STL---Codeforces675D Tree Construction(二叉树节点的父亲节点)

Description

During the programming classes Vasya was assigned a difficult problem. However, he doesn't know how to code and was unable to find the solution in the Internet, so he asks you to help.

You are given a sequence a, consisting of ndistinct integers, that is used to construct the binary search tree. Below is the formal description of the construction process.

1. First element a₁ becomes the root of the tree.
2. Elements a₂, a₃, ..., a_n are added one by one. To add element a_i one needs to traverse the tree starting from the root and using the following rules:
   1. The pointer to the current node is set to the root.
   2. If a_i is greater than the value in the current node, then its right child becomes the current node. Otherwise, the left child of the current node becomes the new current node.
   3. If at some point there is no required child, the new node is created, it is assigned value a_i and becomes the corresponding child of the current node.

Input

The first line of the input contains a single integer n (2 ≤ n ≤ 100 000) — the length of the sequence a.

The second line contains n distinct integers a_i (1 ≤ a_i ≤ 10⁹) — the sequence a itself.

Output

Output n - 1 integers. For all i > 1 print the value written in the node that is the parent of the node with value a_i in it.

Sample Input

Input

3
1 2 3

Output

1 2

Input

5
4 2 3 1 6

Output

4 2 2 4

题意：将n个节点依次插入到二叉树中，第一个数作为根节点，输出2~n节点的父亲节点的值；

思路：对于插入的节点i，它的父节点一定是距离它最近的点，比它大的那个点和比它小的点出现最晚的就是父亲节点，可以用pos[]数组来记录个点出现的次序；

代码如下：

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <set>
#include <map>
using namespace std;
set<int>t;
map<int,int>pos;

int main()
{
    int n,x;
    int times=;
    cin>>n;
    scanf("%d",&x);
    t.insert();
    t.insert(x);
    pos[x]=times++;
    while(times<=n)
    {
        scanf("%d",&x);
        set<int>::iterator it = t.lower_bound(x);
        set<int>::iterator it1=it--;
        set<int>::iterator it2=it;
        if(pos[*it1]<pos[*it2])
            printf("%d ",*it2);
        else printf("%d ",*it1);
        t.insert(x);
        pos[x]=times++;
    }
    return ;
}