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.
Mikhail walks on a Cartesian plane. He starts at the point (0, 0), and in one move he can go to any of eight adjacent points. For example, if Mikhail is currently at the point (0, 0), he can go to any of the following points in one move:
* (1, 0);
* (1, 1);
* (0, 1);
* (-1, 1);
* (-1, 0);
* (-1, -1);
* (0, -1);
* (1, -1).
If Mikhail goes from the point (x1, y1) to the point (x2, y2) in one move, and x1 ≠ x2 and y1 ≠ y2, then such a move is called a diagonal move.
Mikhail has q queries. For the i-th query Mikhail's target is to go to the point (n_i, m_i) from the point (0, 0) in exactly k_i moves. Among all possible movements he want to choose one with the maximum number of diagonal moves. Your task is to find the maximum number of diagonal moves or find that it is impossible to go from the point (0, 0) to the point (n_i, m_i) in k_i moves.
Note that Mikhail can visit any point any number of times (even the destination point!).
Input
The first line of the input contains one integer q (1 ≤ q ≤ 10^4) — the number of queries.
Then q lines follow. The i-th of these q lines contains three integers n_i, m_i and k_i (1 ≤ n_i, m_i, k_i ≤ 10^{18}) — x-coordinate of the destination point of the query, y-coordinate of the destination point of the query and the number of moves in the query, correspondingly.
Output
Print q integers. The i-th integer should be equal to -1 if Mikhail cannot go from the point (0, 0) to the point (n_i, m_i) in exactly k_i moves described above. Otherwise the i-th integer should be equal to the the maximum number of diagonal moves among all possible movements.
Example
Input
3
2 2 3
4 3 7
10 1 9
Output
1
6
-1
Note
One of the possible answers to the first test case: (0, 0) → (1, 0) → (1, 1) → (2, 2).
One of the possible answers to the second test case: (0, 0) → (0, 1) → (1, 2) → (0, 3) → (1, 4) → (2, 3) → (3, 2) → (4, 3).
In the third test case Mikhail cannot reach the point (10, 1) in 9 moves.
Submitted Solution:
```
n = int(input())
for i in range(n):
x, y, k = map(lambda x: abs(int(x)), input().split(" "))
x, y = min(x, y), max(x, y)
minstep = y
ans = 0
if k < minstep:
ans = -1
else:
ans += x
k -= x
if x == y:
if k%2 == 1:
ans -= 1
k -= 1
ans += k
else:
if (y-x) % 2 == 1:
y -= 1
k -= 1
if (y-x)%2 == 1:
k -= 1
ans += k
print(ans)
``` | instruction | 0 | 27,409 | 15 | 54,818 |
No | output | 1 | 27,409 | 15 | 54,819 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mikhail walks on a Cartesian plane. He starts at the point (0, 0), and in one move he can go to any of eight adjacent points. For example, if Mikhail is currently at the point (0, 0), he can go to any of the following points in one move:
* (1, 0);
* (1, 1);
* (0, 1);
* (-1, 1);
* (-1, 0);
* (-1, -1);
* (0, -1);
* (1, -1).
If Mikhail goes from the point (x1, y1) to the point (x2, y2) in one move, and x1 ≠ x2 and y1 ≠ y2, then such a move is called a diagonal move.
Mikhail has q queries. For the i-th query Mikhail's target is to go to the point (n_i, m_i) from the point (0, 0) in exactly k_i moves. Among all possible movements he want to choose one with the maximum number of diagonal moves. Your task is to find the maximum number of diagonal moves or find that it is impossible to go from the point (0, 0) to the point (n_i, m_i) in k_i moves.
Note that Mikhail can visit any point any number of times (even the destination point!).
Input
The first line of the input contains one integer q (1 ≤ q ≤ 10^4) — the number of queries.
Then q lines follow. The i-th of these q lines contains three integers n_i, m_i and k_i (1 ≤ n_i, m_i, k_i ≤ 10^{18}) — x-coordinate of the destination point of the query, y-coordinate of the destination point of the query and the number of moves in the query, correspondingly.
Output
Print q integers. The i-th integer should be equal to -1 if Mikhail cannot go from the point (0, 0) to the point (n_i, m_i) in exactly k_i moves described above. Otherwise the i-th integer should be equal to the the maximum number of diagonal moves among all possible movements.
Example
Input
3
2 2 3
4 3 7
10 1 9
Output
1
6
-1
Note
One of the possible answers to the first test case: (0, 0) → (1, 0) → (1, 1) → (2, 2).
One of the possible answers to the second test case: (0, 0) → (0, 1) → (1, 2) → (0, 3) → (1, 4) → (2, 3) → (3, 2) → (4, 3).
In the third test case Mikhail cannot reach the point (10, 1) in 9 moves.
Submitted Solution:
```
n=int(input())
for j in range(n) :
a,b,c=map(int,input().split())
ost=min(a,b)
ost1=max(a,b)-ost
if ost+ost1>c :
print(-1)
continue
if abs(a-b)%2!=0 :
print(c-1)
else :
if c%2==0 :
print(c)
else :
print(c-2)
``` | instruction | 0 | 27,410 | 15 | 54,820 |
No | output | 1 | 27,410 | 15 | 54,821 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mikhail walks on a Cartesian plane. He starts at the point (0, 0), and in one move he can go to any of eight adjacent points. For example, if Mikhail is currently at the point (0, 0), he can go to any of the following points in one move:
* (1, 0);
* (1, 1);
* (0, 1);
* (-1, 1);
* (-1, 0);
* (-1, -1);
* (0, -1);
* (1, -1).
If Mikhail goes from the point (x1, y1) to the point (x2, y2) in one move, and x1 ≠ x2 and y1 ≠ y2, then such a move is called a diagonal move.
Mikhail has q queries. For the i-th query Mikhail's target is to go to the point (n_i, m_i) from the point (0, 0) in exactly k_i moves. Among all possible movements he want to choose one with the maximum number of diagonal moves. Your task is to find the maximum number of diagonal moves or find that it is impossible to go from the point (0, 0) to the point (n_i, m_i) in k_i moves.
Note that Mikhail can visit any point any number of times (even the destination point!).
Input
The first line of the input contains one integer q (1 ≤ q ≤ 10^4) — the number of queries.
Then q lines follow. The i-th of these q lines contains three integers n_i, m_i and k_i (1 ≤ n_i, m_i, k_i ≤ 10^{18}) — x-coordinate of the destination point of the query, y-coordinate of the destination point of the query and the number of moves in the query, correspondingly.
Output
Print q integers. The i-th integer should be equal to -1 if Mikhail cannot go from the point (0, 0) to the point (n_i, m_i) in exactly k_i moves described above. Otherwise the i-th integer should be equal to the the maximum number of diagonal moves among all possible movements.
Example
Input
3
2 2 3
4 3 7
10 1 9
Output
1
6
-1
Note
One of the possible answers to the first test case: (0, 0) → (1, 0) → (1, 1) → (2, 2).
One of the possible answers to the second test case: (0, 0) → (0, 1) → (1, 2) → (0, 3) → (1, 4) → (2, 3) → (3, 2) → (4, 3).
In the third test case Mikhail cannot reach the point (10, 1) in 9 moves.
Submitted Solution:
```
def er(a, b, c):
if (c < max(a, b)):
return -1
if ((a + b) % 2 == 1):
return (c - 1)
if ((a + b) % 2 == 0):
if ((a+b - 2*c) % 4 == 0):
return c
else:
return (c - 2)
a = int(input())
for i in range(a):
s = input().split()
print(er(int(s[0]), int(s[1]), int(s[2])))
``` | instruction | 0 | 27,411 | 15 | 54,822 |
No | output | 1 | 27,411 | 15 | 54,823 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mikhail walks on a Cartesian plane. He starts at the point (0, 0), and in one move he can go to any of eight adjacent points. For example, if Mikhail is currently at the point (0, 0), he can go to any of the following points in one move:
* (1, 0);
* (1, 1);
* (0, 1);
* (-1, 1);
* (-1, 0);
* (-1, -1);
* (0, -1);
* (1, -1).
If Mikhail goes from the point (x1, y1) to the point (x2, y2) in one move, and x1 ≠ x2 and y1 ≠ y2, then such a move is called a diagonal move.
Mikhail has q queries. For the i-th query Mikhail's target is to go to the point (n_i, m_i) from the point (0, 0) in exactly k_i moves. Among all possible movements he want to choose one with the maximum number of diagonal moves. Your task is to find the maximum number of diagonal moves or find that it is impossible to go from the point (0, 0) to the point (n_i, m_i) in k_i moves.
Note that Mikhail can visit any point any number of times (even the destination point!).
Input
The first line of the input contains one integer q (1 ≤ q ≤ 10^4) — the number of queries.
Then q lines follow. The i-th of these q lines contains three integers n_i, m_i and k_i (1 ≤ n_i, m_i, k_i ≤ 10^{18}) — x-coordinate of the destination point of the query, y-coordinate of the destination point of the query and the number of moves in the query, correspondingly.
Output
Print q integers. The i-th integer should be equal to -1 if Mikhail cannot go from the point (0, 0) to the point (n_i, m_i) in exactly k_i moves described above. Otherwise the i-th integer should be equal to the the maximum number of diagonal moves among all possible movements.
Example
Input
3
2 2 3
4 3 7
10 1 9
Output
1
6
-1
Note
One of the possible answers to the first test case: (0, 0) → (1, 0) → (1, 1) → (2, 2).
One of the possible answers to the second test case: (0, 0) → (0, 1) → (1, 2) → (0, 3) → (1, 4) → (2, 3) → (3, 2) → (4, 3).
In the third test case Mikhail cannot reach the point (10, 1) in 9 moves.
Submitted Solution:
```
def m():
[x, y, k] = [int(i) for i in input().split()]
d=min(x, y)
x-=d
y-=d
k-=d
if k-x-y<0:
print(-1)
else:
x+=y
d+=x
k-=x
if x%2==1:
d-=1
d+=k-k%2
if k%2==1:
d-=1
print(d)
n=int(input())
for i in range(n):
m()
``` | instruction | 0 | 27,412 | 15 | 54,824 |
No | output | 1 | 27,412 | 15 | 54,825 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a board of size n × n, where n is odd (not divisible by 2). Initially, each cell of the board contains one figure.
In one move, you can select exactly one figure presented in some cell and move it to one of the cells sharing a side or a corner with the current cell, i.e. from the cell (i, j) you can move the figure to cells:
* (i - 1, j - 1);
* (i - 1, j);
* (i - 1, j + 1);
* (i, j - 1);
* (i, j + 1);
* (i + 1, j - 1);
* (i + 1, j);
* (i + 1, j + 1);
Of course, you can not move figures to cells out of the board. It is allowed that after a move there will be several figures in one cell.
Your task is to find the minimum number of moves needed to get all the figures into one cell (i.e. n^2-1 cells should contain 0 figures and one cell should contain n^2 figures).
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 200) — the number of test cases. Then t test cases follow.
The only line of the test case contains one integer n (1 ≤ n < 5 ⋅ 10^5) — the size of the board. It is guaranteed that n is odd (not divisible by 2).
It is guaranteed that the sum of n over all test cases does not exceed 5 ⋅ 10^5 (∑ n ≤ 5 ⋅ 10^5).
Output
For each test case print the answer — the minimum number of moves needed to get all the figures into one cell.
Example
Input
3
1
5
499993
Output
0
40
41664916690999888 | instruction | 0 | 27,568 | 15 | 55,136 |
Tags: math
Correct Solution:
```
for _ in range(int(input())):
n=int(input())
print( (n**2-1)*n//3)
``` | output | 1 | 27,568 | 15 | 55,137 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a board of size n × n, where n is odd (not divisible by 2). Initially, each cell of the board contains one figure.
In one move, you can select exactly one figure presented in some cell and move it to one of the cells sharing a side or a corner with the current cell, i.e. from the cell (i, j) you can move the figure to cells:
* (i - 1, j - 1);
* (i - 1, j);
* (i - 1, j + 1);
* (i, j - 1);
* (i, j + 1);
* (i + 1, j - 1);
* (i + 1, j);
* (i + 1, j + 1);
Of course, you can not move figures to cells out of the board. It is allowed that after a move there will be several figures in one cell.
Your task is to find the minimum number of moves needed to get all the figures into one cell (i.e. n^2-1 cells should contain 0 figures and one cell should contain n^2 figures).
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 200) — the number of test cases. Then t test cases follow.
The only line of the test case contains one integer n (1 ≤ n < 5 ⋅ 10^5) — the size of the board. It is guaranteed that n is odd (not divisible by 2).
It is guaranteed that the sum of n over all test cases does not exceed 5 ⋅ 10^5 (∑ n ≤ 5 ⋅ 10^5).
Output
For each test case print the answer — the minimum number of moves needed to get all the figures into one cell.
Example
Input
3
1
5
499993
Output
0
40
41664916690999888 | instruction | 0 | 27,569 | 15 | 55,138 |
Tags: math
Correct Solution:
```
N = 5*(10**5)+1
dp = [0 for i in range(N)]
dp[1] = 0
for i in range(3,N,2):
dp[i] = dp[i-2]+8*(i//2)*(i//2)
t = int(input())
for y in range(t):
n = int(input())
print(dp[n])
``` | output | 1 | 27,569 | 15 | 55,139 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a board of size n × n, where n is odd (not divisible by 2). Initially, each cell of the board contains one figure.
In one move, you can select exactly one figure presented in some cell and move it to one of the cells sharing a side or a corner with the current cell, i.e. from the cell (i, j) you can move the figure to cells:
* (i - 1, j - 1);
* (i - 1, j);
* (i - 1, j + 1);
* (i, j - 1);
* (i, j + 1);
* (i + 1, j - 1);
* (i + 1, j);
* (i + 1, j + 1);
Of course, you can not move figures to cells out of the board. It is allowed that after a move there will be several figures in one cell.
Your task is to find the minimum number of moves needed to get all the figures into one cell (i.e. n^2-1 cells should contain 0 figures and one cell should contain n^2 figures).
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 200) — the number of test cases. Then t test cases follow.
The only line of the test case contains one integer n (1 ≤ n < 5 ⋅ 10^5) — the size of the board. It is guaranteed that n is odd (not divisible by 2).
It is guaranteed that the sum of n over all test cases does not exceed 5 ⋅ 10^5 (∑ n ≤ 5 ⋅ 10^5).
