there are multiple answers, output any of them. In particular, if no ship can be placed on the field, you can output any cell.
Examples
Input
4 3
#..#
#.#.
....
.###
Output
3 2
Input
10 4
#....##...
.#...#....
..#..#..#.
...#.#....
.#..##.#..
.....#...#
...#.##...
.#...#.#..
.....#..#.
...#.#...#
Output
6 1
Input
19 6
##..............###
#......#####.....##
.....#########.....
....###########....
...#############...
..###############..
.#################.
.#################.
.#################.
.#################.
#####....##....####
####............###
####............###
#####...####...####
.#####..####..#####
...###........###..
....###########....
.........##........
#.................#
Output
1 8
Note
The picture below shows the three possible locations of the ship that contain the cell (3, 2) in the first sample.
<image> | instruction | 0 | 28,812 | 15 | 57,624 |
Tags: implementation
Correct Solution:
```
# get inputs
map_size, ship_size = list(map(int, input().split()))
map = [[0 for _ in range(map_size)] for _ in range(map_size)]
for i in range(map_size):
tmpRow = list(input())
for j in range(map_size):
if tmpRow[j] == '.':
map[i][j] = 1
results = [[0 for _ in range(map_size)] for _ in range(map_size)]
max_i, max_j = 0, 0
max_value = 0
for i in range(map_size):
for j in range(map_size):
if map[i][j]==1:
tmp_cnt = 0
for k in range(j, min(j+ship_size,map_size)):
if map[i][k] == 1:
tmp_cnt += 1
if tmp_cnt == ship_size:
for p in range(j, min(j+ship_size,map_size)):
results[i][p] += 1
tmp_cnt = 0
for k in range(i, min(i + ship_size, map_size)):
if map[k][j] == 1:
tmp_cnt += 1
if tmp_cnt == ship_size:
for p in range(i, min(i + ship_size, map_size)):
results[p][j] += 1
for i in range(map_size):
for j in range(map_size):
if results[i][j] > max_value:
max_value = results[i][j]
max_i = i
max_j = j
print(max_i+1, max_j+1)
``` | output | 1 | 28,812 | 15 | 57,625 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Arkady is playing Battleship. The rules of this game aren't really important.
There is a field of n Γ n cells. There should be exactly one k-decker on the field, i. e. a ship that is k cells long oriented either horizontally or vertically. However, Arkady doesn't know where it is located. For each cell Arkady knows if it is definitely empty or can contain a part of the ship.
Consider all possible locations of the ship. Find such a cell that belongs to the maximum possible number of different locations of the ship.
Input
The first line contains two integers n and k (1 β€ k β€ n β€ 100) β the size of the field and the size of the ship.
The next n lines contain the field. Each line contains n characters, each of which is either '#' (denotes a definitely empty cell) or '.' (denotes a cell that can belong to the ship).
Output
Output two integers β the row and the column of a cell that belongs to the maximum possible number of different locations of the ship.
If there are multiple answers, output any of them. In particular, if no ship can be placed on the field, you can output any cell.
Examples
Input
4 3
#..#
#.#.
....
.###
Output
3 2
Input
10 4
#....##...
.#...#....
..#..#..#.
...#.#....
.#..##.#..
.....#...#
...#.##...
.#...#.#..
.....#..#.
...#.#...#
Output
6 1
Input
19 6
##..............###
#......#####.....##
.....#########.....
....###########....
...#############...
..###############..
.#################.
.#################.
.#################.
.#################.
#####....##....####
####............###
####............###
#####...####...####
.#####..####..#####
...###........###..
....###########....
.........##........
#.................#
Output
1 8
Note
The picture below shows the three possible locations of the ship that contain the cell (3, 2) in the first sample.
<image> | instruction | 0 | 28,813 | 15 | 57,626 |
Tags: implementation
Correct Solution:
```
n, k = map(int, input().split())
a = []
for i in range(n):
a.append(input())
maxans = 0
xans = 0
yans = 0
for i in range(n):
for j in range(n):
curans = 0
l = 0
while l < k and i - l >= 0 and a[i - l][j] == '.':
l += 1
r = 0
while r < k and i + r < n and a[i + r][j] == '.':
r += 1
curans += max(0, r + l - k)
if k != 1:
l = 0
while l < k and j - l >= 0 and a[i][j - l] == '.':
l += 1
r = 0
while r < k and j + r < n and a[i][j + r] == '.':
r += 1
curans += max(0, r + l - k)
if curans > maxans:
maxans = curans
xans = i
yans = j
print("{} {}".format(xans + 1, yans + 1))
``` | output | 1 | 28,813 | 15 | 57,627 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Arkady is playing Battleship. The rules of this game aren't really important.
There is a field of n Γ n cells. There should be exactly one k-decker on the field, i. e. a ship that is k cells long oriented either horizontally or vertically. However, Arkady doesn't know where it is located. For each cell Arkady knows if it is definitely empty or can contain a part of the ship.
Consider all possible locations of the ship. Find such a cell that belongs to the maximum possible number of different locations of the ship.
Input
The first line contains two integers n and k (1 β€ k β€ n β€ 100) β the size of the field and the size of the ship.
The next n lines contain the field. Each line contains n characters, each of which is either '#' (denotes a definitely empty cell) or '.' (denotes a cell that can belong to the ship).
Output
Output two integers β the row and the column of a cell that belongs to the maximum possible number of different locations of the ship.
If there are multiple answers, output any of them. In particular, if no ship can be placed on the field, you can output any cell.
Examples
Input
4 3
#..#
#.#.
....
.###
Output
3 2
Input
10 4
#....##...
.#...#....
..#..#..#.
...#.#....
.#..##.#..
.....#...#
...#.##...
.#...#.#..
.....#..#.
...#.#...#
Output
6 1
Input
19 6
##..............###
#......#####.....##
.....#########.....
....###########....
...#############...
..###############..
.#################.
.#################.
.#################.
.#################.
#####....##....####
####............###
####............###
#####...####...####
.#####..####..#####
...###........###..
....###########....
.........##........
#.................#
Output
1 8
Note
The picture below shows the three possible locations of the ship that contain the cell (3, 2) in the first sample.
<image> | instruction | 0 | 28,814 | 15 | 57,628 |
Tags: implementation
Correct Solution:
```
def go():
keep = {}
n, l = (int(i) for i in input().split(' '))
matrix = [None] * n
for i in range(n):
matrix[i] = [i for i in input()]
for i in range(n):
for j in range(n):
if matrix[i][j] == '.':
h = 0
v = 0
for k in range(l):
if j + k < n and matrix[i][j + k] == '.':
h += 1
if i + k < n and matrix[i + k][j] == '.':
v += 1
if h == l:
for k in range(l):
keep.setdefault((i, j + k), 0)
keep[(i, j + k)] += 1
if v == l:
for k in range(l):
keep.setdefault((i + k, j), 0)
keep[(i + k, j)] += 1
if len(keep) != 0:
import operator
result = max(keep.items(), key=operator.itemgetter(1))[0]
return '{} {}'.format(result[0] + 1, result[1] + 1)
return '1 1'
print(go())
``` | output | 1 | 28,814 | 15 | 57,629 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Arkady is playing Battleship. The rules of this game aren't really important.
There is a field of n Γ n cells. There should be exactly one k-decker on the field, i. e. a ship that is k cells long oriented either horizontally or vertically. However, Arkady doesn't know where it is located. For each cell Arkady knows if it is definitely empty or can contain a part of the ship.
Consider all possible locations of the ship. Find such a cell that belongs to the maximum possible number of different locations of the ship.
Input
The first line contains two integers n and k (1 β€ k β€ n β€ 100) β the size of the field and the size of the ship.
The next n lines contain the field. Each line contains n characters, each of which is either '#' (denotes a definitely empty cell) or '.' (denotes a cell that can belong to the ship).
Output
Output two integers β the row and the column of a cell that belongs to the maximum possible number of different locations of the ship.
If there are multiple answers, output any of them. In particular, if no ship can be placed on the field, you can output any cell.
Examples
Input
4 3
#..#
#.#.
....
.###
Output
3 2
Input
10 4
#....##...
.#...#....
..#..#..#.
...#.#....
.#..##.#..
.....#...#
...#.##...
.#...#.#..
.....#..#.
...#.#...#
Output
6 1
Input
19 6
##..............###
#......#####.....##
.....#########.....
....###########....
...#############...
..###############..
.#################.
.#################.
.#################.
.#################.
#####....##....####
####............###
####............###
#####...####...####
.#####..####..#####
...###........###..
....###########....
.........##........
#.................#
Output
1 8
Note
The picture below shows the three possible locations of the ship that contain the cell (3, 2) in the first sample.
<image> | instruction | 0 | 28,815 | 15 | 57,630 |
Tags: implementation
Correct Solution:
```
n, k = map(int, input().split())
def func(i):
return -1 if i == '#' else 0
l = []
for i in range(n):
l.append(list(map(func, input())))
for i in range(n):
for j in range(n - k + 1):
if -1 not in l[i][j : j+k]:
for t in range(k):
l[i][j+t] += 1
l = list(zip(*l))
for i in range(n):
l[i] = list(l[i])
for i in range(n):
for j in range(n - k + 1):
if -1 not in l[i][j : j+k]:
for t in range(k):
l[i][j+t] += 1
l = list(zip(*l))
x, y = 0, 0
m = -1
for i in range(n):
for j in range(n):
if l[i][j] > m:
x = i
y = j
m = l[i][j]
print(x+1, y+1)
``` | output | 1 | 28,815 | 15 | 57,631 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Arkady is playing Battleship. The rules of this game aren't really important.
There is a field of n Γ n cells. There should be exactly one k-decker on the field, i. e. a ship that is k cells long oriented either horizontally or vertically. However, Arkady doesn't know where it is located. For each cell Arkady knows if it is definitely empty or can contain a part of the ship.
Consider all possible locations of the ship. Find such a cell that belongs to the maximum possible number of different locations of the ship.
Input
The first line contains two integers n and k (1 β€ k β€ n β€ 100) β the size of the field and the size of the ship.
The next n lines contain the field. Each line contains n characters, each of which is either '#' (denotes a definitely empty cell) or '.' (denotes a cell that can belong to the ship).
Output
Output two integers β the row and the column of a cell that belongs to the maximum possible number of different locations of the ship.
If there are multiple answers, output any of them. In particular, if no ship can be placed on the field, you can output any cell.
Examples
Input
4 3
#..#
#.#.
....
.###
Output
3 2
Input
10 4
#....##...
.#...#....
..#..#..#.
...#.#....
.#..##.#..
.....#...#
...#.##...
.#...#.#..
.....#..#.
...#.#...#
Output
6 1
Input
19 6
##..............###
#......#####.....##
.....#########.....
