AcWing 796. 子矩阵的和

# 子矩阵的和 - 二维前缀和

#include <iostream>

const int N = 1010;

int n, m, 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;
}

#include <iostream>

const int N = 1010;

int n, m, 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]; // 求前缀和
            
    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]); // 算子矩阵的和
    }
    
    
    return 0;
}