西交校赛 F. GZP and Poker

F. GZP and Poker

GZP often plays games with his friends.Today they went to a board game.There are n players(including GZP) and all of them have some virtual money on the table. ith of them has ai yuan.

Each player can double his virtual wealth any number of times and triple his virtual wealth any number of times.The game has a big prize for making wealth of all players equal.Is it possible for GZP and his friends to win the big prize?

Input

The input consists of several test cases.

First line of input contains an integer n(2≤n≤10^5),the number of players.

The second line contains n integer numbers a1,a2,?,an(1≤ai≤10^9)-the virtual money of players.

Output

For each test case, print a line.

"Yes"(without the quotes)if players can make their wealth equal, or "No" otherwise.

Sample Input

4
75 150 75 50
3
100 150 250

Sample Output

Yes
No

题意：

问这些数能否经过变成同一个数,变换为乘2或者3,乘的次数任意;

思路：

temp=a[i]*2^x*3*y;可以发现a[i]*2^x*3*y=a[j];这里的x和y为整数;那么就可以搜索了;

AC代码：

#include <bits/stdc++.h>
using namespace std;
#define Riep(n) for(int i=1;i<=n;i++)
#define Riop(n) for(int i=0;i<n;i++)
#define Rjep(n) for(int j=1;j<=n;j++)
#define Rjop(n) for(int j=0;j<n;j++)
#define mst(ss,b) memset(ss,b,sizeof(ss));
typedef long long LL;
const LL mod=1e9+7;
const double PI=acos(-1.0);
const int inf=0x3f3f3f3f;
const int N=1e5+5;
int n;
LL a[N];
map<LL,int>mp;
int dfs(LL x)
{
    mp[x]=1;
    if(2*x<=3e9+7&&!mp[2*x])
    {
        dfs(2*x);
    }
    if(3*x<=3e9+7&&!mp[3*x])
    {
        dfs(3*x);
    }
    if(x%2==0&&!mp[x/2])
    {
        dfs(x/2);
    }
    if(x%3==0&&!mp[x/3])
    {
        dfs(x/3);
    }
}
int solve()
{
    for(int i=2;i<=n;i++)
    {
        if(!mp[a[i]])
        {
            printf("No\n");
            return 0;
        }
    }
    printf("Yes\n");
    return 0;
}
int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        mp.clear();
        Riep(n)
        {
            scanf("%lld",&a[i]);
        }
        dfs(a[1]);
        solve();

    }

    return 0;
}