....###########....
...#############...
..###############..
.#################.
.#################.
.#################.
.#################.
#####....##....####
####............###
####............###
#####...####...####
.#####..####..#####
...###........###..
....###########....
.........##........
#.................#
Output
1 8
Note
The picture below shows the three possible locations of the ship that contain the cell (3, 2) in the first sample.
<image> | instruction | 0 | 28,816 | 15 | 57,632 |
Tags: implementation
Correct Solution:
```
n, k = map(int, input().split())
m = []
for i in range(n):
u = str(input())
m.append(u)
v = [[0] * n for _ in range(n)]
for x in range(n - k + 1):
for y in range(n):
s = m[y][x:x+k]
if not '#' in s:
for i in range(k):
v[y][x + i] += 1
for x in range(n):
for y in range(n - k + 1):
o = m[y:y+k]
s = [o[q][x] for q in range(k)]
if not '#' in s:
for i in range(k):
v[y + i][x] += 1
mx = 0
my = 0
m = v[0][0]
for i in range(n):
for q in range(n):
if v[i][q] > m:
mx = q
my = i
m = v[i][q]
print(my + 1, mx + 1)
``` | output | 1 | 28,816 | 15 | 57,633 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Arkady is playing Battleship. The rules of this game aren't really important.
There is a field of n Γ n cells. There should be exactly one k-decker on the field, i. e. a ship that is k cells long oriented either horizontally or vertically. However, Arkady doesn't know where it is located. For each cell Arkady knows if it is definitely empty or can contain a part of the ship.
Consider all possible locations of the ship. Find such a cell that belongs to the maximum possible number of different locations of the ship.
Input
The first line contains two integers n and k (1 β€ k β€ n β€ 100) β the size of the field and the size of the ship.
The next n lines contain the field. Each line contains n characters, each of which is either '#' (denotes a definitely empty cell) or '.' (denotes a cell that can belong to the ship).
Output
Output two integers β the row and the column of a cell that belongs to the maximum possible number of different locations of the ship.
If there are multiple answers, output any of them. In particular, if no ship can be placed on the field, you can output any cell.
Examples
Input
4 3
#..#
#.#.
....
.###
Output
3 2
Input
10 4
#....##...
.#...#....
..#..#..#.
...#.#....
.#..##.#..
.....#...#
...#.##...
.#...#.#..
.....#..#.
...#.#...#
Output
6 1
Input
19 6
##..............###
#......#####.....##
.....#########.....
....###########....
...#############...
..###############..
.#################.
.#################.
.#################.
.#################.
#####....##....####
####............###
####............###
#####...####...####
.#####..####..#####
...###........###..
....###########....
.........##........
#.................#
Output
1 8
Note
The picture below shows the three possible locations of the ship that contain the cell (3, 2) in the first sample.
<image> | instruction | 0 | 28,817 | 15 | 57,634 |
Tags: implementation
Correct Solution:
```
n, k = map(int, input().split())
arr = []
for i in range(n):
arr.append(list(input()))
up = [[0]*n for i in range(n)]
down = [[0]*n for i in range(n)]
left = [[0]*n for i in range(n)]
right = [[0]*n for i in range(n)]
for i in range(n):
up[0][i] = 1 if arr[0][i] == '.' else 0
left[i][0] = 1 if arr[i][0] == '.' else 0
right[i][n-1] = 1 if arr[i][n-1] == '.' else 0
down[n-1][i] = 1 if arr[n-1][i] == '.' else 0
for i in range(1, n):
for j in range(n):
up[i][j] = up[i-1][j]+1 if arr[i][j] == '.' else 0
left[j][i] = left[j][i-1]+1 if arr[j][i] == '.' else 0
right[j][n-1-i] = right[j][n-i]+1 if arr[j][n-1-i] == '.' else 0
down[n-1-i][j] = down[n-i][j]+1 if arr[n-1-i][j] == '.' else 0
def print_fn(arr):
print("\n".join([' '.join(map(str, ar)) for ar in arr]))
print()
'''
print_fn(up)
print_fn(down)
print_fn(left)
print_fn(right)
'''
mx, mxi = 0, (0, 0)
for i in range(n):
for j in range(n):
horizontal = max(0, min(left[i][j], k) + min(right[i][j], k) - k)
vertical = max(0, min(up[i][j], k) + min(down[i][j], k) - k)
parts = horizontal + vertical
if parts > mx:
mx = max(mx, parts)
mxi = (i, j)
print(mxi[0]+1, mxi[1]+1)
``` | output | 1 | 28,817 | 15 | 57,635 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Arkady is playing Battleship. The rules of this game aren't really important.
There is a field of n Γ n cells. There should be exactly one k-decker on the field, i. e. a ship that is k cells long oriented either horizontally or vertically. However, Arkady doesn't know where it is located. For each cell Arkady knows if it is definitely empty or can contain a part of the ship.
Consider all possible locations of the ship. Find such a cell that belongs to the maximum possible number of different locations of the ship.
Input
The first line contains two integers n and k (1 β€ k β€ n β€ 100) β the size of the field and the size of the ship.
The next n lines contain the field. Each line contains n characters, each of which is either '#' (denotes a definitely empty cell) or '.' (denotes a cell that can belong to the ship).
Output
Output two integers β the row and the column of a cell that belongs to the maximum possible number of different locations of the ship.
If there are multiple answers, output any of them. In particular, if no ship can be placed on the field, you can output any cell.
Examples
Input
4 3
#..#
#.#.
....
.###
Output
3 2
Input
10 4
#....##...
.#...#....
..#..#..#.
...#.#....
.#..##.#..
.....#...#
...#.##...
.#...#.#..
.....#..#.
...#.#...#
Output
6 1
Input
19 6
##..............###
#......#####.....##
.....#########.....
....###########....
...#############...
..###############..
.#################.
.#################.
.#################.
.#################.
#####....##....####
####............###
####............###
#####...####...####
.#####..####..#####
...###........###..
....###########....
.........##........
#.................#
Output
1 8
Note
The picture below shows the three possible locations of the ship that contain the cell (3, 2) in the first sample.
<image> | instruction | 0 | 28,818 | 15 | 57,636 |
Tags: implementation
Correct Solution:
```
import re
R = lambda: map(int, input().split())
def f(i, j):
if a[i][j] == '#': return 0
ii = i
while ii >= 0 and a[ii][j] == '.' and i - ii <= k-1: ii -= 1
n1 = i - ii
ii = i
while ii < n and a[ii][j] == '.' and ii - i <= k-1: ii += 1
n2 = ii - i
t1 = max(n1 + n2 - k, 0)
#print(n1, n2)
jj = j
while jj >= 0 and a[i][jj] == '.' and j - jj <= k-1: jj -= 1
n1 = j - jj
jj = j
while jj < n and a[i][jj] == '.' and jj - j <= k-1: jj += 1
n2 = jj - j
#print(n1, n2)
t2 = max(n1 + n2 - k, 0)
#print(i, j, t1, t2)
return t1 + t2
def func():
maxi = maxj = maxv = 0
for i in range(n):
for j in range(n):
v = f(i,j)
if v > maxv:
maxi = i
maxj = j
maxv = v
#print(maxi, maxj, v)
return maxi + 1, maxj + 1
n, k = R()
a = []
for i in range(n): a.append(input())
i, j = func()
print(i, j)
``` | output | 1 | 28,818 | 15 | 57,637 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Arkady is playing Battleship. The rules of this game aren't really important.
There is a field of n Γ n cells. There should be exactly one k-decker on the field, i. e. a ship that is k cells long oriented either horizontally or vertically. However, Arkady doesn't know where it is located. For each cell Arkady knows if it is definitely empty or can contain a part of the ship.
Consider all possible locations of the ship. Find such a cell that belongs to the maximum possible number of different locations of the ship.
Input
The first line contains two integers n and k (1 β€ k β€ n β€ 100) β the size of the field and the size of the ship.
The next n lines contain the field. Each line contains n characters, each of which is either '#' (denotes a definitely empty cell) or '.' (denotes a cell that can belong to the ship).
Output
Output two integers β the row and the column of a cell that belongs to the maximum possible number of different locations of the ship.
If there are multiple answers, output any of them. In particular, if no ship can be placed on the field, you can output any cell.
Examples
Input
4 3
#..#
#.#.
....
.###
Output
3 2
Input
10 4
#....##...
.#...#....
..#..#..#.
...#.#....
.#..##.#..
.....#...#
...#.##...
.#...#.#..
.....#..#.
...#.#...#
Output
6 1
Input
19 6
##..............###
#......#####.....##
.....#########.....
....###########....
...#############...
..###############..
.#################.
.#################.
.#################.
.#################.
#####....##....####
####............###
####............###
#####...####...####
.#####..####..#####
...###........###..
....###########....
.........##........
#.................#
Output
1 8
Note
The picture below shows the three possible locations of the ship that contain the cell (3, 2) in the first sample.
<image> | instruction | 0 | 28,819 | 15 | 57,638 |
Tags: implementation
Correct Solution:
```
from time import time
try:
start = time()
fs, ss = input().split()
fs = int(fs)
ss = int(ss)
field = [[0] * fs for i in range(fs)]
for i in range(fs):
inp = input()
for j in range(fs):
if(inp[j] == '#'):
field[i][j] = -1
for i in range(fs):
for j in range(fs):
d = 1
r = 1
for k in range(ss):
if(i > fs - ss or field[i + k][j] == -1):
r = 0
if(j > fs - ss or field[i][j + k] == -1):
d = 0
if(r == 0 and d == 0):
break
if(r == 1):
for k in range(ss):
field[i + k][j] += 1
if(d == 1):
for k in range(ss):
field[i][j + k] += 1
max = 0
for i in range(fs):
for j in range(fs):
if(field[i][j] > max):
max = field[i][j]
maxi = i
maxj = j
print(maxi + 1, maxj + 1)
except:
print(1,1)
``` | output | 1 | 28,819 | 15 | 57,639 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Arkady is playing Battleship. The rules of this game aren't really important.
There is a field of n Γ n cells. There should be exactly one k-decker on the field, i. e. a ship that is k cells long oriented either horizontally or vertically. However, Arkady doesn't know where it is located. For each cell Arkady knows if it is definitely empty or can contain a part of the ship.
Consider all possible locations of the ship. Find such a cell that belongs to the maximum possible number of different locations of the ship.
Input
The first line contains two integers n and k (1 β€ k β€ n β€ 100) β the size of the field and the size of the ship.
The next n lines contain the field. Each line contains n characters, each of which is either '#' (denotes a definitely empty cell) or '.' (denotes a cell that can belong to the ship).
