HDU 1788 Chinese remainder theorem again

 1 //1788
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cmath>
 5 #include <iostream>
 6 #define LL long long
 7 
 8 using namespace std ;
 9 
10 LL gcd(LL x,LL y)
11 {
12     return y == 0 ? x : gcd(y,x%y) ;
13 }
14 int main()
15 {
16     int I,a ;
17     while(~scanf("%d %d",&I,&a))
18     {
19         if(I == 0 && a == 0) break ;
20         int x ;
21         LL ans = 1;
22         while(I--)
23         {
24             scanf("%d",&x) ;
25             ans = (ans * x)/gcd(ans,x) ;
26         }
27         printf("%I64d\n",ans-a) ;
28     }
29     return 0 ;
30 }