L1-009N个数求和

#include
#include
typedef struct fraction
{
    long long it;//integer
    long long nmr;//numerator
    long long dnm;//denominator
}fraction;
int main()
{
    long long EUCLID_GCD (long long a,long long b);//GCD=greatest_common_divisor
    long long n,i,gcd=1,it=0,nm=0,sum=0,cd=1,cd0=1;//cd=common_denominator
    scanf("%lld",&n);
    getchar();
    fraction frc[n];
    for(i=0;i)
    {
        scanf("%lld/%lld",&frc[i].nmr,&frc[i].dnm);
        gcd=EUCLID_GCD(abs(frc[i].nmr),abs(frc[i].dnm));
        frc[i].nmr=frc[i].nmr/gcd;
        frc[i].dnm=frc[i].dnm/gcd;
        frc[i].it=frc[i].nmr/frc[i].dnm;
        frc[i].nmr=frc[i].nmr%frc[i].dnm;
        getchar();           　　　　　　　　　　　　　　//simplification
    }
    for(i=0;i)
    {
        cd0=cd;
        gcd=EUCLID_GCD(abs(cd),abs(frc[i].dnm));
        cd=(frc[i].dnm/gcd)*cd;                     //reduction of fractions to a common denominator
        sum=(cd/frc[i].dnm)*frc[i].nmr+(cd/cd0)*sum;//summation

        gcd=EUCLID_GCD(abs(cd),abs(sum));
        sum=sum/gcd;
        cd=cd/gcd;                                  //reduction of a fraction

        it=it+sum/cd+frc[i].it;
        sum=sum%cd;                                 //get integer part
    }

    nm=sum;
    if(it<0&&nm<0)
    {
        printf("%lld %lld/%lld",it,nm,cd);
    }else if(it<0&&nm==0)
    {
        printf("%lld",it);
    }else if(it==0&&nm!=0)
    {
        printf("%lld/%lld",nm,cd);
    }else if(it>0&&nm>0)
    {
        printf("%lld %lld/%lld",it,nm,cd);
    }else
    {
        printf("%lld",it);
    }
    return 0;
}

long long EUCLID_GCD (long long a,long long b)
{
    if(b==0)return a;
    else return EUCLID_GCD(b,a%b);
}