Output
Output two integers β the row and the column of a cell that belongs to the maximum possible number of different locations of the ship.
If there are multiple answers, output any of them. In particular, if no ship can be placed on the field, you can output any cell.
Examples
Input
4 3
#..#
#.#.
....
.###
Output
3 2
Input
10 4
#....##...
.#...#....
..#..#..#.
...#.#....
.#..##.#..
.....#...#
...#.##...
.#...#.#..
.....#..#.
...#.#...#
Output
6 1
Input
19 6
##..............###
#......#####.....##
.....#########.....
....###########....
...#############...
..###############..
.#################.
.#################.
.#################.
.#################.
#####....##....####
####............###
####............###
#####...####...####
.#####..####..#####
...###........###..
....###########....
.........##........
#.................#
Output
1 8
Note
The picture below shows the three possible locations of the ship that contain the cell (3, 2) in the first sample.
<image>
Submitted Solution:
```
n,k=map(int,input().split())
a=[]
for i in range(n):
a.append(input())
m,x,y=-1,0,0
for i in range(n):
for j in range(n):
if a[i][j]=='#':continue
R=0;
while j+R<n:
if a[i][j+R]=='#':break
if R==k:break
R+=1
L=0;
while j-L>=0:
if a[i][j-L]=='#':break
if L==k:break
L+=1
U=0;
while i-U>=0:
if a[i-U][j]=='#':break
if U==k:break
U+=1
D=0;
while i+D<n:
if a[i+D][j]=='#':break
if D==k:break
D+=1
z=min(k,max(0,L+R-k))+min(k,max(0,U+D-k))
if z>m: m,x,y=z,j,i
print(y+1,x+1)
``` | instruction | 0 | 28,820 | 15 | 57,640 |
Yes | output | 1 | 28,820 | 15 | 57,641 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Arkady is playing Battleship. The rules of this game aren't really important.
There is a field of n Γ n cells. There should be exactly one k-decker on the field, i. e. a ship that is k cells long oriented either horizontally or vertically. However, Arkady doesn't know where it is located. For each cell Arkady knows if it is definitely empty or can contain a part of the ship.
Consider all possible locations of the ship. Find such a cell that belongs to the maximum possible number of different locations of the ship.
Input
The first line contains two integers n and k (1 β€ k β€ n β€ 100) β the size of the field and the size of the ship.
The next n lines contain the field. Each line contains n characters, each of which is either '#' (denotes a definitely empty cell) or '.' (denotes a cell that can belong to the ship).
Output
Output two integers β the row and the column of a cell that belongs to the maximum possible number of different locations of the ship.
If there are multiple answers, output any of them. In particular, if no ship can be placed on the field, you can output any cell.
Examples
Input
4 3
#..#
#.#.
....
.###
Output
3 2
Input
10 4
#....##...
.#...#....
..#..#..#.
...#.#....
.#..##.#..
.....#...#
...#.##...
.#...#.#..
.....#..#.
...#.#...#
Output
6 1
Input
19 6
##..............###
#......#####.....##
.....#########.....
....###########....
...#############...
..###############..
.#################.
.#################.
.#################.
.#################.
#####....##....####
####............###
####............###
#####...####...####
.#####..####..#####
...###........###..
....###########....
.........##........
#.................#
Output
1 8
Note
The picture below shows the three possible locations of the ship that contain the cell (3, 2) in the first sample.
<image>
Submitted Solution:
```
# https://codeforces.com/problemset/problem/965/B
n, k = map(int, input().strip().split())
lst = []
lst_invert = []
lst_count = []
for i in range(n):
lst_add = []
for j in range(n):
lst_add.append(0)
lst_count.append(lst_add)
# print(lst_count)
for i in range(n):
lst.append(list(input().strip()))
# print(lst)
for i in range(n):
lst_add = []
for j in range(n):
if (lst[i][j:j + k] == list('.' * k)):
for x in range(k):
lst_count[i][j + x] += 1
lst_add.append(lst[j][i])
lst_invert.append(lst_add)
# print(lst_invert)
for i in range(n):
for j in range(n):
if (lst_invert[i][j:j + k] == list('.' * k)):
for x in range(k):
lst_count[j + x][i] += 1
# print(lst_count)
maximum = 0
for i in lst_count:
if (maximum < max(i)):
maximum = max(i)
# print(maximum)
check = 0
for i in range(n):
if (max(lst_count[i]) == maximum):
for j in range(n):
if (lst_count[i][j] == maximum):
print(i+1, j+1)
check = 1
break
if (check == 1):
break
``` | instruction | 0 | 28,821 | 15 | 57,642 |
Yes | output | 1 | 28,821 | 15 | 57,643 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Arkady is playing Battleship. The rules of this game aren't really important.
There is a field of n Γ n cells. There should be exactly one k-decker on the field, i. e. a ship that is k cells long oriented either horizontally or vertically. However, Arkady doesn't know where it is located. For each cell Arkady knows if it is definitely empty or can contain a part of the ship.
Consider all possible locations of the ship. Find such a cell that belongs to the maximum possible number of different locations of the ship.
Input
The first line contains two integers n and k (1 β€ k β€ n β€ 100) β the size of the field and the size of the ship.
The next n lines contain the field. Each line contains n characters, each of which is either '#' (denotes a definitely empty cell) or '.' (denotes a cell that can belong to the ship).
Output
Output two integers β the row and the column of a cell that belongs to the maximum possible number of different locations of the ship.
If there are multiple answers, output any of them. In particular, if no ship can be placed on the field, you can output any cell.
Examples
Input
4 3
#..#
#.#.
....
.###
Output
3 2
Input
10 4
#....##...
.#...#....
..#..#..#.
...#.#....
.#..##.#..
.....#...#
...#.##...
.#...#.#..
.....#..#.
...#.#...#
Output
6 1
Input
19 6
##..............###
#......#####.....##
.....#########.....
....###########....
...#############...
..###############..
.#################.
.#################.
.#################.
.#################.
#####....##....####
####............###
####............###
#####...####...####
.#####..####..#####
...###........###..
....###########....
.........##........
#.................#
Output
1 8
Note
The picture below shows the three possible locations of the ship that contain the cell (3, 2) in the first sample.
<image>
Submitted Solution:
```
def main():
n , k = [int (x) for x in input().split()]
s = []
for x in range(n):
s.append(list(input()))
temp = [[0]*n for x in range(n)]
#print(s)
for i in range(n):
for j in range(n-k+1):
count = 0
for m in range(j, j+k):
if s[i][m] == ".":
count += 1
if(count == k):
for m in range(j, j+k):
temp[i][m] += 1
for i in range(n):
for j in range(n-k+1):
count = 0
for m in range(j, j+k):
if s[m][i] == ".":
count += 1
if(count == k):
for m in range(j, j+k):
temp[m][i]+= 1
x = y =1
maxi = 0
for i in range(n):
for j in range(n):
if maxi < temp[i][j]:
maxi, x, y = temp[i][j], i+1, j+1
print(x, y)
if __name__ == '__main__':
main()
``` | instruction | 0 | 28,822 | 15 | 57,644 |
Yes | output | 1 | 28,822 | 15 | 57,645 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Arkady is playing Battleship. The rules of this game aren't really important.
There is a field of n Γ n cells. There should be exactly one k-decker on the field, i. e. a ship that is k cells long oriented either horizontally or vertically. However, Arkady doesn't know where it is located. For each cell Arkady knows if it is definitely empty or can contain a part of the ship.
Consider all possible locations of the ship. Find such a cell that belongs to the maximum possible number of different locations of the ship.
Input
The first line contains two integers n and k (1 β€ k β€ n β€ 100) β the size of the field and the size of the ship.
The next n lines contain the field. Each line contains n characters, each of which is either '#' (denotes a definitely empty cell) or '.' (denotes a cell that can belong to the ship).
Output
Output two integers β the row and the column of a cell that belongs to the maximum possible number of different locations of the ship.
If there are multiple answers, output any of them. In particular, if no ship can be placed on the field, you can output any cell.
Examples
Input
4 3
#..#
#.#.
....
.###
Output
3 2
Input
10 4
#....##...
.#...#....
..#..#..#.
...#.#....
.#..##.#..
.....#...#
...#.##...
.#...#.#..
.....#..#.
...#.#...#
Output
6 1
Input
19 6
##..............###
#......#####.....##
.....#########.....
....###########....
...#############...
..###############..
.#################.
.#################.
.#################.
.#################.
#####....##....####
####............###
####............###
#####...####...####
.#####..####..#####
...###........###..
....###########....
.........##........
#.................#
Output
1 8
Note
The picture below shows the three possible locations of the ship that contain the cell (3, 2) in the first sample.
<image>
Submitted Solution:
```
n,k = map(int,input().split())
mapp = [list(input()) for i in range(n)]
res = 0
res = [[0 for i in range(n)] for j in range(n)]
for i in range(n):
count = 0
for j in range(n):
if (mapp[i][j] == '.'):
count += 1
else:
count = 0
if (count >= k):
for t in range(j - k + 1,j + 1):
res[i][t] += 1
for i in range(n):
count = 0
for j in range(n):
if (mapp[j][i] == '.'):
count += 1
else:
count = 0
if (count >= k):
for t in range(j - k + 1,j + 1):
res[t][i] += 1
ress = [0,0]
maxx = -1
for i in range(n):
for j in range(n):
if (maxx < res[i][j]):
maxx = res[i][j]
ress[0],ress[1] = i + 1,j + 1
print(*ress)
``` | instruction | 0 | 28,823 | 15 | 57,646 |
Yes | output | 1 | 28,823 | 15 | 57,647 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Arkady is playing Battleship. The rules of this game aren't really important.
There is a field of n Γ n cells. There should be exactly one k-decker on the field, i. e. a ship that is k cells long oriented either horizontally or vertically. However, Arkady doesn't know where it is located. For each cell Arkady knows if it is definitely empty or can contain a part of the ship.
Consider all possible locations of the ship. Find such a cell that belongs to the maximum possible number of different locations of the ship.
Input
The first line contains two integers n and k (1 β€ k β€ n β€ 100) β the size of the field and the size of the ship.
The next n lines contain the field. Each line contains n characters, each of which is either '#' (denotes a definitely empty cell) or '.' (denotes a cell that can belong to the ship).
Output
Output two integers β the row and the column of a cell that belongs to the maximum possible number of different locations of the ship.
If there are multiple answers, output any of them. In particular, if no ship can be placed on the field, you can output any cell.
Examples
Input
4 3
#..#
#.#.
....
.###
Output
3 2
Input
10 4
#....##...
.#...#....
..#..#..#.
...#.#....
.#..##.#..
.....#...#
...#.##...
.#...#.#..
.....#..#.
