1081 Rational Sum (20 分)

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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 (≤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 <bits/stdc++.h>
using namespace std;
typedef long long ll;
ll gcd(ll a,ll b)
{
    return b==0?a:gcd(b,a%b);
}
struct st
{
    ll up;
    ll down;
};
st reduction(st result)
{
    if(result.down<0)
    {
        result.down=-result.down;
        result.up=-result.up;
    }
    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;
}
st add(st f1,st f2)
{
    st result;
    result.up=f1.up*f2.down+f2.up*f1.down;
    result.down=f1.down*f2.down;
    return reduction(result);
}
void show(st r)
{
    reduction(r);
    if(r.down==1)printf("%lld\n",r.up);
    else if(abs(r.up)>abs(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);
    st 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;
}