无脑操作题，由于n<=100,直接枚举所有的子矩阵即可。

#include 
    
     #include 
     
      using namespace std;//Mystery_Sky//#define M 1000int f[M][M], a[M][M];int n, ans, maxn;int main() {    scanf("%d", &n);    for(int i = 1; i <= n; i++)         for(int j = 1; j <= n; j++) {            scanf("%d", &a[i][j]);            f[i][j] = f[i][j-1] + a[i][j];                  }    for(int i = 1; i <= n; i++) {        for(int j = i; j <= n; j++) {            maxn = 0;            for(int k = 1; k <= n; k++) {                maxn += f[k][j] - f[k][i-1];                ans = max(maxn, ans);                if(maxn < 0) maxn = 0;            }        }    }    printf("%d\n", ans);    return 0;}