1081 Rational Sum（20 分）

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 (≤), 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

#include<cstdio>
#include<algorithm>
using namespace std;
typedef long long ll;
struct Fraction{
    ll up,dowm;
};

ll gcd(ll a,ll b){
    return b == 0 ? a : gcd(b,a%b);
}

Fraction reduction(Fraction result){
    if(result.dowm < 0){
        result.up = - result.up;
        result.dowm = - result.dowm;
    }
    if(result.up == 0){
        result.dowm = 1;
    }else{
        int d = gcd(abs(result.dowm),result.up);
        result.dowm /= d;
        result.up /= d;
    }
    return result;
}

Fraction add(Fraction f1,Fraction f2){
    Fraction result;
    result.up = f1.dowm*f2.up + f2.dowm*f1.up;
    result.dowm = f1.dowm*f2.dowm;
    return reduction(result);
}

void showResult(Fraction r){
    reduction(r);
    if(r.dowm == 1) printf("%lld",r.up);
    else if(abs(r.up) > abs(r.dowm)){
        printf("%lld %lld/%lld",r.up/r.dowm,r.up%r.dowm,r.dowm);
    }else{
        printf("%lld/%lld",r.up,r.dowm);
    }
}

int main(){
    int n;
    scanf("%d",&n);
    Fraction sum,temp;
    sum.up = 0, sum.dowm = 1;
    for(int i = 0; i < n; i++){
        scanf("%lld/%lld",&temp.up,&temp.dowm);
        sum = add(sum,temp);
    }
    showResult(sum);
    return 0;
}