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
作者: CHEN, Yue
单位: 浙江大学
时间限制: 400 ms
内存限制: 64 MB

#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); }
int main()
{
	ll n, a, b, up = 0, down = 1;;
	cin >> n;
	for (int i = 0; i < n; i++){
		scanf("%lld/%lld", &a, &b);
		up = up * b + a * down;
		down = down * b;
		ll d = gcd(up, down);
		up /= d, down /= d;
	}
	ll k = up / down;
	up %= down;
	if (k != 0 && up != 0)
		printf("%lld %lld/%lld", k, up, down);
	else if (k == 0 && up != 0)
		printf("%lld/%lld", up, down);
	else
		printf("%lld", k);
	return 0;
}