题目描述
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
cpp
#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;
}