Suma a 2 fractii ordinale dupa simplificarea fractiilor

/* 25.10.2012

13. un program calculeaza suma a 2 fractii ordinale
afisare dupa simplificarea fractiilor
- obs. la citire numitorul nu are voie sa fie 0
A) numitor si numarator naturale
B) numitor si numarator intregi
*/

#include<iostream>
using namespace std;
void citire(int &m,int &n)
{
    cout<<"m="; cin>>m;
    do
    {
        cout<<"n="; cin>>n;
    }while(n==0);
}
int cmmdc(int a,int b)
{
    int r;
    if(b==0)
        return 0;
    else
    {
        int r=a%b;
        while(r)
        {
            a=b;
            b=r;
            r=a%b;
        }
        return b;
    }
}
void afisare(int m,int n)
{
    if(m*n<0)
    {
        m=-m;
        cout<<m;
    }
    else
        cout<<m;
    if(n!=1)
        cout<<"/"<<n<<" ";
}
void suma(int am,int an,int bm,int bn,int &Sm,int &Sn)
{
    Sm=bn*am+bm*an;// numarator
    Sn=bn*an; // numitor
    Sm=Sm/cmmdc(Sm,Sn);// simplific numaratorul
    Sn=Sn/cmmdc(Sm,Sn);// simplific numitorul
}
int main()
{
    int am,an,bm,bn,Sn,Sm;
    citire(am,an);
    citire(bm,bn);
    afisare(am,an);
    afisare(bm,bn);
    cout<<endl<<endl;
    cout<<"Suma fractiilor este ";
    suma(am,an,bm,bn,Sm,Sn);
    return 0;
}