C++

$\color{#cc33ff}{— > 算法基础课题解}$

$\color{gold}{— > 蓝桥杯辅导课题解}$

思路：

二维前缀和

$\color{blue}{模板}$

S[i, j] = 第i行j列格子左上部分所有元素的和
以(x1, y1)为左上角，(x2, y2)为右下角的子矩阵的和为：
S[x2, y2] - S[x1 - 1, y2] - S[x2, y1 - 1] + S[x1 - 1, y1 - 1]

s[i][j] = s[i - 1][j] + s[i][j - 1] - s[i - 1][j - 1]+ a[i][j]; // 求前缀和
s[x2][y2] - s[x1 - 1][y2] - s[x2][y1 - 1] + s[x1 - 1][y1 - 1]);//算子矩阵的和

$求二维前缀和：图解$

$求子矩阵和：图解$

$二维前缀和：$

$时间复杂度：O(nm)$

#include<iostream>
#include<algorithm>
#include<cstring>

using namespace std;

const int N = 1e3 + 10;

int n, m, q;
int a[N][N], s[N][N];

int main() {
    cin >> n >> m >> q;

    for (int i = 1; i <= n; i ++)
        for (int j = 1; j <= m; j ++) {
            cin >> a[i][j];
            s[i][j] = s[i - 1][j] + s[i][j - 1] - s[i - 1][j - 1] + a[i][j]; // 前缀和
        }

    while (q --) {
        int x1, y1, x2, y2;
        cin >> x1 >> y1 >> x2 >> y2;
        cout << s[x2][y2] - s[x1 -  1][y2] - s[x2][y1 - 1] + s[x1 -1][y1 - 1] << endl; // 子矩阵
    }    

    return 0;
}