### HUGE GCD

below given code is for hg spoj or huge gcd spoj.in this question simple logic is that gcd of two number whose products are given is the gcd of each pair of products.

eg:

**2 x 3 x 5=30**;

**4 x 5=20;**

*gcd***(20,30)=10**;

this can be obtained as number of common prime terms in both products

we can write above as

**2 x 3 x 5=30;**

**2 x 2 x 5=20;**

now common terms are

**2 x 5 =10**;

this is logic of gcd studied in 5 or 6th standerd

#include <stdio.h> #include <iostream> using namespace std; long gcd(long a,long b) { return b==0?a:gcd(b,a%b); } long a[1009]; long b[1009]; long store[1000000]; int main() { int n,m; scanf("%d",&n); for(int i=0;i<n;i++) scanf("%ld",&a[i]); scanf("%d",&m); for(int i=0;i<m;i++) scanf("%ld",&b[i]); int k=0; long g; for(int i=0;i<n;i++) { for(int j=0;j<m;j++) { g=gcd(a[i],b[j]); if(b[j]!=1 && g!=1){ store[k++]=g; a[i]=a[i]/g; b[j]=b[j]/g; } } } long long temp,carry; long total=1; int flage=0; int res[10]={0}; res[9]=1; for(int i=0;i<k;i++) { carry=0; total*=store[i]; if(total>999999999){ flage=1; } for(int j=9;j>=1;j--) { temp=res[j]*store[i]; res[j]=(temp + carry)%10; carry=(temp + carry)/10; } } if(flage) for(int i=1;i<=9;i++) printf("%d",res[i]); else printf("%ld",total); printf("\n"); return 0; }