[AcWing 796] 子矩阵的和

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#include<iostream>
using namespace std;
const int N = 1e3 + 10;
int a[N][N], s[N][N];
int main()
{
    int n, m, q;
    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;
}

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1. 前缀和 s[ i ][ j ] 的计算公式，s[ i ][ j ] = s[ i - 1][ j ] + s[ i ][ j - 1 ] -s[ i - 1 ][ j - 1 ] + a[ i ][ j ]；
2. ( x2, y2 ) 和 ( x1, y1 ) 之间的矩阵和等于 s[ x2 ][ y2 ] - s[ x1 - 1 ][ y2 ] - s[ x2 ][ y1 - 1 ] + s[ x1 - 1 ][ y1 - 1 ]；
3. 可以画图，用表格表示，便于理解；