Divide by three, multiply by two CodeForces - 977D（思维+map）

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本文介绍了一个特定条件下的数列排序算法，目标是将一个包含特定数学操作（乘以2或除以3）的数列重新排序，使得每个元素都是前一个元素的两倍或三分之一。该算法首先找到数列中的最小值作为起点，然后通过递归地查找可乘以2或可被3整除的数来构建排序后的数列。

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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.

Examples
inputCopy
6
4 8 6 3 12 9
outputCopy
9 3 6 12 4 8 
inputCopy
4
42 28 84 126
outputCopy
126 42 84 28 
inputCopy
2
1000000000000000000 3000000000000000000
outputCopy
3000000000000000000 1000000000000000000 

Note

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

题意：给出一串长为n的数列，现给出一个排序方案 ai是ai+1的3倍或者ai+1是ai的2倍，保证给出的数列一定可以排序。

思路：因为最小值只有一种方案，所以现找出最小值，然后由最小值向两边加点就可以了（这里用map来判断一个数是否是给出的数列中的数）

#include "iostream"
#include "algorithm"
#include "map"
using namespace std;
typedef long long ll;
ll ans[205],Min=1e19,a;
map<ll,bool> mp;
int main()
{
    ios::sync_with_stdio(false);
    mp.clear();
    int n,t=1,l,r;
    cin>>n;
    for(int i=0;i<n;i++){cin>>a;mp[a]=1;Min=min(Min,a);}
    l=r=100;//有中间向两边加点
    ans[100]=Min;
    while(1){//向左边加点
        if(mp[ans[l]*3]) l--,ans[l]=ans[l+1]*3;
        else if(ans[l]%2==0&&mp[ans[l]/2]) l--,ans[l]=ans[l+1]/2;
        else break;
    }
    while(1&&r-l<n){//向右边加点
        if(mp[ans[r]*2]) r++,ans[r]=ans[r-1]*2;
        else if(ans[r]%3==0&&mp[ans[r]/3]) r++,ans[r]=ans[r-1]/3;
        else break;
    }
    for(int i=l;i<=r;i++){
        cout<<ans[i];
        if(i!=r)
            cout<<" ";
        else
            cout<<endl;
    }
    return 0;
}