Can you fill in these boxes? (Part II)

#include <iostream> #include <list> #include <cmath> using namespace std ;
bool prime(int a){   if(a==2) return true ; if(!(a%2) || a==1) return false ;
    for(int i = 3 ; i < floor(sqrt(a))+1 ; i+= 2) if(!(a%i)) return false ; return true ; }
// ^ junk condensed ^

int done ; //how long the permutation is, 10 here

void loop(list<int> used, list<int> current){ //does permutations
//works by taking what is used and what is left to be organized
    int count = int(current.size()) ;
    for(int i = 0 ; i < count ; i ++){
        list<int>::iterator a = current.begin();
        advance(a, i) ;
        used.push_back(int(*a)) ;
        if(used.size() == done){
            int b[10] ;
            int c = 0 ;
            for(list<int>::const_iterator it = used.begin(); it != used.end(); it++, c++) b[c] = *it ;
            //check conditions and display
            if((double)(b[0]*1000+b[1]*100+b[2]*10+b[3])/(double)(b[4]*10+b[5])==(b[6]-b[8])*10+b[7]-b[9] && b[0]+b[1]+b[2]+b[3] == b[4]+b[5] && prime(b[4]*10+b[5])){
                cout << b[0] <<b[1]<<b[2]<<b[3]<<"/"<<b[4]<<b[5]<<" = "<<b[6]<<b[7]<<" - "<<b[8]<<b[9]<<" = "<<(b[6]-b[8])*10+b[7]-b[9]<< endl ;
            }
        }
        list<int> toPass = current ;
        toPass.remove(int(used.back())) ;
        loop(used, toPass) ;
        used.pop_back() ;
    }
}

int main(){
    int pass1[] = {0,1,2,3,4,5,6,7,8,9} ;
    list<int> pass ;
    pass.assign(pass1, pass1+sizeof(pass1)/sizeof(int)) ;
    done = int(pass.size()) ; //10
    list<int> a ;
    loop(a, pass) ; //pass in a blank list and the one containing 0-9
    return 0;
}