Output
For each test case print the answer — the minimum number of moves needed to get all the figures into one cell.
Example
Input
3
1
5
499993
Output
0
40
41664916690999888 | instruction | 0 | 27,570 | 15 | 55,140 |
Tags: math
Correct Solution:
```
t = int(input())
for z in range(t):
n = int(input())
a = 3
b = 1
s = 0
for i in range(1,n//2+1):
s += i*(a+b)
a = a+2
b = b+2
print(2*s)
``` | output | 1 | 27,570 | 15 | 55,141 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a board of size n × n, where n is odd (not divisible by 2). Initially, each cell of the board contains one figure.
In one move, you can select exactly one figure presented in some cell and move it to one of the cells sharing a side or a corner with the current cell, i.e. from the cell (i, j) you can move the figure to cells:
* (i - 1, j - 1);
* (i - 1, j);
* (i - 1, j + 1);
* (i, j - 1);
* (i, j + 1);
* (i + 1, j - 1);
* (i + 1, j);
* (i + 1, j + 1);
Of course, you can not move figures to cells out of the board. It is allowed that after a move there will be several figures in one cell.
Your task is to find the minimum number of moves needed to get all the figures into one cell (i.e. n^2-1 cells should contain 0 figures and one cell should contain n^2 figures).
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 200) — the number of test cases. Then t test cases follow.
The only line of the test case contains one integer n (1 ≤ n < 5 ⋅ 10^5) — the size of the board. It is guaranteed that n is odd (not divisible by 2).
It is guaranteed that the sum of n over all test cases does not exceed 5 ⋅ 10^5 (∑ n ≤ 5 ⋅ 10^5).
Output
For each test case print the answer — the minimum number of moves needed to get all the figures into one cell.
Example
Input
3
1
5
499993
Output
0
40
41664916690999888 | instruction | 0 | 27,571 | 15 | 55,142 |
Tags: math
Correct Solution:
```
t=int(input());
while t>0:
n=int(input());
sum=0;
for i in range(0,int(n/2)):
sum += 8*(i+1)*(i+1);
print(sum);
t=t-1;
``` | output | 1 | 27,571 | 15 | 55,143 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a board of size n × n, where n is odd (not divisible by 2). Initially, each cell of the board contains one figure.
In one move, you can select exactly one figure presented in some cell and move it to one of the cells sharing a side or a corner with the current cell, i.e. from the cell (i, j) you can move the figure to cells:
* (i - 1, j - 1);
* (i - 1, j);
* (i - 1, j + 1);
* (i, j - 1);
* (i, j + 1);
* (i + 1, j - 1);
* (i + 1, j);
* (i + 1, j + 1);
Of course, you can not move figures to cells out of the board. It is allowed that after a move there will be several figures in one cell.
Your task is to find the minimum number of moves needed to get all the figures into one cell (i.e. n^2-1 cells should contain 0 figures and one cell should contain n^2 figures).
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 200) — the number of test cases. Then t test cases follow.
The only line of the test case contains one integer n (1 ≤ n < 5 ⋅ 10^5) — the size of the board. It is guaranteed that n is odd (not divisible by 2).
It is guaranteed that the sum of n over all test cases does not exceed 5 ⋅ 10^5 (∑ n ≤ 5 ⋅ 10^5).
Output
For each test case print the answer — the minimum number of moves needed to get all the figures into one cell.
Example
Input
3
1
5
499993
Output
0
40
41664916690999888 | instruction | 0 | 27,572 | 15 | 55,144 |
Tags: math
Correct Solution:
```
import sys
input = sys.stdin.readline
t = int(input())
for _ in range(t):
n = int(input())
ans = 0
for i in range(n//2+1):
ans+=i*4*2*i
print(ans)
``` | output | 1 | 27,572 | 15 | 55,145 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a board of size n × n, where n is odd (not divisible by 2). Initially, each cell of the board contains one figure.
In one move, you can select exactly one figure presented in some cell and move it to one of the cells sharing a side or a corner with the current cell, i.e. from the cell (i, j) you can move the figure to cells:
* (i - 1, j - 1);
* (i - 1, j);
* (i - 1, j + 1);
* (i, j - 1);
* (i, j + 1);
* (i + 1, j - 1);
* (i + 1, j);
* (i + 1, j + 1);
Of course, you can not move figures to cells out of the board. It is allowed that after a move there will be several figures in one cell.
Your task is to find the minimum number of moves needed to get all the figures into one cell (i.e. n^2-1 cells should contain 0 figures and one cell should contain n^2 figures).
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 200) — the number of test cases. Then t test cases follow.
The only line of the test case contains one integer n (1 ≤ n < 5 ⋅ 10^5) — the size of the board. It is guaranteed that n is odd (not divisible by 2).
It is guaranteed that the sum of n over all test cases does not exceed 5 ⋅ 10^5 (∑ n ≤ 5 ⋅ 10^5).
Output
For each test case print the answer — the minimum number of moves needed to get all the figures into one cell.
Example
Input
3
1
5
499993
Output
0
40
41664916690999888 | instruction | 0 | 27,573 | 15 | 55,146 |
Tags: math
Correct Solution:
```
t = int(input())
while(t):
n=int(input())
ans = 0
ck= n//2+1
for i in range(1,ck+1):
num = 4*(n-2*(i-1))-4
num*=(abs(i-n//2-1))*1
ans+=num
print(ans)
t-=1
``` | output | 1 | 27,573 | 15 | 55,147 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a board of size n × n, where n is odd (not divisible by 2). Initially, each cell of the board contains one figure.
In one move, you can select exactly one figure presented in some cell and move it to one of the cells sharing a side or a corner with the current cell, i.e. from the cell (i, j) you can move the figure to cells:
* (i - 1, j - 1);
* (i - 1, j);
* (i - 1, j + 1);
* (i, j - 1);
* (i, j + 1);
* (i + 1, j - 1);
* (i + 1, j);
* (i + 1, j + 1);
Of course, you can not move figures to cells out of the board. It is allowed that after a move there will be several figures in one cell.
Your task is to find the minimum number of moves needed to get all the figures into one cell (i.e. n^2-1 cells should contain 0 figures and one cell should contain n^2 figures).
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 200) — the number of test cases. Then t test cases follow.
The only line of the test case contains one integer n (1 ≤ n < 5 ⋅ 10^5) — the size of the board. It is guaranteed that n is odd (not divisible by 2).
It is guaranteed that the sum of n over all test cases does not exceed 5 ⋅ 10^5 (∑ n ≤ 5 ⋅ 10^5).
Output
For each test case print the answer — the minimum number of moves needed to get all the figures into one cell.
Example
Input
3
1
5
499993
Output
0
40
41664916690999888 | instruction | 0 | 27,574 | 15 | 55,148 |
Tags: math
Correct Solution:
```
t=int(input())
while(t):
i=1
n=int(input())
sum=0
while(i<n):
sum = sum + ((i + 2) * (i + 2) - (i * i)) * ((i + 1) / 2)
i=i+2
t-=1
print('%d' % sum)
``` | output | 1 | 27,574 | 15 | 55,149 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a board of size n × n, where n is odd (not divisible by 2). Initially, each cell of the board contains one figure.
In one move, you can select exactly one figure presented in some cell and move it to one of the cells sharing a side or a corner with the current cell, i.e. from the cell (i, j) you can move the figure to cells:
* (i - 1, j - 1);
* (i - 1, j);
* (i - 1, j + 1);
* (i, j - 1);
* (i, j + 1);
* (i + 1, j - 1);
* (i + 1, j);
* (i + 1, j + 1);
Of course, you can not move figures to cells out of the board. It is allowed that after a move there will be several figures in one cell.
Your task is to find the minimum number of moves needed to get all the figures into one cell (i.e. n^2-1 cells should contain 0 figures and one cell should contain n^2 figures).
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 200) — the number of test cases. Then t test cases follow.
The only line of the test case contains one integer n (1 ≤ n < 5 ⋅ 10^5) — the size of the board. It is guaranteed that n is odd (not divisible by 2).
It is guaranteed that the sum of n over all test cases does not exceed 5 ⋅ 10^5 (∑ n ≤ 5 ⋅ 10^5).
Output
For each test case print the answer — the minimum number of moves needed to get all the figures into one cell.
Example
Input
3
1
5
499993
Output
0
40
41664916690999888 | instruction | 0 | 27,575 | 15 | 55,150 |
Tags: math
Correct Solution:
```
t = int(input())
for _ in range(t):
n = int(input())
dp = [0, 0]
for i in range(2, n+1):
if i%2 == 0:
dp.append(0)
else:
dp.append((4*i-4)*(i//2)+dp[-2])
print(dp[-1])
``` | output | 1 | 27,575 | 15 | 55,151 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a board of size n × n, where n is odd (not divisible by 2). Initially, each cell of the board contains one figure.
In one move, you can select exactly one figure presented in some cell and move it to one of the cells sharing a side or a corner with the current cell, i.e. from the cell (i, j) you can move the figure to cells:
* (i - 1, j - 1);
* (i - 1, j);
* (i - 1, j + 1);
* (i, j - 1);
* (i, j + 1);
* (i + 1, j - 1);
* (i + 1, j);
* (i + 1, j + 1);
Of course, you can not move figures to cells out of the board. It is allowed that after a move there will be several figures in one cell.
Your task is to find the minimum number of moves needed to get all the figures into one cell (i.e. n^2-1 cells should contain 0 figures and one cell should contain n^2 figures).
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 200) — the number of test cases. Then t test cases follow.
The only line of the test case contains one integer n (1 ≤ n < 5 ⋅ 10^5) — the size of the board. It is guaranteed that n is odd (not divisible by 2).
It is guaranteed that the sum of n over all test cases does not exceed 5 ⋅ 10^5 (∑ n ≤ 5 ⋅ 10^5).
Output
For each test case print the answer — the minimum number of moves needed to get all the figures into one cell.
Example
Input
3
1
5
499993
Output
0
40
41664916690999888
Submitted Solution:
```
t = int(input())
for ii in range(0, t):
n = int(input())
summa = 0
s_n = 0
k_n = 1
while k_n <= n:
summa = summa + 4 * (k_n - 1) * s_n
k_n += 2
s_n += 1
print(summa)
``` | instruction | 0 | 27,576 | 15 | 55,152 |
Yes | output | 1 | 27,576 | 15 | 55,153 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a board of size n × n, where n is odd (not divisible by 2). Initially, each cell of the board contains one figure.
In one move, you can select exactly one figure presented in some cell and move it to one of the cells sharing a side or a corner with the current cell, i.e. from the cell (i, j) you can move the figure to cells:
* (i - 1, j - 1);
* (i - 1, j);
* (i - 1, j + 1);
* (i, j - 1);
* (i, j + 1);
* (i + 1, j - 1);
* (i + 1, j);
* (i + 1, j + 1);
Of course, you can not move figures to cells out of the board. It is allowed that after a move there will be several figures in one cell.
Your task is to find the minimum number of moves needed to get all the figures into one cell (i.e. n^2-1 cells should contain 0 figures and one cell should contain n^2 figures).
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 200) — the number of test cases. Then t test cases follow.
The only line of the test case contains one integer n (1 ≤ n < 5 ⋅ 10^5) — the size of the board. It is guaranteed that n is odd (not divisible by 2).
It is guaranteed that the sum of n over all test cases does not exceed 5 ⋅ 10^5 (∑ n ≤ 5 ⋅ 10^5).
Output
For each test case print the answer — the minimum number of moves needed to get all the figures into one cell.
Example
Input
3
1
5
499993
Output
0
40
41664916690999888
Submitted Solution:
```
t=int(input())
while t:
n=int(input())
l=(n-1)//2
res=0
if n==1:
print(0)
else:
while l>0:
res+=(n+n-2)*l*2
n-=2
l-=1
print(res)
t-=1
``` | instruction | 0 | 27,577 | 15 | 55,154 |
Yes | output | 1 | 27,577 | 15 | 55,155 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a board of size n × n, where n is odd (not divisible by 2). Initially, each cell of the board contains one figure.
In one move, you can select exactly one figure presented in some cell and move it to one of the cells sharing a side or a corner with the current cell, i.e. from the cell (i, j) you can move the figure to cells:
* (i - 1, j - 1);
* (i - 1, j);
* (i - 1, j + 1);
* (i, j - 1);
* (i, j + 1);
* (i + 1, j - 1);
* (i + 1, j);
* (i + 1, j + 1);
Of course, you can not move figures to cells out of the board. It is allowed that after a move there will be several figures in one cell.
Your task is to find the minimum number of moves needed to get all the figures into one cell (i.e. n^2-1 cells should contain 0 figures and one cell should contain n^2 figures).
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 200) — the number of test cases. Then t test cases follow.
The only line of the test case contains one integer n (1 ≤ n < 5 ⋅ 10^5) — the size of the board. It is guaranteed that n is odd (not divisible by 2).
It is guaranteed that the sum of n over all test cases does not exceed 5 ⋅ 10^5 (∑ n ≤ 5 ⋅ 10^5).
Output
For each test case print the answer — the minimum number of moves needed to get all the figures into one cell.
Example
Input
3
1
5
499993
Output
0
40
41664916690999888
Submitted Solution:
```
h = lambda x: 8 * x * (x - 1) * (x - 2) // 3 + 12 * x * (x - 1) + 8 * x
def solve():
n = int(input())
print(h(n // 2))
t = int(input())
for _ in range(t):
solve()
``` | instruction | 0 | 27,578 | 15 | 55,156 |
Yes | output | 1 | 27,578 | 15 | 55,157 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a board of size n × n, where n is odd (not divisible by 2). Initially, each cell of the board contains one figure.
In one move, you can select exactly one figure presented in some cell and move it to one of the cells sharing a side or a corner with the current cell, i.e. from the cell (i, j) you can move the figure to cells:
* (i - 1, j - 1);
* (i - 1, j);
* (i - 1, j + 1);
* (i, j - 1);
* (i, j + 1);
* (i + 1, j - 1);
* (i + 1, j);
* (i + 1, j + 1);
Of course, you can not move figures to cells out of the board. It is allowed that after a move there will be several figures in one cell.
Your task is to find the minimum number of moves needed to get all the figures into one cell (i.e. n^2-1 cells should contain 0 figures and one cell should contain n^2 figures).
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 200) — the number of test cases. Then t test cases follow.
