题目大意：的加强版，求最大子矩阵和，不过矩阵是可以循环的，矩阵到结尾时可以循环到开头。开始听纠结的，想着难道要分情况讨论吗？！就去网上搜，看到可以通过补全进行处理，也是，通过补全一个相同的，问题就迎刃而解了，所以把n*n的矩阵扩展成2n*2n的矩阵就好了。

1 #include 
      
        2 #include 
       
         3 #define MAXN 160 4  5 int a[MAXN][MAXN], sum[MAXN][MAXN]; 6  7 int main() 8 { 9 #ifdef LOCAL10     freopen("in", "r", stdin);11 #endif12     int T;13     scanf("%d", &T);14     while (T--)15     {16         int n;17         scanf("%d", &n);18         memset(sum, 0, sizeof(sum));19         for (int i = 1; i <= n; i++)20         {21             for (int j = 1; j <= n; j++)22                 scanf("%d", &a[i][j]);23             for (int j = n+1; j <= 2*n; j++)24                 a[i][j] = a[i][j-n];25         }26         for (int i = n+1; i <= 2*n; i++)27             for (int j = 1; j <= 2*n; j++)28                 a[i][j] = a[i-n][j];29         for (int i = 1; i <= 2*n; i++)30             for (int j = 1; j <= 2*n; j++)31                 sum[i][j] = sum[i][j-1] + sum[i-1][j] - sum[i-1][j-1] + a[i][j];32         int max = a[1][1];33         for (int i = 1; i <= n; i++)34             for (int p = 0; p < n; p++)35                 for (int j = 1; j <= n; j++)36                     for (int q = 0; q < n; q++)37                     {38                         int t = sum[i+p][j+q] - sum[i+p][j-1] - sum[i-1][j+q] + sum[i-1][j-1];39                         if (t > max)   max = t;40                     }41         printf("%d\n", max);42     }43     return 0;44 }45 46