文章目录
[C - Orac and LCM](https://vjudge.net/contest/381079#problem/C)题意解题思路代码
C - Orac and LCM
题意
给你 N 个数，让你求着 N个数任意两个数的lcm组成的 ** N*N/2 **个数的 gcd是多少；
解题思路
代码
#include<bits/stdc++.h>
using namespace std
;
#define ll long long
#define fast ios::sync_with_stdio(0)
const int mx
=100100;
ll g
[mx
];
ll a
[mx
];
int main()
{
fast
;
int n
;
cin
>>n
;
for(int i
=1;i
<=n
;i
++) cin
>>a
[i
];
for(int i
=n
;i
;i
--){
g
[i
]=__gcd(a
[i
],g
[i
+1]);
}
ll ans
=0;
for(int i
=1;i
<=n
;i
++){
ans
=__gcd(ans
,a
[i
]*g
[i
+1]/__gcd(a
[i
],g
[i
+1]));
}
cout
<<ans
<<"\n";
return 0;
}
