AtCoder-3920

We have a 3×3 grid. A number c_i,j is written in the square (i,j), where (i,j) denotes the square at the i-th row from the top and the j-th column from the left.
According to Takahashi, there are six integers a₁,a₂,a₃,b₁,b₂,b₃ whose values are fixed, and the number written in the square (i,j) is equal to a_i+b_j.
Determine if he is correct.

Constraints

• c_i,j (1≤i≤3,1≤j≤3) is an integer between 0 and 100 (inclusive).

Input

Input is given from Standard Input in the following format:

c1,1 c1,2 c1,3
c2,1 c2,2 c2,3
c3,1 c3,2 c3,3

Output

If Takahashi's statement is correct, print Yes; otherwise, print No.

Sample Input 1

1 0 1
2 1 2
1 0 1

Sample Output 1

Yes

Takahashi is correct, since there are possible sets of integers such as:a₁=0,a₂=1,a₃=0,b₁=1,b₂=0,b₃=1.

Sample Input 2

2 2 2
2 1 2
2 2 2

Sample Output 2

No

Takahashi is incorrect in this case.

Sample Input 3

0 8 8
0 8 8
0 8 8

Sample Output 3

Yes

Sample Input 4

1 8 6
2 9 7
0 7 7

Sample Output 4

No

题解：这一题由于给的数据范围不大，所以可根据关系式，遍历每个整数A；看是否找到使关系式成立的A。如有，则输出Yes，否则输出No

AC代码为：

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;

int a[4][4];
int main()
{
for (int i = 1; i <= 3; i++)
{
for (int j = 1; j <= 3; j++)
{
cin >> a[i][j];
}
}
int flag = 1;
for (int k = -1000; k <= 1000; k++)
{
int a2, a3, b1, b2, b3;
int a1 = k;
b1 = a[1][1] - k;
b2 = a[1][2] - k;
b3 = a[1][3] - k;
a2 = a[2][1] - b1;
a3 = a[3][1] - b1;
if (a[2][2] != a2 + b2) flag = 0;
if (a[2][3] != a2 + b3) flag = 0;
if (a[3][2] != a3 + b2) flag = 0;
if (a[3][3] != a3 + b3) flag = 0;
if (flag)
{
break;
}
}
if (flag)
cout << "Yes" << endl;
else
cout << "No" << endl;
return 0;
}