正题

题目链接:https://www.luogu.com.cn/problem/P3911

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题目大意

给出数列$A$求$$\sum _{i=1}^n\sum _{j=1}^{n}lcm(A_i,A_j)$$

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解题思路

设$c_i$表示$A_j=i$的个数，然后答案就是（下面$n=5e4$）$$\sum _{i=1}^n\sum _{j=1}^{n}lcm(i,j)c_i{c}_j$$
$$\sum _{i=1}^n\sum _{j=1}^n\frac{ij}{gcd(i,j)}{c}_i{c}_j$$
$$\sum _{x=1}^{n}x\sum _{i=1}^{\lfloor \frac{n}{x}\rfloor }\sum _{j=1}^{\lfloor \frac{n}{x}\rfloor }[gcd(i,j)==1]ij{c}_{i\ast x}{c}_{j\ast x}$$
然后反演一下
$$\sum _{x=1}^n\sum _{x|d}\mu (\frac{d}{x})d\sum _{i=1}^{\lfloor \frac{n}{d}\rfloor }\sum _{j=1}^{\lfloor \frac{n}{d}\rfloor }ij{c}_{i\ast d}{c}_{j\ast d}$$
$$\sum _{d=1}^{n}d\sum _{x|d}\mu (\frac{d}{x})\sum _{i=1}^{\lfloor \frac{n}{d}\rfloor }\sum _{j=1}^{\lfloor \frac{n}{d}\rfloor }ij{c}_{i\ast d}{c}_{j\ast d}$$
然后两部分分开处理就好了，时间复杂度$O(n\log n)$

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$code$

#include<cstdio>
#include<cstring>
#include<algorithm>
#define ll long long
using namespace std;
const ll N=5e4+10;
ll n,cnt,pri[N],mu[N],c[N],g[N],f[N],ans;
bool v[N];
void prime(){mu[1]=1;for(ll i=2;i<N;i++){if(!v[i])pri[++cnt]=i,mu[i]=-1;for(ll j=1;j<=cnt&&i*pri[j]<N;j++){v[i*pri[j]]=1;if(i%pri[j]==0)break;mu[i*pri[j]]=mu[i]*mu[pri[j]];}}for(ll i=1;i<N;i++)for(ll j=i;j<N;j+=i)g[j]+=mu[i]*i;return;
}
int main()
{prime();scanf("%lld",&n);for(ll i=1;i<=n;i++){ll x;scanf("%lld",&x);c[x]++;}for(ll i=1;i<N;i++)for(ll j=1;i*j<N;j++)f[i]+=c[i*j]*j;for(ll i=1;i<N;i++)ans+=f[i]*f[i]*i*g[i];printf("%lld\n",ans);return 0;
}