...#.#...#
Output
6 1
Input
19 6
##..............###
#......#####.....##
.....#########.....
....###########....
...#############...
..###############..
.#################.
.#################.
.#################.
.#################.
#####....##....####
####............###
####............###
#####...####...####
.#####..####..#####
...###........###..
....###########....
.........##........
#.................#
Output
1 8
Note
The picture below shows the three possible locations of the ship that contain the cell (3, 2) in the first sample.
<image>
Submitted Solution:
```
n,k=map(int,input().split())
mat=[]
for i in range(n):
mat.append(input())
dp=[0]*(n)
cnt=0
ok=False
for i in range(n):
for j in range(n):
if(mat[i][j]=='.'):
dp[j]+=1
cnt+=1
else:
dp[j]=0
cnt=0
if(cnt>=k or dp[j]>=k):
ok=True
break
if(ok):
break
if(ok):
print(i+1,j+1)
``` | instruction | 0 | 28,824 | 15 | 57,648 |
No | output | 1 | 28,824 | 15 | 57,649 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Arkady is playing Battleship. The rules of this game aren't really important.
There is a field of n Γ n cells. There should be exactly one k-decker on the field, i. e. a ship that is k cells long oriented either horizontally or vertically. However, Arkady doesn't know where it is located. For each cell Arkady knows if it is definitely empty or can contain a part of the ship.
Consider all possible locations of the ship. Find such a cell that belongs to the maximum possible number of different locations of the ship.
Input
The first line contains two integers n and k (1 β€ k β€ n β€ 100) β the size of the field and the size of the ship.
The next n lines contain the field. Each line contains n characters, each of which is either '#' (denotes a definitely empty cell) or '.' (denotes a cell that can belong to the ship).
Output
Output two integers β the row and the column of a cell that belongs to the maximum possible number of different locations of the ship.
If there are multiple answers, output any of them. In particular, if no ship can be placed on the field, you can output any cell.
Examples
Input
4 3
#..#
#.#.
....
.###
Output
3 2
Input
10 4
#....##...
.#...#....
..#..#..#.
...#.#....
.#..##.#..
.....#...#
...#.##...
.#...#.#..
.....#..#.
...#.#...#
Output
6 1
Input
19 6
##..............###
#......#####.....##
.....#########.....
....###########....
...#############...
..###############..
.#################.
.#################.
.#################.
.#################.
#####....##....####
####............###
####............###
#####...####...####
.#####..####..#####
...###........###..
....###########....
.........##........
#.................#
Output
1 8
Note
The picture below shows the three possible locations of the ship that contain the cell (3, 2) in the first sample.
<image>
Submitted Solution:
```
import sys
input = sys.stdin.readline
n,k = list(map(int,input().split()))
m = [None]*n
d1 = [[0,1],[0,-1]]
d2 = [[1,0],[-1,0]]
#ans = [[0]*n for _ in range(n)]
for i in range(n):
m[i] = input()
max_num = 0
res = [0,0]
for i in range(n):
for j in range(n):
if m[i][j]!='.':
continue
tmp = 0
cnt = 1
for x in d1:
for z in range(1,k):
if 0<=i+x[0]*z<n and 0<=j+x[1]*z<n and m[i+x[0]*z][j+x[1]*z]=='.':
cnt += 1
else:
break
tmp += max(cnt-k+1,0)
#ans[i][j] += max(cnt-k+1,0)
cnt = 1
for x in d2:
for z in range(1,k):
if 0<=i+x[0]*z<n and 0<=j+x[1]*z<n and m[i+x[0]*z][j+x[1]*z]=='.':
cnt += 1
else:
break
tmp += max(cnt-k+1,0)
#ans[i][j] += max(cnt-k+1,0)
if tmp > max_num:
res = [i+1,j+1]
max_num = tmp
print(*res)
``` | instruction | 0 | 28,825 | 15 | 57,650 |
No | output | 1 | 28,825 | 15 | 57,651 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Arkady is playing Battleship. The rules of this game aren't really important.
There is a field of n Γ n cells. There should be exactly one k-decker on the field, i. e. a ship that is k cells long oriented either horizontally or vertically. However, Arkady doesn't know where it is located. For each cell Arkady knows if it is definitely empty or can contain a part of the ship.
Consider all possible locations of the ship. Find such a cell that belongs to the maximum possible number of different locations of the ship.
Input
The first line contains two integers n and k (1 β€ k β€ n β€ 100) β the size of the field and the size of the ship.
The next n lines contain the field. Each line contains n characters, each of which is either '#' (denotes a definitely empty cell) or '.' (denotes a cell that can belong to the ship).
Output
Output two integers β the row and the column of a cell that belongs to the maximum possible number of different locations of the ship.
If there are multiple answers, output any of them. In particular, if no ship can be placed on the field, you can output any cell.
Examples
Input
4 3
#..#
#.#.
....
.###
Output
3 2
Input
10 4
#....##...
.#...#....
..#..#..#.
...#.#....
.#..##.#..
.....#...#
...#.##...
.#...#.#..
.....#..#.
...#.#...#
Output
6 1
Input
19 6
##..............###
#......#####.....##
.....#########.....
....###########....
...#############...
..###############..
.#################.
.#################.
.#################.
.#################.
#####....##....####
####............###
####............###
#####...####...####
.#####..####..#####
...###........###..
....###########....
.........##........
#.................#
Output
1 8
Note
The picture below shows the three possible locations of the ship that contain the cell (3, 2) in the first sample.
<image>
Submitted Solution:
```
n, k = map(int, input().split())
s = [input() for i in range(n)]
count = [[0 for j in range(n)] for i in range(n)]
for i in range(n):
for j in range(n):
if j+k < n:
if all(list(map(lambda x: x=='.', s[i][j:j+k]))) :
#print(i,j)
#print(s[i][j:j+k])
count[i][j:j+k] = list(map(lambda x: x+1, count[i][j:j+k]))
if i+k < n:
if all(list(map(lambda x: x=='.', [s[i+m][j] for m in range(k)]))) :
for m in range(k):
count[i+m][j] += 1
m, p, max = 1, 1, -1
for i in range(n):
for j in range(n):
#print(count[i][j], end = " ")
if count[i][j]>max:
max = count[i][j]
m, p = i+1, j+1
#print()
print(m,p)
``` | instruction | 0 | 28,826 | 15 | 57,652 |
No | output | 1 | 28,826 | 15 | 57,653 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Arkady is playing Battleship. The rules of this game aren't really important.
There is a field of n Γ n cells. There should be exactly one k-decker on the field, i. e. a ship that is k cells long oriented either horizontally or vertically. However, Arkady doesn't know where it is located. For each cell Arkady knows if it is definitely empty or can contain a part of the ship.
Consider all possible locations of the ship. Find such a cell that belongs to the maximum possible number of different locations of the ship.
Input
The first line contains two integers n and k (1 β€ k β€ n β€ 100) β the size of the field and the size of the ship.
The next n lines contain the field. Each line contains n characters, each of which is either '#' (denotes a definitely empty cell) or '.' (denotes a cell that can belong to the ship).
Output
Output two integers β the row and the column of a cell that belongs to the maximum possible number of different locations of the ship.
If there are multiple answers, output any of them. In particular, if no ship can be placed on the field, you can output any cell.
Examples
Input
4 3
#..#
#.#.
....
.###
Output
3 2
Input
10 4
#....##...
.#...#....
..#..#..#.
...#.#....
.#..##.#..
.....#...#
...#.##...
.#...#.#..
.....#..#.
...#.#...#
Output
6 1
Input
19 6
##..............###
#......#####.....##
.....#########.....
....###########....
...#############...
..###############..
.#################.
.#################.
.#################.
.#################.
#####....##....####
####............###
####............###
#####...####...####
.#####..####..#####
...###........###..
....###########....
.........##........
#.................#
Output
1 8
Note
The picture below shows the three possible locations of the ship that contain the cell (3, 2) in the first sample.
<image>
Submitted Solution:
```
n, k = [int(x) for x in input().split()]
arr = list()
for k in range(0,n):
arr.append(input().strip())
def se(i,j):
global arr, n, k
x = [0,0]
pos = -1
for m in [1,-1]:
a = i
pos += 1
while 0 <= a < n:
if arr[a][j] == '.':
x[pos] += 1
else:
break
a += m
y = [0,0]
pos = -1
for m in [1,-1]:
a = j
pos += 1
while 0 <= a < n:
if arr[i][a] == '.':
y[pos] += 1
else:
break
a += m
ans = 0
def f(v,k):
if sum(v) - 1 >= k:
return min(min(v),k)
return 0
return f(x,k) + f(y,k)
ans = [0,0,0]
for i in range(0,n):
for j in range(0,n):
if arr[i][j] == '.':
t = se(i,j)
if t >= ans[2]:
ans = [i, j, t]
print(ans[0] + 1, ans[1] + 1)
``` | instruction | 0 | 28,827 | 15 | 57,654 |
No | output | 1 | 28,827 | 15 | 57,655 |
Provide tags and a correct Python 3 solution for this coding contest problem.
[3R2 as DJ Mashiro - Happiness Breeze](https://open.spotify.com/track/2qGqK8GRS65Wlf20qUBEak)
[Ice - DJ Mashiro is dead or alive](https://soundcloud.com/iceloki/dj-mashiro-is-dead-or-alive)
NEKO#Ξ¦ΟΞ¦ has just got a new maze game on her PC!
The game's main puzzle is a maze, in the forms of a 2 Γ n rectangle grid. NEKO's task is to lead a Nekomimi girl from cell (1, 1) to the gate at (2, n) and escape the maze. The girl can only move between cells sharing a common side.
However, at some moments during the game, some cells may change their state: either from normal ground to lava (which forbids movement into that cell), or vice versa (which makes that cell passable again). Initially all cells are of the ground type.
After hours of streaming, NEKO finally figured out there are only q such moments: the i-th moment toggles the state of cell (r_i, c_i) (either from ground to lava or vice versa).
Knowing this, NEKO wonders, after each of the q moments, whether it is still possible to move from cell (1, 1) to cell (2, n) without going through any lava cells.
Although NEKO is a great streamer and gamer, she still can't get through quizzes and problems requiring large amount of Brain Power. Can you help her?
Input
The first line contains integers n, q (2 β€ n β€ 10^5, 1 β€ q β€ 10^5).
The i-th of q following lines contains two integers r_i, c_i (1 β€ r_i β€ 2, 1 β€ c_i β€ n), denoting the coordinates of the cell to be flipped at the i-th moment.
It is guaranteed that cells (1, 1) and (2, n) never appear in the query list.
Output
For each moment, if it is possible to travel from cell (1, 1) to cell (2, n), print "Yes", otherwise print "No". There should be exactly q answers, one after every update.
