51Nod 1227 平均最小公倍数（杜教筛）

题目链接：哆啦A梦传送门

题意：

求$\begin{array}{r}\sum _{i=a}^b\frac{1}{i}\sum _{j=1}^{i}lcm(j,i)\end{array}$

我们设：

$\begin{array}{rl}ans& =\sum _{i=1}^n\frac{1}{i}\sum _{j=1}^{i}lcm(j,i)\\ & =\sum _{i=1}^n\frac{1}{i}\sum _{j=1}^i\frac{i\ast j}{gcd(i,j)}\\ & =\sum _{d=1}^{n}d\sum _{i=1}^{\frac{n}{d}}\frac{1}{i}\ast i\sum _{j=1}^i[gcd(i,j)=1]j\\ & =\sum _{d=1}^{n}d\sum _{i=1}^{\frac{n}{d}}\frac{i\ast \phi (i)+[n==1]}{2}\\ & =(\frac{1}{2}\sum _{d=1}^{n}d\sum _{i=1}^{\frac{n}{d}}i\ast \phi (i))+\frac{n}{2}\end{array}$

现在我们看到这个式子，只需分块计算，但我们还得快速得出$\begin{array}{r}\sum _{i=1}^{n}i\ast \phi (i)\end{array}$

我们知道有一个关于$\phi (i)$的性质：
$n={\sum }_{d|n}\phi (d)$

那么我们设$h(x)=x\ast \phi (x)$，$S(n)={\sum }_{i=1}^{n}h(i)$

我们知道：${\sum }_{i=1}^n{i}^2=\frac{n\ast (n+1)\ast (2\ast n+1)}{6}$
即：

$\begin{array}{rl}\sum _{i=1}^n{i}^2& =\sum _{d=1}^n\sum _{i|d}\phi (i)\ast d\\ & =\sum _{d=1}^n\sum _{i|d}h(i)\ast \frac{d}{i}\\ & =\sum _{d=1}^n\sum _{i=1}^{\frac{n}{d}}h(i)\ast \frac{(d\ast i)}{i}\\ & =\sum _{d=1}^{n}d\sum _{i=1}^{\frac{n}{d}}h(i)\end{array}$

即：
$\begin{array}{r}\frac{n\ast (n+1)\ast (2\ast n+1)}{6}=S(n)+\sum _{d=2}^{n}d\sum _{i=1}^{\frac{n}{d}}h(i)\end{array}$
$\begin{array}{r}S(n)=\frac{n\ast (n+1)\ast (2\ast n+1)}{6}-\sum _{d=2}^{n}d\sum _{i=1}^{\frac{n}{d}}h(i)\end{array}$

故我们可以预处理前6000000现$x\ast \phi (x)$和。再分块计算就好了。

#include
#include
#include
#include

using namespace std;
typedef long long LL;
const LL mod=1000000007;
const int N=6e6;

LL sum[N];///存放S的前N项
bool vis[N+10];
LL phi[N+10];
int pri[N+10],tot;

tr1::unordered_map w;

LL inv2,inv4,inv6;

void init()
{


    phi[1]=1;

    for(int i=2;i<=N;i++)
    {
        if(!vis[i]){
            phi[i]=i-1;
            pri[++tot]=i;
        }
        for(int j=1;j<=tot&&i*pri[j]<=N;j++)
        {
            int t=pri[j];
            vis[i*t]=1;
            if(i%t==0){
                phi[i*t]=phi[i]*t;break;
            }
            phi[i*t]=phi[i]*(t-1);
        }
    }

    ///预处理 \sum_{i=1}^{N}i*phi[i]
    for(int i=1;i<=N;i++)
    {
        sum[i]=(sum[i-1]+1LL*phi[i]%mod*i%mod)%mod;
    }

}

LL fast_pow(LL a1,LL n)
{
    LL ans=1;
    while(n)
    {
        if(n&1) ans=ans*a1%mod;
        a1=a1*a1%mod;
        n>>=1;
    }
    return ans;
}

LL s2(LL x) ///\sum_{i=1}^{x}i*i
{
    LL t=x%mod;
    return t*(t+1)%mod*(2*t+1)%mod*inv6%mod;
}

LL s1(LL x) ///\sum_{i=1}^{x}i
{
    x=x%mod;
    return x*(x+1)%mod*inv2%mod;
}


LL djs(LL x)
{
    if(x<=N) return sum[x];
    if(w[x]) return w[x];

    LL t=x%mod;
    LL ans=s2(x);
    for(LL l=2,r;l<=x;l=r+1)
    {
        r=x/(x/l);
        ans=(ans-(s1(r)-s1(l-1)+mod)%mod*djs(x/l)+mod)%mod;
    }

    return w[x]=ans;


}

LL solve(LL n)
{

    LL ans=n%mod;
    for(LL l=1,r;l<=n;l=r+1)
    {
        r=n/(n/l);
        ans=(ans+(r-l+1+mod)%mod*djs(n/l)%mod)%mod;
    }

    return ans*inv2%mod;
}



int main()
{

    LL n;

     inv2=fast_pow((LL)2,mod-2);
     inv4=fast_pow((LL)4,mod-2);
     inv6=fast_pow((LL)6,mod-2);

    init();
    LL a,b;

    scanf("%lld%lld",&a,&b);

    LL ans=0;

    ans=(solve(b)-solve(a-1)+mod)%mod;


    printf("%lld\n",ans);

    return 0;
}

\begin{aligned}\end{aligned}