[杜教筛] 51nod1238. 最小公倍数之和 V3

令

A(n)=n∑i=1ni(i,n)

=n∑d|n∑i=1ndi[(i,nd)=1]

=n∑d|ndφ(d)+n

那么

Ans=2A(n)−∑i=1ni

=∑i=1ni∑d|idφ(d)

=∑d=1nd2φ(d)∑i=1⌊nd⌋i

∑nd=1d2φ(d) 这个东西可以用杜教筛推

p(n)=∑i=1ni2φ(i)

∑i=1n∑d|id2φ(d)(id)2=∑i=1ni3=∑i=1ni2p(⌊ni⌋)

杜教筛一下就好了

#include 
#include 
#include 
#include 

using namespace std;

typedef long long ll;

const int N=10000010,P=1e9+7;

int p[N],phi[N];
ll S[N];

inline void Pre(){
  phi[1]=1;
  for(int i=2;i<=10000000;i++){
    if(!p[i]) p[++*p]=i,phi[i]=i-1;
    for(int j=1;j<=*p && 1LL*i*p[j]<=10000000;j++){
      p[i*p[j]]=1;
      if(i%p[j]) phi[i*p[j]]=phi[i]*(p[j]-1);
      else{
    phi[i*p[j]]=phi[i]*p[j];
    break;
      }
    }
  }
  for(int i=1;i<=10000000;i++)
    S[i]=(S[i-1]+1LL*i*i%P*phi[i])%P;
}

const ll inv2=P+1>>1,inv=(P+1)/6;

inline ll SS(ll l,ll r){
  return (r-l+1)%P*((l+r)%P)%P*inv2%P;
}

inline ll Sqr(ll n){
  n%=P;
  return n*(n+1)%P*(2*n+1)%P*inv%P;
}

inline ll SqrS(ll l,ll r){
  return (Sqr(r)-Sqr(l-1))%P;
}

map M;

inline ll Cube(ll n){
  return SS(1,n)*SS(1,n)%P;
}

ll Sum(ll n){
  if(n<=10000000) return S[n];
  if(M.count(n)) return M[n];
  ll ret=Cube(n);
  for(ll i=2,j;i<=n;i=j+1){
    j=n/(n/i);
    ret=(ret-SqrS(i,j)*Sum(n/i))%P;
  }
  return M[n]=ret;
}

int main(){
  ll n,ans=0;
  freopen("1.in","r",stdin);
  freopen("1.out","w",stdout);
  scanf("%lld",&n); Pre();
  for(ll i=1,j,lst=0;i<=n;i=j+1){
    j=n/(n/i); ll cur=Sum(j);
    ans=(ans+SS(1,n/i)*(cur-lst))%P;
    lst=cur;
  }
  printf("%lld\n",(ans+P)%P);
  return 0;
}