[luogu P3911] 最小公倍数之和

我丢 : https://www.luogu.com.cn/problem/P3911

思路

这题的关键是在$A_i$的范围这里，说明这题肯定是和值域有关的
涉及到$lcm$肯定和$gcd$有关，所以应该是要莫比乌斯反演

题解

$记a[i]为i出现的次数$
$\begin{array}{rl}& \sum _{i=1}^n\sum _{j=1}^{n}lcm(i,j)\times a_i\times a_j\\ =& \sum _{i=1}^n\sum _{j=1}^n\frac{i\times j\times a_i\times a_j}{\gcd (i,j)}\\ =& \sum _{d=1}^n\sum _{i=1}^{\lfloor n/d\rfloor }\sum _{j=1}^{\lfloor n/d\rfloor }[\gcd (i,j)=1]d\times i\times j\times a_{id}\times a_{jd}(枚举d=gcd)\\ =& \sum _{d=1}^n\sum _{i=1}^{\lfloor n/d\rfloor }\sum _{j=1}^{\lfloor n/d\rfloor }\sum _{k|\gcd (i,j)}\mu (k)\times d\times i\times j\times a_{id}\times a_{jd}\\ =& \sum _{d=1}^n\sum _{k=1}^{\lfloor n/d\rfloor }\sum _{i=1}^{\lfloor n/kd\rfloor }\sum _{j=1}^{\lfloor n/kd\rfloor }\mu (k)\times d\times ik\times jk\times a_{idk}\times a_{jdk}\\ =& \sum _{T=1}^{n}T\times (\sum _{i=1}^{\lfloor n/T\rfloor }i\times a_{iT}{)}^2\sum _{k|T}\mu (k)\times k(设T=kd,k\ast k\ast d=T\ast k)\end{array}$

$显然{\sum }_{k|T}\mu (k)\times k是可以预处理的$
code:

#include<bits/stdc++.h>
#define N 1000005
#define int long long
using namespace std;
int prime[N], vis[N], mu[N], s[N], sz;
void init() {
	mu[1] = 1;
	for(int i = 2; i < N; i ++) {
		if(!vis[i]) {
			prime[++ sz] = i;
			mu[i] = -1;
		}
		for(int j = 1; j <= sz && prime[j] * i < N; j ++) {
			vis[prime[j] * i] = 1;
			if(i % prime[j] == 0) break;
			else mu[i * prime[j]] = - mu[i];
		}
	}	
	for(int i = 1; i < N; i ++)
		for(int j = i; j < N; j += i)
			s[j] += mu[i] * i;
}
int n, a[N];
signed main() {
	init();
	scanf("%lld", &n);
	for(int i = 1, x = 0; i <= n; i ++) scanf("%lld", &x), a[x] ++;
	int ans = 0;
	for(int T = 1; T < N; T ++) {
		int ha = 0;
		for(int i = 1; i < N / T; i ++) ha += i * a[i * T];
		ans += T * ha * ha * s[T];
	}
	printf("%lld", ans);	
	return 0;
}