The only line of the test case contains one integer n (1 ≤ n < 5 ⋅ 10^5) — the size of the board. It is guaranteed that n is odd (not divisible by 2).
It is guaranteed that the sum of n over all test cases does not exceed 5 ⋅ 10^5 (∑ n ≤ 5 ⋅ 10^5).
Output
For each test case print the answer — the minimum number of moves needed to get all the figures into one cell.
Example
Input
3
1
5
499993
Output
0
40
41664916690999888
Submitted Solution:
```
for z in range(int(input())):
n=int(input())
sum=0
for i in range(1,n+1,2):
sum+=int(i//2*(4*i-4))
print(sum)
``` | instruction | 0 | 27,579 | 15 | 55,158 |
Yes | output | 1 | 27,579 | 15 | 55,159 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a board of size n × n, where n is odd (not divisible by 2). Initially, each cell of the board contains one figure.
In one move, you can select exactly one figure presented in some cell and move it to one of the cells sharing a side or a corner with the current cell, i.e. from the cell (i, j) you can move the figure to cells:
* (i - 1, j - 1);
* (i - 1, j);
* (i - 1, j + 1);
* (i, j - 1);
* (i, j + 1);
* (i + 1, j - 1);
* (i + 1, j);
* (i + 1, j + 1);
Of course, you can not move figures to cells out of the board. It is allowed that after a move there will be several figures in one cell.
Your task is to find the minimum number of moves needed to get all the figures into one cell (i.e. n^2-1 cells should contain 0 figures and one cell should contain n^2 figures).
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 200) — the number of test cases. Then t test cases follow.
The only line of the test case contains one integer n (1 ≤ n < 5 ⋅ 10^5) — the size of the board. It is guaranteed that n is odd (not divisible by 2).
It is guaranteed that the sum of n over all test cases does not exceed 5 ⋅ 10^5 (∑ n ≤ 5 ⋅ 10^5).
Output
For each test case print the answer — the minimum number of moves needed to get all the figures into one cell.
Example
Input
3
1
5
499993
Output
0
40
41664916690999888
Submitted Solution:
```
t = int(input())
for x in range(t):
n = int(input())
n = n-1
ans = int((n*(n+1)/2)*4)
print(ans)
``` | instruction | 0 | 27,580 | 15 | 55,160 |
No | output | 1 | 27,580 | 15 | 55,161 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a board of size n × n, where n is odd (not divisible by 2). Initially, each cell of the board contains one figure.
In one move, you can select exactly one figure presented in some cell and move it to one of the cells sharing a side or a corner with the current cell, i.e. from the cell (i, j) you can move the figure to cells:
* (i - 1, j - 1);
* (i - 1, j);
* (i - 1, j + 1);
* (i, j - 1);
* (i, j + 1);
* (i + 1, j - 1);
* (i + 1, j);
* (i + 1, j + 1);
Of course, you can not move figures to cells out of the board. It is allowed that after a move there will be several figures in one cell.
Your task is to find the minimum number of moves needed to get all the figures into one cell (i.e. n^2-1 cells should contain 0 figures and one cell should contain n^2 figures).
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 200) — the number of test cases. Then t test cases follow.
The only line of the test case contains one integer n (1 ≤ n < 5 ⋅ 10^5) — the size of the board. It is guaranteed that n is odd (not divisible by 2).
It is guaranteed that the sum of n over all test cases does not exceed 5 ⋅ 10^5 (∑ n ≤ 5 ⋅ 10^5).
Output
For each test case print the answer — the minimum number of moves needed to get all the figures into one cell.
Example
Input
3
1
5
499993
Output
0
40
41664916690999888
Submitted Solution:
```
n=int(input())
l=[0]*n
for i in range(n):
l[i]=int(input())
ans=0
for i in range(n):
for j in range((l[i]+1)//2):
ans+=j*((2*j+1)**2 - (2*j-1)**2)
print(ans)
``` | instruction | 0 | 27,581 | 15 | 55,162 |
No | output | 1 | 27,581 | 15 | 55,163 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a board of size n × n, where n is odd (not divisible by 2). Initially, each cell of the board contains one figure.
In one move, you can select exactly one figure presented in some cell and move it to one of the cells sharing a side or a corner with the current cell, i.e. from the cell (i, j) you can move the figure to cells:
* (i - 1, j - 1);
* (i - 1, j);
* (i - 1, j + 1);
* (i, j - 1);
* (i, j + 1);
* (i + 1, j - 1);
* (i + 1, j);
* (i + 1, j + 1);
Of course, you can not move figures to cells out of the board. It is allowed that after a move there will be several figures in one cell.
Your task is to find the minimum number of moves needed to get all the figures into one cell (i.e. n^2-1 cells should contain 0 figures and one cell should contain n^2 figures).
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 200) — the number of test cases. Then t test cases follow.
The only line of the test case contains one integer n (1 ≤ n < 5 ⋅ 10^5) — the size of the board. It is guaranteed that n is odd (not divisible by 2).
It is guaranteed that the sum of n over all test cases does not exceed 5 ⋅ 10^5 (∑ n ≤ 5 ⋅ 10^5).
Output
For each test case print the answer — the minimum number of moves needed to get all the figures into one cell.
Example
Input
3
1
5
499993
Output
0
40
41664916690999888
Submitted Solution:
```
t = int(input())
for _ in range(t):
n = int(input())
if n == 3:
print(8)
else:
res = 0
start = 8
for i in range(1, n-2):
res += start * i
start += 8
print(res)
``` | instruction | 0 | 27,582 | 15 | 55,164 |
No | output | 1 | 27,582 | 15 | 55,165 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a board of size n × n, where n is odd (not divisible by 2). Initially, each cell of the board contains one figure.
In one move, you can select exactly one figure presented in some cell and move it to one of the cells sharing a side or a corner with the current cell, i.e. from the cell (i, j) you can move the figure to cells:
* (i - 1, j - 1);
* (i - 1, j);
* (i - 1, j + 1);
* (i, j - 1);
* (i, j + 1);
* (i + 1, j - 1);
* (i + 1, j);
* (i + 1, j + 1);
Of course, you can not move figures to cells out of the board. It is allowed that after a move there will be several figures in one cell.
Your task is to find the minimum number of moves needed to get all the figures into one cell (i.e. n^2-1 cells should contain 0 figures and one cell should contain n^2 figures).
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 200) — the number of test cases. Then t test cases follow.
The only line of the test case contains one integer n (1 ≤ n < 5 ⋅ 10^5) — the size of the board. It is guaranteed that n is odd (not divisible by 2).
It is guaranteed that the sum of n over all test cases does not exceed 5 ⋅ 10^5 (∑ n ≤ 5 ⋅ 10^5).
Output
For each test case print the answer — the minimum number of moves needed to get all the figures into one cell.
Example
Input
3
1
5
499993
Output
0
40
41664916690999888
Submitted Solution:
```
testcases = int(input())
while(testcases>0):
testcases = testcases - 1
n = int(input())
print((n*(n+1)*(2*n+1)*8)//6)
``` | instruction | 0 | 27,583 | 15 | 55,166 |
No | output | 1 | 27,583 | 15 | 55,167 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Artem is building a new robot. He has a matrix a consisting of n rows and m columns. The cell located on the i-th row from the top and the j-th column from the left has a value a_{i,j} written in it.
If two adjacent cells contain the same value, the robot will break. A matrix is called good if no two adjacent cells contain the same value, where two cells are called adjacent if they share a side.
Artem wants to increment the values in some cells by one to make a good.
More formally, find a good matrix b that satisfies the following condition —
* For all valid (i,j), either b_{i,j} = a_{i,j} or b_{i,j} = a_{i,j}+1.
For the constraints of this problem, it can be shown that such a matrix b always exists. If there are several such tables, you can output any of them. Please note that you do not have to minimize the number of increments.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10). Description of the test cases follows.
The first line of each test case contains two integers n, m (1 ≤ n ≤ 100, 1 ≤ m ≤ 100) — the number of rows and columns, respectively.
The following n lines each contain m integers. The j-th integer in the i-th line is a_{i,j} (1 ≤ a_{i,j} ≤ 10^9).
Output
For each case, output n lines each containing m integers. The j-th integer in the i-th line is b_{i,j}.
Example
Input
3
3 2
1 2
4 5
7 8
2 2
1 1
3 3
2 2
1 3
2 2
Output
1 2
5 6
7 8
2 1
4 3
2 4
3 2
Note
In all the cases, you can verify that no two adjacent cells have the same value and that b is the same as a with some values incremented by one. | instruction | 0 | 27,633 | 15 | 55,266 |
Tags: 2-sat, chinese remainder theorem, constructive algorithms, fft, flows
Correct Solution:
```
for _ in range(int(input())):
n, m = map(int, input().split())
mat = []
for i in range(n):
row = list(map(int, input().split()))
mat.append(row)
for i in range(n):
for j in range(m):
if ((i%2==0 and j%2==0) or (i%2==1 and j%2==1)) and mat[i][j]%2==1:
mat[i][j]+=1
elif ((i%2==1 and j%2==0) or (i%2==0 and j%2==1)) and mat[i][j]%2==0:
mat[i][j]+=1
for i in range(n):
print(*(mat[i]))
``` | output | 1 | 27,633 | 15 | 55,267 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Artem is building a new robot. He has a matrix a consisting of n rows and m columns. The cell located on the i-th row from the top and the j-th column from the left has a value a_{i,j} written in it.
If two adjacent cells contain the same value, the robot will break. A matrix is called good if no two adjacent cells contain the same value, where two cells are called adjacent if they share a side.
Artem wants to increment the values in some cells by one to make a good.
More formally, find a good matrix b that satisfies the following condition —
* For all valid (i,j), either b_{i,j} = a_{i,j} or b_{i,j} = a_{i,j}+1.
For the constraints of this problem, it can be shown that such a matrix b always exists. If there are several such tables, you can output any of them. Please note that you do not have to minimize the number of increments.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10). Description of the test cases follows.
The first line of each test case contains two integers n, m (1 ≤ n ≤ 100, 1 ≤ m ≤ 100) — the number of rows and columns, respectively.
The following n lines each contain m integers. The j-th integer in the i-th line is a_{i,j} (1 ≤ a_{i,j} ≤ 10^9).
Output
For each case, output n lines each containing m integers. The j-th integer in the i-th line is b_{i,j}.
Example
Input
3
3 2
1 2
4 5
7 8
2 2
1 1
3 3
2 2
1 3
2 2
Output
1 2
5 6
7 8
2 1
4 3
2 4
3 2
Note
In all the cases, you can verify that no two adjacent cells have the same value and that b is the same as a with some values incremented by one. | instruction | 0 | 27,634 | 15 | 55,268 |
Tags: 2-sat, chinese remainder theorem, constructive algorithms, fft, flows
Correct Solution:
```
import time,math as mt,bisect as bs,sys
from sys import stdin,stdout
from collections import deque
from collections import defaultdict
from fractions import Fraction
from collections import Counter
from collections import OrderedDict
pi=3.14159265358979323846264338327950
def II(): # to take integer input
return int(stdin.readline())
def IP(): # to take tuple as input
return map(int,stdin.readline().split())
def L(): # to take list as input
return list(map(int,stdin.readline().split()))
def P(x): # to print integer,list,string etc..
return stdout.write(str(x)+"\n")
def PI(x,y): # to print tuple separatedly
return stdout.write(str(x)+" "+str(y)+"\n")
def lcm(a,b): # to calculate lcm
return (a*b)//gcd(a,b)
def gcd(a,b): # to calculate gcd
if a==0:
return b
elif b==0:
return a
if a>b:
return gcd(a%b,b)
else:
return gcd(a,b%a)
def bfs(adj,v): # a schema of bfs
visited=[False]*(v+1)
q=deque()
while q:
pass
def sieve():
li=[True]*(2*(10**5)+5)
li[0],li[1]=False,False
for i in range(2,len(li),1):
if li[i]==True:
for j in range(i*i,len(li),i):
li[j]=False
prime=[]
for i in range((2*(10**5)+5)):
if li[i]==True:
prime.append(i)
return prime
def setBit(n):
count=0
while n!=0:
n=n&(n-1)
count+=1
return count
mx=10**7
spf=[mx]*(mx+1)
def SPF():
spf[1]=1
for i in range(2,mx+1):
if spf[i]==mx:
spf[i]=i
for j in range(i*i,mx+1,i):
if i<spf[j]:
spf[j]=i
return
def readTree(n,e): # to read tree
adj=[set() for i in range(n+1)]
for i in range(e):
u1,u2=IP()
adj[u1].add(u2)
return adj
#####################################################################################
mod=10**9+7
def solve():
n,m=IP()
li=[]
for i in range(n):
temp=L()
li.append(temp)
for i in range(n):
for j in range(m):
if (i+j)%2 and (li[i][j]%2==0):
li[i][j]+=1
elif (i+j)%2==0 and li[i][j]%2==1:
li[i][j]+=1
for ele in li:
print(*ele)
t=II()
for i in range(t):
solve()
#######
#
#
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# `````````1¶¶¶¶¶¶¶¶¶¶¶0¶¶¶00000000000000000¶11___1_`````
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# ``````````¶¶¶¶¶¶¶¶0¶¶¶¶¶¶¶¶¶0000000000000000__`````11``
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# ```````0¶¶¶¶¶¶¶¶¶000000000000000000000000000¶01___`````
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# ``````0¶¶0¶¶¶0¶¶0000000000000000000000000000¶01`1_11_``
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# ``````¶¶¶¶0¶¶¶¶00000000000000000000000000000001```_1_11
``` | output | 1 | 27,634 | 15 | 55,269 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Artem is building a new robot. He has a matrix a consisting of n rows and m columns. The cell located on the i-th row from the top and the j-th column from the left has a value a_{i,j} written in it.
If two adjacent cells contain the same value, the robot will break. A matrix is called good if no two adjacent cells contain the same value, where two cells are called adjacent if they share a side.
Artem wants to increment the values in some cells by one to make a good.
More formally, find a good matrix b that satisfies the following condition —
* For all valid (i,j), either b_{i,j} = a_{i,j} or b_{i,j} = a_{i,j}+1.
For the constraints of this problem, it can be shown that such a matrix b always exists. If there are several such tables, you can output any of them. Please note that you do not have to minimize the number of increments.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10). Description of the test cases follows.
