【CF】【318div2C】【整数表示】【gcd】_c. bear and poker time limit per test2 seconds mem-优快云博客

本文探讨了熊Limak与他的朋友们在赌场中通过调整赌注大小，利用2倍和3倍的倍增操作，是否有可能使所有赌注相等，从而赢得 jackpot 的可能性。

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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

因为任何一个整数都可以表示成素因子的积。所以x = 2^k1 * 3^k2 * 5^k3 * 7^k4

然后这些数通过*2 *3最终的结果一样所以等价于每个数的k3,k4.....kn 都一样。因为前面的2和3的幂次可以取得非常大。比如2^11111111111 * 3^1111111111。

所以希望 gcd(x,y) = 2^(min(kx1,ky1)) * 3^(min(kx2,ky2) * 5^k3 * 5^k4.

然后等价于x / gcd(x,y) 的因子只有2,3。

#include <iostream>
#include <cstring>
#include <cmath>
#include <queue>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <string>
#include <cstdlib>
#include <cstdio>
#include <algorithm>
using namespace std;
    
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define mp push_back

int gcd(int a,int b)
{
	return a==0?b:gcd(b%a,a);
}

int arr[100010];
int main()
{
	int n;
	while(scanf("%d",&n) != EOF)
	{
		int gg = 0;
		for(int i=0;i<n;i++)
		{
			scanf("%d",&arr[i]);
			gg = gcd(gg,arr[i]);
		}
		bool ok = true;
	
		for(int i=0;i<n;i++)
		{
			arr[i] =  arr[i] / gg;

			while(arr[i] % 2 ==0) arr[i] /= 2;
			while(arr[i] % 3 ==0) arr[i] /= 3;
			if(arr[i] != 1)
			{
				ok = false;
				break;
			}
		}
		if(ok) printf("Yes\n");
		else printf("No\n");
	
	}	
    return 0;
}