1081. 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

代碼實現

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;

long long gcd(long long a,long long b)  //求最大公約數
{
    return b==0 ? a:gcd(b,a%b);
}
struct fraction{
  long long up,down;
};
fraction reduction(fraction result)    //化簡
{
    if(result.down<0)
    {
        result.up=-result.up;
        result.down= - result.down;
    }
    if(result.up == 0)
    {
        result.down = 1;
    }
    else
    {
        int d=gcd(abs(result.up),abs(result.down));
        result.up/=d;
        result.down/=d;
    }
    return result;
}
fraction add(fraction f1,fraction f2)
{
    fraction result;
    result.up= f1.up*f2.down + f2.up*f1.down;
    result.down=f1.down*f2.down;
    return reduction(result);
}
void show(fraction r)
{
    r=reduction(r);
    if(r.down==1) printf("%lld\n",r.up);
    else if(abs(r.up)>r.down)
    {
        printf("%lld %lld/%lld\n",r.up/r.down,abs(r.up)%r.down,r.down);
    }
    else
    {
        printf("%lld/%lld\n",r.up,r.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;
}