The first line of each test case contains two integers n, m (1 ≤ n ≤ 100, 1 ≤ m ≤ 100) — the number of rows and columns, respectively.
The following n lines each contain m integers. The j-th integer in the i-th line is a_{i,j} (1 ≤ a_{i,j} ≤ 10^9).
Output
For each case, output n lines each containing m integers. The j-th integer in the i-th line is b_{i,j}.
Example
Input
3
3 2
1 2
4 5
7 8
2 2
1 1
3 3
2 2
1 3
2 2
Output
1 2
5 6
7 8
2 1
4 3
2 4
3 2
Note
In all the cases, you can verify that no two adjacent cells have the same value and that b is the same as a with some values incremented by one. | instruction | 0 | 27,635 | 15 | 55,270 |
Tags: 2-sat, chinese remainder theorem, constructive algorithms, fft, flows
Correct Solution:
```
import sys
def input():
return sys.stdin.readline()
dif = [(-1, 0), (1, 0), (0, 1), (0, -1)]
def solve():
m, n = map(int, input().split())
matrix = [list(map(int, input().split())) for i in range(m)]
for i in range(m):
for j in range(n):
which = (i+j) % 2
if which == 0:
if matrix[i][j] % 2 != 0:
matrix[i][j] += 1
else:
if matrix[i][j] % 2 == 0:
matrix[i][j] += 1
for i in range(m):
print(*matrix[i])
return
def main():
t = int(input())
for i in range(t):
solve()
return
if __name__ == "__main__":
main()
``` | output | 1 | 27,635 | 15 | 55,271 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Artem is building a new robot. He has a matrix a consisting of n rows and m columns. The cell located on the i-th row from the top and the j-th column from the left has a value a_{i,j} written in it.
If two adjacent cells contain the same value, the robot will break. A matrix is called good if no two adjacent cells contain the same value, where two cells are called adjacent if they share a side.
Artem wants to increment the values in some cells by one to make a good.
More formally, find a good matrix b that satisfies the following condition —
* For all valid (i,j), either b_{i,j} = a_{i,j} or b_{i,j} = a_{i,j}+1.
For the constraints of this problem, it can be shown that such a matrix b always exists. If there are several such tables, you can output any of them. Please note that you do not have to minimize the number of increments.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10). Description of the test cases follows.
The first line of each test case contains two integers n, m (1 ≤ n ≤ 100, 1 ≤ m ≤ 100) — the number of rows and columns, respectively.
The following n lines each contain m integers. The j-th integer in the i-th line is a_{i,j} (1 ≤ a_{i,j} ≤ 10^9).
Output
For each case, output n lines each containing m integers. The j-th integer in the i-th line is b_{i,j}.
Example
Input
3
3 2
1 2
4 5
7 8
2 2
1 1
3 3
2 2
1 3
2 2
Output
1 2
5 6
7 8
2 1
4 3
2 4
3 2
Note
In all the cases, you can verify that no two adjacent cells have the same value and that b is the same as a with some values incremented by one. | instruction | 0 | 27,636 | 15 | 55,272 |
Tags: 2-sat, chinese remainder theorem, constructive algorithms, fft, flows
Correct Solution:
```
from __future__ import division, print_function
import os,sys
from io import BytesIO, IOBase
if sys.version_info[0] < 3:
from __builtin__ import xrange as range
from future_builtins import ascii, filter, hex, map, oct, zip
def so(): return int(input())
def st(): return input()
def mj(): return map(int,input().strip().split(" "))
def msj(): return map(str,input().strip().split(" "))
def le(): return list(map(int,input().split()))
def dmain():
sys.setrecursionlimit(1000000)
threading.stack_size(1024000)
thread = threading.Thread(target=main)
thread.start()
def decimalToBinary(n): return bin(n).replace("0b","")
def read():
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
def tr(n):
return n*(n+1)//2
def main():
import math
for i in range(so()):
n,q=mj()
d=[]
for i in range(n):
d.append(le())
for i in range(n):
for j in range(q):
if(d[i][j]%2!=(i+j)%2):
d[i][j]=d[i][j]+1
for i in range(len(d)):
print(*d[i])
# region fastio
# template taken from https://github.com/cheran-senthil/PyRival/blob/master/templates/template.py
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")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
#read()
main()
#dmain()
# Comment Read()
``` | output | 1 | 27,636 | 15 | 55,273 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Artem is building a new robot. He has a matrix a consisting of n rows and m columns. The cell located on the i-th row from the top and the j-th column from the left has a value a_{i,j} written in it.
If two adjacent cells contain the same value, the robot will break. A matrix is called good if no two adjacent cells contain the same value, where two cells are called adjacent if they share a side.
Artem wants to increment the values in some cells by one to make a good.
More formally, find a good matrix b that satisfies the following condition —
* For all valid (i,j), either b_{i,j} = a_{i,j} or b_{i,j} = a_{i,j}+1.
For the constraints of this problem, it can be shown that such a matrix b always exists. If there are several such tables, you can output any of them. Please note that you do not have to minimize the number of increments.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10). Description of the test cases follows.
The first line of each test case contains two integers n, m (1 ≤ n ≤ 100, 1 ≤ m ≤ 100) — the number of rows and columns, respectively.
The following n lines each contain m integers. The j-th integer in the i-th line is a_{i,j} (1 ≤ a_{i,j} ≤ 10^9).
Output
For each case, output n lines each containing m integers. The j-th integer in the i-th line is b_{i,j}.
Example
Input
3
3 2
1 2
4 5
7 8
2 2
1 1
3 3
2 2
1 3
2 2
Output
1 2
5 6
7 8
2 1
4 3
2 4
3 2
Note
In all the cases, you can verify that no two adjacent cells have the same value and that b is the same as a with some values incremented by one. | instruction | 0 | 27,637 | 15 | 55,274 |
Tags: 2-sat, chinese remainder theorem, constructive algorithms, fft, flows
Correct Solution:
```
t=int(input())
ad=[(1,0),(0,1),(-1,0),(0,-1)]
def aze(x,y):
if x<n and x>-1 and y<m and y>-1:
return(True)
return(False)
print()
def dfs(k):
q=[k]
while q:
x=q.pop()
i,j=x//m,x%m
for w in ad:
ip,jp=i+w[0],j+w[1]
if not aze(ip,jp):continue
if mat[i][j]!=mat[ip][jp]:continue
q.append(jp+ip*m)
if vi[jp+ip*m]:
mat[ip][jp]-=1
vi[jp+ip*m]=False
else:
mat[ip][jp]+=1
vi[jp+ip*m]=True
for _ in range(t):
n,m=map(int,input().split())
mat=[[] for i in range(n)]
vi=[False for i in range(n*m)]
for o in range(n):
mat[o]=list(map(int,input().split()))
for i in range(n):
for j in range(m):
dfs(j+i*m)
for i in range(n):
print(*mat[i])
print()
``` | output | 1 | 27,637 | 15 | 55,275 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Artem is building a new robot. He has a matrix a consisting of n rows and m columns. The cell located on the i-th row from the top and the j-th column from the left has a value a_{i,j} written in it.
If two adjacent cells contain the same value, the robot will break. A matrix is called good if no two adjacent cells contain the same value, where two cells are called adjacent if they share a side.
Artem wants to increment the values in some cells by one to make a good.
More formally, find a good matrix b that satisfies the following condition —
* For all valid (i,j), either b_{i,j} = a_{i,j} or b_{i,j} = a_{i,j}+1.
For the constraints of this problem, it can be shown that such a matrix b always exists. If there are several such tables, you can output any of them. Please note that you do not have to minimize the number of increments.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10). Description of the test cases follows.
The first line of each test case contains two integers n, m (1 ≤ n ≤ 100, 1 ≤ m ≤ 100) — the number of rows and columns, respectively.
The following n lines each contain m integers. The j-th integer in the i-th line is a_{i,j} (1 ≤ a_{i,j} ≤ 10^9).
Output
For each case, output n lines each containing m integers. The j-th integer in the i-th line is b_{i,j}.
Example
Input
3
3 2
1 2
4 5
7 8
2 2
1 1
3 3
2 2
1 3
2 2
Output
1 2
5 6
7 8
2 1
4 3
2 4
3 2
Note
In all the cases, you can verify that no two adjacent cells have the same value and that b is the same as a with some values incremented by one. | instruction | 0 | 27,638 | 15 | 55,276 |
Tags: 2-sat, chinese remainder theorem, constructive algorithms, fft, flows
Correct Solution:
```
from math import *
from sys import *
t=int(stdin.readline())
for _ in range(t):
n, k = map(int, stdin.readline().split())
a=[]
for i in range(n):
m = list(map(int, stdin.readline().split()))
a.append(m)
f=0
for i in range(n):
for j in range(k):
if f==0:
if j%2==0:
if a[i][j]%2==1:
a[i][j]+=1
else:
if a[i][j]%2==0:
a[i][j]+=1
else:
if j%2==0:
if a[i][j]%2==0:
a[i][j]+=1
else:
if a[i][j]%2==1:
a[i][j]+=1
f+=1
f=f%2
for i in range(n):
print(*a[i])
``` | output | 1 | 27,638 | 15 | 55,277 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Artem is building a new robot. He has a matrix a consisting of n rows and m columns. The cell located on the i-th row from the top and the j-th column from the left has a value a_{i,j} written in it.
If two adjacent cells contain the same value, the robot will break. A matrix is called good if no two adjacent cells contain the same value, where two cells are called adjacent if they share a side.
Artem wants to increment the values in some cells by one to make a good.
More formally, find a good matrix b that satisfies the following condition —
* For all valid (i,j), either b_{i,j} = a_{i,j} or b_{i,j} = a_{i,j}+1.
For the constraints of this problem, it can be shown that such a matrix b always exists. If there are several such tables, you can output any of them. Please note that you do not have to minimize the number of increments.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10). Description of the test cases follows.
The first line of each test case contains two integers n, m (1 ≤ n ≤ 100, 1 ≤ m ≤ 100) — the number of rows and columns, respectively.
The following n lines each contain m integers. The j-th integer in the i-th line is a_{i,j} (1 ≤ a_{i,j} ≤ 10^9).
Output
For each case, output n lines each containing m integers. The j-th integer in the i-th line is b_{i,j}.
Example
Input
3
3 2
1 2
4 5
7 8
2 2
1 1
3 3
2 2
1 3
2 2
Output
1 2
5 6
7 8
2 1
4 3
2 4
3 2
Note
In all the cases, you can verify that no two adjacent cells have the same value and that b is the same as a with some values incremented by one. | instruction | 0 | 27,639 | 15 | 55,278 |
Tags: 2-sat, chinese remainder theorem, constructive algorithms, fft, flows
Correct Solution:
```
from sys import stdin, stdout
t=int(stdin.readline())
for _ in range(t):
n,m=map(int,stdin.readline().split())
arr=[]
for _ in range(n):
arr.append(list(map(int,stdin.readline().split())))
for i in range(n):
for j in range(m):
if (i+j)%2==0:
if arr[i][j]%2==0:
continue
else:
arr[i][j]+=1
else:
if arr[i][j]%2!=0:
continue
else:
arr[i][j]+=1
for i in range(n):
for j in range(m):
print(arr[i][j],end=" ")
print("")
``` | output | 1 | 27,639 | 15 | 55,279 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Artem is building a new robot. He has a matrix a consisting of n rows and m columns. The cell located on the i-th row from the top and the j-th column from the left has a value a_{i,j} written in it.
If two adjacent cells contain the same value, the robot will break. A matrix is called good if no two adjacent cells contain the same value, where two cells are called adjacent if they share a side.
Artem wants to increment the values in some cells by one to make a good.
More formally, find a good matrix b that satisfies the following condition —
* For all valid (i,j), either b_{i,j} = a_{i,j} or b_{i,j} = a_{i,j}+1.
For the constraints of this problem, it can be shown that such a matrix b always exists. If there are several such tables, you can output any of them. Please note that you do not have to minimize the number of increments.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10). Description of the test cases follows.
The first line of each test case contains two integers n, m (1 ≤ n ≤ 100, 1 ≤ m ≤ 100) — the number of rows and columns, respectively.
The following n lines each contain m integers. The j-th integer in the i-th line is a_{i,j} (1 ≤ a_{i,j} ≤ 10^9).
Output
For each case, output n lines each containing m integers. The j-th integer in the i-th line is b_{i,j}.
Example
Input
3
3 2
1 2
4 5
7 8
2 2
1 1
3 3
2 2
1 3
2 2
Output
1 2
5 6
7 8
2 1
4 3
2 4
3 2
Note
In all the cases, you can verify that no two adjacent cells have the same value and that b is the same as a with some values incremented by one. | instruction | 0 | 27,640 | 15 | 55,280 |
Tags: 2-sat, chinese remainder theorem, constructive algorithms, fft, flows
Correct Solution:
```
for _ in range(int(input())):
n, m = list(map(int, input().split()))
a = []
for i in range(n):
a.append(list(map(int, input().split())))
for i in range(n):
for r in range(m):
if (i + r) % 2 != a[i][r] % 2:
a[i][r] += 1
for i in range(n):
print(*a[i])
``` | output | 1 | 27,640 | 15 | 55,281 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Artem is building a new robot. He has a matrix a consisting of n rows and m columns. The cell located on the i-th row from the top and the j-th column from the left has a value a_{i,j} written in it.
If two adjacent cells contain the same value, the robot will break. A matrix is called good if no two adjacent cells contain the same value, where two cells are called adjacent if they share a side.
Artem wants to increment the values in some cells by one to make a good.
More formally, find a good matrix b that satisfies the following condition —
* For all valid (i,j), either b_{i,j} = a_{i,j} or b_{i,j} = a_{i,j}+1.
For the constraints of this problem, it can be shown that such a matrix b always exists. If there are several such tables, you can output any of them. Please note that you do not have to minimize the number of increments.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10). Description of the test cases follows.
The first line of each test case contains two integers n, m (1 ≤ n ≤ 100, 1 ≤ m ≤ 100) — the number of rows and columns, respectively.
The following n lines each contain m integers. The j-th integer in the i-th line is a_{i,j} (1 ≤ a_{i,j} ≤ 10^9).
Output
For each case, output n lines each containing m integers. The j-th integer in the i-th line is b_{i,j}.
