Divide by three, multiply by two

最新推荐文章于 2023-10-09 02:02:59 发布

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最新推荐文章于 2023-10-09 02:02:59 发布

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分类专栏： code force

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本文介绍了一个算法问题，即如何通过重新排序一个包含特定数学操作（除以3或乘以2）的结果序列，来还原可能的操作流程。该问题要求输入序列能够按照特定规则重新组织，使得每个数都是前一个数的两倍或是三分之一。

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

divide the number $x$ by $3$ ( $x$ must be divisible by $3$ );
multiply the number $x$ by $2$ .

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

You are given a sequence of length $n$ — 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 $n$ ( $2 \leq n \leq 100$ ) — the number of the elements in the sequence. The second line of the input contains $n$ integer numbers $a_{1}, a_{2}, \dots, a_{n}$ ( $1 \leq a_{i} \leq 3 \cdot 10^{18}$ ) — rearranged (reordered) sequence that Polycarp can wrote down on the board.

Output

Print $n$ 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.

这道题其实就是按能膜3的数目由高到低排列因为只有两种方式一种是除以3一种是乘以2 所以能膜的3的数目肯定是依次递减的

并且如果相邻的两者膜3的数目相等那肯定是数小的在前面数大的在后面选择的是乘以2的方式

所以pair先比较除以3的数目再比较乘以2的数目。

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
typedef long long ll;
ll n,x;
pair<ll,ll>a[105];

int main(void)
{
    ll i,c,t;
    cin>>n;
    for(i=0;i<n;++i)
    {
        cin>>x;
        for(t=x,c=0;t%3==0;t/=3)c++;
        a[i]={-c,x};

    }
    sort(a,a+n);

    for(int i=0;i<n;++i)
    {
        cout<<a[i].first<<'*';
        cout<<a[i].second<<' ';
    }
    return 0;



}