You can print the words in any case (either lowercase, uppercase or mixed).
Example
Input
5 5
2 3
1 4
2 4
2 3
1 4
Output
Yes
No
No
No
Yes
Note
We'll crack down the example test here:
* After the first query, the girl still able to reach the goal. One of the shortest path ways should be: (1,1) β (1,2) β (1,3) β (1,4) β (1,5) β (2,5).
* After the second query, it's impossible to move to the goal, since the farthest cell she could reach is (1, 3).
* After the fourth query, the (2, 3) is not blocked, but now all the 4-th column is blocked, so she still can't reach the goal.
* After the fifth query, the column barrier has been lifted, thus she can go to the final goal again. | instruction | 0 | 29,235 | 15 | 58,470 |
Tags: data structures, dsu, implementation
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
blocked_pairs = 0
def solution(arr, r, c, n):
global blocked_pairs
delta = 1 if (not arr[r][c]) else - 1
arr[r][c] = not arr[r][c]
for dy in range(-1, 2):
if c + dy < 0 or c + dy >= n:
continue
if arr[1 - r][c + dy]:
blocked_pairs += delta
write("NO" if blocked_pairs >= 1 else "YES")
def main():
n, q = r_array()
arr = [[False] * n for _ in range(2)]
for _ in range(q):
r, c = r_array()
solution(arr, r - 1, c - 1, n)
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
def input():
return sys.stdin.readline().rstrip("\r\n")
def write(*args, end='\n'):
for x in args:
sys.stdout.write(str(x) + ' ')
sys.stdout.write(end)
def r_array():
return [int(x) for x in input().split()]
def r_int():
return int(input())
if __name__ == "__main__":
main()
``` | output | 1 | 29,235 | 15 | 58,471 |
Provide tags and a correct Python 3 solution for this coding contest problem.
[3R2 as DJ Mashiro - Happiness Breeze](https://open.spotify.com/track/2qGqK8GRS65Wlf20qUBEak)
[Ice - DJ Mashiro is dead or alive](https://soundcloud.com/iceloki/dj-mashiro-is-dead-or-alive)
NEKO#Ξ¦ΟΞ¦ has just got a new maze game on her PC!
The game's main puzzle is a maze, in the forms of a 2 Γ n rectangle grid. NEKO's task is to lead a Nekomimi girl from cell (1, 1) to the gate at (2, n) and escape the maze. The girl can only move between cells sharing a common side.
However, at some moments during the game, some cells may change their state: either from normal ground to lava (which forbids movement into that cell), or vice versa (which makes that cell passable again). Initially all cells are of the ground type.
After hours of streaming, NEKO finally figured out there are only q such moments: the i-th moment toggles the state of cell (r_i, c_i) (either from ground to lava or vice versa).
Knowing this, NEKO wonders, after each of the q moments, whether it is still possible to move from cell (1, 1) to cell (2, n) without going through any lava cells.
Although NEKO is a great streamer and gamer, she still can't get through quizzes and problems requiring large amount of Brain Power. Can you help her?
Input
The first line contains integers n, q (2 β€ n β€ 10^5, 1 β€ q β€ 10^5).
The i-th of q following lines contains two integers r_i, c_i (1 β€ r_i β€ 2, 1 β€ c_i β€ n), denoting the coordinates of the cell to be flipped at the i-th moment.
It is guaranteed that cells (1, 1) and (2, n) never appear in the query list.
Output
For each moment, if it is possible to travel from cell (1, 1) to cell (2, n), print "Yes", otherwise print "No". There should be exactly q answers, one after every update.
You can print the words in any case (either lowercase, uppercase or mixed).
Example
Input
5 5
2 3
1 4
2 4
2 3
1 4
Output
Yes
No
No
No
Yes
Note
We'll crack down the example test here:
* After the first query, the girl still able to reach the goal. One of the shortest path ways should be: (1,1) β (1,2) β (1,3) β (1,4) β (1,5) β (2,5).
* After the second query, it's impossible to move to the goal, since the farthest cell she could reach is (1, 3).
* After the fourth query, the (2, 3) is not blocked, but now all the 4-th column is blocked, so she still can't reach the goal.
* After the fifth query, the column barrier has been lifted, thus she can go to the final goal again. | instruction | 0 | 29,236 | 15 | 58,472 |
Tags: data structures, dsu, implementation
Correct Solution:
```
def is_blocked(board, c):
if board[c][0] == 1 and board[c][1] == 1:
return True
if board[c][0] == 1:
if c > 0 and board[c - 1][1] == 1:
return True
if c < len(board) - 1 and board[c + 1][1] == 1:
return True
if board[c][1] == 1:
if c > 0 and board[c - 1][0] == 1:
return True
if c < len(board) - 1 and board[c + 1][0] == 1:
return True
return False
a = list(map(int, input().split()))
n = a[0]
q = a[1]
board = []
for i in range(n):
board.append([0, 0])
blocked = set()
for i in range(q):
rc = list(map(int, input().split()))
col = rc[1] - 1
row = rc[0] - 1
if board[col][row] == 0:
board[col][row] = 1
if board[col][1 - row] == 1:
blocked.add(col)
if col > 0 and board[col - 1][1 - row] == 1:
blocked.add(col)
blocked.add(col - 1)
if col < n - 1 and board[col + 1][1 - row] == 1:
blocked.add(col)
blocked.add(col + 1)
else:
board[col][row] = 0
if not is_blocked(board, col) and col in blocked:
blocked.remove(col)
if col > 0 and col - 1 in blocked and not is_blocked(board, col - 1):
blocked.remove(col - 1)
if col < n - 1 and col + 1 in blocked and not is_blocked(board, col + 1):
blocked.remove(col + 1)
if len(blocked) == 0:
print("Yes")
else:
print("No")
``` | output | 1 | 29,236 | 15 | 58,473 |
Provide tags and a correct Python 3 solution for this coding contest problem.
[3R2 as DJ Mashiro - Happiness Breeze](https://open.spotify.com/track/2qGqK8GRS65Wlf20qUBEak)
[Ice - DJ Mashiro is dead or alive](https://soundcloud.com/iceloki/dj-mashiro-is-dead-or-alive)
NEKO#Ξ¦ΟΞ¦ has just got a new maze game on her PC!
The game's main puzzle is a maze, in the forms of a 2 Γ n rectangle grid. NEKO's task is to lead a Nekomimi girl from cell (1, 1) to the gate at (2, n) and escape the maze. The girl can only move between cells sharing a common side.
However, at some moments during the game, some cells may change their state: either from normal ground to lava (which forbids movement into that cell), or vice versa (which makes that cell passable again). Initially all cells are of the ground type.
After hours of streaming, NEKO finally figured out there are only q such moments: the i-th moment toggles the state of cell (r_i, c_i) (either from ground to lava or vice versa).
Knowing this, NEKO wonders, after each of the q moments, whether it is still possible to move from cell (1, 1) to cell (2, n) without going through any lava cells.
Although NEKO is a great streamer and gamer, she still can't get through quizzes and problems requiring large amount of Brain Power. Can you help her?
Input
The first line contains integers n, q (2 β€ n β€ 10^5, 1 β€ q β€ 10^5).
The i-th of q following lines contains two integers r_i, c_i (1 β€ r_i β€ 2, 1 β€ c_i β€ n), denoting the coordinates of the cell to be flipped at the i-th moment.
It is guaranteed that cells (1, 1) and (2, n) never appear in the query list.
Output
For each moment, if it is possible to travel from cell (1, 1) to cell (2, n), print "Yes", otherwise print "No". There should be exactly q answers, one after every update.
You can print the words in any case (either lowercase, uppercase or mixed).
Example
Input
5 5
2 3
1 4
2 4
2 3
1 4
Output
Yes
No
No
No
Yes
Note
We'll crack down the example test here:
* After the first query, the girl still able to reach the goal. One of the shortest path ways should be: (1,1) β (1,2) β (1,3) β (1,4) β (1,5) β (2,5).
* After the second query, it's impossible to move to the goal, since the farthest cell she could reach is (1, 3).
* After the fourth query, the (2, 3) is not blocked, but now all the 4-th column is blocked, so she still can't reach the goal.
* After the fifth query, the column barrier has been lifted, thus she can go to the final goal again. | instruction | 0 | 29,237 | 15 | 58,474 |
Tags: data structures, dsu, implementation
Correct Solution:
```
n,q=map(int,input().split())
ans=[[0 for i in range(n)]for i in range(2)]
bilaspur=0
for _ in range(q):
x,y=map(int,input().split())
x-=1
y-=1
t=1 if ans[x][y]==0 else -1
ans[x][y]=1-ans[x][y]
for i in range(-1,2):
if y+i>=0 and y+i<n and ans[1-x][y+i]:
bilaspur+=t
if bilaspur==0:
print("YES")
else:
print("NO")
``` | output | 1 | 29,237 | 15 | 58,475 |
Provide tags and a correct Python 3 solution for this coding contest problem.
[3R2 as DJ Mashiro - Happiness Breeze](https://open.spotify.com/track/2qGqK8GRS65Wlf20qUBEak)
[Ice - DJ Mashiro is dead or alive](https://soundcloud.com/iceloki/dj-mashiro-is-dead-or-alive)
NEKO#Ξ¦ΟΞ¦ has just got a new maze game on her PC!
The game's main puzzle is a maze, in the forms of a 2 Γ n rectangle grid. NEKO's task is to lead a Nekomimi girl from cell (1, 1) to the gate at (2, n) and escape the maze. The girl can only move between cells sharing a common side.
However, at some moments during the game, some cells may change their state: either from normal ground to lava (which forbids movement into that cell), or vice versa (which makes that cell passable again). Initially all cells are of the ground type.
After hours of streaming, NEKO finally figured out there are only q such moments: the i-th moment toggles the state of cell (r_i, c_i) (either from ground to lava or vice versa).
Knowing this, NEKO wonders, after each of the q moments, whether it is still possible to move from cell (1, 1) to cell (2, n) without going through any lava cells.
Although NEKO is a great streamer and gamer, she still can't get through quizzes and problems requiring large amount of Brain Power. Can you help her?
Input
The first line contains integers n, q (2 β€ n β€ 10^5, 1 β€ q β€ 10^5).
The i-th of q following lines contains two integers r_i, c_i (1 β€ r_i β€ 2, 1 β€ c_i β€ n), denoting the coordinates of the cell to be flipped at the i-th moment.
It is guaranteed that cells (1, 1) and (2, n) never appear in the query list.
Output
For each moment, if it is possible to travel from cell (1, 1) to cell (2, n), print "Yes", otherwise print "No". There should be exactly q answers, one after every update.
You can print the words in any case (either lowercase, uppercase or mixed).
