D. Divide by three, multiply by two
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?1 operations 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.
题意：
给你n个数，让你给他们排序，后一个数应该是前一个数的二倍 或 三分之一，输出任意一种结果，数据保证有解。
思路：
深搜，n个点，如果点 j 是 点 i 的二倍 或 三分之一， 那么点 i 到 点 j 有一条单向边 ，最终的排序结果，也就是遍历所有点的任意一种方式。
参考代码：
#include <iostream>
#include <algorithm>
#include <vector>
#include <cstdio>
#include <cstring>using namespace std;
long long a[110];
vector <int> b[110];
int vis[110];
bool flag = 0;
int n;void dfs(int i, long long ans[], int k) {ans[k] = a[i];if(flag) return ;if(k >= n) {for(int i=1; i<=n; i++)cout << ans[i] << " ";cout << endl;flag = true;}int len = b[i].size();for(int j=0; j<len; j++) {if(vis[b[i][j]])continue;vis[b[i][j]] = 1;dfs(b[i][j], ans, k+1);vis[b[i][j]] = 0;}
}int main() {flag = false;cin >> n;for(int i=1; i<=n; i++) {cin >> a[i];}for(int i=1; i<=n; i++) {for(int j=1; j<=n; j++) {if(a[j]== a[i]*2 || a[i] == a[j]*3) {b[i].push_back(j);}}}/* for(int i=1; i<=n; i++) {int len = b[i].size();for(int j=0; j<len; j++) {cout << a[b[i][j]] << " ";}cout << endl;}*/long long ans[110];for(int i=1; i<=n; i++) {memset(vis, 0, sizeof(vis));vis[i] = 1;dfs(i, ans, 1);if(flag) {break;}}return 0;
}