AcWing 796. 子矩阵的和

这个题的重点是仿照一维的数组，所以a[N][N]也是从1索引开始的。画个图举个例子就非常清晰了

之所以不好理解是因为没画格子，一个格子代表一个点，就很好理解了。

java代码：

import java.io.*;
public class Main{
    static int N = 1010;
    static int[][] arr = new int[N][N];
    static int[][] s = new int[N][N];
    public static void main(String[] args) throws IOException{
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        String[] str1 = br.readLine().split(" ");
        int n = Integer.parseInt(str1[0]);
        int m = Integer.parseInt(str1[1]);
        int q = Integer.parseInt(str1[2]);
        for(int i = 1; i <= n; i++){
            String[] str2 = br.readLine().split(" ");
            for(int j = 1; j <= m; j++){
                arr[i][j] = Integer.parseInt(str2[j-1]);
            }
        }
        for(int i = 1; i <= n; i++){
            for(int j = 1; j <= m;j++){
                s[i][j] = arr[i][j] + s[i-1][j] + s[i][j-1] - s[i-1][j-1];
            }
        }
        for(int i = 0 ; i < q ; i++){
        String[] str3 = br.readLine().split(" ");
        int x1 = Integer.parseInt(str3[0]);
        int y1 = Integer.parseInt(str3[1]);
        int x2 = Integer.parseInt(str3[2]);
        int y2 = Integer.parseInt(str3[3]);
        System.out.println(s[x2][y2] - s[x1 - 1][y2] - s[x2][y1 - 1] + s[x1 - 1][y1 - 1]);
        }
    }
}