CodeForces 687B(剩余定理)

问题描述:

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.

Example

Input

4 52 3 5 12

Output

Yes

Input

2 72 3

Output

No

Note

In the first sample, Arya can understand  because 5 is one of the ancient numbers.

In the second sample, Arya can't be sure what  is. For example 1 and 7 have the same remainders after dividing by 2 and 3, but they differ in remainders after dividing by 7.

问题描述:给你n个数(c1,c2,,,,cn) 我们知道X%ci=ai 每个ci对应的ai，问能否知道X%k等于多少？

题目分析:参考大神:点击打开链接

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m1*m2...*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^k1*p2^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

这个逻辑，数论菜鸡表示方！

代码如下:

#include<iostream>#include<cstdio>#include<cmath>#include<cstring>#define ll long longusing namespace std;ll Gcd(ll a,ll b){    if (b==0) return a;    else return Gcd(b,a%b);}ll Lcm(ll a,ll b){    return a/Gcd(a,b)*b;}int main(){    int n,k;    while (scanf("%d%d",&n,&k)!=EOF) {        ll lcm=1;        for (int i=1;i<=n;i++) {            ll c;            scanf("%lld",&c);            lcm=Lcm(lcm,c)%k;        }        if (lcm%k==0)            puts("Yes");        else            puts("No");    }    return 0;}