Codeforces Round #318 [RussianCodeCup Thanks-Round] (Div. 2) C 模擬

連結：戳這裡

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.

題意：

n個人在打牌，給出n個人的籌碼，如果這n個人的籌碼可以通過無限的擴大兩倍或者三倍使得他們全部相等

輸出Yes否則輸出No

思路：

先求出所有的數的最大公約數併除去，然後統計是否是2或3的倍數

程式碼：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<string>
#include<vector>
#include <ctime>
#include<queue>
#include<set>
#include<map>
#include<stack>
#include<iomanip>
#include<cmath>
#define mst(ss,b) memset((ss),(b),sizeof(ss))
#define maxn 0x3f3f3f3f
#define MAX 1000100
///#pragma comment(linker, "/STACK:102400000,102400000")
typedef long long ll;
typedef unsigned long long ull;
#define INF (1ll<<60)-1
using namespace std;
int n;
ll a[200100];
ll gcd(ll a,ll b){
    return b==0?a:gcd(b,a%b);
}
ll b[200100];
int main(){
    scanf("%d",&n);
    for(int i=1;i<=n;i++) scanf("%I64d",&a[i]);
    ll tmp=a[1];
    for(int i=2;i<=n;i++){
        tmp=gcd(tmp,a[i]);
    }
    for(int i=1;i<=n;i++) b[i]=a[i]/tmp;
    for(int i=1;i<=n;i++){
        while(b[i]%3==0){
            b[i]/=3;
        }
        while(b[i]%2==0){
            b[i]/=2;
        }
        if(b[i]!=1){
            cout<<"No"<<endl;
            return 0;
        }
    }
    cout<<"Yes"<<endl;
    return 0;
}