Example
Input
3
3 2
1 2
4 5
7 8
2 2
1 1
3 3
2 2
1 3
2 2
Output
1 2
5 6
7 8
2 1
4 3
2 4
3 2
Note
In all the cases, you can verify that no two adjacent cells have the same value and that b is the same as a with some values incremented by one.
Submitted Solution:
```
t=int(input())
for i in range(t):
n,m=map(int,input().split())
a=[[0 for k in range(m)] for j in range(n)]
ans=[[0 for k in range(m)] for j in range(n)]
for l in range(n):
v=list(map(int,input().split()))
for u in range(m):
a[l][u]=v[u]
for y in range(n):
for p in range(m):
if (p+y)%2==0 and a[y][p]%2!=0:
ans[y][p]=a[y][p]+1
elif (p+y)%2!=0 and a[y][p]%2==0:
ans[y][p]=a[y][p]+1
else:
ans[y][p]=a[y][p]
for h in range(n):
print(*ans[h],sep=" ")
``` | instruction | 0 | 27,641 | 15 | 55,282 |
Yes | output | 1 | 27,641 | 15 | 55,283 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Artem is building a new robot. He has a matrix a consisting of n rows and m columns. The cell located on the i-th row from the top and the j-th column from the left has a value a_{i,j} written in it.
If two adjacent cells contain the same value, the robot will break. A matrix is called good if no two adjacent cells contain the same value, where two cells are called adjacent if they share a side.
Artem wants to increment the values in some cells by one to make a good.
More formally, find a good matrix b that satisfies the following condition —
* For all valid (i,j), either b_{i,j} = a_{i,j} or b_{i,j} = a_{i,j}+1.
For the constraints of this problem, it can be shown that such a matrix b always exists. If there are several such tables, you can output any of them. Please note that you do not have to minimize the number of increments.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10). Description of the test cases follows.
The first line of each test case contains two integers n, m (1 ≤ n ≤ 100, 1 ≤ m ≤ 100) — the number of rows and columns, respectively.
The following n lines each contain m integers. The j-th integer in the i-th line is a_{i,j} (1 ≤ a_{i,j} ≤ 10^9).
Output
For each case, output n lines each containing m integers. The j-th integer in the i-th line is b_{i,j}.
Example
Input
3
3 2
1 2
4 5
7 8
2 2
1 1
3 3
2 2
1 3
2 2
Output
1 2
5 6
7 8
2 1
4 3
2 4
3 2
Note
In all the cases, you can verify that no two adjacent cells have the same value and that b is the same as a with some values incremented by one.
Submitted Solution:
```
t = int(input())
for _ in range(t):
n,m = map(int,input().split())
for i in range(n):
a = list(map(int,input().split()))
print(*[a[j]+(a[j]+i+j&1) for j in range(m)])
``` | instruction | 0 | 27,642 | 15 | 55,284 |
Yes | output | 1 | 27,642 | 15 | 55,285 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Artem is building a new robot. He has a matrix a consisting of n rows and m columns. The cell located on the i-th row from the top and the j-th column from the left has a value a_{i,j} written in it.
If two adjacent cells contain the same value, the robot will break. A matrix is called good if no two adjacent cells contain the same value, where two cells are called adjacent if they share a side.
Artem wants to increment the values in some cells by one to make a good.
More formally, find a good matrix b that satisfies the following condition —
* For all valid (i,j), either b_{i,j} = a_{i,j} or b_{i,j} = a_{i,j}+1.
For the constraints of this problem, it can be shown that such a matrix b always exists. If there are several such tables, you can output any of them. Please note that you do not have to minimize the number of increments.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10). Description of the test cases follows.
The first line of each test case contains two integers n, m (1 ≤ n ≤ 100, 1 ≤ m ≤ 100) — the number of rows and columns, respectively.
The following n lines each contain m integers. The j-th integer in the i-th line is a_{i,j} (1 ≤ a_{i,j} ≤ 10^9).
Output
For each case, output n lines each containing m integers. The j-th integer in the i-th line is b_{i,j}.
Example
Input
3
3 2
1 2
4 5
7 8
2 2
1 1
3 3
2 2
1 3
2 2
Output
1 2
5 6
7 8
2 1
4 3
2 4
3 2
Note
In all the cases, you can verify that no two adjacent cells have the same value and that b is the same as a with some values incremented by one.
Submitted Solution:
```
for _ in range(int(input())):
n,m=[int(c) for c in input().split()]
arr =[[0 for i in range(m+1)]]
for i in range(n):
ar =[int(c) for c in input().split()]
arr.append([0]+ar)
for i in range(n+1):
for j in range(m+1):
if (i+j)&1 != arr[i][j]&1:
arr[i][j]+=1
for i in range(1,n+1):
print(*arr[i][1:])
``` | instruction | 0 | 27,643 | 15 | 55,286 |
Yes | output | 1 | 27,643 | 15 | 55,287 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Artem is building a new robot. He has a matrix a consisting of n rows and m columns. The cell located on the i-th row from the top and the j-th column from the left has a value a_{i,j} written in it.
If two adjacent cells contain the same value, the robot will break. A matrix is called good if no two adjacent cells contain the same value, where two cells are called adjacent if they share a side.
Artem wants to increment the values in some cells by one to make a good.
More formally, find a good matrix b that satisfies the following condition —
* For all valid (i,j), either b_{i,j} = a_{i,j} or b_{i,j} = a_{i,j}+1.
For the constraints of this problem, it can be shown that such a matrix b always exists. If there are several such tables, you can output any of them. Please note that you do not have to minimize the number of increments.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10). Description of the test cases follows.
The first line of each test case contains two integers n, m (1 ≤ n ≤ 100, 1 ≤ m ≤ 100) — the number of rows and columns, respectively.
The following n lines each contain m integers. The j-th integer in the i-th line is a_{i,j} (1 ≤ a_{i,j} ≤ 10^9).
Output
For each case, output n lines each containing m integers. The j-th integer in the i-th line is b_{i,j}.
Example
Input
3
3 2
1 2
4 5
7 8
2 2
1 1
3 3
2 2
1 3
2 2
Output
1 2
5 6
7 8
2 1
4 3
2 4
3 2
Note
In all the cases, you can verify that no two adjacent cells have the same value and that b is the same as a with some values incremented by one.
Submitted Solution:
```
# 3 rows, 2 columns
#
# 1 2
# 4 5
# 7 8
#
#
# 1 1 -> 1 2
# 3 3 4 3
#
# x 2
# 3[2] <- stuck
for _ in range(int(input())):
n,m = map(int,input().split())
for k in range(n):
l = list(map(int,input().split()))
# if k % 2 == 0: # row
for i in range(m):
# if i % 2 == 0:
# if l[i] % 2 != 0:
# l[i] += 1
# else:
# if l[i] % 2 == 0:
# l[i] += 1
# else:
# for i in range(m):
# if i % 2 != 0:
# if l[i] % 2 == 0:
# l[i] += 1
# else:
# if l[i] % 2 != 0:
# l[i] += 1
#
if (k+i+l[i]) % 2 == 0:
l[i] += 1
print(*l)
``` | instruction | 0 | 27,644 | 15 | 55,288 |
Yes | output | 1 | 27,644 | 15 | 55,289 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Artem is building a new robot. He has a matrix a consisting of n rows and m columns. The cell located on the i-th row from the top and the j-th column from the left has a value a_{i,j} written in it.
If two adjacent cells contain the same value, the robot will break. A matrix is called good if no two adjacent cells contain the same value, where two cells are called adjacent if they share a side.
Artem wants to increment the values in some cells by one to make a good.
More formally, find a good matrix b that satisfies the following condition —
* For all valid (i,j), either b_{i,j} = a_{i,j} or b_{i,j} = a_{i,j}+1.
For the constraints of this problem, it can be shown that such a matrix b always exists. If there are several such tables, you can output any of them. Please note that you do not have to minimize the number of increments.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10). Description of the test cases follows.
The first line of each test case contains two integers n, m (1 ≤ n ≤ 100, 1 ≤ m ≤ 100) — the number of rows and columns, respectively.
The following n lines each contain m integers. The j-th integer in the i-th line is a_{i,j} (1 ≤ a_{i,j} ≤ 10^9).
Output
For each case, output n lines each containing m integers. The j-th integer in the i-th line is b_{i,j}.
Example
Input
3
3 2
1 2
4 5
7 8
2 2
1 1
3 3
2 2
1 3
2 2
Output
1 2
5 6
7 8
2 1
4 3
2 4
3 2
Note
In all the cases, you can verify that no two adjacent cells have the same value and that b is the same as a with some values incremented by one.
Submitted Solution:
```
import sys
import math
import bisect
from sys import stdin, stdout
from math import gcd, floor, sqrt, log2, ceil
from collections import defaultdict as dd
from bisect import bisect_left as bl, bisect_right as br
from bisect import insort
from collections import Counter
from collections import deque
from heapq import heappush,heappop,heapify
from itertools import permutations,combinations
mod = int(1e9)+7
ip = lambda : int(stdin.readline())
inp = lambda: map(int,stdin.readline().split())
ips = lambda: stdin.readline().rstrip()
out = lambda x : stdout.write(str(x)+"\n")
t = ip()
for _ in range(t):
n,m = inp()
arr = []
for i in range(n):
ch = list(inp())
#ch = [[i,i+1] for i in ch]
arr.append(ch)
xor = 0
for i in range(n):
if xor == 0:
for j in range(m):
if j%2 == 0:
if arr[i][j]%2 != 0:
arr[i][j] += 1
else:
if ch[j]%2 == 0:
arr[i][j] += 1
else:
for j in range(m):
if j%2 == 1:
if arr[i][j]%2 != 0:
arr[i][j] += 1
else:
if arr[i][j]%2 == 0:
arr[i][j] += 1
xor ^= 1
for i in arr:
print(*i)
``` | instruction | 0 | 27,645 | 15 | 55,290 |
No | output | 1 | 27,645 | 15 | 55,291 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Artem is building a new robot. He has a matrix a consisting of n rows and m columns. The cell located on the i-th row from the top and the j-th column from the left has a value a_{i,j} written in it.
If two adjacent cells contain the same value, the robot will break. A matrix is called good if no two adjacent cells contain the same value, where two cells are called adjacent if they share a side.
Artem wants to increment the values in some cells by one to make a good.
More formally, find a good matrix b that satisfies the following condition —
* For all valid (i,j), either b_{i,j} = a_{i,j} or b_{i,j} = a_{i,j}+1.
For the constraints of this problem, it can be shown that such a matrix b always exists. If there are several such tables, you can output any of them. Please note that you do not have to minimize the number of increments.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10). Description of the test cases follows.
The first line of each test case contains two integers n, m (1 ≤ n ≤ 100, 1 ≤ m ≤ 100) — the number of rows and columns, respectively.
The following n lines each contain m integers. The j-th integer in the i-th line is a_{i,j} (1 ≤ a_{i,j} ≤ 10^9).
Output
For each case, output n lines each containing m integers. The j-th integer in the i-th line is b_{i,j}.
Example
Input
3
3 2
1 2
4 5
7 8
2 2
1 1
3 3
2 2
1 3
2 2
Output
1 2
5 6
7 8
2 1
4 3
2 4
3 2
Note
In all the cases, you can verify that no two adjacent cells have the same value and that b is the same as a with some values incremented by one.
Submitted Solution:
```
import sys
# import math
# import re
# from heapq import *
# from collections import defaultdict as dd
# from collections import OrderedDict as odict
# from collections import Counter as cc
# from collections import deque
# sys.setrecursionlimit(3*(10**5)+1)#thsis is must
# mod = 10**9+7; md = 998244353
# m = 2**32
input = lambda: sys.stdin.readline().strip()
inp = lambda: list(map(int,sys.stdin.readline().strip().split()))
# def C(n,r,mod):
# if r>n:
# return 0
# num = den = 1
# for i in range(r):
# num = (num*(n-i))%mod
# den = (den*(i+1))%mod
# return (num*pow(den,mod-2,mod))%mod
# def pfcs(M = 100000):
# pfc = [0]*(M+1)
# for i in range(2,M+1):
# if pfc[i]==0:
# for j in range(i,M+1,i):
# pfc[j]+=1
# return pfc
#______________________________________________________
for _ in range(int(input())):
n,m = list(map(int,input().split()))
b = [list(map(int,input().split())) for i in range(n)]
# b = [[0 for i in range(m)] for j in range(n)]
d = [[1,0],[-1,0],[0,1],[0,-1]]
def dir(i,j):
res = []
for l,r in d:
row = i+l
col = j+r
if row<0 or row>=n or col<0 or col>=m:
continue
# print(row,col,"lol")s
res.append(b[row][col])
return res
for i in range(n):
for j in range(m):
tar = b[i][j]
res = dir(i,j)
count = res.count(tar)
count2 = res.count(tar+1)
if count2==0 and count==1:
b[i][j] = tar+1
else:
b[i][j] = tar
for i in b:
print(*i)
``` | instruction | 0 | 27,646 | 15 | 55,292 |
No | output | 1 | 27,646 | 15 | 55,293 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Artem is building a new robot. He has a matrix a consisting of n rows and m columns. The cell located on the i-th row from the top and the j-th column from the left has a value a_{i,j} written in it.
If two adjacent cells contain the same value, the robot will break. A matrix is called good if no two adjacent cells contain the same value, where two cells are called adjacent if they share a side.
Artem wants to increment the values in some cells by one to make a good.
More formally, find a good matrix b that satisfies the following condition —
* For all valid (i,j), either b_{i,j} = a_{i,j} or b_{i,j} = a_{i,j}+1.
For the constraints of this problem, it can be shown that such a matrix b always exists. If there are several such tables, you can output any of them. Please note that you do not have to minimize the number of increments.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10). Description of the test cases follows.
The first line of each test case contains two integers n, m (1 ≤ n ≤ 100, 1 ≤ m ≤ 100) — the number of rows and columns, respectively.
The following n lines each contain m integers. The j-th integer in the i-th line is a_{i,j} (1 ≤ a_{i,j} ≤ 10^9).
Output
For each case, output n lines each containing m integers. The j-th integer in the i-th line is b_{i,j}.
Example
Input
3
3 2
1 2
4 5
7 8
2 2
1 1
3 3
2 2
1 3
2 2
Output
1 2
5 6
7 8
2 1
4 3
2 4
3 2
Note
In all the cases, you can verify that no two adjacent cells have the same value and that b is the same as a with some values incremented by one.
