#13. 井字棋

井字棋

见pdf文档

#include <cstring>
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cassert>
using namespace std;
const int N = 4 + 5;
char s[N][N];
int a[N][N];

int main()
{
    int T;
    scanf("%d", &T);
    while (T--)
    {	
        scanf("%s%s%s", s[1], s[2], s[3]);
        for (int i = 1; i <= 3; i++)
            for (int j = 1; j <= 3; j++)
                a[i][j] = s[i][j - 1] - '0';

    /*
        a[i][j]存储第i行第j个数 0/1/2
        已处理T组数据,每组输出一行"tie"(不包含引号)


    */
        printf("tie\n");

    }
    return 0;
}

`