題面

Description

給出一個數字N，求\(\sum\limits_{i=1}^n\varphi(i)\)i,1<=i<=N

Input

正整數N。N<=2*10^9

Output

輸出答案。

Sample Input

10

Sample Output

32

題目分析

杜教篩模板題。

由\((1*\varphi)=Id\)，取\(g(x)=1\)。
\[ S(n)=\frac {n \cdot (n+1)}2-\sum_{i=2}^nS(\frac ni) \]

代碼實現

#include<iostream>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<cstdio>
#include<iomanip>
#include<cstdlib>
#include<map>
#define MAXN 0x7fffffff
typedef long long LL;
const int N=1e7+5,M=1e7;
using namespace std;
inline int Getint(){register int x=0,f=1;register char ch=getchar();while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}while(isdigit(ch)){x=x*10+ch-'0';ch=getchar();}return x*f;}
bool vis[N];
int prime[N];
LL phi[N];
map<LL,LL>sphi;
LL Sphi(int x){
    if(x<=M)return phi[x];
    if(sphi[x])return sphi[x];
    LL ret=1ll*x*(x+1)/2;
    for(int l=2,r;l<=x;l=r+1){
        r=x/(x/l);
        ret-=(r-l+1)*Sphi(x/l);
    }
    return sphi[x]=ret;
} 
int main(){
    phi[1]=1;
    for(int i=2;i<=M;i++){
        if(!vis[i])prime[++prime[0]]=i,phi[i]=i-1;
        for(int j=1;j<=prime[0]&&1ll*prime[j]*i<=M;j++){
            vis[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(int i=2;i<=M;i++)phi[i]+=phi[i-1];
    int n=Getint(); 
    cout<<Sphi(n); 
    return 0;
}
 

【BZOJ4805】歐拉函數求和