# 算法基础：子矩阵的和()-其他

## 算法基础：子矩阵的和()

### 算法：子矩阵的和

``````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]

#include <bits/stdc++.h>

using namespace std;

const int N = 1e3 + 10;
int a[N][N], s[N][N];  //s[n][n]二维矩阵前缀和

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

for(int i = 1; i <= n; i++){
for(int j = 1; j <= m; j++){
cin >> 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]; //二维矩阵前缀和的预处理公式
}
}

while(q--){
int x1, y1, x2, y2;
cin >> x1 >> y1 >> x2 >> y2;

cout << s[x2][y2] - s[x2][y1 - 1] - s[x1 -1][y2] + s[x1 -1][y1-1] << endl;
}

return 0;
}

``````
————————

### 算法：子矩阵的和

``````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]

#include <bits/stdc++.h>

using namespace std;

const int N = 1e3 + 10;
int a[N][N], s[N][N];  //s[n][n]二维矩阵前缀和

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

for(int i = 1; i <= n; i++){
for(int j = 1; j <= m; j++){
cin >> 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]; //二维矩阵前缀和的预处理公式
}
}

while(q--){
int x1, y1, x2, y2;
cin >> x1 >> y1 >> x2 >> y2;

cout << s[x2][y2] - s[x2][y1 - 1] - s[x1 -1][y2] + s[x1 -1][y1-1] << endl;
}

return 0;
}

``````