luogu3911 最小公倍数之和(莫比乌斯反演)

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给定(A_1,A_2,dots,A_N)，求(sum_{i=1}^Nsum_{j=1}^Nlcm(A_i,A_j))

(1le Nle 50000;1le A_ile 50000)

为了推式子方便我们设：

(n=50000) (a_i=sum_{j=1}^N[A_j=i])

答案就是(sum_{i=1}^nsum_{j=1}^na_ia_jlcm(i,j))

(sum_{i=1}^nsum_{j=1}^na_ia_jlcm(i,j))

(=sum_{i=1}^nsum_{j=1}^na_ia_jfrac{ij}{gcd(i,j)})

(=sum_{p=1}^nfrac1psum_{i=1}^nsum_{j=1}^na_ia_jij[gcd(i,j)=p])

(=sum_{p=1}^npsum_{i=1}^{n/p}sum_{j=1}^{n/p}a_{ip}a_{jp}ij[gcd(i,j)=1])

(=sum_{p=1}^npsum_{i=1}^{n/p}sum_{j=1}^{n/p}a_{ip}a_{jp}ijsum_{d|i,d|j}mu(d))

(=sum_{p=1}^npsum_{d=1}^nmu(d)d^2sum_{i=1}^{n/dp}sum_{j=1}^{n/dp}a_{idp}a_{jdp}ij)

(=sum_{q=1}^nqsum_{d|q}dmu(d)sum_{i=1}^{n/q}sum_{j=1}^{n/q}a_{iq}a_{jq}ij)

(=sum_{q=1}^nqsum_{d|q}dmu(d)left(sum_{i=1}^{n/q}a_{iq}i ight)^2)

(sum_{i=1}^{n/q}a_{iq}i)对于每个(q)求值，总复杂度为(frac n 1+frac n 2+frac n 3+dots+frac nn)复杂度为调和级数(O(nlog n))

前半部分筛(d)后枚举(d)及其倍数，复杂度还是调和级数(O(nlog n))

然后直接求和就行了，貌似不用打数论分块

41行一遍A。。。真不用打数论分块，复杂度nlogn

#include <cstdio>
using namespace std;

int prime[50010], mu[50010], tot, fuck = 50000;
bool vis[50010];
long long sum1[50010], sum2[50010];
int bucket[50010];

int main()
{
	mu[1] = 1;
	for (int i = 2; i <= fuck; i++)
	{
		if (vis[i] == false) prime[++tot] = i, mu[i] = -1;
		for (int j = 1; j <= tot && i * prime[j] <= fuck; j++)
		{
			vis[i * prime[j]] = true;
			if (i % prime[j] == 0) break;
			mu[i * prime[j]] = -mu[i];
		}
		mu[i] *= i;
	}
	for (int d = 1; d <= fuck; d++)
		for (int q = d; q <= fuck; q += d)
			sum1[q] += mu[d];
	for (int i = 1; i <= fuck; i++)
		sum1[i] *= i;
	int n; scanf("%d", &n);
	for (int x, i = 1; i <= n; i++)
		scanf("%d", &x), bucket[x]++;
	long long ans = 0;
	for (int q = 1; q <= fuck; q++)
	{
		int sb = fuck / q;
		for (int i = 1; i <= sb; i++)
			sum2[q] += bucket[i * q] * (long long)i;
		ans += sum2[q] * sum1[q] * sum2[q];
	}
	printf("%lld
", ans);
	return 0;
}