I think the algorithm is pretty straightforward. First calculate all the primes between 1 and 10000, then for each of the 10 number, records the number of different primes which are divisors of the number. Then use the formula (num1+1)*(num2+1)*...*(numN+1) where numi is the number of occurences of the ith prime in [1, 10000].

My code:
===========================================================
#include <iostream>
#include <cstdlib>
#include <vector>
using namespace std;
int main()
{
    vector<int> primes;
    bool e[10001];
    for(int i = 2; i <= 10000; i++)
       e[i] = true;
    for(int i = 2; i <= 10000; i++)
       if(e[i])
          for(int j = 2; j <= 10000/i; j++)
                  e[j*i] = false;
    for(int i = 2; i <= 10000; i++)
       if(e[i])
          primes.push_back(i);

    int num[10001];
    for(int i = 0; i < primes.size(); i++)
            num[i] = 0;
    for(int i = 0; i < 10; i++)
    {
       int n;
       cin >> n;
       for(int j = 0; j < primes.size(); j++)
          if(!(n%primes[j]))
            num[j]++;
    }
    int total = 1;
    for(int i = 0; i < primes.size(); i++)
            total = total*(num[i]+1)%10;
    cout << (total%10) << endl;
    //system("PAUSE");
    return 0;
}

I didn't look through your solution, but I can give you test, at which your program give wrong answer:
10
20
30
40
50
60
70
80
90
100
wright answer is 0 and your's is 8 (
(because number of divisors is 2340)

Thank you very much. I got AC.

Number of divisor is 2470 :) .