Codeforces Round #318 [RussianCodeCup Thanks-Round] (Div. 1) A. Bear and Poker 分解-白红宇

Codeforces Round #318 [RussianCodeCup Thanks-Round] (Div. 1) A. Bear and Poker 分解

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发布时间：2019-06-22

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A. Bear and Poker

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/573/problem/A

Description

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 Input

75 150 75 50

Sample Output

Yes

HINT

题意

给你n个数，你可以把操作任意一个数，使他翻倍或者翻三倍无数次，问你是否可以让所有数都相同

题解：

假设最后变成了K，那么K=2^x+3^y+tmp

所以把所有数都按照2和3分解之后，只要剩下的数都是tmp，那就是yes

代码：

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                    typedef long long ll;using namespace std;//freopen("D.in","r",stdin);//freopen("D.out","w",stdout);#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)#define maxn 200051#define mod 10007#define eps 1e-9int Num;//const int inf=0x7fffffff; //нчоч╢Сconst int inf=0x3f3f3f3f;inline ll read(){ ll x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){ if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f;}//**************************************************************************************ll a[maxn];int check(ll x){ if(x==1) return 1; if(x==2) return 1; if(x==3) return 1; if(x%2==0) return check(x/2); if(x%3==0) return check(x/3); return 0;}int n;ll gcd(ll a,ll b){ return b==0?a:gcd(b,a%b);}ll lcm(ll a,ll b){ return a*b/gcd(a,b);}int main(){ n=read(); for(int i=0;i

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