Example
Input
5 5
2 3
1 4
2 4
2 3
1 4
Output
Yes
No
No
No
Yes
Note
We'll crack down the example test here:
* After the first query, the girl still able to reach the goal. One of the shortest path ways should be: (1,1) β (1,2) β (1,3) β (1,4) β (1,5) β (2,5).
* After the second query, it's impossible to move to the goal, since the farthest cell she could reach is (1, 3).
* After the fourth query, the (2, 3) is not blocked, but now all the 4-th column is blocked, so she still can't reach the goal.
* After the fifth query, the column barrier has been lifted, thus she can go to the final goal again. | instruction | 0 | 29,238 | 15 | 58,476 |
Tags: data structures, dsu, implementation
Correct Solution:
```
from collections import Counter
from collections import defaultdict
import math
n,q=list(map(int,input().split()))
d=defaultdict(lambda:0)
c=0
for _ in range(0,q):
i,j=list(map(int,input().split()))
if(d[(i,j)]==0):
d[(i,j)]=1
if(d[(i-1,j-1)]>0):
c=c+1
if(d[(i-1,j)]>0):
c=c+1
if(d[(i-1,j+1)]>0):
c=c+1
if(d[(i+1,j-1)]>0):
c=c+1
if(d[(i+1,j)]>0):
c=c+1
if(d[(i+1,j+1)]>0):
c=c+1
else:
d[(i,j)]=0
if(d[(i-1,j-1)]>0):
c=c-1
if(d[(i-1,j)]>0):
c=c-1
if(d[(i-1,j+1)]>0):
c=c-1
if(d[(i+1,j-1)]>0):
c=c-1
if(d[(i+1,j)]>0):
c=c-1
if(d[(i+1,j+1)]>0):
c=c-1
if(c==0):
print("Yes")
else:
print("No")
``` | output | 1 | 29,238 | 15 | 58,477 |
Provide tags and a correct Python 3 solution for this coding contest problem.
[3R2 as DJ Mashiro - Happiness Breeze](https://open.spotify.com/track/2qGqK8GRS65Wlf20qUBEak)
[Ice - DJ Mashiro is dead or alive](https://soundcloud.com/iceloki/dj-mashiro-is-dead-or-alive)
NEKO#Ξ¦ΟΞ¦ has just got a new maze game on her PC!
The game's main puzzle is a maze, in the forms of a 2 Γ n rectangle grid. NEKO's task is to lead a Nekomimi girl from cell (1, 1) to the gate at (2, n) and escape the maze. The girl can only move between cells sharing a common side.
However, at some moments during the game, some cells may change their state: either from normal ground to lava (which forbids movement into that cell), or vice versa (which makes that cell passable again). Initially all cells are of the ground type.
After hours of streaming, NEKO finally figured out there are only q such moments: the i-th moment toggles the state of cell (r_i, c_i) (either from ground to lava or vice versa).
Knowing this, NEKO wonders, after each of the q moments, whether it is still possible to move from cell (1, 1) to cell (2, n) without going through any lava cells.
Although NEKO is a great streamer and gamer, she still can't get through quizzes and problems requiring large amount of Brain Power. Can you help her?
Input
The first line contains integers n, q (2 β€ n β€ 10^5, 1 β€ q β€ 10^5).
The i-th of q following lines contains two integers r_i, c_i (1 β€ r_i β€ 2, 1 β€ c_i β€ n), denoting the coordinates of the cell to be flipped at the i-th moment.
It is guaranteed that cells (1, 1) and (2, n) never appear in the query list.
Output
For each moment, if it is possible to travel from cell (1, 1) to cell (2, n), print "Yes", otherwise print "No". There should be exactly q answers, one after every update.
You can print the words in any case (either lowercase, uppercase or mixed).
Example
Input
5 5
2 3
1 4
2 4
2 3
1 4
Output
Yes
No
No
No
Yes
Note
We'll crack down the example test here:
* After the first query, the girl still able to reach the goal. One of the shortest path ways should be: (1,1) β (1,2) β (1,3) β (1,4) β (1,5) β (2,5).
* After the second query, it's impossible to move to the goal, since the farthest cell she could reach is (1, 3).
* After the fourth query, the (2, 3) is not blocked, but now all the 4-th column is blocked, so she still can't reach the goal.
* After the fifth query, the column barrier has been lifted, thus she can go to the final goal again. | instruction | 0 | 29,239 | 15 | 58,478 |
Tags: data structures, dsu, implementation
Correct Solution:
```
n, q = map(int, input().strip().split())
nodes = [[0 for _ in range(n+2)], [0 for _ in range(n+2)]]
ans = 0
for i in range(q):
x, y = map(int, input().strip().split())
x = x - 1
pos = (x + 1) % 2
if nodes[x][y] == 1:
nodes[x][y] = 0
ans = ans - nodes[pos][y-1] - nodes[pos][y] - nodes[pos][y+1]
else:
nodes[x][y] = 1
ans = ans + nodes[pos][y-1] + nodes[pos][y] + nodes[pos][y+1]
if ans == 0:
print('Yes')
else:
print('No')
``` | output | 1 | 29,239 | 15 | 58,479 |
Provide tags and a correct Python 3 solution for this coding contest problem.
[3R2 as DJ Mashiro - Happiness Breeze](https://open.spotify.com/track/2qGqK8GRS65Wlf20qUBEak)
[Ice - DJ Mashiro is dead or alive](https://soundcloud.com/iceloki/dj-mashiro-is-dead-or-alive)
NEKO#Ξ¦ΟΞ¦ has just got a new maze game on her PC!
The game's main puzzle is a maze, in the forms of a 2 Γ n rectangle grid. NEKO's task is to lead a Nekomimi girl from cell (1, 1) to the gate at (2, n) and escape the maze. The girl can only move between cells sharing a common side.
However, at some moments during the game, some cells may change their state: either from normal ground to lava (which forbids movement into that cell), or vice versa (which makes that cell passable again). Initially all cells are of the ground type.
After hours of streaming, NEKO finally figured out there are only q such moments: the i-th moment toggles the state of cell (r_i, c_i) (either from ground to lava or vice versa).
Knowing this, NEKO wonders, after each of the q moments, whether it is still possible to move from cell (1, 1) to cell (2, n) without going through any lava cells.
Although NEKO is a great streamer and gamer, she still can't get through quizzes and problems requiring large amount of Brain Power. Can you help her?
Input
The first line contains integers n, q (2 β€ n β€ 10^5, 1 β€ q β€ 10^5).
The i-th of q following lines contains two integers r_i, c_i (1 β€ r_i β€ 2, 1 β€ c_i β€ n), denoting the coordinates of the cell to be flipped at the i-th moment.
It is guaranteed that cells (1, 1) and (2, n) never appear in the query list.
Output
For each moment, if it is possible to travel from cell (1, 1) to cell (2, n), print "Yes", otherwise print "No". There should be exactly q answers, one after every update.
You can print the words in any case (either lowercase, uppercase or mixed).
Example
Input
5 5
2 3
1 4
2 4
2 3
1 4
Output
Yes
No
No
No
Yes
Note
We'll crack down the example test here:
* After the first query, the girl still able to reach the goal. One of the shortest path ways should be: (1,1) β (1,2) β (1,3) β (1,4) β (1,5) β (2,5).
* After the second query, it's impossible to move to the goal, since the farthest cell she could reach is (1, 3).
* After the fourth query, the (2, 3) is not blocked, but now all the 4-th column is blocked, so she still can't reach the goal.
* After the fifth query, the column barrier has been lifted, thus she can go to the final goal again. | instruction | 0 | 29,240 | 15 | 58,480 |
Tags: data structures, dsu, implementation
Correct Solution:
```
n,q = map(int,input().split())
l = set()
close = set()
for i in range(q):
x,y = map(int,input().split())
# l = []
if (x,y) not in l:
l.add((x,y))
if x == 1:
if (x+1,y) in l:
close.add(((x,y),(x+1,y)))
if (x+1,y+1) in l:
close.add(((x,y),(x+1,y+1)))
if (x+1,y-1) in l:
close.add(((x,y),(x+1,y-1)))
elif x == 2:
if (x-1,y) in l:
close.add(((x-1,y),(x,y)))
if (x-1,y+1) in l:
close.add(((x-1,y+1),(x,y)))
if (x-1,y-1) in l:
close.add(((x-1,y-1),(x,y)))
else:
l.remove((x,y))
if x == 1:
if (x+1,y) in l:
close.remove(((x,y),(x+1,y)))
if (x+1,y+1) in l:
close.remove(((x,y),(x+1,y+1)))
if (x+1,y-1) in l:
close.remove(((x,y),(x+1,y-1)))
elif x == 2:
if (x-1,y) in l:
close.remove(((x-1,y),(x,y)))
if (x-1,y+1) in l:
close.remove(((x-1,y+1),(x,y)))
if (x-1,y-1) in l:
close.remove(((x-1,y-1),(x,y)))
if len(close):
print("NO")
else:
print("YES")
``` | output | 1 | 29,240 | 15 | 58,481 |
Provide tags and a correct Python 3 solution for this coding contest problem.
[3R2 as DJ Mashiro - Happiness Breeze](https://open.spotify.com/track/2qGqK8GRS65Wlf20qUBEak)
[Ice - DJ Mashiro is dead or alive](https://soundcloud.com/iceloki/dj-mashiro-is-dead-or-alive)
NEKO#Ξ¦ΟΞ¦ has just got a new maze game on her PC!
The game's main puzzle is a maze, in the forms of a 2 Γ n rectangle grid. NEKO's task is to lead a Nekomimi girl from cell (1, 1) to the gate at (2, n) and escape the maze. The girl can only move between cells sharing a common side.
However, at some moments during the game, some cells may change their state: either from normal ground to lava (which forbids movement into that cell), or vice versa (which makes that cell passable again). Initially all cells are of the ground type.
After hours of streaming, NEKO finally figured out there are only q such moments: the i-th moment toggles the state of cell (r_i, c_i) (either from ground to lava or vice versa).
Knowing this, NEKO wonders, after each of the q moments, whether it is still possible to move from cell (1, 1) to cell (2, n) without going through any lava cells.
Although NEKO is a great streamer and gamer, she still can't get through quizzes and problems requiring large amount of Brain Power. Can you help her?
Input
The first line contains integers n, q (2 β€ n β€ 10^5, 1 β€ q β€ 10^5).
The i-th of q following lines contains two integers r_i, c_i (1 β€ r_i β€ 2, 1 β€ c_i β€ n), denoting the coordinates of the cell to be flipped at the i-th moment.
It is guaranteed that cells (1, 1) and (2, n) never appear in the query list.
