1081. Rational Sum (20)

the problem is from PAT,which website is http://pat.zju.edu.cn/contests/pat-a-practise/1081

the code is as followed:

#include<stdio.h>
#include<math.h>
long int gongyue(long int num1, long int num2)
{
    long int gcd=0;
    if (num1==num2)
    {
        gcd = num1;
    }
    if (num1>num2)
    {
        long int tmp = num1;
        num1 = num2;
        num2 = tmp;
    }
    if (num2 % num1 == 0)
    {
        gcd = num1;
    }
    else
    {
        long int tmp = num1;
        num1 = num2 % num1;
        num2 = tmp;
        gcd = gongyue(num1, num2);
    }
    return gcd;
}
long int gongbei(long int x, long int y)
{
    return x * y / gongyue(x,y);
}

int main()
{
    long int fenzi,fenmu;
    long int tempzi,tempmu;
    int n;
    //printf("%d",gongbei(3,8));
    scanf("%d",&n);
    scanf("%ld/%ld",&fenzi,&fenmu);
    n -= 1;
    long int temp = fenzi;
    if (fenzi == 0)
    {
        fenmu = 1;
    }
    else
    {
        fenzi = fenzi / gongyue(abs(temp), fenmu);
        fenmu = fenmu / gongyue(abs(temp), fenmu);
    }
    while (n--)
    {

        scanf("%ld/%ld",&tempzi,&tempmu);
        fenzi = fenzi * gongbei(fenmu,tempmu)/fenmu + tempzi * gongbei(fenmu,tempmu)/tempmu;
        fenmu = gongbei(fenmu,tempmu);
        long int tempfenzi = fenzi;
        if (fenzi == 0)
        {
            fenmu = 1;
        }
        else
        {
           fenzi = fenzi / gongyue(abs(tempfenzi), fenmu);
           fenmu = fenmu / gongyue(abs(tempfenzi), fenmu);
        }

    }
    if (abs(fenzi)>=fenmu)
    {
        printf("%ld",fenzi/fenmu);
        if (fenzi%fenmu != 0)
        {
            printf(" %ld/%ld",abs(fenzi)%fenmu,fenmu);
        }
        printf("\n");
    }
    else
    {
        if (fenzi == 0)
        {
            printf("0");
        }
        else
        {
            printf("%ld/%ld",fenzi,fenmu);
        }
        printf("\n");
    }
}

the time complexity is O(n) .