hdu 1788 Chinese remainder theorem again((数学:简单题)

#include <stdio.h>
#define LL long long

LL gcd(LL a, LL b) {
return b==0 ? a : gcd(b, a%b);
}

int main(void) {
LL n, a, m, ans;
while(scanf("%I64d%I64d", &n, &a) && (n+a)) {
ans = 1;
while(n--) {
scanf("%I64d", &m);
ans = ans*m/gcd(ans, m);
}
printf("%I64d\n", ans-a);
}
}