AntiSudoku CodeForces 1335D 多解

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 99 elements of this field (i.e. choose some 1≤i,j≤91≤i,j≤9 and change the number at the position (i,j)(i,j) to any other number) to make it anti-sudoku. The anti-sudoku is the 9×99×9 field, in which:

• Each row contains at least two equal elements;
• each column contains at least two equal elements;
• each 3×33×3 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 tt independent test cases.

Input

The first line of the input contains one integer tt (1≤t≤1041≤t≤104) — the number of test cases. Then tt test cases follow.

Each test case consists of 99 lines, each line consists of 99 characters from 11 to 99 without any whitespaces — the correct solution of the sudoku puzzle.

Output

For each test case, print the answer — the initial field with at most 99 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

ACcode＃１

#include<bits/stdc++.h>
using namespace std;

const int N = 2e5+7;
int cnt[N];

int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);

    int t;
    cin>>t;

    while (t--) {
        for (int i=0; i<9; i++) {
            string s;
            cin>>s;

            for (char &c: s)
                if (c == '1')//随便一个1~9的数字变成1~9的另一个数字即可
                    c = '2';
            cout<<s<<endl;
        }
    }
}

ACcode#2

#include <bits/stdc++.h>
typedef long long ll;
using namespace std;
const ll inf = 1e18;
const int mod = 1000000007;
const int mx = 100; //check the limits, dummy
typedef pair<int, int> pa;
const double PI = acos(-1);
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
#define swa(a,b) a^=b^=a^=b
#define re(i,a,b) for(int i=(a),_=(b);i<_;i++)
#define rb(i,a,b) for(int i=(b),_=(a);i>=_;i--)
#define clr(a) memset(a, 0, sizeof(a))
#define lowbit(x) ((x)&(x-1))
#define mkp make_pair
void sc(int& x) { scanf("%d", &x); }void sc(int64_t& x) { scanf("%lld", &x); }void sc(double& x) { scanf("%lf", &x); }void sc(char& x) { scanf(" %c", &x); }void sc(char* x) { scanf("%s", x); }
ll  m, n,t,x,k,ans=0,sum=0;
int main()
{
    ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    cin >> t;
    while (t--) {
        vector<string> s(9);
        for (int i = 0; i < 9; i++) {
            cin >> s[i];
        }
        vector<int> a = { 0, 4, 8, 1, 5, 6, 2, 3, 7 };
        for (int i = 0; i < 9; i++) {
            int j = a[i];
            s[i][j] = (s[i][j] == '9' ? '1' : s[i][j] + 1);
        }
        for (int i = 0; i < 9; i++) {
            cout << s[i] << '\n';
        }
    }
}

ACcode#3

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;

int main()
{
    int t;
    scanf("%d",&t);
    vector<array<int,2>> v;
    for(int i=0;i<3;i++)
    {
        int r=3*i;
        int c=i;
        for(int j=0;j<3;j++) v.push_back({r+j,c+3*j});
    }
    while(t--)
    {
        char s[9][10];
        for(int o=0;o<9;o++) scanf("%s",s[o]);
        for(auto [a,b]:v)
        {
            if(s[a][b]=='1') s[a][b]='2';
            else s[a][b]='1';
        }
        for(int o=0;o<9;o++) printf("%s\n",s[o]);
    }
    return 0;
}

ACcode#4

#include<bits/stdc++.h>
#define MAXN 100005
#define INF 1000000000
#define MOD 1000000007
#define F first
#define S second
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
int t,n,a[MAXN];
char mp[15][15];
bool valid[10];
int main()
{
    scanf("%d",&t);
    while(t--)
    {
        for(int i=0;i<9;i++) scanf("%s",mp[i]);
        int tot=0,j=0;
        for(int i=0;i<9;i++)
        {
            if(i%3==0)
            {
                j=tot;
                tot++;
            }
            else j+=3;
            if(mp[i][j]=='1') mp[i][j]='2'; else mp[i][j]='1';
        }
        for(int i=0;i<9;i++) printf("%s\n",mp[i]);
    }
    return 0;
}