My solution is very intelligent!
After every read number, we make factorization of it.
Every number is smaller than 10000.
We have array fact with 10000 elements and this array is global, because every elements are zero when we declare it.
When we make factorization of every divisor p (for factorization), we increase the p-th element in the array.
In the end of reading in array fact we have how many times is used i-th element in factorization - we have factorization of the whole product.
With one looking the array we calculate the amount of divisors with the so famous formula:
N=p[1]k[1]*p[2]k[2]*p[3]*k[3]*..., when p[i] is prime number , k[i] is amount of p[i] in factorization of N.
The amount of divisors is:(k[1]+1)*(k[2]+1)*(k[3]+1)*... .
See it!
//Brave-balloonists
#include<iostream>
using namespace std;
int fact[10000];
int main () {
int a;
for (int i=0;i<10;i++) {
int a;
cin>>a;
if (a!=1)
   {
int p=1,st=1;
while (a!=1)           {
p+=st;
while (a%p==0) {
a/=p;
fact[p]++;
               }
if (p==2) {p=1;st=2;}
                       }
  }
}
int br_del=1;
for (int i=0;i<10000;i++) br_del*=(fact[i]+1);
printf("%d\n",br_del%10);
return 0;
}

you must delete it.