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) to make it anti-sudoku. The anti-sudoku is the 9×9 field, in which:
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≤104) — 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
思路:最多改9次,一共有9个块,这样的话,就每个块都改一个就行了。
代码如下:
#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int maxx=10;
int s[maxx][maxx];
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
for(int i=0;i<9;i++)
for(int j=0;j<9;j++) scanf("%1d",&s[i][j]);
s[0][0]=(s[0][0]+1)%10;
s[1][3]=(s[1][3]+1)%10;
s[2][6]=(s[2][6]+1)%10;
s[3][1]=(s[3][1]+1)%10;
s[4][4]=(s[4][4]+1)%10;
s[5][7]=(s[5][7]+1)%10;
s[6][2]=(s[6][2]+1)%10;
s[7][5]=(s[7][5]+1)%10;
s[8][8]=(s[8][8]+1)%10;
for(int i=0;i<9;i++)
{
for(int j=0;j<9;j++) printf("%1d",s[i][j]?s[i][j]:1);
cout<<endl;
}
}
return 0;
}
努力加油a啊,(o)/~