Submitted Solution:
```
for i in range(int(input())):
n,m = map(int,input().split())
a = []
for i in range(n):
l = list(map(int,input().split()))
for k in range(m):
a.append(l[k])
c = [0]*(n*m)
for i in range(n*m):
if i == n*m - 1:
break
if i+m > n*m - 1:
if a[i] == a[i+1]:
if c[i] == -1:
a[i+1] += 1
c[i+1] = -1
else:
a[i] += 1
c[i] = -1
elif (i+1) % m == 0:
if a[i] == a[i+m]:
if c[i] == -1:
a[i+m] += 1
c[i+m] = - 1
else:
a[i] += 1
c[i] = -1
else:
if a[i] == a[i+m]:
if c[i] == -1:
a[i+m] += 1
c[i+m] = - 1
else:
a[i] += 1
c[i] = -1
if a[i] == a[i+1]:
if c[i] == -1:
a[i+1] += 1
c[i+1] = -1
else:
a[i] += 1
c[i] = -1
if a[i] == a[i+1]:
if c[i] == -1:
a[i+1] += 1
c[i+1] = -1
else:
a[i] += 1
c[i] = -1
if a[i] == a[i+m]:
if c[i] == -1:
a[i+m] += 1
c[i+m] = - 1
else:
a[i] += 1
c[i] = -1
for i in range(n*m):
if (i+1) % m == 0:
print(a[i])
else:
print(a[i], end = " ")
``` | instruction | 0 | 27,647 | 15 | 55,294 |
No | output | 1 | 27,647 | 15 | 55,295 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Artem is building a new robot. He has a matrix a consisting of n rows and m columns. The cell located on the i-th row from the top and the j-th column from the left has a value a_{i,j} written in it.
If two adjacent cells contain the same value, the robot will break. A matrix is called good if no two adjacent cells contain the same value, where two cells are called adjacent if they share a side.
Artem wants to increment the values in some cells by one to make a good.
More formally, find a good matrix b that satisfies the following condition —
* For all valid (i,j), either b_{i,j} = a_{i,j} or b_{i,j} = a_{i,j}+1.
For the constraints of this problem, it can be shown that such a matrix b always exists. If there are several such tables, you can output any of them. Please note that you do not have to minimize the number of increments.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10). Description of the test cases follows.
The first line of each test case contains two integers n, m (1 ≤ n ≤ 100, 1 ≤ m ≤ 100) — the number of rows and columns, respectively.
The following n lines each contain m integers. The j-th integer in the i-th line is a_{i,j} (1 ≤ a_{i,j} ≤ 10^9).
Output
For each case, output n lines each containing m integers. The j-th integer in the i-th line is b_{i,j}.
Example
Input
3
3 2
1 2
4 5
7 8
2 2
1 1
3 3
2 2
1 3
2 2
Output
1 2
5 6
7 8
2 1
4 3
2 4
3 2
Note
In all the cases, you can verify that no two adjacent cells have the same value and that b is the same as a with some values incremented by one.
Submitted Solution:
```
import io
import os
from collections import Counter, defaultdict, deque
from heapq import heappush, heappop, heapify
def solve(R, C, grid):
if False:
print(R, C)
for row in grid:
print(row)
drdc = [(-1, 0), (0, -1), (0, 1), (1, 0)]
heap = []
for r in range(R):
for c in range(C):
heappush(heap, (-grid[r][c], r + c, r, c))
while heap:
neg, _, r, c = heappop(heap)
bad = False
for dr, dc in drdc:
nr = r + dr
nc = c + dc
if 0 <= nr < R and 0 <= nc < C:
if grid[nr][nc] == grid[r][c] + 1:
bad = True
break
if not bad:
grid[r][c] += 1
return "\n".join(" ".join(map(str, row)) for row in grid)
if __name__ == "__main__":
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
TC = int(input())
for tc in range(1, TC + 1):
N, M = [int(x) for x in input().split()]
grid = [[int(x) for x in input().split()] for i in range(N)]
ans = solve(N, M, grid)
print(ans)
``` | instruction | 0 | 27,648 | 15 | 55,296 |
No | output | 1 | 27,648 | 15 | 55,297 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Elephant loves magic squares very much.
A magic square is a 3 × 3 table, each cell contains some positive integer. At that the sums of integers in all rows, columns and diagonals of the table are equal. The figure below shows the magic square, the sum of integers in all its rows, columns and diagonals equals 15.
<image>
The Little Elephant remembered one magic square. He started writing this square on a piece of paper, but as he wrote, he forgot all three elements of the main diagonal of the magic square. Fortunately, the Little Elephant clearly remembered that all elements of the magic square did not exceed 105.
Help the Little Elephant, restore the original magic square, given the Elephant's notes.
Input
The first three lines of the input contain the Little Elephant's notes. The first line contains elements of the first row of the magic square. The second line contains the elements of the second row, the third line is for the third row. The main diagonal elements that have been forgotten by the Elephant are represented by zeroes.
It is guaranteed that the notes contain exactly three zeroes and they are all located on the main diagonal. It is guaranteed that all positive numbers in the table do not exceed 105.
Output
Print three lines, in each line print three integers — the Little Elephant's magic square. If there are multiple magic squares, you are allowed to print any of them. Note that all numbers you print must be positive and not exceed 105.
It is guaranteed that there exists at least one magic square that meets the conditions.
Examples
Input
0 1 1
1 0 1
1 1 0
Output
1 1 1
1 1 1
1 1 1
Input
0 3 6
5 0 5
4 7 0
Output
6 3 6
5 5 5
4 7 4 | instruction | 0 | 27,732 | 15 | 55,464 |
Tags: brute force, implementation
Correct Solution:
```
def main():
def is_valid(grid):
if sum(grid[0]) != sum(grid[1]) or sum(grid[1]) != sum(grid[2]):
return False
if grid[0][0] + grid[1][1] + grid[2][2] != grid[0][2] + grid[1][1] + grid[2][0]:
return False
return True
grid = []
for i in range(3):
grid.append(list(map(int, input().split(' '))))
m = [sum(grid[i]) for i in range(3)]
for i in range(int(1e6)):
grid[0][0] = i - sum(grid[0])
grid[1][1] = i - sum(grid[1])
grid[2][2] = i - sum(grid[2])
if is_valid(grid) and grid[0][0] != 0:
break
grid[0][0] = 0
grid[1][1] = 0
grid[2][2] = 0
for i in range(3):
print(' '.join(list(map(str,grid[i]))))
main()
``` | output | 1 | 27,732 | 15 | 55,465 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Pavel loves grid mazes. A grid maze is an n × m rectangle maze where each cell is either empty, or is a wall. You can go from one cell to another only if both cells are empty and have a common side.
Pavel drew a grid maze with all empty cells forming a connected area. That is, you can go from any empty cell to any other one. Pavel doesn't like it when his maze has too little walls. He wants to turn exactly k empty cells into walls so that all the remaining cells still formed a connected area. Help him.
Input
The first line contains three integers n, m, k (1 ≤ n, m ≤ 500, 0 ≤ k < s), where n and m are the maze's height and width, correspondingly, k is the number of walls Pavel wants to add and letter s represents the number of empty cells in the original maze.
Each of the next n lines contains m characters. They describe the original maze. If a character on a line equals ".", then the corresponding cell is empty and if the character equals "#", then the cell is a wall.
Output
Print n lines containing m characters each: the new maze that fits Pavel's requirements. Mark the empty cells that you transformed into walls as "X", the other cells must be left without changes (that is, "." and "#").
It is guaranteed that a solution exists. If there are multiple solutions you can output any of them.
Examples
Input
3 4 2
#..#
..#.
#...
Output
#.X#
X.#.
#...
Input
5 4 5
#...
#.#.
.#..
...#
.#.#
Output
#XXX
#X#.
X#..
...#
.#.# | instruction | 0 | 27,784 | 15 | 55,568 |
Tags: dfs and similar
Correct Solution:
```
from collections import deque,namedtuple
dx = [0,0,1,-1]
dy = [1,-1,0,0]
def main():
n,m,limit = map(int,input().split())
a,sx,sy,tot = [],0,0,0
for i in range(n):
s,t = input(),[]
for j,c in enumerate(s):
if c == '.':
sx,sy = i,j
tot += 1
c = 'X'
t.append(c)
a.append(t[:])
dq,done = deque([(sx,sy)]),0
while len(dq) > 0:
now = dq.popleft()
if done >= tot - limit:
break
if(a[now[0]][now[1]] == 'X'):
done += 1
else:
continue
a[now[0]][now[1]] = '.'
for i in range(4):
tx,ty = dx[i] + now[0],dy[i] + now[1]
if 0 <= tx < n and 0 <= ty < m and a[tx][ty] == 'X':
dq.append((tx,ty))
for i in a:
print("".join(i))
if __name__ == "__main__":
main()
``` | output | 1 | 27,784 | 15 | 55,569 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Pavel loves grid mazes. A grid maze is an n × m rectangle maze where each cell is either empty, or is a wall. You can go from one cell to another only if both cells are empty and have a common side.
Pavel drew a grid maze with all empty cells forming a connected area. That is, you can go from any empty cell to any other one. Pavel doesn't like it when his maze has too little walls. He wants to turn exactly k empty cells into walls so that all the remaining cells still formed a connected area. Help him.
Input
The first line contains three integers n, m, k (1 ≤ n, m ≤ 500, 0 ≤ k < s), where n and m are the maze's height and width, correspondingly, k is the number of walls Pavel wants to add and letter s represents the number of empty cells in the original maze.
Each of the next n lines contains m characters. They describe the original maze. If a character on a line equals ".", then the corresponding cell is empty and if the character equals "#", then the cell is a wall.
Output
Print n lines containing m characters each: the new maze that fits Pavel's requirements. Mark the empty cells that you transformed into walls as "X", the other cells must be left without changes (that is, "." and "#").
It is guaranteed that a solution exists. If there are multiple solutions you can output any of them.
Examples
Input
3 4 2
#..#
..#.
#...
Output
#.X#
X.#.
#...
Input
5 4 5
#...
#.#.
.#..
...#
.#.#
Output
#XXX
#X#.
X#..
...#
.#.# | instruction | 0 | 27,785 | 15 | 55,570 |
Tags: dfs and similar
Correct Solution:
```
class Queue:
def __init__(self):
self.list = []
def push(self, value):
self.list.append(value)
def pop(self):
return self.list.pop(0)
def top(self):
return self.list[0]
def empty(self):
return len(self.list) == 0
class Cell:
def __init__(self, row, column):
self.row = row
self.column = column
def neighbors(self):
return [Cell(self.row + 1, self.column ), # South
Cell(self.row - 1, self.column ), # North
Cell(self.row , self.column + 1), # West
Cell(self.row , self.column - 1)] # East
rows, columns, k = map(int, input().split())
top_row = ''.join(['#' for i in range(2 + columns)])
bottom_row = ''.join(['#' for i in range(2 + columns)])
matrix = [top_row] + ['#' + input() + '#' for i in range(rows)] + [bottom_row]
matrix = [list(i) for i in matrix]
emptyCellNum = 0
firstEmptyCell = False
for i in range(rows):
for j in range(columns):
if matrix[i+1][j+1] == '.':
emptyCellNum += 1
if not firstEmptyCell:
startCell = Cell(i+1, j+1)
firstEmptyCell = True
# Number of neighrbors that the vertex has
degree = [[0 for i in range(2 + columns)] for j in range(2 + rows)]
visited = [[False for i in range(2 + columns)] for j in range(2 + rows)]
def BFS(start):
counter = 1
q = Queue()
q.push(start)
visited[start.row][start.column]=True
while not q.empty():
cell = q.pop()
for neighbor in cell.neighbors():
if matrix[neighbor.row][neighbor.column] == '.':
if not visited[neighbor.row][neighbor.column]:
if counter == emptyCellNum - k:
return
visited[neighbor.row][neighbor.column]= True
counter += 1
q.push(neighbor)
BFS(startCell)
for i in range(1, 1 + rows):
for j in range(1, 1 + columns):
if matrix[i][j] == '.' and visited[i][j] == False:
matrix[i][j] = 'X'
for i in range(2 + rows):
matrix[i] = ''.join(matrix[i][1:-1])
print("\n".join(matrix[1:-1]))
``` | output | 1 | 27,785 | 15 | 55,571 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Pavel loves grid mazes. A grid maze is an n × m rectangle maze where each cell is either empty, or is a wall. You can go from one cell to another only if both cells are empty and have a common side.
Pavel drew a grid maze with all empty cells forming a connected area. That is, you can go from any empty cell to any other one. Pavel doesn't like it when his maze has too little walls. He wants to turn exactly k empty cells into walls so that all the remaining cells still formed a connected area. Help him.
Input
The first line contains three integers n, m, k (1 ≤ n, m ≤ 500, 0 ≤ k < s), where n and m are the maze's height and width, correspondingly, k is the number of walls Pavel wants to add and letter s represents the number of empty cells in the original maze.
Each of the next n lines contains m characters. They describe the original maze. If a character on a line equals ".", then the corresponding cell is empty and if the character equals "#", then the cell is a wall.
Output
Print n lines containing m characters each: the new maze that fits Pavel's requirements. Mark the empty cells that you transformed into walls as "X", the other cells must be left without changes (that is, "." and "#").
It is guaranteed that a solution exists. If there are multiple solutions you can output any of them.
Examples
Input
3 4 2
#..#
..#.
#...
Output
#.X#
X.#.
#...
Input
5 4 5
#...
#.#.
.#..
...#
.#.#
Output
#XXX
#X#.
X#..
