题目：

1. 第一类指令形如“C l r d”，表示把数列中第l~r个数都加d。

2. 第二类指令形如“Q X”，表示询问数列中第x个数的值。

分析树状数组

1. 修改某一个元素
2. 求前缀和

差分

可以用差分的思想，完成题目的求法。
差分： b1 + b2 + … bn = an
-> b1 = a1 - a0
-> b2 = a2 - a1
…

import java.io.*;
class Main{
    static BufferedReader read = new BufferedReader(new InputStreamReader(System.in));
    static BufferedWriter log  = new BufferedWriter(new OutputStreamWriter(System.out));
    static int[] tr, a;
    static int n, m;
    public static void main(String[] args) throws Exception{
        String[] ss = read.readLine().split(" ");
        n = Integer.valueOf(ss[0]); m = Integer.valueOf(ss[1]);
        tr = new int[n + 1]; a = new int[n + 1];
        ss = read.readLine().split(" ");
        for(int i = 1; i <= n; i++) {
            a[i] = Integer.valueOf(ss[i - 1]);
            add(i, a[i] - a[i - 1]);
        }
        while(m -- > 0){
            ss = read.readLine().split(" ");
            switch(ss[0]){
                case "C":
                    int l = Integer.valueOf(ss[1]);
                    int r = Integer.valueOf(ss[2]);
                    int c = Integer.valueOf(ss[3]);
                    add(l, c); add(r + 1, -c);
                    break;
                case "Q":
                    int x = Integer.valueOf(ss[1]);
                    log.write(sum(x) + "\n");
                    break;
            }
        }
        log.flush();

    }

    public static void add(int x, int c){
        for(int i = x; i <= n; i += lowBit(i)) {
            tr[i] += c;
        }
    }

    public static long sum(int r){
        long res = 0;
        for(int i = r; i > 0; i -= lowBit(i)) res += tr[i];
        return res;
    }

    public static int lowBit(int x){
        return x & - x;
    }
}