题面
题解
话说……就一个杜教筛……刚才那道拿过来改几行就行了……
//minamoto
#include<bits/stdc++.h>
#define R register
#define ll long long
#define IT map<ll,int>::iterator
#define fp(i,a,b) for(R int i=a,I=b+1;i<I;++i)
#define fd(i,a,b) for(R int i=a,I=b-1;i>I;--i)
#define go(u) for(int i=head[u],v=e[i].v;i;i=e[i].nx,v=e[i].v)
using namespace std;
const int N=5e6+5,P=1e9+7,inv2=500000004;
inline int add(R int x,R int y){return x+y>=P?x+y-P:x+y;}
inline int dec(R int x,R int y){return x-y<0?x-y+P:x-y;}
inline int mul(R int x,R int y){return 1ll*x*y-1ll*x*y/P*P;}
inline int calc(R int x){return (1ll*x*(x+1)>>1)%P;}
bitset<N>vis;int p[N],phi[N],m,sqr,res;ll n;map<ll,int>mp;IT it;
void init(int n){
phi[1]=1;
fp(i,2,n){
if(!vis[i])p[++m]=i,phi[i]=i-1;
for(R int j=1;j<=m&&1ll*i*p[j]<=n;++j){
vis[i*p[j]]=1;
if(i%p[j]==0){phi[i*p[j]]=phi[i]*p[j];break;}
phi[i*p[j]]=phi[i]*(p[j]-1);
}
}
fp(i,2,n)phi[i]=add(phi[i],phi[i-1]);
}
int Phi(ll n){
if(n<=sqr)return phi[n];
it=mp.find(n);
if(it!=mp.end())return it->second;
int res=calc(n%P);
for(R ll i=2,j;i<=n;i=j+1)
j=n/(n/i),res=dec(res,mul((j-i+1)%P,Phi(n/i)));
return mp[n]=res;
}
int main(){
// freopen("testdata.in","r",stdin);
scanf("%lld",&n),init(sqr=N-5);
printf("%d\n",Phi(n));
return 0;
}