Codeforces 977D Divide by three, multiply by two（拓扑排序）

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
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: $[9, 3, 6, 12, 4, 8]$ . It can match possible Polycarp's game which started with $x = 9$ .

题目大意：给定的数组按照以下要求排序：后一个数是前一个数的三分之一，或者是前一个数的二倍。

思路：如果a[v]是a[u]的三分之一或者二倍，就给u->v加一条有向边，然后跑一遍拓扑排序就行了，注意得到的拓扑数组是下标

代码：

#include<cstdio>
#include<iostream>
#include<string>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<set>
#include<vector>
#include<map>
typedef long long ll;
using namespace std;
const int maxn = 200000 + 100;
int G[100 +10][100 + 10];
int c[maxn];
ll topo[maxn], t;
int n;

bool dfs(int u){
    c[u] = -1;
    for(int i = 0; i < n; i++)if(G[u][i]){
    	if(c[i] < 0)return false;
        else if(!c[i] && !dfs(i))return false;
	}
	c[u] = 1;topo[--t] = u;
	return true;
}
bool toposort(){
	t = n;
	memset(c, 0, sizeof(c));
	for(int i = 0; i < n; i++)if(!c[i]){
		if(!dfs(i)) return false;
	}
	return true;
}
int main(){
	scanf("%d", &n);
	ll a[maxn];
	memset(G, 0, sizeof(G));
	for(int i = 0; i < n; i++){
		scanf("%lld", &a[i]);
		
	}
	sort(a, a+n);
	for(int i = 0; i < n ; i++){
		int it = lower_bound(a, a+n, a[i]/3)-a;
		int itt = lower_bound(a, a+n, a[i]*2)-a;
		if(a[i]%3==0&&a[it]==a[i]/3)G[i][it] = 1;
		if(a[i]*2==a[itt])G[i][itt] = 1;
		
	}
	
	toposort();
	for(int i = 0; i < n; i++)printf("%lld ", a[topo[i]]);
}