BZOJ4805: 欧拉函数求和(杜教筛)

4805: 欧拉函数求和

Time Limit: 15 Sec  Memory Limit: 256 MB
Submit: 614  Solved: 342
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Description

给出一个数字N，求sigma(phi(i)),1<=i<=N

Input

正整数N。N<=2*10^9

Output

输出答案。

 

Sample Input

10

Sample Output

32

HINT

 

Source

By FancyCoder

 

直接大力杜教筛

$\sum_{i=1}^{n}\varphi(i) = \frac{n\times(n+1)}{2} - \sum_{d=2}^{n}\sum_{i=1}^{\lfloor\frac{n}{d}\rfloor}\varphi(i)$

 

#include<cstdio>

#include<map>

#include<ext/pb_ds/assoc_container.hpp>

#include<ext/pb_ds/hash_policy.hpp>

#define LL long long

using namespace std;

using namespace __gnu_pbds;

const int MAXN=;

int N,limit=,tot=,vis[MAXN],prime[MAXN];

LL phi[MAXN];

gp_hash_table<int,LL>Aphi;

void GetPhi()

{

    vis[]=;phi[]=;

    for(int i=;i<=limit;i++)

    {

        if(!vis[i]) prime[++tot]=i,phi[i]=i-;

        for(int j=;j<=tot&&i*prime[j]<=limit;j++)

        {

            vis[i*prime[j]]=;

            if(i%prime[j]==) {phi[i*prime[j]]=phi[i]*prime[j];break;}

            else phi[i*prime[j]]=phi[i]*(prime[j]-);

        }

    }

    for(int i=;i<=limit;i++) phi[i]+=phi[i-];

}

LL SolvePhi(LL n)

{

    if(n<=limit) return phi[n];

    if(Aphi[n]) return Aphi[n];

    LL tmp=n*(n+)/;

    for(int i=,nxt;i<=n;i=nxt+)

    {

        nxt=min(n,n/(n/i));

        tmp-=SolvePhi(n/i)*(LL)(nxt-i+);

    }

    return Aphi[n]=tmp;

}

int main()

{

    GetPhi();

    scanf("%lld",&N);

    printf("%lld",SolvePhi(N));

    return ;

}