心得

code 半小时 debug 两小时。。
Java配合bufferedreader写的是心累啊。。。
需要注意的有两点：
1. 输入如果是Scanner，则时间不够用，类似地，矩阵前缀和也是那样，需要用到BufferReader
2. System.out.println也不可以作为输出，即使一样也会报错。这个不知道是不是bug，而改成BufferWriter则可以AC

算法1

C++ 代码

import java.io.*;

public class Main {

    static void insert(int i1, int j1, int i2, int j2, int x, int[][] b) {
        b[i1][j1] += x;
        b[i1][j2 + 1] -= x;
        b[i2 + 1][j1] -= x;
        b[i2 + 1][j2 + 1] += x;
    }
    public static void main(String[] args) throws IOException {
        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        BufferedWriter writer = new BufferedWriter(new OutputStreamWriter(System.out));
        String[] str1 = reader.readLine().split(" ");
        int n = Integer.parseInt(str1[0]);
        int m = Integer.parseInt(str1[1]);
        int q = Integer.parseInt(str1[2]);
        int N = 1010;
        int[][] b = new int[N][N];

        for (int i = 1; i <= n; i++) {
            String[] str2 = reader.readLine().split(" ");
            for (int j = 1; j <= m; j++) {
                int x = Integer.parseInt(str2[j - 1]);
                insert(i, j, i, j, x, b);
            }
        }

        while (q > 0) {
            String[] str3 = reader.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]);
            int c = Integer.parseInt(str3[4]);
            insert(x1, y1, x2, y2, c, b);
            q--;
        }

        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= m; j++) {
                b[i][j] += b[i - 1][j] + b[i][j - 1] - b[i - 1][j - 1]; 
                writer.write(b[i][j] + " ");
                //和求S一样，因为b[i - 1][j]和b[i][j - 1]之前已经被算过了，都表示前缀和，b[i][j]还没有被算过   
            }
            writer.write("\n");
        }
        writer.flush();
    }
}