Codeforces977D——Divide by three, multiply by two

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≤n≤100) — the number of the elements in the sequence. The second line of the input contains n integer numbers a1,a2,…,an (1≤ai≤3⋅1018) — 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
Input
6
4 8 6 3 12 9
Output
9 3 6 12 4 8
Input
4
42 28 84 126
Output
126 42 84 28
Input
2
1000000000000000000 3000000000000000000
Output
3000000000000000000 1000000000000000000

这题感觉挺好的，数的变化，给出两种变换的规律，/3 或者 *2 然后给出一列数，求真正变换的序列
dfs一波
用map记录每个数出现的次数，可以在bfs中判断，然后就是枚举最后一个数了，然后dfs递推进去，有条件不满足就退出，当所有条件都满足了，就从底层输出。

代码：

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <map>
using namespace std;
const int N=105;
long long a[N];
map<int,long long> t;
map<long long,int> r;
int n;
int flag;
bool dfs(long long a,int t){
    //printf("%lld %d\n",a,t);
    r[a]--;
    if(t==n){
        if(r[a]>=0){
            //printf("%lld ",a);
            return true;
        }
        else{
            return false;
        }
    }
    else{
        int fa=0;
        int fb=0;
        long long ta=3*a;
        if(r[ta]==0){
            fa=1;
        }
        else{
            if(dfs(ta,t+1)){
                printf("%lld ",ta);
                return true;
            }
            else{
                r[ta]++;
                return false;
            }
        }
        if(a%2!=0){
            return false;
        }
        ta=a/2;
        if(r[ta]==0){
            fb=1;
        }
        else{
            if(dfs(ta,t+1)){
                printf("%lld ",ta);
                return true;
            }
            else{
                r[ta]++;
                return false;
            }
        }
        if(fa && fb){
            return false;
        }
    }
}
int main(void){
    //freopen("data.txt","r",stdin);
    scanf("%d",&n);
    for(int i=0;i<n;i++){
        scanf("%lld",&a[i]);
        t[i]=a[i];
        r[a[i]]++;
    }
    // if(dfs(a[5],1)){
   
    // printf("%lld\n",a[1]);
    //
    // }
    for(int i=0;i<n;i++){
        if(dfs(a[i],1)){
            printf("%lld\n",a[i]);
            break;
        }
        r[a[i]]++;
    }
    return 0;
}