5.3 分数的四则运算：A1081 Rational Sum

A1081 Rational Sum

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<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
using namespace std;
typedef long long LL;
LL gcd(LL a,LL b){
    return b==0 ? a:gcd(b,a%b);
}
struct Fraction{
    LL up,down;
};
Fraction reduction(Fraction 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;
}
Fraction add(Fraction f1,Fraction f2){
    Fraction result;
    result.up=f1.up*f2.down+f1.down*f2.up;
    result.down=f1.down*f2.down;
    return reduction(result);
}
void showResult(Fraction f){
    f=reduction(f);
    if(f.down==1) printf("%lld\n",f.up);
    else if(abs(f.up)>abs(f.down)){
        printf("%lld %lld/%lld\n",f.up/f.down,abs(f.up)%f.down,f.down);
    }
    else{
        printf("%lld/%lld\n",f.up,f.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);
    }
    showResult(sum);
    return 0;
}