首先将树状数组中的数都置为1，代表还没有数被选
然后从后往前遍历，去找处在a[i] + 1位置上且没被选的牛是哪一个，过程用二分维护即可

#include <bits/stdc++.h>

using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> PII;

const int N = 100010;

int tr[N];
int n;
int a[N], ans[N], idx;

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

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

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

int main()
{
    ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    cin >> n;

    for (int i = 2; i <= n; i ++) cin >> a[i];

    for (int i = 1; i <= n; i ++) add(i, 1);

    for (int i = n; i >= 1; i --)
    {
        int y = a[i] + 1;
        int l = 1, r = n;
        while (l < r)
        {
            int mid = l + r >> 1;
            if (sum(mid) >= y) r = mid;
            else l = mid + 1;
        }
        add(l, -1);
        ans[++ idx] = l;
    }

    for (int i = n; i >= 1; i --) cout << ans[i] << endl;

    return 0;
}