【CF-1350 C

Orac and LCM

题意

有一个数组s，相关定义如下

1. (gcd(s))是最大的一个整数x，s中的所有数字都可以整除x
2. (lcm(s))是最小的一个整数x，x可以整除s中的所有数字

给出一个有n个数字的数组a，根据数组a，得到另一个数组(t={lcm(a_i,a_j)|i<j}),

让求出(gcd(t))

思路

[egin{aligned} gcd(t)=&gcd{\ &gcd{lcm(a_1,a_2),lcm(a_1,a_3),lcm(a_1,a_4)...lcm(a_1,a_n)},\ &gcd{lcm(a_2,a_3),lcm(a_2,a_4)...lcm(a_2,a_n)},\ &...\ &gcd{lcm(a_{n-2},a_{n-1}),lcm(a_{n-2},a_n)},\ &gcd{lcm(a_{n-1},a_n)}\ } end{aligned} ]

单看(gcd{lcm(a_1,a_2),lcm(a_1,a_3),lcm(a_1,a_4)...lcm(a_1,a_n)})

可以知道(lcm(a,b)=a*frac{b}{gcd(a,b)})，那么

[egin{aligned} &gcd{lcm(a_1,a_2),lcm(a_1,a_3),lcm(a_1,a_4)...lcm(a_1,a_n)}\ &=gcd{a_1*frac{a_2}{gcd(a_1,a_2)},a_1*frac{a_3}{gcd(a_1,a_3)},a_1*frac{a_4}{gcd(a_1,a_4)}...a_1*frac{a_n}{gcd(a_1,a_n)}}\ &=a_1*gcd{frac{a_2}{gcd(a_1,a_2)},frac{a_3}{gcd(a_1,a_3)},frac{a_4}{gcd(a_1,a_4)},...,frac{a_n}{gcd(a_1,a_m)}}\ &=a_1*{frac{gcd{a_2,a_3,a_4,..,a_n}}{gcd{a_1,gcd{a_2,a_3,a_4,..,a_n}}}}\ &=a_1*{frac{gcd{a_2,a_3,a_4,..,a_n}} {gcd{a_1,a_2,a_3,a_4,..,a_n}}}\ end{aligned} ]

定义(suf[i]=gcd{a_i,a_{i+1},a_{i+2},...,a_{n-1},a_n}),那么

[egin{aligned} gcd(t)=gcd{\ &a_1*frac{suf[2]} {suf[1]},\ &a_2*frac{suf[3]} {suf[2]},\ &...\ &a_{n-2}*frac{suf[n-1]} {suf[n-2]},\ &a_{n-1}*frac{suf[n]} {suf[n-1]},\ } end{aligned} ]

最后顺序求一遍(a_i*frac{suf[i+1]}{suf[i]})的gcd就可以了

代码

/*Gts2m ranks first in the world*/
#include<bits/stdc++.h>
using namespace std;
const int N=2e5+10;
typedef long long ll;

ll suf[N],arr[N],rel[N];
int main()
{
    ll n;
    scanf("%lld",&n);
    for(ll i=1;i<=n;i++)
        scanf("%lld",&arr[i]);
    suf[n+1]=arr[n];
    for(ll i=n;i;i--)
    {
        suf[i]=__gcd(suf[i+1],arr[i]);
        rel[i]=1LL*suf[i+1]/suf[i]*arr[i];
    }
    ll ans=rel[n-1];
    for(ll i=1;i<n-1;i++) ans=__gcd(ans,rel[i]);
    printf("%lld
",ans);
    return 0;
}