Orac and LCM cf地址

For the multiset of positive integers s={s1,s2,…,sk}, define the Greatest Common Divisor (GCD) and Least Common Multiple (LCM) of s as follow:

gcd(s) is the maximum positive integer x, such that all integers in s are divisible on x.

lcm(s) is the minimum positive integer x, that divisible on all integers from s.
For example, gcd({8,12})=4,gcd({12,18,6})=6 and lcm({4,6})=12. Note that for any positive integer x, gcd({x})=lcm({x})=x.

Orac has a sequence a with length n. He come up with the multiset t={lcm({ai,aj}) | i<j}, and asked you to find the value of gcd(t) for him. In other words, you need to calculate the GCD of LCMs of all pairs of elements in the given sequence.

Input
The first line contains one integer n (2≤n≤100000).

The second line contains n integers, a1,a2,…,an (1≤ai≤200000).

Output
Print one integer: gcd({lcm({ai,aj}) | i<j}).

Examples
inputCopy
2
1 1
outputCopy
1
inputCopy
4
10 24 40 80
outputCopy
40
inputCopy
10
540 648 810 648 720 540 594 864 972 648
outputCopy
54
Note
For the first example, t={lcm({1,1})}={1}, so gcd(t)=1.

For the second example, t={120,40,80,120,240,80}, and it’s not hard to see that gcd(t)=40.

题目大意： 两两求最小公倍数，然后得到集合，求最大公因数。

思路： 求质因数次小的指数。
注意：当有某个质因数的个数为n的的时候，应该any*=次小指数
当为n-1的时候 应该为any*=最小指数。

		直接把0算进去应该不用考虑这两种情况了

我的代码:

#include <cstdio>
#include <cstring>
#include <string>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <queue>
#include <cstdlib>
#include <stack>
#include <vector>
#include <set>
#include <map>
#include <bitset>
#define INF 0x3f3f3f3f3f3f3f3f
#define FILL(a,b) (memset(a,b,sizeof(a)))
#define re register
#define lson rt<<1
#define rson rt<<1|1
#define lowbit(a) ((a)&-(a))
#define ios std::ios::sync_with_stdio(false);std::cin.tie(0);std::cout.tie(0);
#define fi first
#define rep(i,n) for(int i=0;(i)<(n);i++)
#define rep1(i,n) for(int i=1;(i)<=(n);i++)
#define se secondusing namespace std;
typedef long long  ll;
typedef unsigned long long  ull;
typedef pair<int,int > pii;
const ll mod=104857601;
const int N =2e5+10;
const double eps = 1e-6;
const double pi=acos(-1);
int gcd(int a,int b){return !b?a:gcd(b,a%b);}
int dx[4]={-1,0,1,0} , dy[4] = {0,1,0,-1};
int a[N];
int p[N], cnt;
bool st[N];void get_primes(int n)
{for (int i = 2; i <= n; i ++ ){if (!st[i]) p[cnt ++ ] = i;for (ll j = 0; p[j] <= n / i; j ++ ){st[p[j] * i] = true;if (i % p[j] == 0) break;}}
}int pow1(int a,int b)
{ll any=1;while(b){if(b&1) any*=a;a*=a;b>>=1;}return any;
}int d1[N];
int d2[N];
int d3[N];
int main()
{get_primes(N);int n;scanf("%d",&n);for(int i=1;i<=n;i++){scanf("%d",&a[i]);}if(n==2){   printf("%lld",ll(a[1]/gcd(a[1],a[2]))*a[2]);return 0;}for(int i=1;i<=n;i++){for(int j=0;j<cnt;j++){if(a[i]<p[j]||a[i]<p[j]*p[j]) break;if(a[i]%p[j]==0){int x=0;d3[p[j]]++;while(a[i]%p[j]==0){a[i]/=p[j];x++;}if(x<=d1[p[j]]||d1[p[j]]==0){d2[p[j]]=d1[p[j]];d1[p[j]]=x;}else if(x<d2[p[j]]||d2[p[j]]==0){d2[p[j]]=x;}}}if(a[i]>1){d3[a[i]]++;int x=1;if(x<=d1[a[i]]||d1[a[i]]==0){d2[a[i]]=d1[a[i]];d1[a[i]]=x;}else if(x<d2[a[i]]||d2[a[i]]==0){d2[a[i]]=x;}}}ll any=1;for(int i=0;i<200000;i++){if(d3[i]==n) any*=pow1(i,d2[i]);else if(d3[i]==n-1) any*=pow1(i,d1[i]);}printf("%lld\n",any);return 0;
}

大神代码：优美的求次小指数，在最小指数的基础上求。

#include <bits/stdc++.h>using namespace std;typedef long long ll;ll n,d[100007],a,b;int main()
{cin>>n;cin>>d[1]>>d[2];a=__gcd(d[1],d[2]);b=d[1]*d[2];for(int i=3;i<=n;i++){cin>>d[i];b=__gcd(a*d[i],b);a=__gcd(a,d[i]);}cout<<b/a;return 0;
}