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:

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 ==  ? a : gcd(b,a%b);

}

Fraction reduction(Fraction result){

    if(result.dowm < ){

        result.up = - result.up;

        result.dowm = - result.dowm;

    }

    if(result.up == ){

        result.dowm = ;

    }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 == ) 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 = , sum.dowm = ;

    for(int i = ; i < n; i++){

        scanf("%lld/%lld",&temp.up,&temp.dowm);

        sum = add(sum,temp);

    }

    showResult(sum);

    return ;

}

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