第一問是來搞笑的。由歐拉函數的計算公式容易發現φ(i²)=iφ(i)。那麽可以發現φ(n²)*id(n)=Σd*φ(d)*(n/d)=nΣφ(d)=n² 。這樣就有了杜教篩所要求的容易算前綴和的兩個函數。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<‘0‘||c>‘9‘)) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
    int x=0,f=1;char c=getchar();
    while (c<‘0‘||c>‘9
 
‘) {if (c==‘-‘) f=-1;c=getchar();}
    while (c>=‘0‘&&c<=‘9‘) x=(x<<1)+(x<<3)+(c^48),c=getchar();
    return x*f;
}
int n,phi[N],iphi[N],prime[N],cnt,inv6=166666668;
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+1)>>1)%P;}
int sumtwo(int x){return 1ll*x*(x+1)%P*(x<<1|1)%P*inv6%P;}
int work(int x)
{
    if (x<=min(n,N-10)) return iphi[x];
    if (f.find(x)!=f.end()) return f[x];
    int s=sumtwo(x);
    for (int i=2;i<=x;i++)
    {
        int t=x/(x/i);
        inc(s,P-1ll*(sumone(t)-sumone(i-1)+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<<1<<endl;
    flag[1]=1;phi[1]=1;
    for (int i=2;i<=min(n,N-10);i++)
    {
        if (!flag[i]) prime[++cnt]=i,phi[i]=i-1;
        for (int j=1;j<=cnt&&prime[j]*i<=min(n,N-10);j++)
        {
            flag[prime[j]*i]=1;
            if (i%prime[j]==0) {phi[prime[j]*i]=phi[i]*prime[j];break;}
            phi[prime[j]*i]=phi[i]*(prime[j]-1);
        }
    }
    for (int i=1;i<=min(n,N-10);i++) iphi[i]=1ll*i*phi[i]%P;
    for (int i=1;i<=min(n,N-10);i++) inc(iphi[i],iphi[i-1]);
    cout<<work(n);
    return 0;
}

BZOJ4916 神犇和蒟蒻（歐拉函數+杜教篩）