Codeforces Round #634 (Div. 3) D.Anti-Sudoku

Codeforces Round #634 (Div. 3) D.Anti-Sudoku

题目链接
You are given a correct solution of the sudoku puzzle. If you don’t know what is the sudoku, you can read about it here.

The picture showing the correct sudoku solution:

Blocks are bordered with bold black color.

Your task is to change at most 9 elements of this field (i.e. choose some 1≤i,j≤9 and change the number at the position (i,j) to any other number in range [1;9]) to make it anti-sudoku. The anti-sudoku is the 9×9 field, in which:

• Any number in this field is in range [1;9];
• each row contains at least two equal elements;
• each column contains at least two equal elements;
• each 3×3 block (you can read what is the block in the link above) contains at least two equal elements.

It is guaranteed that the answer exists.

You have to answer t independent test cases.

Input

The first line of the input contains one integer t (1≤t≤1e4) — the number of test cases. Then t test cases follow.

Each test case consists of 9 lines, each line consists of 9 characters from 1 to 9 without any whitespaces — the correct solution of the sudoku puzzle.

Output

For each test case, print the answer — the initial field with at most 9 changed elements so that the obtained field is anti-sudoku. If there are several solutions, you can print any. It is guaranteed that the answer exists.

Example

input

1
154873296
386592714
729641835
863725149
975314628
412968357
631457982
598236471
247189563

output

154873396
336592714
729645835
863725145
979314628
412958357
631457992
998236471
247789563

思维题，我们知道每一个位置的元素都锁定一行一列，那么我们九个点要覆盖所有行所有列，我们考虑如下点，也可以不按我的方法，满足上述条件即可：
$$\left[\begin{array}{ccccccccc}1& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 1& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 1& 0& 0\\ 0& 1& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 1& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 1& 0\\ 0& 0& 1& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 1& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 1\end{array}\right]$$
1的位置代表我选择的点，只需把点的位置数改成其他任意一个数即可，AC代码如下：

#include
using namespace std;
typedef long long ll;

int main(){
    int t;
    cin>>t;
    while(t--){
        char a[15][15];
        for(int i=0;i<9;i++){
            for(int j=0;j<9;j++){
                cin>>a[i][j];
            }
        }
        char b[10]={'1','2','3','4','5','6','7','8','9'};
        for(int i=0;i<10;i++) if(b[i]!=a[0][0]) {a[0][0]=b[i];break;}
        for(int i=0;i<10;i++) if(b[i]!=a[1][3]) {a[1][3]=b[i];break;}
        for(int i=0;i<10;i++) if(b[i]!=a[2][6]) {a[2][6]=b[i];break;}
        for(int i=0;i<10;i++) if(b[i]!=a[3][1]) {a[3][1]=b[i];break;}
        for(int i=0;i<10;i++) if(b[i]!=a[4][4]) {a[4][4]=b[i];break;}
        for(int i=0;i<10;i++) if(b[i]!=a[5][7]) {a[5][7]=b[i];break;}
        for(int i=0;i<10;i++) if(b[i]!=a[6][2]) {a[6][2]=b[i];break;}
        for(int i=0;i<10;i++) if(b[i]!=a[7][5]) {a[7][5]=b[i];break;}
        for(int i=0;i<10;i++) if(b[i]!=a[8][8]) {a[8][8]=b[i];break;}
        for(int i=0;i<9;i++){
            for(int j=0;j<9;j++){
                cout<<a[i][j];
            }
            puts("");
        }
    }
    return 0;
}