Output
For each moment, if it is possible to travel from cell (1, 1) to cell (2, n), print "Yes", otherwise print "No". There should be exactly q answers, one after every update.
You can print the words in any case (either lowercase, uppercase or mixed).
Example
Input
5 5
2 3
1 4
2 4
2 3
1 4
Output
Yes
No
No
No
Yes
Note
We'll crack down the example test here:
* After the first query, the girl still able to reach the goal. One of the shortest path ways should be: (1,1) β (1,2) β (1,3) β (1,4) β (1,5) β (2,5).
* After the second query, it's impossible to move to the goal, since the farthest cell she could reach is (1, 3).
* After the fourth query, the (2, 3) is not blocked, but now all the 4-th column is blocked, so she still can't reach the goal.
* After the fifth query, the column barrier has been lifted, thus she can go to the final goal again. | instruction | 0 | 29,241 | 15 | 58,482 |
Tags: data structures, dsu, implementation
Correct Solution:
```
n, q = map(int, input().split())
a = [[0 for i in range(n)] for j in range(2)]
cnt = 0
for i in range(q):
x, y = map(int, input().split())
if a[x - 1][y - 1]:
k = -1
else:
k = 1
i = 1 - x + 1
for j in range(max(0, y - 2), min(n, y + 1)):
if a[i][j] == 1:
cnt += k
a[x - 1][y - 1] = 1 - a[x - 1][y - 1]
if cnt or a[0][0] or a[1][n - 1]:
print('No')
else:
print('Yes')
``` | output | 1 | 29,241 | 15 | 58,483 |
Provide tags and a correct Python 3 solution for this coding contest problem.
[3R2 as DJ Mashiro - Happiness Breeze](https://open.spotify.com/track/2qGqK8GRS65Wlf20qUBEak)
[Ice - DJ Mashiro is dead or alive](https://soundcloud.com/iceloki/dj-mashiro-is-dead-or-alive)
NEKO#Ξ¦ΟΞ¦ has just got a new maze game on her PC!
The game's main puzzle is a maze, in the forms of a 2 Γ n rectangle grid. NEKO's task is to lead a Nekomimi girl from cell (1, 1) to the gate at (2, n) and escape the maze. The girl can only move between cells sharing a common side.
However, at some moments during the game, some cells may change their state: either from normal ground to lava (which forbids movement into that cell), or vice versa (which makes that cell passable again). Initially all cells are of the ground type.
After hours of streaming, NEKO finally figured out there are only q such moments: the i-th moment toggles the state of cell (r_i, c_i) (either from ground to lava or vice versa).
Knowing this, NEKO wonders, after each of the q moments, whether it is still possible to move from cell (1, 1) to cell (2, n) without going through any lava cells.
Although NEKO is a great streamer and gamer, she still can't get through quizzes and problems requiring large amount of Brain Power. Can you help her?
Input
The first line contains integers n, q (2 β€ n β€ 10^5, 1 β€ q β€ 10^5).
The i-th of q following lines contains two integers r_i, c_i (1 β€ r_i β€ 2, 1 β€ c_i β€ n), denoting the coordinates of the cell to be flipped at the i-th moment.
It is guaranteed that cells (1, 1) and (2, n) never appear in the query list.
Output
For each moment, if it is possible to travel from cell (1, 1) to cell (2, n), print "Yes", otherwise print "No". There should be exactly q answers, one after every update.
You can print the words in any case (either lowercase, uppercase or mixed).
Example
Input
5 5
2 3
1 4
2 4
2 3
1 4
Output
Yes
No
No
No
Yes
Note
We'll crack down the example test here:
* After the first query, the girl still able to reach the goal. One of the shortest path ways should be: (1,1) β (1,2) β (1,3) β (1,4) β (1,5) β (2,5).
* After the second query, it's impossible to move to the goal, since the farthest cell she could reach is (1, 3).
* After the fourth query, the (2, 3) is not blocked, but now all the 4-th column is blocked, so she still can't reach the goal.
* After the fifth query, the column barrier has been lifted, thus she can go to the final goal again. | instruction | 0 | 29,242 | 15 | 58,484 |
Tags: data structures, dsu, implementation
Correct Solution:
```
n,q = map(int, input().split())
G = [[0]*n for j in range(2)]
ans = 0
c = {}
for i in range(q):
x,y = map(int, input().split())
G[x-1][y-1] = 1 - G[x-1][y-1]
if G[x-1][y-1] == 1:
if y-1>=0 and G[2-x][y-1] == 1:
ans += 1
if y<n and G[2-x][y]==1:
ans +=1
if y-2>=0 and G[2-x][y-2]==1:
ans +=1
else:
if y - 1 >= 0 and G[2 - x][y - 1] == 1:
ans -= 1
if y < n and G[2 - x][y] == 1:
ans -= 1
if y - 2 >= 0 and G[2 - x][y - 2] == 1:
ans -= 1
if ans>0:
print("No")
else:
print("Yes")
``` | output | 1 | 29,242 | 15 | 58,485 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
[3R2 as DJ Mashiro - Happiness Breeze](https://open.spotify.com/track/2qGqK8GRS65Wlf20qUBEak)
[Ice - DJ Mashiro is dead or alive](https://soundcloud.com/iceloki/dj-mashiro-is-dead-or-alive)
NEKO#Ξ¦ΟΞ¦ has just got a new maze game on her PC!
The game's main puzzle is a maze, in the forms of a 2 Γ n rectangle grid. NEKO's task is to lead a Nekomimi girl from cell (1, 1) to the gate at (2, n) and escape the maze. The girl can only move between cells sharing a common side.
However, at some moments during the game, some cells may change their state: either from normal ground to lava (which forbids movement into that cell), or vice versa (which makes that cell passable again). Initially all cells are of the ground type.
After hours of streaming, NEKO finally figured out there are only q such moments: the i-th moment toggles the state of cell (r_i, c_i) (either from ground to lava or vice versa).
Knowing this, NEKO wonders, after each of the q moments, whether it is still possible to move from cell (1, 1) to cell (2, n) without going through any lava cells.
Although NEKO is a great streamer and gamer, she still can't get through quizzes and problems requiring large amount of Brain Power. Can you help her?
Input
The first line contains integers n, q (2 β€ n β€ 10^5, 1 β€ q β€ 10^5).
The i-th of q following lines contains two integers r_i, c_i (1 β€ r_i β€ 2, 1 β€ c_i β€ n), denoting the coordinates of the cell to be flipped at the i-th moment.
It is guaranteed that cells (1, 1) and (2, n) never appear in the query list.
Output
For each moment, if it is possible to travel from cell (1, 1) to cell (2, n), print "Yes", otherwise print "No". There should be exactly q answers, one after every update.
You can print the words in any case (either lowercase, uppercase or mixed).
Example
Input
5 5
2 3
1 4
2 4
2 3
1 4
Output
Yes
No
No
No
Yes
Note
We'll crack down the example test here:
* After the first query, the girl still able to reach the goal. One of the shortest path ways should be: (1,1) β (1,2) β (1,3) β (1,4) β (1,5) β (2,5).
* After the second query, it's impossible to move to the goal, since the farthest cell she could reach is (1, 3).
* After the fourth query, the (2, 3) is not blocked, but now all the 4-th column is blocked, so she still can't reach the goal.
* After the fifth query, the column barrier has been lifted, thus she can go to the final goal again.
Submitted Solution:
```
# import collections
# import random
# import math
from collections import defaultdict
# import itertools
# from sys import stdin, stdout
#import math
import sys
# import operator
# from decimal import Decimal
# sys.setrecursionlimit(10**6)
p2D = lambda x: print(*x, sep="\n")
def II(): return int(sys.stdin.buffer.readline())
def MI(): return map(int, sys.stdin.buffer.readline().split())
def LI(): return list(map(int, sys.stdin.buffer.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def BI(): return sys.stdin.buffer.readline().rstrip()
def SI(): return sys.stdin.buffer.readline().rstrip().decode()
def li(): return [int(i) for i in input().split()]
def lli(rows): return [li() for _ in range(rows)]
def si(): return input()
def ii(): return int(input())
def ins(): return input().split()
def solve():
n,q = LI()
p = 0
a = [[0]*n for _ in range(2)]
for _ in range(q):
t = LI()
if t == [2,n]: ans = "NO"
else:
a[t[0] - 1][t[1] - 1] ^= 1
sad = 1 if a[t[0] - 1][t[1] - 1] ==1 else -1
if a[2 - t[0]][t[1] - 1] == 1: p += sad
if t[1] > 1 and a[2 - t[0]][t[1] - 2] == 1: p += sad
if t[1] < n and a[2 - t[0]][t[1]] == 1: p += sad
if p>0: ans = "NO"
else: ans = "YES"
sys.stdout.write(ans + "\n")
return
def main():
#n, q = li()
#for _ in range(q):
solve()
#sys.stdout.write(str(solve()) + "\n")
# z += str(ans) + '\n'
# print(len(ans), ' '.join(map(str, ans)), sep='\n')
# stdout.write(z)
# for interactive problems
# print("? {} {}".format(l,m), flush=True)
# or print this after each print statement
# sys.stdout.flush()
if __name__ == "__main__":
main()
``` | instruction | 0 | 29,243 | 15 | 58,486 |
Yes | output | 1 | 29,243 | 15 | 58,487 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
[3R2 as DJ Mashiro - Happiness Breeze](https://open.spotify.com/track/2qGqK8GRS65Wlf20qUBEak)
[Ice - DJ Mashiro is dead or alive](https://soundcloud.com/iceloki/dj-mashiro-is-dead-or-alive)
NEKO#Ξ¦ΟΞ¦ has just got a new maze game on her PC!
The game's main puzzle is a maze, in the forms of a 2 Γ n rectangle grid. NEKO's task is to lead a Nekomimi girl from cell (1, 1) to the gate at (2, n) and escape the maze. The girl can only move between cells sharing a common side.
However, at some moments during the game, some cells may change their state: either from normal ground to lava (which forbids movement into that cell), or vice versa (which makes that cell passable again). Initially all cells are of the ground type.
After hours of streaming, NEKO finally figured out there are only q such moments: the i-th moment toggles the state of cell (r_i, c_i) (either from ground to lava or vice versa).
Knowing this, NEKO wonders, after each of the q moments, whether it is still possible to move from cell (1, 1) to cell (2, n) without going through any lava cells.
Although NEKO is a great streamer and gamer, she still can't get through quizzes and problems requiring large amount of Brain Power. Can you help her?
Input
The first line contains integers n, q (2 β€ n β€ 10^5, 1 β€ q β€ 10^5).
The i-th of q following lines contains two integers r_i, c_i (1 β€ r_i β€ 2, 1 β€ c_i β€ n), denoting the coordinates of the cell to be flipped at the i-th moment.
