[51Nod1238]最小公倍数之和-V3

题意

  给定$n$，求$\sum_{i=1}^n\sum_{j=1}^n lcm(i,j)$

解法

  本题有两种化简式子的方法，虽然最后的复杂度都是O($n^\frac{2}{3}$)，但是代码难度却截然不同
  壹：$\sum_{i=1}^n\sum_{j=1}^n lcm(i,j)=\sum_{d=1}^n d^{-1} \sum_{i=1}^n\sum_{j=1}^n ij*[gcd(i,j)=1]$

$$=\sum_{d=1}^n d^{-1}*d^2*\sum_{i=1}^{\lfloor \frac{n}{d} \rfloor}\sum_{j=1}^{\lfloor \frac{n}{d} \rfloor} ij*[gcd(i,j)=1]$$
$$=\sum_{d=1}^n d \sum_{p=1}^{\lfloor \frac{n}{d} \rfloor}\mu(p)*p^2*\sum_{i=1}^{\lfloor \frac{n}{dp} \rfloor}\sum_{j=1}^{\lfloor \frac{n}{dp} \rfloor} ij$$
  令$w=dp$，$F(n)=\sum_{i=1}^n\sum_{j=1}^n ij$，则有：
$$原式=\sum_{w=1}^n F(\lfloor \frac{n}{w} \rfloor)*w*\sum_{d|w}\mu(\dfrac{w}{d})*\dfrac{w}{d}=\sum_{w=1}^n F(\lfloor \frac{n}{w} \rfloor)*w*\sum_{d|w}\mu(d)*d$$
  只看后半部分：
$$\sum_{w=1}^n w\sum_{d|w}\mu(d)*d=\sum_{w=1}^n\sum_{d|w}\mu(d)*d*w=\sum_{p=1}^n\sum_{d=1}^{\lfloor \frac{n}{p} \rfloor}p*d^2*\mu(d)=\sum_{d=1}^nd^2*\mu(d)\sum_{p=1}^{\lfloor \frac{n}{d} \rfloor}p$$
  令$S(n)=\sum_{i=1}^n i$，有：$原式=\sum_{d=1}^n S(\lfloor \frac{n}{d} \rfloor)*d^2\mu(d)$
  然后还是只看后半部分：
$$∵[n=1]=\sum_{d|n}\mu(d)$$
$$∴\sum_{w=1}^n\sum_{d|w}d^2*\mu(d)*(\dfrac{w}{d})^2=\sum_{w=1}^n w^2*[\sum_{d|w}\mu(d)]=\sum_{w=1}^n w^2*[w=1]=1$$
$$又∵\sum_{w=1}^n\sum_{d|w}d^2*\mu(d)*(\dfrac{w}{d})^2=\sum_{k=1}^{n}k^2\sum_{d=1}^{\left \lfloor \frac{n}{k} \right \rfloor}d^2\mu(d)$$
$$∴1^2*\sum_{d=1}^nd^2\mu(d)=1-\sum_{k=2}^{n}k^2\sum_{d=1}^{\left \lfloor \frac{n}{k} \right \rfloor}d^2\mu(d)$$
  然后再一层一层套回去，很难写……
  贰：$\sum_{i=1}^n\sum_{j=1}^n lcm(i,j)=\sum_{d=1}^n d^{-1} \sum_{i=1}^n\sum_{j=1}^n ij*[gcd(i,j)=1]$
$$=\sum_{d=1}^n d*\sum_{i=1}^{\lfloor \frac{n}{d} \rfloor}\sum_{j=1}^{\lfloor \frac{n}{d} \rfloor} ij*[gcd(i,j)=1]$$
  只看后面：
$$\sum_{i=1}^{\lfloor \frac{n}{d} \rfloor}\sum_{j=1}^{\lfloor \frac{n}{d} \rfloor} ij*[gcd(i,j)=1]=\sum_{i=1}^{\lfloor \frac{n}{d} \rfloor}i\sum_{j=1}^{\lfloor \frac{n}{d} \rfloor} j*[gcd(i,j)=1]=2*\sum_{i=1}^{\lfloor \frac{n}{d} \rfloor} i \sum_{j=1}^{i} j*[gcd(i,j)=1]-1$$
$$=2*\sum_{i=1}^{\lfloor \frac{n}{d} \rfloor} i*\dfrac{i*\varphi(i)+[i=1]}{2}-1=\sum_{i=1}^{\lfloor \frac{n}{d} \rfloor} i^2*\varphi(i)=\sum_{i=1}^{\lfloor \frac{n}{d} \rfloor}\sum_{p|i}p^2*\varphi(p)*(\dfrac{i}{p})^2$$
$$=\sum_{p=1}^{\lfloor \frac{n}{d} \rfloor} p^2*\sum_{t=1}^{\lfloor \frac{n}{dp} \rfloor}t^2*\varphi(t)=\dfrac{w^2*(w+1)^2}{4}-\sum_{k=2}^w k^2*\sum_{i=1}^{\lfloor \frac{w}{k} \rfloor} i^2\varphi(i)，w=\lfloor \frac{n}{d} \rfloor$$

复杂度

O($n^\frac{2}{3}$)

代码

#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<map>
#define Lint long long int
using namespace std;
const int m2=500000004;
const int m6=166666668;
const int N=10001000;
const int M=1e9+7;
bool vis[N];
Lint phi[N],n;
int pri[N/10],cnt;
int ans;
map<Lint,int> f;
int F(Lint n)
{
    n%=M;n=1ll*n*(n+1)%M;
    return 1ll*n*m2%M;
}
int G(Lint n)
{
    n%=M;int x=1ll*(n+1)*(2*n+1)%M;
    return n*x%M*m6%M;
}
int H(Lint n)   { return 1ll*phi[n]*(n%M*(n%M)%M)%M; }
void Prepare()
{
    Lint L=min( 1ll*N,n+1 );
    phi[1]=1;
    for(int i=2;i<L;i++)
    {
        if( !vis[i] )   pri[++cnt]=i,phi[i]=i-1;
        for(int j=1;j<=cnt;j++)
        {
            int x=pri[j]*i;
            if( x>=L )   break ;
            vis[x]=1;
            if( i%pri[j] )   phi[x]=phi[i]*phi[pri[j]];
            else
            {
                phi[x]=phi[i]*pri[j];
                break ;
            }
        }
    }
    for(int i=1;i<L;i++)   phi[i]=(phi[i-1]+H(i))%M;
}
int cal(Lint n)
{
    if( n<N )   return phi[n];
    if( f.count(n) )   return f[n];
    int ret=1ll*F(n)*F(n)%M;
    Lint x=2;
    while( x<=n )
    {
        Lint y=n/(n/x);
        ret=(ret-1ll*(G(y)-G(x-1)+M)%M*cal(n/x)%M+M)%M;
        x=y+1;
    }
    return f[n]=ret;
}
int main()
{
    Lint x=1;
    scanf("%lld",&n);
    Prepare();
    while( x<=n )
    {
        Lint y=n/(n/x);
        ans+=1ll*(y-x+1)%M*(y%M+x%M)%M*m2%M*cal(n/x)%M;
        ans%=M,x=y+1;
    }
    printf("%d\n",ans);
    return 0;
}