D. Remainders Game

题目连接：

http://www.codeforces.com/contest/688/problem/D

Description

Today Pari and Arya are playing a game called Remainders.

Pari chooses two positive integer x and k, and tells Arya k but not x. Arya have to find the value . There are n ancient numbers c1, c2, …, cn and Pari has to tell Arya if Arya wants. Given k and the ancient values, tell us if Arya has a winning strategy independent of value of x or not. Formally, is it true that Arya can understand the value for any positive integer x?

Note, that means the remainder of x after dividing it by y.

Input

The first line of the input contains two integers n and k (1 ≤ n,  k ≤ 1 000 000) — the number of ancient integers and value k that is chosen by Pari.

The second line contains n integers c1, c2, …, cn (1 ≤ ci ≤ 1 000 000).

Output

Print “Yes” (without quotes) if Arya has a winning strategy independent of value of x, or “No” (without quotes) otherwise.

Sample Input

4 5
2 3 5 12

Sample Output

Yes

Hint

题意

给你k和n

你现在想要知道x，但是人家不告诉你

给你n个ci，表示你可以知道x%ci

问你能不能唯一确定x

题解

首先，根据剩余定理，如果我们想知道x%m等于多少，当且仅当我们知道x%m1,x%m2..x%mr分别等于多少，其中m1m2…mr=m，并且mi相互互质，即构成独立剩余系。令m的素数分解为m=p1^k1p2^k2…pr^kr，如果任意i，都有pi^ki的倍数出现在集合中，那么m就能被猜出来。
这个问题等价于问LCM(ci)%m是否等于0

所以只要求出LCM(ci)即可，不过要边求lcm，边和m取gcd，防止爆int

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e6+6;
int n;
long long k,c[maxn];
long long gcd(long long a,long long b){
    if(b==0)return a;
    return gcd(b,a%b);
}
long long lcm(long long a,long long b){
    return a*b/gcd(a,b);
}
int main(){
    scanf("%d",&n);
    scanf("%lld",&k);
    long long tmp = 1;
    for(int i=1;i<=n;i++)
    {
        scanf("%lld",&c[i]);
        tmp=lcm(tmp,c[i]);
        tmp=gcd(tmp,k);
        if(tmp==k){
            printf("Yes\n");
            return 0;
        }
    }
    printf("No\n");
}