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

谜一样的牛 https://www.acwing.com/problem/content/245/

从后往前做，若最后一头牛前有n头牛比他矮，则最后一头牛排第n+1，再推倒数第二头…
1.从剩余的数中，找出第k小的数
2.删除这个数 （1和2是平衡树的基本操作）

1.a1～an初始为1，表示每个身高可以用一次
2.用树状数组维护前缀和，sum(x)表示1～x中剩余多少数
3.从剩余的数中找到第k小的数 -> 找到一个最小的x，使得sum(x) = k
4.sum是单调的，可以用二分找

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

const int N = 1e5 + 10;
int h[N], ans[N], tr[N], n;

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;
}

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

int main()
{
    cin >> n;
    for ( int i = 2; i <= n; i ++ ) cin >> h[i];
    for ( int i = 1; i <= n; i ++ ) tr[i] = lowbit(i);  //每个ai都是1，tr[i] = lowbit(i)
    for ( int i = n; i; i -- )
    {
        int k = h[i] + 1;   //前面有h[i]个牛比他矮，则他高h[i]+1
        int l = 1, r = n;
        while ( l < r )
        {
            int mid = l + r >> 1;
            if ( sum(mid) >= k ) r = mid;
            else l = mid + 1;
        }
        ans[i] = r;
        add(r, -1);
    }
    for ( int i = 1; i <= n; i ++ ) cout << ans[i] << ' ' ;
    return 0;
}