BZOJ4916 神犇和蒟蒻（欧拉函数+杜教筛）

　　第一问是来搞笑的。由欧拉函数的计算公式容易发现φ(i²)=iφ(i)。那么可以发现φ(n²)*id(n)（此处为卷积）=Σd*φ(d)*(n/d)=nΣφ(d)=n² 。这样就有了杜教筛所要求的容易算前缀和的两个函数。一通套路即可。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<map>
using namespace std;
#define ll long long
#define P 1000000007
#define N 1000010
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<''||c>'')) c=getchar();return c;}
int gcd(int n,int m){return m==?n:gcd(m,n%m);}
int read()
{
    int x=,f=;char c=getchar();
    while (c<''||c>'') {if (c=='-') f=-;c=getchar();}
    while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),c=getchar();
    return x*f;
}
int n,phi[N],iphi[N],prime[N],cnt,inv6=;
map<int,int> f;
bool flag[N];
inline void inc(int &x,int y){x+=y;if (x>=P) x-=P;}
int sumone(int x){return (1ll*x*(x+)>>)%P;}
int sumtwo(int x){return 1ll*x*(x+)%P*(x<<|)%P*inv6%P;}
int work(int x)
{
    if (x<=min(n,N-)) return iphi[x];
    if (f.find(x)!=f.end()) return f[x];
    int s=sumtwo(x);
    for (int i=;i<=x;i++)
    {
        int t=x/(x/i);
        inc(s,P-1ll*(sumone(t)-sumone(i-)+P)*work(x/i)%P);
        i=t;
    }
    f[x]=s;return s;
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj4916.in","r",stdin);
    freopen("bzoj4916.out","w",stdout);
    const char LL[]="%I64d\n";
#else
    const char LL[]="%lld\n";
#endif
    n=read();cout<<<<endl;
    flag[]=;phi[]=;
    for (int i=;i<=min(n,N-);i++)
    {
        if (!flag[i]) prime[++cnt]=i,phi[i]=i-;
        for (int j=;j<=cnt&&prime[j]*i<=min(n,N-);j++)
        {
            flag[prime[j]*i]=;
            if (i%prime[j]==) {phi[prime[j]*i]=phi[i]*prime[j];break;}
            phi[prime[j]*i]=phi[i]*(prime[j]-);
        }
    }
    for (int i=;i<=min(n,N-);i++) iphi[i]=1ll*i*phi[i]%P;
    for (int i=;i<=min(n,N-);i++) inc(iphi[i],iphi[i-]);
    cout<<work(n);
    return ;
}