51nod1238 最小公倍数之和 V3

题解

　　感觉柿子画起来有点费劲，我太蒻了。

$$ans=\sum_{i=1}^n\sum_{j=1}^n\frac{ij}{(i,j)}\\ =\sum_{d=1}^n\frac{1}{d}\sum_{d|i}^n\sum_{d|j}^nij[(i,j)=1]\\ =\sum_{d=1}^nd\sum_{i=1}^{\lfloor\frac{n}{d}\rfloor}\sum_{j=1}^{\lfloor\frac{n}{d}\rfloor}ij[(i,j)=1]\\ =\sum_{d=1}^nd\times(2\sum_{i=1}^{\lfloor\frac{n}{d}\rfloor}i(\sum_{j=1}^ij[(i,j)=1]))-1)\\ =\sum_{d=1}^nd(2\sum_{i=1}^{\lfloor\frac{n}{d}\rfloor}i\frac{i\varphi(i)+[i=1]}{2}-1)\\ =\sum_{d=1}^nd\sum_{i=1}^{\lfloor\frac{n}{d}\rfloor}i^2\varphi(i)$$
　　令 $f(i)=i^2\varphi(i)$， $s(n)=\sum_{i=1}^nf(i)$
$$ans=\sum_{d=1}^nd\times s(\lfloor\frac{n}{d}\rfloor)$$
　　 $id_2(n)=n^2$
$$\sum_{d|n}d^2\varphi(i)id_2(\frac{n}{d})\\ =\sum_{d|n}d^2\varphi(i)\frac{n^2}{d^2}\\ =n^3$$
　　要画出杜教筛，要试着对卷积求个前缀和
$$\sum_{i=1}^n\sum_{d|i}f(d)id_2(\frac{i}{d})\\ =\sum_{\frac{i}{d}=1}^n\sum_{d=1}^{\left\lfloor\frac{n}{\frac{i}{d}}\right\rfloor}f(d)id_2(\frac{i}{d})\\ =\sum_{t=1}^n\sum_{d=1}^{\lfloor\frac{n}{t}\rfloor}f(d)id_2(t)\\ =\sum_{t=1}^nid_2(t)s(\lfloor\frac{n}{t}\rfloor)$$
　　因为强迫症，我把 $t$都换成 $i$。（和之前的 $i$没有任何关系）
$$\sum_{i=1}^nid_2(i)s(\lfloor\frac{n}{i}\rfloor)=\sum_{i=1}^ni^3\\ s(n)=\sum_{i=1}^{n}i^3-\sum_{i=2}^nid_2(i)s(({\lfloor\frac{n}{i}\rfloor})$$
　　好了

代码

//杜教筛 
#include <cstdio>
#include <algorithm>
#define maxn 4700000
#define mod 1000000007ll
using namespace std;
typedef long long ll;
ll f[maxn+10], phi[maxn+10], _2, _6, N;
int prime[maxn+10];
bool mark[maxn+10];
void shai()
{
    ll i, j;
    phi[1]=1;
    for(i=2;i<maxn;i++)
    {
        if(!mark[i])prime[++*prime]=i, phi[i]=i-1;
        for(j=1;i*prime[j]<maxn;j++)
        {
            mark[i*prime[j]]=1;
            if(i%prime[j]==0){phi[i*prime[j]]=phi[i]*prime[j];break;}
            phi[i*prime[j]]=phi[i]*phi[prime[j]];
        }
    }
    for(i=1;i<maxn;i++)phi[i]=(phi[i]*i%mod*i%mod+phi[i-1])%mod;
}
ll sqr(ll x){if(x>mod)x%=mod;return x*x%mod;}
ll s1(ll x){if(x>mod)x%=mod;return x*(1+x)%mod*_2%mod;}
ll s2(ll x){if(x>mod)x%=mod;return x*(1+x)%mod*(2*x+1)%mod*_6%mod;}
ll s3(ll x){if(x>mod)x%mod;return sqr(s1(x));}
ll getf(ll x){return x<maxn?phi[x]:f[N/x];}
void djs(ll n)
{
    if(n<maxn or getf(n))return;
    ll i, last, t=N/n;
    f[t]=s3(n);
    for(i=2;i<=n;i=last+1)
    {
        last=n/(n/i);
        djs(n/i);
        f[t]=(f[t]-(s2(last)-s2(i-1))*getf(n/i)%mod)%mod;
    }
}
int main()
{
    ll i, last, ans=0;
    _2=500000004;
    _6=166666668;
    scanf("%lld",&N);
    shai();
    djs(N);
    for(i=1;i<=N;i=last+1)
    {
        last=N/(N/i);
        ans=(ans+getf(N/i)*(s1(last)-s1(i-1)))%mod;
    }
    printf("%lld",(ans+mod)%mod);
    return 0;
}