C - Orac and LCM

Codeforces Round #641 (Div. 2)
2021/11/12

题目大意：

给一组数 $a[n]$ ,求 $gcd(\{lcm(a_i,a_j)|i<j\})$

思路：

不失一般性，假设 a[n] = {a,b,c};
那么考虑 i = a；
则 $gcd(lcm(a,b),lcm(a,c))=lcm(a,gcd(b,c))$
由此可知：
设 $g_1=lcm(a,gcd(b,c,d...a[n]))$
$g_2=lcm(b,gcd(c,d....,a[n]))$

$ans=gcd(g_1,g_2....,g_n-1)$

用后缀数组处理下gcd；

Code

// Problem: C. Orac and LCM
// Contest: Codeforces - Codeforces Round #641 (Div. 2)
// URL: https://codeforces.com/problemset/problem/1350/C
// Memory Limit: 256 MB
// Time Limit: 3000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include
using namespace std;
#define _orz ios::sync_with_stdio(false),cin.tie(0)
#define mem(str,num) memset(str,num,sizeof(str))
#define forr(i,a,b) for(int i = a;i <= b;i++)
#define forn(i,n) for(int i = 0; i < n; i++)
#define all(a) (a.begin(),a.end())
#define dbg() cout << "0k!" << endl;
//#define _DEBUG
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int N = 1e5+10;
const ll MOD = 1e9+7;
ll gcd(ll a,ll b){
	return b?gcd(b,a%b):a;
}
ll lcm(ll a, ll b){
	ll res = a/gcd(a,b)*b;
	return res;
}

ll suf[N];
int a[N];
void so1ve(){
	int n; cin >> n;
	forr(i,1,n)cin >> a[i];
	suf[n] = a[n];
	for(int i = n-1; i >= 1; i--){
		suf[i] = gcd(suf[i+1],a[i]);
	}
	ll ans = lcm(a[1],suf[2]);
	forr(i,2,n-1) ans = gcd(ans,lcm(a[i],suf[i+1]));
	cout << ans << endl;
}
int main()
{
#ifdef _DEBUG
    //freopen("input.txt","r",stdin);
    freopen("out.txt","w",stdout);
#endif
    int t=1;;
    while(t--) so1ve();
    return 0;
}