bzoj4916 神犇和蒟蒻

Description

很久很久以前，有一只神犇叫yzy;

很久很久之后，有一只蒟蒻叫lty; 

Input

请你读入一个整数N;1<=N<=1E9,A、B模1E9+7;

Output

请你输出一个整数A=\sum_{i=1}^N{\mu (i^2)};

请你输出一个整数B=\sum_{i=1}^N{\varphi (i^2)};

Sample Input

1

Sample Output

1
1

 

正解：杜教筛。

第一问答案是$1$。

第二问，先给个结论：$\varphi (n^{2})=n\varphi (n)$，于是我们要求$F(n)=\sum_{i=1}^{n}i\varphi (i)$。

设$f(n)=n\varphi (n)$，考虑$f$与$id$函数的狄利克雷卷积，$id*f(n)=\sum_{d|n}id(d)f(\frac{n}{d})$

$id*f(n)=n\sum_{d|n}\varphi (\frac{n}{d})=n^{2}$，那么$\sum_{i=1}^{n}id*f(i)=\frac{n(n+1)(2n+1)}{6}$

又$\sum_{i=1}^{n}id*f(i)=\sum_{i=1}^{n}\sum_{d|i}id(d)f(\frac{i}{d})=\sum_{ij\leq n}id(i)f(j)=\sum_{i=1}^{n}id(i)F(\left \lfloor \frac{n}{i} \right \rfloor)$

于是$F(n)=id(1)F(n)=\sum_{i=1}^{n}id*f(i)-\sum_{i=2}^{n}id(i)F(\left \lfloor \frac{n}{i} \right \rfloor)=\frac{n(n+1)(2n+1)}{6}-\sum_{i=2}^{n}id(i)F(\left \lfloor \frac{n}{i} \right \rfloor)$

然后直接用杜教筛的套路：数论分块+记忆化搜索就行了。

 

//It is made by wfj_2048~
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <vector>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#define rhl (1000000007)
#define N (3000010)
#define inf (1<<30)
#define il inline
#define RG register
#define ll long long
#define File(s) freopen(s".in","r",stdin),freopen(s".out","w",stdout)

using namespace std;

int vis[N],phi[N],prime[N],n,maxn,cnt;
ll f[N],In2,In6;

map <ll,ll> F,vi;

il int gi(){
    RG int x=0,q=1; RG char ch=getchar();
    while ((ch<'0' || ch>'9') && ch!='-') ch=getchar();
    if (ch=='-') q=-1,ch=getchar();
    while (ch>='0' && ch<='9') x=x*10+ch-48,ch=getchar();
    return q*x;
}

il ll qpow(RG ll a,RG ll b){
    RG ll ans=1;
    while (b){
    if (b&1) ans=ans*a%rhl;
    a=a*a%rhl,b>>=1;
    }
    return ans;
}

il void sieve(){
    phi[1]=f[1]=1;
    for (RG int i=2;i<=maxn;++i){
    if (!vis[i]) prime[++cnt]=i,phi[i]=i-1;
    for (RG int j=1,k;j<=cnt;++j){
        k=i*prime[j]; if (k>maxn) break; vis[k]=1;
        if (i%prime[j]) phi[k]=phi[i]*phi[prime[j]];
        else{ phi[k]=phi[i]*prime[j]; break; }
    }
    }
    for (RG int i=2;i<=maxn;++i) f[i]=(f[i-1]+(ll)i*(ll)phi[i])%rhl; return;
}

il ll du(RG ll n){
    if (n<=maxn) return f[n]; if (vi[n]) return F[n];
    RG ll ans=n*(n+1)%rhl*(2*n+1)%rhl*In6%rhl,pos; vi[n]=1;
    for (RG ll i=2;i<=n;i=pos+1){
    pos=n/(n/i);
    ans-=(i+pos)*(pos-i+1)%rhl*du(n/i)%rhl*In2%rhl;
    if (ans<0) ans+=rhl;
    }
    return F[n]=ans;
}

il void work(){
    n=gi(),puts("1"),maxn=min(3000000,n),sieve();
    In2=qpow(2,rhl-2),In6=qpow(6,rhl-2);
    printf("%lld\n",du(n)); return;
}

int main(){
    File("phi");
    work();
    return 0;
}