51nod 1220 约数之和【莫比乌斯反演+杜教筛】

首先由这样一个式子：\( d(ij)=\sum_{p|i}\sum_{q|j}[gcd(p,q)==1]\frac{pj}{q} \)大概感性证明一下吧我不会证

然后开始推：

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

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

\[\sum_{p=1}^{n}p\sum_{q=1}^{n}[gcd(p,q)==1]\left \lfloor \frac{n}{p} \right \rfloor\sum_{j=1}^{\left \lfloor \frac{n}{q} \right \rfloor}j
\]

方便起见设\( f(n)=\sum_{i=1}^{n}i \)

\[\sum_{p=1}^{n}p\sum_{q=1}^{n}\sum_{k|p,k|q}\mu(k)\left \lfloor \frac{n}{p} \right \rfloor f(\left \lfloor \frac{n}{q} \right \rfloor)
\]

\[\sum_{k=1}^{n}\mu(k)\sum_{k|p}p\left \lfloor \frac{n}{p} \right \rfloor\sum_{k|q}f(\left \lfloor \frac{n}{q} \right \rfloor)
\]

\[\sum_{k=1}^{n}\mu(k)\sum_{i=1}^{\left \lfloor \frac{n}{k} \right \rfloor}ik\left \lfloor \frac{n}{ik} \right \rfloor\sum_{j=1}^{\left \lfloor \frac{n}{k} \right \rfloor}f(\left \lfloor \frac{n}{jk} \right \rfloor)
\]

\[\sum_{k=1}^{n}\mu(k)k\sum_{i=1}^{\left \lfloor \frac{n}{k} \right \rfloor}i\left \lfloor \frac{n}{ik} \right \rfloor\sum_{j=1}^{\left \lfloor \frac{n}{k} \right \rfloor}f(\left \lfloor \frac{n}{jk} \right \rfloor)
\]

这个样子显然可以用杜教筛了，但是注意到后面有两个求和式，可能会增大常数（但是也不会T啦），所以考虑这两个求和式的关系：

\[\sum_{i=1}^{n}f(\left \lfloor \frac{n}{i} \right \rfloor)
\]

\[=\sum_{i=1}^{n}\sum_{j=1}^{\left \lfloor \frac{n}{i} \right \rfloor}j
\]

\[=\sum_{i=1}^{n}\sum_{j=1}^{\left \lfloor \frac{n}{i} \right \rfloor}j
\]

\[=\sum_{j=1}^{n}j\left \lfloor \frac{n}{j} \right \rfloor
\]

所以这两个式子是一样的！于是就变成了：

\[\sum_{k=1}^{n}\mu(k)k(\sum_{j=1}^{\left \lfloor \frac{n}{k} \right \rfloor}f(\left \lfloor \frac{n}{jk} \right \rfloor))^2
\]

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
const int N=1000005,inv2=500000004,mod=1e9+7;
int m,mb[N],q[N],tot;
long long n,s[N],ans,ha[N];
bool v[N];
long long slv(long long n)
{
	return n*(n+1)%mod*inv2%mod;
}
long long wk(long long x)
{
	if(x<=m)
		return s[x];//cout<<x<<endl;
	if(ha[n/x])
		return ha[n/x];
	long long re=1ll;
	for(int i=2,la;i<=x;i=la+1)
	{
		la=x/(x/i);
		re=(re-(slv(la)-slv(i-1))*wk(x/i)%mod)%mod;
	}
	return ha[n/x]=re;
}
long long clc(long long n)
{
	long long re=0ll;
	for(int i=1,la;i<=n;i=la+1)
	{
		la=n/(n/i);
		re=(re+(la-i+1)*slv(n/i)%mod)%mod;
	}
	return re;
}
int main()
{
	scanf("%lld",&n);
	m=(int)ceil(pow((int)n,2.0/3));
	mb[1]=1;
	for(int i=2;i<=m;i++)
	{
		if(!v[i])
		{
			q[++tot]=i;
			mb[i]=-1;
		}
		for(int j=1;j<=tot&&q[j]*i<=m;j++)
		{
			int k=i*q[j];
			v[k]=1;
			if(i%q[j]==0)
			{
				mb[k]=0;
				break;
			}
			mb[k]=-mb[i];
		}
	}
	for(int i=1;i<=m;i++)
		s[i]=(s[i-1]+i*mb[i])%mod;
	//cout<<wk(102)<<" "<<wk(101)<<endl;
	for(int i=1,la;i<=n;i=la+1)
	{
		la=n/(n/i);
		long long ml=clc(n/i);//if(i!=1)cout<<i-1<<" "<<n/(i-1)<<endl<<la<<" "<<n/la<<endl;
		ans=(ans+(wk(la)-wk(i-1))*ml%mod*ml%mod)%mod;
	}
	printf("%lld",(ans%mod+mod)%mod);
	return 0;
}