题目描述

blablabla

样例

#include <iostream>

using namespace std;

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]);

    // 初始化前缀和数组
    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];

    // 询问
    while (q -- )
    {
        int x1, y1, x2, y2;
        scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
        printf("%d\n", s[x2][y2] - s[x2][y1 - 1] - s[x1 - 1][y2] + s[x1 - 1][y1 - 1]);
    }

    return 0;
}

算法1

(暴力枚举) $O(n^2)$

blablabla

时间复杂度分析：blablabla

C++ 代码

import java.lang.reflect.Array;
import java.math.BigInteger;
import java.util.*;


public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt(), m = sc.nextInt(), q = sc.nextInt() ;
        int[][] matrix = new int[n+1][m+1];
        int[][] sum = new int[n+1][m+1];
        for(int i = 1;i<=n;i++)
            for(int j = 1; j <=m; j++){
                matrix[i][j]=sc.nextInt();
                sum[i][j] = sum[i-1][j]+sum[i][j-1]-sum[i-1][j-1]+matrix[i][j];
            }
        while(q-->0){
            int x1 = sc.nextInt();
            int y1 = sc.nextInt();
            int x2 = sc.nextInt();
            int y2 = sc.nextInt();
            System.out.println(sum[x2][y2]-sum[x1-1][y2]-sum[x2][y1-1]+sum[x1-1][y1-1]);
        }
    }

}

算法2

(暴力枚举) $O(n^2)$

blablabla

时间复杂度分析：blablabla

C++ 代码

blablabla