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.

題目大意： 兩兩求最小公倍數，然后得到集合，求最大公因數。

思路： 求質因數次小的指數。
注意：當有某個質因數的個數為n的的時候，應該any*=次小指數
當為n-1的時候 應該為any*=最小指數。

直接把0算進去應該不用考慮這兩種情況了

我的代碼:

#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; }

大神代碼：優美的求次小指數，在最小指數的基礎上求。

#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; }

總結

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