It is guaranteed that cells (1, 1) and (2, n) never appear in the query list.
Output
For each moment, if it is possible to travel from cell (1, 1) to cell (2, n), print "Yes", otherwise print "No". There should be exactly q answers, one after every update.
You can print the words in any case (either lowercase, uppercase or mixed).
Example
Input
5 5
2 3
1 4
2 4
2 3
1 4
Output
Yes
No
No
No
Yes
Note
We'll crack down the example test here:
* After the first query, the girl still able to reach the goal. One of the shortest path ways should be: (1,1) β (1,2) β (1,3) β (1,4) β (1,5) β (2,5).
* After the second query, it's impossible to move to the goal, since the farthest cell she could reach is (1, 3).
* After the fourth query, the (2, 3) is not blocked, but now all the 4-th column is blocked, so she still can't reach the goal.
* After the fifth query, the column barrier has been lifted, thus she can go to the final goal again.
Submitted Solution:
```
def binarySearch(arr,x):
l=0
r=len(arr)-1
while l <= r:
mid = (l + r)//2;
if arr[mid] == x:
return mid
elif arr[mid] < x:
l = mid + 1
else:
r = mid - 1
return -1
def js(arr,x):
l=0
r=len(arr)-1
ans=-1
while(l<=r):
m=(l+r)//2
if(arr[m]<=x):
ans=m
l=m+1
else:
r=m-1
return ans
def jg(arr,x):
l=0
r=len(arr)-1
ans=-1
while(l<=r):
m=(l+r)//2
if(arr[m]>=x):
ans=m
r=m-1
else:
l=m+1
return ans
from math import *
n,q = map(int,input().split())
dp=[]
dp.append([0]*n)
dp.append([0]*n)
block=0
for i in range(q):
a,b=map(int,input().split())
a-=1
b-=1
if(a==0):
if(dp[a][b]==0):
dp[a][b]=1
block+=dp[a+1][b]
if(b==0 or b<n-1):
block+=dp[a+1][b+1]
if(b==n-1 or b>0):
block+=dp[a+1][b-1]
else:
dp[a][b]=0
block-=dp[a+1][b]
if(b==0 or b<n-1):
block-=dp[a+1][b+1]
if(b==n-1 or b>0):
block-=dp[a+1][b-1]
else:
if(dp[a][b]==0):
dp[a][b]=1
block+=dp[a-1][b]
if(b==0 or b<n-1):
block+=dp[a-1][b+1]
if(b==n-1 or b>0):
block+=dp[a-1][b-1]
else:
dp[a][b]=0
block-=dp[a-1][b]
if(b==0 or b<n-1):
block-=dp[a-1][b+1]
if(b==n-1 or b>0):
block-=dp[a-1][b-1]
# print(block,dp,end=" ")
if(block>0):
print("No")
else:
print("Yes")
``` | instruction | 0 | 29,244 | 15 | 58,488 |
Yes | output | 1 | 29,244 | 15 | 58,489 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
[3R2 as DJ Mashiro - Happiness Breeze](https://open.spotify.com/track/2qGqK8GRS65Wlf20qUBEak)
[Ice - DJ Mashiro is dead or alive](https://soundcloud.com/iceloki/dj-mashiro-is-dead-or-alive)
NEKO#Ξ¦ΟΞ¦ has just got a new maze game on her PC!
The game's main puzzle is a maze, in the forms of a 2 Γ n rectangle grid. NEKO's task is to lead a Nekomimi girl from cell (1, 1) to the gate at (2, n) and escape the maze. The girl can only move between cells sharing a common side.
However, at some moments during the game, some cells may change their state: either from normal ground to lava (which forbids movement into that cell), or vice versa (which makes that cell passable again). Initially all cells are of the ground type.
After hours of streaming, NEKO finally figured out there are only q such moments: the i-th moment toggles the state of cell (r_i, c_i) (either from ground to lava or vice versa).
Knowing this, NEKO wonders, after each of the q moments, whether it is still possible to move from cell (1, 1) to cell (2, n) without going through any lava cells.
Although NEKO is a great streamer and gamer, she still can't get through quizzes and problems requiring large amount of Brain Power. Can you help her?
Input
The first line contains integers n, q (2 β€ n β€ 10^5, 1 β€ q β€ 10^5).
The i-th of q following lines contains two integers r_i, c_i (1 β€ r_i β€ 2, 1 β€ c_i β€ n), denoting the coordinates of the cell to be flipped at the i-th moment.
It is guaranteed that cells (1, 1) and (2, n) never appear in the query list.
Output
For each moment, if it is possible to travel from cell (1, 1) to cell (2, n), print "Yes", otherwise print "No". There should be exactly q answers, one after every update.
You can print the words in any case (either lowercase, uppercase or mixed).
Example
Input
5 5
2 3
1 4
2 4
2 3
1 4
Output
Yes
No
No
No
Yes
Note
We'll crack down the example test here:
* After the first query, the girl still able to reach the goal. One of the shortest path ways should be: (1,1) β (1,2) β (1,3) β (1,4) β (1,5) β (2,5).
* After the second query, it's impossible to move to the goal, since the farthest cell she could reach is (1, 3).
* After the fourth query, the (2, 3) is not blocked, but now all the 4-th column is blocked, so she still can't reach the goal.
* After the fifth query, the column barrier has been lifted, thus she can go to the final goal again.
Submitted Solution:
```
n,q = map(int,input().split())
cells = {}
bad_nei = 0
for i in range(q):
row,col = map(int,input().split())
was_forbidden = cells.get((row,col),False)
r = (row%2)+1
for c in range(col-1,col+2):
if r==row:
continue
if r==row and c==col:
continue
if not (1<=r<=2 and 1<=c<=n):
continue
if (r,c) in cells:
if was_forbidden:
bad_nei-=1
else:
bad_nei+=1
if (row,col) in cells:
del cells[row,col]
else:
cells[row,col] = True
if bad_nei>=1:
print("No")
else:
print("Yes")
``` | instruction | 0 | 29,245 | 15 | 58,490 |
Yes | output | 1 | 29,245 | 15 | 58,491 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
[3R2 as DJ Mashiro - Happiness Breeze](https://open.spotify.com/track/2qGqK8GRS65Wlf20qUBEak)
[Ice - DJ Mashiro is dead or alive](https://soundcloud.com/iceloki/dj-mashiro-is-dead-or-alive)
NEKO#Ξ¦ΟΞ¦ has just got a new maze game on her PC!
The game's main puzzle is a maze, in the forms of a 2 Γ n rectangle grid. NEKO's task is to lead a Nekomimi girl from cell (1, 1) to the gate at (2, n) and escape the maze. The girl can only move between cells sharing a common side.
However, at some moments during the game, some cells may change their state: either from normal ground to lava (which forbids movement into that cell), or vice versa (which makes that cell passable again). Initially all cells are of the ground type.
After hours of streaming, NEKO finally figured out there are only q such moments: the i-th moment toggles the state of cell (r_i, c_i) (either from ground to lava or vice versa).
Knowing this, NEKO wonders, after each of the q moments, whether it is still possible to move from cell (1, 1) to cell (2, n) without going through any lava cells.
Although NEKO is a great streamer and gamer, she still can't get through quizzes and problems requiring large amount of Brain Power. Can you help her?
Input
The first line contains integers n, q (2 β€ n β€ 10^5, 1 β€ q β€ 10^5).
The i-th of q following lines contains two integers r_i, c_i (1 β€ r_i β€ 2, 1 β€ c_i β€ n), denoting the coordinates of the cell to be flipped at the i-th moment.
It is guaranteed that cells (1, 1) and (2, n) never appear in the query list.
Output
For each moment, if it is possible to travel from cell (1, 1) to cell (2, n), print "Yes", otherwise print "No". There should be exactly q answers, one after every update.
You can print the words in any case (either lowercase, uppercase or mixed).
Example
Input
5 5
2 3
1 4
2 4
2 3
1 4
Output
Yes
No
No
No
Yes
Note
We'll crack down the example test here:
* After the first query, the girl still able to reach the goal. One of the shortest path ways should be: (1,1) β (1,2) β (1,3) β (1,4) β (1,5) β (2,5).
* After the second query, it's impossible to move to the goal, since the farthest cell she could reach is (1, 3).
* After the fourth query, the (2, 3) is not blocked, but now all the 4-th column is blocked, so she still can't reach the goal.
* After the fifth query, the column barrier has been lifted, thus she can go to the final goal again.
Submitted Solution:
```
from sys import stdin, stdout
from math import floor, gcd, fabs, factorial, fmod, sqrt, inf, log
from collections import defaultdict as dd, deque
from heapq import merge, heapify, heappop, heappush, nsmallest
from bisect import bisect_left as bl, bisect_right as br, bisect
mod = pow(10, 9) + 7
mod2 = 998244353
def inp(): return stdin.readline().strip()
def iinp(): return int(inp())
def out(var, end="\n"): stdout.write(str(var)+"\n")
def outa(*var, end="\n"): stdout.write(' '.join(map(str, var)) + end)
def lmp(): return list(mp())
def mp(): return map(int, inp().split())
def smp(): return map(str, inp().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [l1d(m, val) for j in range(n)]
def remadd(x, y): return 1 if x%y else 0
def ceil(a,b): return (a+b-1)//b
S1 = 'abcdefghijklmnopqrstuvwxyz'
S2 = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
def isprime(x):
if x<=1: return False
if x in (2, 3): return True
if x%2 == 0: return False
for i in range(3, int(sqrt(x))+1, 2):
if x%i == 0: return False
return True
n, q = mp()
md = dd(int)
ml = l2d(2, n)
prev = dd(int)
s = set()
for i in range(q):
x, y = mp()
x -= 1
y -= 1
if ml[x][y]:
ml[x][y] = 0
if x:
if y>0:
if (y-1, y) in s: s.remove((y-1, y))
if y<n-1:
if (y+1, y) in s: s.remove((y+1, y))
else:
if y>0:
if (y, y-1) in s: s.remove((y, y-1))
if y<n-1:
if (y, y+1) in s: s.remove((y, y+1))
if (y, y) in s: s.remove((y, y))
else:
ml[x][y] = 1
if x:
if y>0:
if ml[0][y-1]: s.add((y-1, y))
if y<n-1:
if ml[0][y+1]: s.add((y+1, y))
else:
if y>0:
if ml[1][y-1]: s.add((y, y-1))
if y<n-1:
if ml[1][y+1]: s.add((y, y+1))
if ml[1-x][y]: s.add((y, y))
print("No" if len(s) else "Yes")
``` | instruction | 0 | 29,246 | 15 | 58,492 |
Yes | output | 1 | 29,246 | 15 | 58,493 |
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