HDU1788 Chinese remainder theorem again 最小公倍数

#include <iostream>
#include <cstdio>
using namespace std;
typedef long long LL;
LL gcd(LL a,LL b)
{
return !b?a:gcd(b,a%b);
}
int main()
{
int n,k,t;
while(cin>>n>>k)
{
if(n==0&&k==0) break;
LL ans=1;
for(int i=0;i<n;i++)
{
cin>>t;
ans=t/gcd(ans,t)*ans;
}
cout<<ans-k<<endl;
}
return 0;
}