Codeforces Round #318 [RussianCodeCup Thanks-Round] (Div. 2)C. Bear and Poker

                                              C. Bear and Poker

                                                                 time limit per test 

2 seconds

                                                                 memory limit per test 

256 megabytes

 

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.

Sample test(s)

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.

题意：给你n个数，每个数可以乘2或3，问你是否能让所有数相同；

题解：先找出最大公约数，再不断除2，3，判断是否是1就好了

#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <queue>
#include <typeinfo>
#include <map>
#include <stack>
typedef __int64 ll;
#define inf 0x7fffffff
using namespace std;
inline ll read()
{
    ll x=,f=;
    char ch=getchar();
    while(ch<''||ch>'')
    {
        if(ch=='-')f=-;
        ch=getchar();
    }
    while(ch>=''&&ch<='')
    {
        x=x*+ch-'';
        ch=getchar();
    }
    return x*f;
}
//**************************************************************************************

ll  gcd(ll a,ll b)
{
    if(b==)return a;
    else return gcd(b,a%b);
}
int A[];
int main()
{
    int n=read();
    ll a=read();
    int cnt=;
    A[++cnt]=a;
    for(int i=;i<n;i++)
    {
       int  b=read();
       A[++cnt]=b;
       a=gcd(a,b);
    }
    int flag=;
    for(int i=;i<=n;i++)
    {
        A[i]/=a;
        while(A[i]%==)A[i]=A[i]/;
        while(A[i]%==)A[i]=A[i]/;
        if(A[i]!=){
            flag=;
            cout<<"NO"<<endl;
            break;
        }
    }
    if(!flag)cout<<"YES"<<endl;
    return ;
}

代码