【CF】【318div2C】【整数表示】【gcd】

C. Bear and Poker

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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

475 150 75 50

output

Yes

input

3100 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_backint 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;}