AtCoder Beginner Contest 088 C Takahashi's Information

Problem Statement

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

枚举。
代码：

#include <bits/stdc++.h>
using namespace std;
int s[3][3];
int check()
{
    if(s[0][1] - s[0][0] != s[1][1] - s[1][0] || s[0][1] - s[0][0] != s[2][1] - s[2][0])return 0;
    if(s[0][2] - s[0][0] != s[1][2] - s[1][0] || s[0][2] - s[0][0] != s[2][2] - s[2][0])return 0;
    if(s[0][1] - s[0][2] != s[1][1] - s[1][2] || s[0][1] - s[0][2] != s[2][1] - s[2][2])return 0;
    if(s[0][0] - s[1][0] != s[0][1] - s[1][1] || s[0][0] - s[1][0] != s[0][2] - s[1][2])return 0;
    if(s[0][0] - s[2][0] != s[0][1] - s[2][1] || s[0][0] - s[2][0] != s[0][2] - s[2][2])return 0;
    if(s[2][0] - s[1][0] != s[2][1] - s[1][1] || s[2][0] - s[1][0] != s[2][2] - s[1][2])return 0;
    return 1;
}
int main()
{
    for(int i = 0;i < 3;i ++)
    {
        for(int j = 0;j < 3;j ++)
        {
            cin>>s[i][j];
        }
    }
    if(check())cout<<"Yes";
    else cout<<"No";
}