A1081. Rational Sum (20)

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

#include <stdio.h>
#include <stdlib.h>
#include <iostream>
#include <string.h>
#include <string>
#include <math.h>
#include <algorithm>
using namespace std;
//求最大公约数
int gcd(long long  a,long long b) 
{
    if(b==0)return a;
    else return gcd(b,a%b); 
}

struct Fraction{
    long long up;
    long long down;
};
Fraction  reduction(Fraction r)
{
    if(r.down<0)
    {
        r.up*=(-1);
        r.down*=(-1); 
    }
    if(r.up==0)
    {
        r.down=1;
    }else
    {
        int d=gcd(r.up,r.down);
        r.up=r.up/d;
        r.down/=d;
    }
    return r;
}

Fraction add(Fraction a,Fraction b)
{
    Fraction r;
    r.down=a.down*b.down;
    r.up=a.up*b.down+a.down*b.up;
    return reduction(r);
} 

void show(Fraction a)
{
    if(a.down==1)
    printf("%lld",a.up);
    else if(abs(a.up)>a.down)
    {
        printf("%lld %lld/%lld",a.up/a.down,abs(a.up)%a.down,a.down);
    }
    else
    {
        printf("%lld/%lld",a.up,a.down);
    }
}
int main(){
    int n;
    scanf("%d",&n);
    Fraction sum,temp;
    sum.up=0;sum.down=1;
    for(int i=0;i<n;i++)
    {
        scanf("%lld/%lld",&temp.up,&temp.down);
        sum=add(sum,temp);
    } 
    show(sum);
    return 0;
}