796. 子矩阵的和(暴力)

#include<iostream>
using namespace std;
const int N = 1010;
int s[N][N];
int n,m,q;

int main(){

    scanf("%d%d%d",&n,&m,&q);
    for(int i = 1 ; i <= n ; i ++){
        for(int j = 1 ; j <= m ; j ++){
            scanf("%d",&s[i][j]);
        }
    }
    //求出每个矩阵和
    for(int i = 1 ; i <= n ; i ++){
        for(int j = 1 ; j <= m ; j ++){
            s[i][j] += s[i-1][j] + s[i][j-1] - s[i-1][j-1];
        }
    }
    //大矩阵 删除左边 和上边 然后再把删除的交集补上
    int x1,x2,y1,y2;
    while(q --){
        scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
        printf("%d\n",s[x2][y2] - s[x2][y1-1] - s[x1-1][y2] + s[x1-1][y1-1]);
    }



    return 0;
}