[51Nod1238]最小公倍数之和-V3

题意

  给定n$n$，求∑ni=1∑nj=1lcm(i,j)$\sum _{i=1}^n\sum _{j=1}^{n}lcm(i,j)$

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(n23$n^{\frac{2}{3}}$)，但是代码难度却截然不同
  壹：∑ni=1∑nj=1lcm(i,j)=∑nd=1d−1∑ni=1∑nj=1ij∗[gcd(i,j)=1]$\sum _{i=1}^n\sum _{j=1}^{n}lcm(i,j)=\sum _{d=1}^n{d}^{-1}\sum _{i=1}^n\sum _{j=1}^{n}ij\ast [gcd(i,j)=1]$

=∑d=1nd−1∗d2∗∑i=1⌊nd⌋∑j=1⌊nd⌋ij∗[gcd(i,j)=1]$$=\sum _{d=1}^n{d}^{-1}\ast d^2\ast \sum _{i=1}^{\lfloor \frac{n}{d}\rfloor }\sum _{j=1}^{\lfloor \frac{n}{d}\rfloor }ij\ast [gcd(i,j)=1]$$

=∑d=1nd∑p=1⌊nd⌋μ(p)∗p2∗∑i=1⌊ndp⌋∑j=1⌊ndp⌋ij$$=\sum _{d=1}^{n}d\sum _{p=1}^{\lfloor \frac{n}{d}\rfloor }\mu (p)\ast p^2\ast \sum _{i=1}^{\lfloor \frac{n}{dp}\rfloor }\sum _{j=1}^{\lfloor \frac{n}{dp}\rfloor }ij$$

  令w=dp$w=dp$，F(n)=∑ni=1∑nj=1ij$F(n)=\sum _{i=1}^n\sum _{j=1}^{n}ij$，则有：

原式=∑w=1nF(⌊nw⌋)∗w∗∑d|wμ(wd)∗wd=∑w=1nF(⌊nw⌋)∗w∗∑d|wμ(d)∗d$$原式=\sum _{w=1}^{n}F(\lfloor \frac{n}{w}\rfloor )\ast w\ast \sum _{d|w}\mu ({\displaystyle \frac{w}{d}})\ast {\displaystyle \frac{w}{d}}=\sum _{w=1}^{n}F(\lfloor \frac{n}{w}\rfloor )\ast w\ast \sum _{d|w}\mu (d)\ast d$$

  只看后半部分：

∑w=1nw∑d|wμ(d)∗d=∑w=1n∑d|wμ(d)∗d∗w=∑p=1n∑d=1⌊np⌋p∗d2∗μ(d)=∑d=1nd2∗μ(d)∑p=1⌊nd⌋p$$\sum _{w=1}^{n}w\sum _{d|w}\mu (d)\ast d=\sum _{w=1}^n\sum _{d|w}\mu (d)\ast d\ast w=\sum _{p=1}^n\sum _{d=1}^{\lfloor \frac{n}{p}\rfloor }p\ast d^2\ast \mu (d)=\sum _{d=1}^n{d}^2\ast \mu (d)\sum _{p=1}^{\lfloor \frac{n}{d}\rfloor }p$$

  令S(n)=∑ni=1i$S(n)=\sum _{i=1}^{n}i$，有：原式=∑nd=1S(⌊nd⌋)∗d2μ(d)$原式=\sum _{d=1}^{n}S(\lfloor \frac{n}{d}\rfloor )\ast d^2\mu (d)$
  然后还是只看后半部分：

∵[n=1]=∑d|nμ(d)$$\because [n=1]=\sum _{d|n}\mu (d)$$

∴∑w=1n∑d|wd2∗μ(d)∗(wd)2=∑w=1nw2∗[∑d|wμ(d)]=∑w=1nw2∗[w=1]=1$$\therefore \sum _{w=1}^n\sum _{d|w}{d}^2\ast \mu (d)\ast ({\displaystyle \frac{w}{d}}{)}^2=\sum _{w=1}^n{w}^2\ast [\sum _{d|w}\mu (d)]=\sum _{w=1}^n{w}^2\ast [w=1]=1$$

