B1081 Rational Sum (分数计算)

B1081 Rational Sum (20分)

晓得了，原来gcd是可以处理负数的啊~~

#include<bits/stdc++.h>
#include<cstring>
using namespace std;
const int MAX=110;
int gcd(int a,int b)
{
   return !b ? a:gcd(b,a%b);
}
int main()
{
    int n;
    int a,b;
    cin>>n;
    scanf("%d/%d",&a,&b);
    int t=gcd(a,b);
    a/=t,b/=t;
    for(int i=1;i<n;i++)
    {
        int c,d;
        scanf("%d/%d",&c,&d);
        a=a*d+b*c;
        b=b*d;
        t=gcd(a,b);
        a/=t;
        b/=t;
    }
    if(a%b==0)
        printf("%d",a/b);
    else
    {
        if(abs(a)>abs(b))
        {
            if(a>0)
                printf("%d %d/%d",a/b,a%b,b);
            else
            {
                a=-a;
                printf("-%d %d/%d",a/b,a%b,b);
            }
        }
        else
            printf("%d/%d",a,b);
    }
    return 0;
}