51nod1244 莫比乌斯函数之和

杜教筛模板

给$mu$找个函数1，因为$mu * 1=I$，其中I是元函数，I(x)=[x==1]。

那就$sum_{i=1}^{n}sum_{d|i}mu(d)=sum_{k=1}^{n}1sum_{d=1}^{left lfloor frac{n}{k} ight floor}mu(d)=sum_{k=1}^{n}S(left lfloor frac{n}{k} ight floor)$

然后$S(n)=sum_{i=1}^{n}sum_{d|i}mu(d)-sum_{k=2}^{n}S(left lfloor frac{n}{k} ight floor)=1-S(left lfloor frac{n}{k} ight floor)$。

搞定！

#include<string.h>
#include<stdlib.h>
#include<stdio.h>
#include<math.h>
//#include<assert.h>
#include<algorithm> 
//#include<iostream>
//#include<bitset>
using namespace std;

#define LL long long
LL n,m;
#define maxn 5000011
int miu[maxn],summiu[maxn],prime[maxn/10],lp; bool notprime[maxn];
void pre(int n)
{
    lp=0; miu[1]=1; summiu[1]=1;
    for (int i=2;i<=n;i++)
    {
        if (!notprime[i]) {prime[++lp]=i; miu[i]=-1;}
        summiu[i]=summiu[i-1]+miu[i];
        for (int j=1,tmp;j<=lp && 1ll*prime[j]*i<=n;j++)
        {
            notprime[tmp=i*prime[j]]=1;
            if (i%prime[j]) miu[tmp]=-miu[i];
            else {miu[tmp]=0; break;}
        }
    }
}

struct Edge{LL to; int v,next;};
#define maxh 1000007
struct Hash
{
    int first[maxh],le;Edge edge[maxn];
    Hash() {le=2;}
    void insert(LL y,int v)
    {int x=y%maxh; Edge &e=edge[le]; e.to=y; e.v=v; e.next=first[x]; first[x]=le++;}
    int find(LL y)
    {
        int x=y%maxh;
        for (int i=first[x];i;i=edge[i].next) if (edge[i].to==y) return edge[i].v;
        return -1;
    }
}h;

int du(LL n)
{
    if (n<=m) return summiu[n];
    int tmp=h.find(n); if (tmp!=-1) return tmp;
    LL tot=0;
    for (LL i=2,last;i<=n;i=last+1)
    {
        last=n/(n/i);
        tot+=(last-i+1)*du(n/i);
    }
    LL ans=1-tot;
    h.insert(n,ans);
    return ans;
}

int main()
{
    LL left;scanf("%lld%lld",&left,&n);
    m=(LL)pow(pow(n,1.0/3),2); pre(m);
    printf("%d
",du(n)-du(left-1));
    return 0;
}