树状数组详解
https://blog.csdn.net/sjystone/article/details/115396746

简单的整数问题I https://www.acwing.com/problem/content/248/

树状数组拓展
将某一区间的每个数加上c，求a[x] -> 差分
-> 树状数组中存储a[i]的差分

构造差分:
b[i] = a[i] - a[i - 1]

#include <iostream>
#include <cstring>
using namespace std;

typedef long long LL;
const int N = 1e5 + 10;
int a[N], tr[N], n, m;

int lowbit(int x)
{
    return x & -x;
}

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

LL sum(int x)
{
    LL res = 0;
    for ( int i = x; i; i -= lowbit(i) ) res += tr[i];
    return res;
}

int main()
{
    cin >> n >> m;
    for ( int i = 1; i <= n; i ++ ) scanf("%d", &a[i]);
    for ( int i = 1; i <= n; i ++ ) add(i, a[i] - a[i - 1]); //树状数组中存a[i]的差分
    while ( m -- )
    {
        char op[2];
        scanf("%s", op);
        if ( *op == 'C' ) 
        {
            int l, r, d;
            scanf("%d%d%d", &l, &r, &d);
            add(l, d), add(r + 1, -d);  //差分日常操作
        }
        else 
        {
            int x;
            scanf("%d",&x);
            printf("%lld\n", sum(x));     //差分数组的前缀和便是a[i]
        }
    }
    return 0;
}