题目描述

blablabla

样例

#include <iostream>
const int N = 1010;
int m, n, q;
int a[N][N], s[N][N];
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", &a[i][j]);

    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] + a[i][j];

    while(q--){
        int x1, y1, x2, y2;
        scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
        printf("%d\n", s[x2][y2] - s[x1-1][y2] - s[x2][y1-1] + s[x1-1][y1-1]);
    }
    return 0;
}

算法1

(暴力枚举) $O(n^2)$

blablabla

时间复杂度

参考文献

C++ 代码

blablabla

算法2

(暴力枚举) $O(n^2)$

blablabla

时间复杂度

参考文献

C++ 代码

blablabla