树状数组

lowbit(x)=2^k
int lowbit(int t)
    {
        return t & (-t);
    }

单点更新
x是更改的位置，y是更改的值
void add(int x, int y)
    {
        for (int i = x; i <= n; i += lowbit(i))
            tree[i] += y;
    }

求A数组中前x项的和 
int getsum(int x)
    {
        int ans = 0;
        for (int i = x; i > 0; i -= lowbit(i))
            ans += tree[i];
        return ans;
    }

大佬的解释

https://www.cnblogs.com/hsd-/p/6139376.html