A1081

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<iostream>
#include<string>
#include<vector>
#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<cstdio>
#include<algorithm>
using namespace std;
long long gcd(long long a, long long b) {return b == 0 ? abs(a) : gcd(b, a % b);}
int main() {
    long long n, a, b, suma = 0, sumb = 1, gcdvalue;
    scanf("%lld", &n);
    for(int i = 0; i < n; i++) {
        scanf("%lld/%lld", &a, &b);
        gcdvalue = gcd(a, b);
        a = a / gcdvalue;
        b = b / gcdvalue;
        suma = a * sumb + suma * b;
        sumb = b * sumb;
        gcdvalue = gcd(suma, sumb);
        sumb = sumb / gcdvalue;
        suma = suma / gcdvalue;
    }
    long long integer = suma / sumb;
    suma = suma - (sumb * integer);
    if(integer != 0) {
        printf("%lld", integer);
        if(suma != 0) printf(" ");
    }
    if(suma != 0)
        printf("%lld/%lld", suma, sumb);
    if(integer == 0 && suma == 0)
        printf("0");
    return 0;
}