题目描述

“蓝桥杯”练习系统 (lanqiao.cn)

题目分析 

思路：

1.去重排序将其进行预处理

2.用gcd得到最简比值

3.用gcd_sub分别计算分子、分母的指数最大公约数 

#include<bits/stdc++.h>
using namespace std;
const int N = 110;
typedef long long ll;
ll n, cnt, a[N], x[N], y[N];
ll gcd(ll a, ll b)
{
	return b ? gcd(b , a % b) : a;
}
ll gcd_sub(ll a, ll b)
{
	if(a < b)swap(a, b);
	if(b == 1)return a;
	return gcd_sub(b, a / b);
}
int main()
{
	cin >> n;
	for(int i = 1; i <= n; i ++)
	{
		cin >> a[i];
	}
	sort(a + 1, a + 1 + n);
	for(int i = 2; i <= n; i ++)
	{
		if(a[i] != a[i - 1])
		{
			ll d = gcd(a[i], a[1]);
			cnt ++; 
			x[cnt] = a[i] / d;//分子 
			y[cnt] = a[1] / d;//分母 
		}
	}
	ll u = x[1], d = y[1];
	for(int i = 2; i <= cnt; i ++)
	{
		u = gcd_sub(u, x[i]);
		d = gcd_sub(d, y[i]);
	}
	cout << u << '/' << d;
	return 0;
}