...#
.#.# | instruction | 0 | 27,786 | 15 | 55,572 |
Tags: dfs and similar
Correct Solution:
```
#: Author - Soumya Saurav
import sys,io,os,time
from collections import defaultdict
from collections import Counter
from collections import deque
from itertools import combinations
from itertools import permutations
import bisect,math,heapq
alphabet = "abcdefghijklmnopqrstuvwxyz"
input = sys.stdin.readline
########################################
n , m , k = map(int , input().split())
arr = []
empty = []
visited = [[False]*m for i in range(n)]
for i in range(n):
temp = list(input())[:-1]
arr.append(temp)
for j in range(len(temp)):
if temp[j] == ".":
empty = [i , j]
order = []
visited[empty[0]][empty[1]] = True
q = deque([empty])
while q:
curx , cury = q.popleft()
order.append([curx,cury])
for dx, dy in [[0,1], [1,0], [0,-1], [-1,0]]:
x = dx + curx
y = dy + cury
if not(0 <= x <= n-1) or not(0 <= y <= m-1):
continue
if not visited[x][y] and arr[x][y] == ".":
visited[x][y] = True
q.append([x,y])
order = order[::-1]
for i in range(k):
arr[order[i][0]][order[i][1]] = "X"
for i in range(n):
print("".join(arr[i]))
'''
# Wrap solution for more recursion depth
#submit as python3
import collections,sys,threading
sys.setrecursionlimit(10**9)
threading.stack_size(10**8)
threading.Thread(target=solve).start()
'''
``` | output | 1 | 27,786 | 15 | 55,573 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Pavel loves grid mazes. A grid maze is an n × m rectangle maze where each cell is either empty, or is a wall. You can go from one cell to another only if both cells are empty and have a common side.
Pavel drew a grid maze with all empty cells forming a connected area. That is, you can go from any empty cell to any other one. Pavel doesn't like it when his maze has too little walls. He wants to turn exactly k empty cells into walls so that all the remaining cells still formed a connected area. Help him.
Input
The first line contains three integers n, m, k (1 ≤ n, m ≤ 500, 0 ≤ k < s), where n and m are the maze's height and width, correspondingly, k is the number of walls Pavel wants to add and letter s represents the number of empty cells in the original maze.
Each of the next n lines contains m characters. They describe the original maze. If a character on a line equals ".", then the corresponding cell is empty and if the character equals "#", then the cell is a wall.
Output
Print n lines containing m characters each: the new maze that fits Pavel's requirements. Mark the empty cells that you transformed into walls as "X", the other cells must be left without changes (that is, "." and "#").
It is guaranteed that a solution exists. If there are multiple solutions you can output any of them.
Examples
Input
3 4 2
#..#
..#.
#...
Output
#.X#
X.#.
#...
Input
5 4 5
#...
#.#.
.#..
...#
.#.#
Output
#XXX
#X#.
X#..
...#
.#.# | instruction | 0 | 27,787 | 15 | 55,574 |
Tags: dfs and similar
Correct Solution:
```
import sys
import threading
n,m,k = map(int,input().split())
matrix = []
for i in range(n):
row = input()
r=[]
for j in range(len(row)):
r.append(row[j])
matrix.append(r)
number_of_empty_cell = 0
global count
count = 0
for i in range(n):
for j in range(m):
if matrix[i][j]==".":
number_of_empty_cell+=1
def isSafe(i,j):
return True if 0<= i < n and 0<= j < m else False
def boarder_check(i,j):
return True if i==0 or j==0 or i==n-1 or j==m-1 else False
def dfs(i,j):
global count
if count < number_of_empty_cell-k:
matrix[i][j]="$"
count+=1
if isSafe(i,j+1) and matrix[i][j+1]==".":
dfs(i,j+1)
if isSafe(i+1,j) and matrix[i+1][j]==".":
dfs(i+1,j)
if isSafe(i,j-1) and matrix[i][j-1]==".":
dfs(i,j-1)
if isSafe(i-1,j) and matrix[i-1][j]==".":
dfs(i-1,j)
def main():
for i in range(n):
for j in range(m):
if matrix[i][j]==".":
dfs(i,j)
for i in range(n):
for j in range(m):
if matrix[i][j]=="$":
matrix[i][j]="."
elif matrix[i][j]==".":
matrix[i][j] = "X"
for i in range(n):
s = "".join(matrix[i])
print(s,end="\n")
if __name__ == "__main__":
sys.setrecursionlimit(10**6)
threading.stack_size(10**8)
t = threading.Thread(target=main)
t.start()
t.join()
``` | output | 1 | 27,787 | 15 | 55,575 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Pavel loves grid mazes. A grid maze is an n × m rectangle maze where each cell is either empty, or is a wall. You can go from one cell to another only if both cells are empty and have a common side.
Pavel drew a grid maze with all empty cells forming a connected area. That is, you can go from any empty cell to any other one. Pavel doesn't like it when his maze has too little walls. He wants to turn exactly k empty cells into walls so that all the remaining cells still formed a connected area. Help him.
Input
The first line contains three integers n, m, k (1 ≤ n, m ≤ 500, 0 ≤ k < s), where n and m are the maze's height and width, correspondingly, k is the number of walls Pavel wants to add and letter s represents the number of empty cells in the original maze.
Each of the next n lines contains m characters. They describe the original maze. If a character on a line equals ".", then the corresponding cell is empty and if the character equals "#", then the cell is a wall.
Output
Print n lines containing m characters each: the new maze that fits Pavel's requirements. Mark the empty cells that you transformed into walls as "X", the other cells must be left without changes (that is, "." and "#").
It is guaranteed that a solution exists. If there are multiple solutions you can output any of them.
Examples
Input
3 4 2
#..#
..#.
#...
Output
#.X#
X.#.
#...
Input
5 4 5
#...
#.#.
.#..
...#
.#.#
Output
#XXX
#X#.
X#..
...#
.#.# | instruction | 0 | 27,788 | 15 | 55,576 |
Tags: dfs and similar
Correct Solution:
```
from collections import deque,namedtuple
dx = [0,0,1,-1]
dy = [1,-1,0,0]
def main():
n,m,limit = map(int,input().split())
a,sx,sy,tot = [],0,0,0
for i in range(n):
s,t = input(),[]
for j,c in enumerate(s):
if c == '.':
sx,sy = i,j
tot += 1
c = 'X'
t.append(c)
a.append(t[:])
dq,done = deque([(sx,sy)]),0
while len(dq) > 0:
now = dq.pop()
if done >= tot - limit:
break
if(a[now[0]][now[1]] == 'X'):
done += 1
a[now[0]][now[1]] = '.'
for i in range(4):
tx,ty = dx[i] + now[0],dy[i] + now[1]
if 0 <= tx < n and 0 <= ty < m and a[tx][ty] == 'X':
dq.append((tx,ty))
for i in a:
print("".join(i))
if __name__ == "__main__":
main()
``` | output | 1 | 27,788 | 15 | 55,577 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Pavel loves grid mazes. A grid maze is an n × m rectangle maze where each cell is either empty, or is a wall. You can go from one cell to another only if both cells are empty and have a common side.
Pavel drew a grid maze with all empty cells forming a connected area. That is, you can go from any empty cell to any other one. Pavel doesn't like it when his maze has too little walls. He wants to turn exactly k empty cells into walls so that all the remaining cells still formed a connected area. Help him.
Input
The first line contains three integers n, m, k (1 ≤ n, m ≤ 500, 0 ≤ k < s), where n and m are the maze's height and width, correspondingly, k is the number of walls Pavel wants to add and letter s represents the number of empty cells in the original maze.
Each of the next n lines contains m characters. They describe the original maze. If a character on a line equals ".", then the corresponding cell is empty and if the character equals "#", then the cell is a wall.
Output
Print n lines containing m characters each: the new maze that fits Pavel's requirements. Mark the empty cells that you transformed into walls as "X", the other cells must be left without changes (that is, "." and "#").
It is guaranteed that a solution exists. If there are multiple solutions you can output any of them.
Examples
Input
3 4 2
#..#
..#.
#...
Output
#.X#
X.#.
#...
Input
5 4 5
#...
#.#.
.#..
...#
.#.#
Output
#XXX
#X#.
X#..
...#
.#.# | instruction | 0 | 27,789 | 15 | 55,578 |
Tags: dfs and similar
Correct Solution:
```
import sys
reader = (line.rstrip() for line in sys.stdin)
input = reader.__next__
def getInts():
return [int(s) for s in input().split()]
def getInt():
return int(input())
def getStrs():
return [s for s in input().split()]
def getStr():
return input()
def listStr():
return list(input())
import collections as c
def solve():
N, M, K = getInts()
maze = []
for n in range(N):
maze.append(listStr())
graph = c.defaultdict(set)
first_flag = False
empty_count = 0
if K == 0:
for n in range(N):
print(''.join(maze[n]))
return
for n in range(N):
for m in range(M):
if maze[n][m] == '#':
continue
empty_count += 1
if not first_flag:
first_n, first_m = n, m
first_flag = True
if n > 0 and maze[n-1][m] != '#':
graph[(n,m)].add((n-1,m))
if n < N-1 and maze[n+1][m] != '#':
graph[(n,m)].add((n+1,m))
if m > 0 and maze[n][m-1] != '#':
graph[(n,m)].add((n,m-1))
if m < M-1 and maze[n][m+1] != '#':
graph[(n,m)].add((n,m+1))
assert(len(graph[(n,m)]) > 0)
#bfs starting from (first_n, first_m), stopping when node count reaches S-K
def bfs(graph,visited,node,queue,visited_dict,max_nodes):
visited.append(node)
visited_dict[node] = 1
count = 0
queue.append(node)
queue = c.deque(queue)
if len(visited) == max_nodes:
return
while queue:
#be careful: this pop(0) can be slow. Consider using a deque
s = queue.popleft()
for neighbour in graph[s]:
if neighbour not in visited_dict:
visited.append(neighbour)
visited_dict[neighbour] = 1
if len(visited) == max_nodes:
return
queue.append(neighbour)
return
visited = []
visited_dict = c.defaultdict(int)
bfs(graph,visited,(first_n,first_m),[],visited_dict,empty_count-K)
for n in range(N):
for m in range(M):
if maze[n][m] != '#':
if (n,m) not in visited_dict:
maze[n][m] = 'X'
for n in range(N):
print(''.join(maze[n]))
return
solve()
``` | output | 1 | 27,789 | 15 | 55,579 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Pavel loves grid mazes. A grid maze is an n × m rectangle maze where each cell is either empty, or is a wall. You can go from one cell to another only if both cells are empty and have a common side.
Pavel drew a grid maze with all empty cells forming a connected area. That is, you can go from any empty cell to any other one. Pavel doesn't like it when his maze has too little walls. He wants to turn exactly k empty cells into walls so that all the remaining cells still formed a connected area. Help him.
Input
The first line contains three integers n, m, k (1 ≤ n, m ≤ 500, 0 ≤ k < s), where n and m are the maze's height and width, correspondingly, k is the number of walls Pavel wants to add and letter s represents the number of empty cells in the original maze.
Each of the next n lines contains m characters. They describe the original maze. If a character on a line equals ".", then the corresponding cell is empty and if the character equals "#", then the cell is a wall.
Output
Print n lines containing m characters each: the new maze that fits Pavel's requirements. Mark the empty cells that you transformed into walls as "X", the other cells must be left without changes (that is, "." and "#").
It is guaranteed that a solution exists. If there are multiple solutions you can output any of them.
Examples
Input
3 4 2
#..#
..#.
#...
Output
#.X#
X.#.
#...
Input
5 4 5
#...
#.#.
.#..
...#
.#.#
Output
#XXX
#X#.
X#..
...#
.#.# | instruction | 0 | 27,790 | 15 | 55,580 |
Tags: dfs and similar
Correct Solution:
```
from sys import stdin
from collections import *
def arr_inp(n):
if n == 1:
return [int(x) for x in stdin.readline().split()]
elif n == 2:
return [float(x) for x in stdin.readline().split()]
else:
return list(stdin.readline()[:-1])
def valid(i, j):
global n, m
return i > -1 and i < n and j > -1 and j < m
dx, dy = [-1, 0, 1, 0], [0, 1, 0, -1]
def print_maze():
for i in range(n):
print(*maze[i], sep='')
def bfs_util(i, j):
queue, visit, q2 = deque([[i, j]]), defaultdict(int, {(i, j): 1}), deque([[i, j]])
while queue:
# dequeue parent vertix
s = queue.popleft()
# enqueue child vertices
for i in range(4):
ni, nj = s[0] + dx[i], s[1] + dy[i]
if valid(ni, nj) and not visit[ni, nj] and maze[ni][nj] != '#':
queue.append([ni, nj])
q2.append([ni, nj])
visit[ni, nj] = 1
for i in range(k):
ix1, ix2 = q2.pop()
maze[ix1][ix2] = 'X'
print_maze()
n, m, k = arr_inp(1)
maze = [arr_inp(3) for i in range(n)]
if k == 0:
print_maze()
else:
for i in range(n):
try:
ix = maze[i].index('.')
if ix != -1:
bfs_util(i, ix)
break
except:
continue
``` | output | 1 | 27,790 | 15 | 55,581 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Pavel loves grid mazes. A grid maze is an n × m rectangle maze where each cell is either empty, or is a wall. You can go from one cell to another only if both cells are empty and have a common side.
Pavel drew a grid maze with all empty cells forming a connected area. That is, you can go from any empty cell to any other one. Pavel doesn't like it when his maze has too little walls. He wants to turn exactly k empty cells into walls so that all the remaining cells still formed a connected area. Help him.
Input
The first line contains three integers n, m, k (1 ≤ n, m ≤ 500, 0 ≤ k < s), where n and m are the maze's height and width, correspondingly, k is the number of walls Pavel wants to add and letter s represents the number of empty cells in the original maze.
Each of the next n lines contains m characters. They describe the original maze. If a character on a line equals ".", then the corresponding cell is empty and if the character equals "#", then the cell is a wall.
Output
Print n lines containing m characters each: the new maze that fits Pavel's requirements. Mark the empty cells that you transformed into walls as "X", the other cells must be left without changes (that is, "." and "#").
It is guaranteed that a solution exists. If there are multiple solutions you can output any of them.
Examples
Input
3 4 2
#..#
..#.
#...
Output
#.X#
X.#.
#...
Input
5 4 5
#...
#.#.
.#..
...#
.#.#
Output
#XXX
#X#.
X#..
