bzoj 4176: Lucas的数论【莫比乌斯反演+杜教筛】

首先由这样一个结论：

[d(ij)=sum_{p|i}sum_{q|j}[gcd(p,q)==1] ]

然后推反演公式：

[sum_{i=1}^{n}sum_{j=1}^{n}sum_{p|i}sum_{q|j}[gcd(p,q)==1] ]

[sum_{p=1}^{n}sum_{q=1}^{n}[gcd(p,q)==1]left lfloor frac{n}{p} ight floorleft lfloor frac{n}{q} ight floor ]

[sum_{p=1}^{n}sum_{q=1}^{n}sum_{k|p,k|q}mu(k)left lfloor frac{n}{p} ight floorleft lfloor frac{n}{q} ight floor ]

[sum_{k=1}^{n}mu(k)(sum_{k|p}left lfloor frac{n}{p} ight floor)^2 ]

[sum_{k=1}^{n}mu(k)(sum_{p=1}^{left lfloor frac{n}{k} ight floor}left lfloor frac{n}{pk} ight floor)^2 ]

然后对于这个递归子问题形式就可以用杜教筛求解了

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const long long N=2000005,m=2000000,P=100005,mod=1e9+7;
long long n,mb[N],q[N],tot,ans,p[P],t;
bool v[N];
long long F(long long n)
{
	long long sum=0;
	for(long long i=1,la;i<=n;i=la+1)
	{
		la=n/(n/i);
		sum=(sum+(la-(i-1))*(n/i)%mod)%mod;
	}
	return sum*sum%mod;
}
long long getp(long long x,long long n)
{
	return (x<=m)?mb[x]:p[n/x];
}
void slv(long long x,long long n)
{
	if(x<=m) 
		return;
	long long i,j=1,t=n/x;
	if(v[t]) 
		return;
	v[t]=1;
	p[t]=1;
	while(j<x)
	{
		i=j+1;
		j=x/(x/i);
		slv(x/i,n);
		p[t]=(p[t]-getp(x/i,n)*(j-i+1)+mod)%mod;
	}
}
long long wk(long long n)
{
	if(n<=m)
		return mb[n];
	memset(v,0,sizeof(v));
	slv(n,n);
	return p[1];
}
int main()
{
	mb[1]=1;
	for(long long i=2;i<=m;i++)
	{
		if(!v[i])
		{
			q[++tot]=i;
			mb[i]=-1;
		}
		for(long long j=1;j<=tot&&i*q[j]<=m;j++)
		{
			long long k=i*q[j];
			v[k]=1;
			if(i%q[j]==0)
			{
				mb[k]=0;
				break;
			}
			mb[k]=-mb[i];
		}
	}
	for(long long i=1;i<=m;i++)
		mb[i]+=mb[i-1];
	scanf("%lld",&n);
	for(long long i=1,la;i<=n;i=la+1)
	{
		la=n/(n/i);//cout<<la<<" "<<i-1<<" "<<wk(la)<<" "<<wk(i-1)<<endl;
		ans=(ans+(wk(la)-wk(i-1)+mod)%mod*F(n/i)%mod)%mod;
	}
	printf("%lld
",ans);
	return 0;
}