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;
}