L1-009 N个数求和

#include <iostream>
#include <algorithm>

using namespace std;

typedef long long ll;
const int N = 105;

typedef struct node {
	ll x, y;
}node;
node a[N];

ll gcd(ll a, ll b)
{
	return b ? gcd(b, a % b) : a;
}

int main()
{
	int n;
	cin >> n;
	for (int i = 0; i < n; i++) scanf("%lld/%lld", &a[i].x, &a[i].y);

	ll ans = a[0].y;
	for (int i = 1; i < n; i++)
	{
		ans = ans / gcd(ans, a[i].y) * a[i].y;
	}
	//printf("%lld\n", ans);
	ll res = 0; //分子和
	for (int i = 0; i < n; i++)
	{
		res = res + ans / a[i].y * a[i].x;
	}
	ll x = res / ans;
	ll b = res % ans;
	ll c = 1;
	if (b != 0)
	{
		c = gcd(b, ans);
		b /= c, ans /= c;
	}
	if (x == 0)
	{
		if (b != 0)printf("%lld/%lld", b, ans);
		else printf("0");
	}
	
	if (b == 0 && x != 0) printf("%lld", x);
	if(x!=0 && b!= 0) printf("%lld %lld/%lld", x, b, ans);
	return 0;
}