...#
.#.# | instruction | 0 | 27,791 | 15 | 55,582 |
Tags: dfs and similar
Correct Solution:
```
def solve():
n,m,k=map(int,input().split())
Maze=[list(input())for row in range(n)]
bfs(Maze,n,m,k)
print("\n".join("".join(row) for row in Maze))
def bfs(M,n,m,k):
r,c=findEmptyCell(M)
visited=[[False for col in range(m)]for row in range(n)]
visited[r][c]=True
queue=deque()
queue.append((r,c))
s=sum(sum((x=="."for x in row),0)for row in M)
toConnect=s-k-1
while queue:
coords=queue.popleft()
for(r,c)in neighbors(coords,n,m):
if not visited[r][c]and M[r][c]==".":
visited[r][c]=True
queue.append((r,c))
toConnect-=1
if toConnect<0:
M[r][c]="X"
def findEmptyCell(M):
for(r, row)in enumerate(M):
for(c, cell)in enumerate(row):
if cell==".":
return(r, c)
def neighbors(coords,n,m):
r,c=coords
if r>0:
yield(r-1,c)
if r<n-1:
yield(r+1,c)
if c>0:
yield(r,c-1)
if c<m-1:
yield(r,c+1)
if __name__ =='__main__':
from collections import deque
solve()
``` | output | 1 | 27,791 | 15 | 55,583 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Pavel loves grid mazes. A grid maze is an n × m rectangle maze where each cell is either empty, or is a wall. You can go from one cell to another only if both cells are empty and have a common side.
Pavel drew a grid maze with all empty cells forming a connected area. That is, you can go from any empty cell to any other one. Pavel doesn't like it when his maze has too little walls. He wants to turn exactly k empty cells into walls so that all the remaining cells still formed a connected area. Help him.
Input
The first line contains three integers n, m, k (1 ≤ n, m ≤ 500, 0 ≤ k < s), where n and m are the maze's height and width, correspondingly, k is the number of walls Pavel wants to add and letter s represents the number of empty cells in the original maze.
Each of the next n lines contains m characters. They describe the original maze. If a character on a line equals ".", then the corresponding cell is empty and if the character equals "#", then the cell is a wall.
Output
Print n lines containing m characters each: the new maze that fits Pavel's requirements. Mark the empty cells that you transformed into walls as "X", the other cells must be left without changes (that is, "." and "#").
It is guaranteed that a solution exists. If there are multiple solutions you can output any of them.
Examples
Input
3 4 2
#..#
..#.
#...
Output
#.X#
X.#.
#...
Input
5 4 5
#...
#.#.
.#..
...#
.#.#
Output
#XXX
#X#.
X#..
...#
.#.#
Submitted Solution:
```
import sys
import time
from collections import deque
#sys.exit()
def read_line(cast=None):
if cast is not None:
return cast(input().strip())
else:
return input().strip()
def read_list(cast=int):
return list(map(cast, read_line().split()))
INTMAX = 1000000007
def solve(n,m,k):
nm = n*m
tab = []
for i in range(n):
l = list(read_line())
tab.append(l)
g = [[] for _ in range(nm)]
#deg = [0 for _ in range(nm)]
d = [(0,1),(0,-1),(1,0),(-1,0)]
sfn = set()
ln=None
for i in range(n):
for j in range(m):
if tab[i][j] == '.':
cn = m*i + j
sfn.add(cn)
if ln is None:
ln = cn
for di,dj in d:
ii,jj=i+di,j+dj
if ii >=0 and ii < n and jj>=0 and jj <m and tab[ii][jj]=='.':
g[cn].append(m*ii + jj)
st = [ln]
vis = set()
while len(sfn) > k: #st_1deg and
cn = st.pop()
vis.add(cn)
if cn in sfn:
sfn.remove(cn)
for nn in g[cn]:
ii, jj= nn // m, nn % m
if tab[ii][jj] == '.' and nn not in vis:#nn in sfn:
st.append(nn)
for cn in sfn:
i, j= cn // m, cn % m
tab[i][j] = 'X'
printTab(n, m, tab)
def printTab(n, m, tab):
v = 0
for i in range(n):
s = ""
for j in range(m):
s+=tab[i][j]
if tab[i][j]=='X':
v+=1
print(s)
#print(v)
n, m, k = read_list()
solve(n,m,k)
``` | instruction | 0 | 27,792 | 15 | 55,584 |
Yes | output | 1 | 27,792 | 15 | 55,585 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Pavel loves grid mazes. A grid maze is an n × m rectangle maze where each cell is either empty, or is a wall. You can go from one cell to another only if both cells are empty and have a common side.
Pavel drew a grid maze with all empty cells forming a connected area. That is, you can go from any empty cell to any other one. Pavel doesn't like it when his maze has too little walls. He wants to turn exactly k empty cells into walls so that all the remaining cells still formed a connected area. Help him.
Input
The first line contains three integers n, m, k (1 ≤ n, m ≤ 500, 0 ≤ k < s), where n and m are the maze's height and width, correspondingly, k is the number of walls Pavel wants to add and letter s represents the number of empty cells in the original maze.
Each of the next n lines contains m characters. They describe the original maze. If a character on a line equals ".", then the corresponding cell is empty and if the character equals "#", then the cell is a wall.
Output
Print n lines containing m characters each: the new maze that fits Pavel's requirements. Mark the empty cells that you transformed into walls as "X", the other cells must be left without changes (that is, "." and "#").
It is guaranteed that a solution exists. If there are multiple solutions you can output any of them.
Examples
Input
3 4 2
#..#
..#.
#...
Output
#.X#
X.#.
#...
Input
5 4 5
#...
#.#.
.#..
...#
.#.#
Output
#XXX
#X#.
X#..
...#
.#.#
Submitted Solution:
```
import sys
import threading
n,m,k=list(map(int,input().split()))
li=[list(input()) for _ in range(n)]
def issafe(r,c):
return 0<=r<n and 0<=c<m
def fun(r,c,cnt):
#print(r,c)
if cnt[0]>=k:
return
li[r][c]="1"
if issafe(r-1,c) and li[r-1][c]==".":
fun(r-1,c,cnt)
if issafe(r+1,c) and li[r+1][c]==".":
fun(r+1,c,cnt)
if issafe(r,c+1) and li[r][c+1]==".":
fun(r,c+1,cnt)
if issafe(r,c-1) and li[r][c-1]==".":
fun(r,c-1,cnt)
if cnt[0]<k:
cnt[0]+=1
li[r][c]="X"
def main():
cnt,i=[0],0
while(i<n):
for j in range(m):
if cnt[0]<k and li[i][j]==".":
fun(i,j,cnt)
if cnt[0]>=k:
break
if cnt[0]>=k:
break
i+=1
for i in range(n):
for j in range(m):
if li[i][j]=="1":
print(".",end="")
else:
print(li[i][j],end="")
print()
if __name__=="__main__":
sys.setrecursionlimit(10**6)
threading.stack_size(10**8)
t = threading.Thread(target=main)
t.start()
t.join()
``` | instruction | 0 | 27,793 | 15 | 55,586 |
Yes | output | 1 | 27,793 | 15 | 55,587 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Pavel loves grid mazes. A grid maze is an n × m rectangle maze where each cell is either empty, or is a wall. You can go from one cell to another only if both cells are empty and have a common side.
Pavel drew a grid maze with all empty cells forming a connected area. That is, you can go from any empty cell to any other one. Pavel doesn't like it when his maze has too little walls. He wants to turn exactly k empty cells into walls so that all the remaining cells still formed a connected area. Help him.
Input
The first line contains three integers n, m, k (1 ≤ n, m ≤ 500, 0 ≤ k < s), where n and m are the maze's height and width, correspondingly, k is the number of walls Pavel wants to add and letter s represents the number of empty cells in the original maze.
Each of the next n lines contains m characters. They describe the original maze. If a character on a line equals ".", then the corresponding cell is empty and if the character equals "#", then the cell is a wall.
Output
Print n lines containing m characters each: the new maze that fits Pavel's requirements. Mark the empty cells that you transformed into walls as "X", the other cells must be left without changes (that is, "." and "#").
It is guaranteed that a solution exists. If there are multiple solutions you can output any of them.
Examples
Input
3 4 2
#..#
..#.
#...
Output
#.X#
X.#.
#...
Input
5 4 5
#...
#.#.
.#..
...#
.#.#
Output
#XXX
#X#.
X#..
...#
.#.#
Submitted Solution:
```
n,m,k=map(int,input().split())
t=[input().replace('.','X') for i in range(n)]
k=n*m-k-sum(i.count('#')for i in t)
t=[list(i) for i in t]
i,p=0,[]
while k:
if 'X' in t[i]:
j=t[i].index('X')
t[i][j],p='.',[(i,j)]
k-=1
break
i+=1
while k:
x,y=p.pop()
for i,j in ((x,y-1),(x,y+1),(x-1,y),(x+1,y)):
if 0<=i<n and 0<=j<m and t[i][j]=='X':
t[i][j]='.'
p.append((i,j))
k-=1
if k==0: break
for i in range(n):
t[i]=''.join(t[i])
print('\n'.join(t))
``` | instruction | 0 | 27,794 | 15 | 55,588 |
Yes | output | 1 | 27,794 | 15 | 55,589 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Pavel loves grid mazes. A grid maze is an n × m rectangle maze where each cell is either empty, or is a wall. You can go from one cell to another only if both cells are empty and have a common side.
Pavel drew a grid maze with all empty cells forming a connected area. That is, you can go from any empty cell to any other one. Pavel doesn't like it when his maze has too little walls. He wants to turn exactly k empty cells into walls so that all the remaining cells still formed a connected area. Help him.
Input
The first line contains three integers n, m, k (1 ≤ n, m ≤ 500, 0 ≤ k < s), where n and m are the maze's height and width, correspondingly, k is the number of walls Pavel wants to add and letter s represents the number of empty cells in the original maze.
Each of the next n lines contains m characters. They describe the original maze. If a character on a line equals ".", then the corresponding cell is empty and if the character equals "#", then the cell is a wall.
Output
Print n lines containing m characters each: the new maze that fits Pavel's requirements. Mark the empty cells that you transformed into walls as "X", the other cells must be left without changes (that is, "." and "#").
It is guaranteed that a solution exists. If there are multiple solutions you can output any of them.
Examples
Input
3 4 2
#..#
..#.
#...
Output
#.X#
X.#.
#...
Input
5 4 5
#...
#.#.
.#..
...#
.#.#
Output
#XXX
#X#.
X#..
...#
.#.#
Submitted Solution:
```
def main():
n, m, k = map(int, input().split())
mn = m * n
l = [_ == '.' for _ in range(n) for _ in input()]
neigh = [[] for _ in range(mn)]
for y in range(n - 1):
for x in range(m):
yx = y * m + x
if l[yx] and l[yx + m]:
neigh[yx].append(yx + m)
neigh[yx + m].append(yx)
for y in range(n):
for x in range(m - 1):
yx = y * m + x
if l[yx] and l[yx + 1]:
neigh[yx].append(yx + 1)
neigh[yx + 1].append(yx)
res = [True] * mn
cur, nxt = set(), {l.index(True)}
cnt = sum(l) - k
while cnt:
cur, nxt = nxt, cur
while cur:
i = cur.pop()
if res[i]:
res[i] = False
cnt -= 1
if not cnt:
break
for j in neigh[i]:
nxt.add(j)
for x, (a, b) in enumerate(zip(res, l)):
if a and b:
l[x] = 2
l = [''.join(('#', '.', 'X')[_] for _ in l[x:x + m]) for x in range(0, mn, m)]
print('\n'.join(l))
if __name__ == '__main__':
main()
``` | instruction | 0 | 27,795 | 15 | 55,590 |
Yes | output | 1 | 27,795 | 15 | 55,591 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Pavel loves grid mazes. A grid maze is an n × m rectangle maze where each cell is either empty, or is a wall. You can go from one cell to another only if both cells are empty and have a common side.
Pavel drew a grid maze with all empty cells forming a connected area. That is, you can go from any empty cell to any other one. Pavel doesn't like it when his maze has too little walls. He wants to turn exactly k empty cells into walls so that all the remaining cells still formed a connected area. Help him.
Input
The first line contains three integers n, m, k (1 ≤ n, m ≤ 500, 0 ≤ k < s), where n and m are the maze's height and width, correspondingly, k is the number of walls Pavel wants to add and letter s represents the number of empty cells in the original maze.
Each of the next n lines contains m characters. They describe the original maze. If a character on a line equals ".", then the corresponding cell is empty and if the character equals "#", then the cell is a wall.
Output
Print n lines containing m characters each: the new maze that fits Pavel's requirements. Mark the empty cells that you transformed into walls as "X", the other cells must be left without changes (that is, "." and "#").
It is guaranteed that a solution exists. If there are multiple solutions you can output any of them.
Examples
Input
3 4 2
#..#
..#.
#...
Output
#.X#
X.#.
#...
Input
5 4 5
#...
#.#.
.#..
...#
.#.#
Output
#XXX
#X#.
X#..
...#
.#.#
Submitted Solution:
```
n,m,k=map(int,input().split())
a=[['' for x in range(m)] for x in range(n)]
dopi=0
for i in range(n):
a[i]=list(input())
dopi+=a[i].count('.')
b=a.copy()
if k!=0:
pos=[-1,-1]
for lin in range(n):
if pos!=[-1,-1]:
break
for i in range(m):
if a[lin][i]=='.':
pos=[i,lin]
break
import collections
k=dopi-k+1
#print (k)
if (pos!=[-1,-1])and(k!=1)and(b[pos[1]][pos[0]]=='.'):
b[pos[1]][pos[0]]=='1'
k-=1
q=collections.deque([pos])
#print (pos)
while (len(q)!=0)and(k>0):
now=q.popleft()
if (now[1]-1>=0)and(b[now[1]-1] [now[0]]=='.'):
k-=1
b[now[1]-1] [now[0]]='1'
q.append([now[0],now[1]-1])
if k==0:
break
if (now[1]+1<n)and(b[now[1]+1] [now[0]]=='.'):
k-=1
b[now[1]+1] [now[0]]='1'
q.append([now[0],now[1]+1])
if k==0:
break
if (now[0]-1>=0)and(b[now[1]] [now[0]-1]=='.'):
k-=1
b[now[1]] [now[0]-1]='1'
q.append([now[0]-1,now[1]])
if k==0:
break
if (now[0]+1<m)and(b[now[1]] [now[0]+1]=='.'):
k-=1
b[now[1]] [now[0]+1]='1'
q.append([now[0]+1,now[1]])
if k==0:
break
#print(k)
#for i in b:
#print (i)
#print('-'*9)
for i in range(n):
for j in range(m):
if b[i][j]=='1':
b[i][j]='.'
elif b[i][j]=='.':
b[i][j]='X'
s=''
for j in b[i]:
s+=str(j)
print(s)
else:
s=''
for j in b[i]:
s+=str(j)
print(s)
``` | instruction | 0 | 27,796 | 15 | 55,592 |
No | output | 1 | 27,796 | 15 | 55,593 |
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