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
");
return 0;
}
}
printf("No
");
}