D. Divide by three, multiply by two(要什么脑子，暴力吖)

最新推荐文章于 2023-09-27 12:16:52 发布

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最新推荐文章于 2023-09-27 12:16:52 发布

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本文介绍了一个有趣的问题：如何重新排列一个包含特定操作后的数值序列，使其满足后一个数要么是前一个数的三分之一，要么是两倍的条件。通过一个简单而巧妙的方法，确保了在任意给定的序列中都能找到正确的顺序。

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D. Divide by three, multiply by two

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Polycarp likes to play with numbers. He takes some integer number xx, writes it down on the board, and then performs with it n−1n−1operations of the two kinds:

divide the number xx by 33 (xx must be divisible by 33);
multiply the number xx by 22.

After each operation, Polycarp writes down the result on the board and replaces xx by the result. So there will be nn numbers on the board after all.

You are given a sequence of length nn — the numbers that Polycarp wrote down. This sequence is given in arbitrary order, i.e. the order of the sequence can mismatch the order of the numbers written on the board.

Your problem is to rearrange (reorder) elements of this sequence in such a way that it can match possible Polycarp's game in the order of the numbers written on the board. I.e. each next number will be exactly two times of the previous number or exactly one third of previous number.

It is guaranteed that the answer exists.

Input

The first line of the input contatins an integer number nn (2≤n≤1002≤n≤100) — the number of the elements in the sequence. The second line of the input contains nn integer numbers a1,a2,…,ana1,a2,…,an (1≤ai≤3⋅10181≤ai≤3⋅1018) — rearranged (reordered) sequence that Polycarp can wrote down on the board.

Output

Print nn integer numbers — rearranged (reordered) input sequence that can be the sequence that Polycarp could write down on the board.

It is guaranteed that the answer exists.

Examples

input

Copy

6
4 8 6 3 12 9

output

Copy

9 3 6 12 4 8

input

Copy

4
42 28 84 126

output

Copy

126 42 84 28

input

Copy

2
1000000000000000000 3000000000000000000

output

Copy

3000000000000000000 1000000000000000000

Note

In the first example the given sequence can be rearranged in the following way: [9,3,6,12,4,8][9,3,6,12,4,8]. It can match possible Polycarp's game which started with x=9x=9.

看题目，给你一个序列，重新排序使得新数列满足以下条件

后一个数要么是前一个数的三分之一，要么是两倍。完。。。

仔细看一遍题，保证一定有解，并且，解唯一（没说多组答案）

然后就好玩了，随便从原数列中抽一个数，向后向前延伸即可，

本来是要dfs的，后来一想，题目这么氵，那就直接写好了

时间复杂度大概emm直接就是O（n）吧

ac代码如下

#include<bits/stdc++.h>
#include<stdio.h>
#include<stdlib.h>
#include<time.h>
#include<string.h>
#include <queue>
using namespace std;
typedef long long ll;
ll cnt[105];
ll num[300];
int n;
map<ll,int> P;
int main()
{
    cin>>n;
    for(int i=0;i<n;i++)
        cin>>cnt[i],P[cnt[i]]++;

    num[105]=cnt[0];

    for(int i=105;;i++)
    {
        ll c=num[i];
        if(c%3==0 && P[c/3])
        {
            P[c/3]--;
            num[i+1]=c/3;
        }
        else if(P[c*2])
        {
            P[c*2]--;
            num[i+1]=c*2;
        }
        if(!num[i+1])
            break;
    }

    for(int i=105;;i--)
    {
        ll c=num[i];
        if(P[c*3])
        {
            P[c*3]--;
            num[i-1]=c*3;
        }
        else if(c%2==0 && P[c/2])
        {
            P[c/2]--;
            num[i-1]=c/2;
        }
        if(!num[i-1])
            break;
    }

    int i=0;

    while(!num[i])i++;

    printf("%lld",num[i++]);

    while(num[i])printf(" %lld",num[i++]);

    printf("\n");


    return 0;
}