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 <cmath>
using namespace std;
int gcd(int a,int b) {
	return b==0?a:gcd(b,a%b);
}
int main(){
	int n;
	cin>>
 
n;
	int a,b;
	scanf("%d/%d",&a,&b);
	int t=gcd(a,b);
	a=a/t,b=b/t;
	for(int i=1;i<n;i++){
		int c,d;
		scanf("%d/%d",&c,&d);
		a=a*d+b*c;
		b=b*d;
		int t=gcd(a,b);
		a=a/t,b=b/t;
	}
	if(a%b==0) printf("%d",a/b);
	else{
		if(abs(a)>abs(b)){
			if(a>0) printf("%d %d/%d",a/b,a-a/b*b,b);
			else{
				a=-a;
				printf("-%d %d/%d",a/b,a-a/b*b,b);
			}
		}else printf("%d/%d",a,b);
	}
	return 0;
}