PAT-A-1081-Rational Sum

Given N rational numbers in the form numerator/denominator, you are supposed to calculate their sum.

Input Specification:

Each input file contains one test case. Each case starts with a positive integer N (≤100), followed in the next line N rational numbers a1/b1 a2/b2 ... where all the numerators and denominators are in the range of long int. If there is a negative number, then the sign must appear in front of the numerator.

Output Specification:

For each test case, output the sum in the simplest form integer numerator/denominator where integer is the integer part of the sum, numerator < denominator, and the numerator and the denominator have no common factor. You must output only the fractional part if the integer part is 0.

Sample Input 1:

5
2/5 4/15 1/30 -2/60 8/3

Sample Output 1:

3 1/3

Sample Input 2:

2
4/3 2/3

Sample Output 2:

2

Sample Input 3:

3
1/3 -1/6 1/8

Sample Output 3:

7/24

Code

#include <iostream>
using namespace std;

typedef long long ll;

ll gcd(ll a,ll b){
    return !b? a:gcd(b,a%b);
}

struct Fraction{
    ll up,down;//分子和分母
};
//化简
void reduction(Fraction& f1){
    //正负的化简
    if(f1.down<0){
        f1.up=-f1.up;
        f1.down=-f1.down;
    }
    //为0的化简
    if(f1.up==0)
        f1.down=1;
    else{
        //约去公约数
        ll d = gcd(abs(f1.up),abs(f1.down));
        f1.up/=d;
        f1.down/=d;
    }
}

void add(Fraction& f1,Fraction& f2){
    f1.up=f1.up*f2.down+f1.down*f2.up;
    f1.down=f1.down*f2.down;
    reduction(f1);  
}

void showFraction(Fraction& f1){
    if(f1.down==1)
        printf("%lld\n",f1.up);
    else if(abs(f1.up)>f1.down)//假分式
        printf("%lld %lld/%lld\n",f1.up/f1.down,abs(f1.up)%f1.down,f1.down);
    else
        printf("%lld/%lld\n",f1.up,f1.down);
    
}

int main(){
	int n;
	scanf("%d",&n);
    Fraction sum,temp;
    sum.up=0,sum.down=1;

    while(n--){
        scanf("%lld/%lld",&temp.up,&temp.down);
        add(sum,temp);
    }

    showFraction(sum);
	return 0;
}