Sorry for my bad English:)

This problem can be solved by brute force answer and counting how many numbers the answer divides.

It's O(R log R)

where R = maximum number in a[] array.

you have to make an array cnt[], which will keep the count of integers x.

cnt[x]=count of number x.

and code:

for (int i=1;i<=R;i++) {
   int k=0;//count of numbers that are divisible by i
   for (int j=i;j<=R;j+=i)
      k+=cnt[j];
   if (k>=2) ans=max(ans,i);//if count of number more than one, 
   //then we have two numbers with a divider i, 
   //and it is not necessarily maximum divisor of these numbers.
}

O(R log R), because R/1 + R/2 + ... + R/R ~ R log R

Sorry because my english ability is not good. I think that this problem can be solved by dynamic programming. We call F[i] is the number of different elements in array a which are the mutiple of i. The result is the largest integer x that F[x] is larger than 1. We can calculate F[i] by following way. First we must initial the value of array F by F[a[i]] = F[a[i]] + 1. i = p1^m1*p2^m2*p3^m3*...*pk^mk. (k <= 7 because a[i] is smaller than 100000) If we can calculate the value of F[i] then F[i/p1] = F[i/p1] + F[i]. F[i/p2] = F[i/p2] + F[i]. F[i/p3] = F[i/p3] + F[i]. ... F[i/pk] = F[i/pk] + F[i]. You can calculate p1, p2, ..., pk in O(1). You will have more details here: http://e-maxx.ru/algo/prime_sieve_linear Because k <= 7, this approach can run in O(7*maxC) with maxC is the largest value in array a. Sorry again because my english ability is bad.

#include <stdio.h>
#include <stdlib.h>  
long  i,k,n,nmax;
long  brojevi[1001];
long nzd(long a, long b)
{
     if(!b) return a;
     a%=b;
     return nzd(b,a);

}
int main()
{
 
   
    scanf("%d",&n);
   
    for(i=1;i<=n;++i)
    {
      scanf("%d",&brojevi[i]);
     
    }
  for(k=1;k<n;++k)
  {
      for(i=k+1;i<=n;++i)
      {
             if(nzd(brojevi[k],brojevi[i])>nmax) nmax=nzd(brojevi[k],brojevi[i]);
      }
  }  
 
    printf("%d\n",nmax);
    system("pause");
    return 0;
}

I think this will be helpful.