Code for calculating number of trailing zeroes after factorial
Code Review
All return paths from a function must return  value. Leaving a function that has a non void type without a return is undefined behavior (google nasal dragons).
int getZeroes(long long n){
    if(n&gt;0){
       // STUFF
       return stuff;
    }
    // BUT HERE there is no return.
    // Any negative value or zero causes undefined behavior.
}

This while loop seems overly cumbersome with a break;
        while(a%5){
            if(a%5 == 0){  // will this ever be true.
                // STUFF   // Since it is just after a test that checks the opposite?
            }else{
                break;
            }
            a = a/5;
        }

This looks exactly the same both sides of the else:
        return (b+getZeroes(n-1));
    }else{
        return (getZeroes(n-1));

The only difference is b (which just needs to be made zero for the second case then) you can yank it out of the conditional.
Why store the results?
x.push_back(getZeroes(cases));

The only thing you do with them is print them. May as well just print them as you generate them!