又∵∑w=1n∑d|wd2∗μ(d)∗(wd)2=∑k=1nk2∑d=1⌊nk⌋d2μ(d)$$又\because \sum _{w=1}^n\sum _{d|w}{d}^2\ast \mu (d)\ast ({\displaystyle \frac{w}{d}}{)}^2=\sum _{k=1}^n{k}^2\sum _{d=1}^{\lfloor \frac{n}{k}\rfloor }{d}^2\mu (d)$$

∴12∗∑d=1nd2μ(d)=1−∑k=2nk2∑d=1⌊nk⌋d2μ(d)$$\therefore 1^2\ast \sum _{d=1}^n{d}^2\mu (d)=1-\sum _{k=2}^n{k}^2\sum _{d=1}^{\lfloor \frac{n}{k}\rfloor }{d}^2\mu (d)$$

  然后再一层一层套回去，很难写……
  贰：∑ni=1∑nj=1lcm(i,j)=∑nd=1d−1∑ni=1∑nj=1ij∗[gcd(i,j)=1]$\sum _{i=1}^n\sum _{j=1}^{n}lcm(i,j)=\sum _{d=1}^n{d}^{-1}\sum _{i=1}^n\sum _{j=1}^{n}ij\ast [gcd(i,j)=1]$

=∑d=1nd∗∑i=1⌊nd⌋∑j=1⌊nd⌋ij∗[gcd(i,j)=1]$$=\sum _{d=1}^{n}d\ast \sum _{i=1}^{\lfloor \frac{n}{d}\rfloor }\sum _{j=1}^{\lfloor \frac{n}{d}\rfloor }ij\ast [gcd(i,j)=1]$$

  只看后面：

∑i=1⌊nd⌋∑j=1⌊nd⌋ij∗[gcd(i,j)=1]=∑i=1⌊nd⌋i∑j=1⌊nd⌋j∗[gcd(i,j)=1]=2∗∑i=1⌊nd⌋i∑j=1ij∗[gcd(i,j)=1]−1$$\sum _{i=1}^{\lfloor \frac{n}{d}\rfloor }\sum _{j=1}^{\lfloor \frac{n}{d}\rfloor }ij\ast [gcd(i,j)=1]=\sum _{i=1}^{\lfloor \frac{n}{d}\rfloor }i\sum _{j=1}^{\lfloor \frac{n}{d}\rfloor }j\ast [gcd(i,j)=1]=2\ast \sum _{i=1}^{\lfloor \frac{n}{d}\rfloor }i\sum _{j=1}^{i}j\ast [gcd(i,j)=1]-1$$

=2∗∑i=1⌊nd⌋i∗i∗φ(i)+[i=1]2−1=∑i=1⌊nd⌋i2∗φ(i)=∑i=1⌊nd⌋∑p|ip2∗φ(p)∗(ip)2$$=2\ast \sum _{i=1}^{\lfloor \frac{n}{d}\rfloor }i\ast {\displaystyle \frac{i\ast \phi (i)+[i=1]}{2}}-1=\sum _{i=1}^{\lfloor \frac{n}{d}\rfloor }{i}^2\ast \phi (i)=\sum _{i=1}^{\lfloor \frac{n}{d}\rfloor }\sum _{p|i}{p}^2\ast \phi (p)\ast ({\displaystyle \frac{i}{p}}{)}^2$$

=∑p=1⌊nd⌋p2∗∑t=1⌊ndp⌋t2∗φ(t)=w2∗(w+1)24−∑k=2wk2∗∑i=1⌊wk⌋i2φ(i)，w=⌊nd⌋$$=\sum _{p=1}^{\lfloor \frac{n}{d}\rfloor }{p}^2\ast \sum _{t=1}^{\lfloor \frac{n}{dp}\rfloor }{t}^2\ast \phi (t)={\displaystyle \frac{w^2\ast (w+1{)}^2}{4}}-\sum _{k=2}^w{k}^2\ast \sum _{i=1}^{\lfloor \frac{w}{k}\rfloor }{i}^2\phi (i)，w=\lfloor \frac{n}{d}\rfloor$$

复杂度

O(n23$n^{\frac{2}{3}}$)

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;
}