#include <iostream>

using namespace std;

int i0, i1, i2, i3, i4, i5;
int d0, d1;
int countD0, countD1, countBoth;

// This assignment is worth 5 points.
// -1 for not having gramatically correct output
// -0.5 if the only grammar error is having 0 with singular verbs
// -2 for failing to count the cases for divisible by both divisors correctly

int main()
{
	cout << "Enter in six integers: ";
	cin >> i0 >> i1 >> i2 >> i3 >> i4 >> i5;
	cout << "Enter in two divisors: ";
	cin >> d0 >> d1;

	countD0 = (i0 % d0 == 0) + (i1 % d0 == 0) + (i2 % d0 == 0)
		+ (i3 % d0 == 0) + (i4 % d0 == 0) + (i5 % d0 == 0);

	countD1 = (i0 % d1 == 0) + (i1 % d1 == 0) + (i2 % d1 == 0)
                + (i3 % d1 == 0) + (i4 % d1 == 0) + (i5 % d1 == 0);

	countBoth = (i0 % d0 == 0 && i0 % d1 == 0)
		  + (i1 % d0 == 0 && i1 % d1 == 0)
		  + (i2 % d0 == 0 && i2 % d1 == 0)
		  + (i3 % d0 == 0 && i3 % d1 == 0)
		  + (i4 % d0 == 0 && i4 % d1 == 0)
		  + (i5 % d0 == 0 && i5 % d1 == 0);

	/* Alternatively, you can do this to count your multiples instead:
	 *
	 * countD0 = !(i0 % d0) + !(i1 % d0) + !(i2 % d0)
	 * 	   + !(i3 % d0) + !(i4 % d0) + !(i5 % d0);
	 *
	 * Because ! would flip the 0 to a 1 and non-zeros to 0.
	 */

	if(countD0 != 1)	// Grammar check
		cout << countD0 << " numbers are divisible by " << d0
		     << '.' << endl;
	else
		cout << countD0 << " number is divisible by " << d0
		     << '.' << endl;

	if(countD1 != 1)
                cout << countD1 << " numbers are divisible by " << d1
                     << '.' << endl;
        else
                cout << countD1 << " number is divisible by " << d1
                     << '.' << endl;

	if(countBoth != 1)
                cout << countBoth << " numbers are divisible by both." << endl;
        else
                cout << countBoth << " number is divisible by both." << endl;

	return 0;
}