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;
}

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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;
}