题干：
Limak is an old brown bear. He often plays poker with his friends. Today they went to a casino. There are n players (including Limak himself) and right now all of them have bids on the table. i-th of them has bid with size ai dollars.
Each player can double his bid any number of times and triple his bid any number of times. The casino has a great jackpot for making all bids equal. Is it possible that Limak and his friends will win a jackpot?
Input
First line of input contains an integer n (2 ≤ n ≤ 105), the number of players.
The second line contains n integer numbers a1, a2, ..., an (1 ≤ ai ≤ 109) — the bids of players.
Output
Print "Yes" (without the quotes) if players can make their bids become equal, or "No" otherwise.
Examples
Input
4
75 150 75 50
Output
Yes
Input
3
100 150 250
Output
No
Note
In the first sample test first and third players should double their bids twice, second player should double his bid once and fourth player should both double and triple his bid.
It can be shown that in the second sample test there is no way to make all bids equal.
题目大意：
最开始每个人都有一个值，任何一个人都可以将自己的值变成原来的两倍或三倍，问能不能让所有人的值都相等。
解题报告：
除了暴力想不到别的方法了，，随便交一发没想到a了。。不过不会算时间复杂度？感觉o(n^2)级别的吧？本来想先把a[i]都存下，然后再o(n)判断一遍的，但是感觉可能会超时？(毕竟不会算复杂度)于是采取边处理便判断的方式，找到NO的就直接输出然后结束程序了，算是一种剪枝？
至于这道题的思路，也挺好想的吧算是。我们分析最终得数，如果最终得数要相同，因为可以乘任意次，也就是说最终得数中可以包含任意个3和2做因子，于是我们把这些因子除去，看剩下的数字是否是一样的，也就可以判断最终的答案是否是一样的。
法2：人选两个人出来，a1,a2,按题意方法要有
AC代码：
//暴力？
#include<bits/stdc++.h>
using namespace std;
const int MAX = 1e5 +5;
int a[MAX];
int main()
{
int n;
cin>>n;
for(int i = 1; i<=n; i++) scanf("%d",a+i);
for(int i = 1; i<=n; i++) {
while(a[i]%6==0) a[i]/=6;
while(a[i]%2==0) a[i]/=2;
while(a[i]%3==0) a[i]/=3;
if(i==1) continue;
if(a[i] != a[i-1]) {
puts("No");return 0;
}
}
puts("Yes");
return 0 ;
}