Codeforces Round #634 D. Anti-Sudoku（构造/水）

题目描述

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

• Any number in this field is in range [1;9] [1; 9] [1;9] ;
• each row contains at least two equal elements;
• each column contains at least two equal elements;
• each 3×3 3 imes 3 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 t t independent test cases.

输入格式

The first line of the input contains one integer t t t ( 1≤t≤104 1 le t le 10^4 1≤t≤104 ) — the number of test cases. Then t t t test cases follow.

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

输出格式

For each test case, print the answer — the initial field with at most 9 9 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.

输入输出样例

输入 #1 复制

1
154873296
386592714
729641835
863725149
975314628
412968357
631457982
598236471
247189563

输出 #1 复制

154873396
336592714
729645835
863725145
979314628
412958357
631457992
998236471
247789563
一开始构造麻烦了。。其实对于其中一个数x，把所有数变成x+1（9的话变成1）即可。因为每一行，每一列，每相邻九格里一定有一个x，把它+1一定会和另一个数重复，满足要求。
代码瞎邒写的，切勿参考。

#include <bits/stdc++.h>
using namespace std;
char a[11][11];
int main()
{
    int t;
    cin>>t;
    getchar();
    while(t--)
    {
        int i,j;
        for(i=0;i<9;i++)
        {
            for(j=0;j<9;j++)
            {
                scanf("%c",&a[i][j]);
            }
            getchar();
        }
        
        for(i=0;i<9;i++)
        {
            int j;
            switch(i)
            {
                case 0:
                    j=0;
                    break;
                case 1:
                    j=4;
                    break;
                case 2:
                    j=8;
                    break;
                case 3:
                    j=1;
                    break;
                case 4:
                    j=5;
                    break;
                case 5:
                    j=6;
                    break;
                case 6:
                    j=2;
                    break;
                case 7:
                    j=3;
                    break;
                case 8:
                    j=7;
                    break;    
            }
            if(a[i][j]!='9')a[i][j]=(int)(a[i][j]-'0'+1)+'0';
            else a[i][j]='1';
        }
        
        
        for(i=0;i<=8;i++)
        {
            for(j=0;j<=8;j++)
            {
                printf("%c",a[i][j]);
            }
            cout<<endl;
        }
    }
    //0 0 0 4 0 8
    //4 0 4 4 0 8
    //8 0 8 4 8 8
    
    return 0;
}