一、思路

和一维差分的思想差不多，但是二维处理却麻烦很多(这题用的是BufferedReader和BufferedWriter，因为用Scanner超时了，真的对Java太不友好了，难受)。s[i] [j]为差分数组，我们这里把原数组也当作只处理当前位置的差分来处理，这样可以方便统一操作。insert函数就是用来处理差分的。最后再还原数组就行了。

二、代码模板

import java.io.*;
public class Main {
    static int[][] s;
    public static void insert(int x1, int y1, int x2, int y2, int num){
        s[x1][y1] += num;
        s[x2 + 1][y1] -= num;
        s[x1][y2 + 1] -= num;
        s[x2 + 1][y2 + 1] += num;
    }
    public static void main(String[] args) throws IOException {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        BufferedWriter out = new BufferedWriter(new OutputStreamWriter(System.out));
        String[] str = in.readLine().split(" ");
        int n = Integer.parseInt(str[0]);
        int m = Integer.parseInt(str[1]);
        int q = Integer.parseInt(str[2]);
        s = new int[n + 2][m + 2];      //差分数组
        for (int i = 1; i <= n; i ++){
            str = in.readLine().split(" "); //读入原数组
            for (int j = 1; j <= m; j++)
                insert(i, j, i, j, Integer.parseInt(str[j - 1]));   //因为str下标为0开始
        }
        for (int i = 0; i < q; i++){
            str = in.readLine().split(" ");
            int x1 = Integer.parseInt(str[0]);
            int y1 = Integer.parseInt(str[1]);
            int x2 = Integer.parseInt(str[2]);
            int y2 = Integer.parseInt(str[3]);
            int num = Integer.parseInt(str[4]);
            insert(x1, y1, x2, y2, num);    //操作差分数组
        }
        for (int i = 1; i <= n; i++){
            for (int j = 1; j <= m;j++){    //还原数组
                s[i][j] = s[i][j] + s[i - 1][j] + s[i][j - 1] - s[i - 1][j - 1];
                out.write(s[i][j] + " ");
            }
            out.write("\n");
        }
        out.flush();    //刷新
    }
}

三、复杂度分析

• 时间复杂度：O(m * n)
• 空间复杂度：O(m * n)