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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>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!