G40【模板】杜教筛

G40 杜教筛_哔哩哔哩_bilibili

 

P4213 【模板】杜教筛 - 洛谷

 

#include <algorithm>
#include <cstdio>
#include <cstring>
#include <map>
using namespace std;
#define LL long long
const int N=2000010;
LL vis[N],pm[N],mu[N],phi[N],cnt;
map<LL,LL> mp_mu, mp_phi;

void init(){
  mu[1]=phi[1]=1;
  for(int i=2; i<N; i++){
    if(!vis[i])pm[++cnt]=i,mu[i]=-1,phi[i]=i-1;
    for(int j=1;i*pm[j]<N;j++){
      int p=pm[j]; vis[i*p]=1;
      if(i%p==0){phi[i*p]=phi[i]*p; break;}
      mu[i*p]=-mu[i]; phi[i*p]=phi[i]*(p-1);
    }
  }
  for(int i=1;i<N;i++)mu[i]+=mu[i-1],phi[i]+=phi[i-1]; 
}
LL Sphi(LL n){
  if(n<N) return phi[n]; //预处理的剪枝
  if(mp_phi[n]) return mp_phi[n]; //记忆化剪枝
  LL ans = n*(n+1)/2;
  for(LL l=2,r; l<=n; l=r+1){
    r=n/(n/l);
    ans-=Sphi(n/l)*(r-l+1);
  }
  return mp_phi[n]=ans;
}
LL Smu(LL n){
  if(n<N) return mu[n];
  if(mp_mu[n]) return mp_mu[n];
  LL ans = 1;
  for(LL l=2,r; l<=n; l=r+1){
    r=n/(n/l);
    ans-=Smu(n/l)*(r-l+1);
  }
  return mp_mu[n]=ans;
}
int main(){
  init();
  LL T, n; scanf("%lld", &T);
  while(T--){
    scanf("%lld", &n);
    printf("%lld %lld\n", Sphi(n), Smu(n));
  }
}

 Luogu P3768 简单的数学题

#include<bits/stdc++.h>
#define N 5000000
#define LL long long
using namespace std;

int vis[N],p[N],phi[N],cnt;
LL P,n,sum[N],inv;
map<LL,LL> mp;

LL qpow(LL a,LL b){
  LL res=1;
  while(b){
    if(b&1) res=res*a%P;
    a=a*a%P;
    b>>=1;
  }
  return res;
}
void init(){
  inv=qpow(6,P-2);
  phi[1]=1;
  for(int i=2;i<N;i++){
    if(!vis[i])p[++cnt]=i,phi[i]=i-1;
    for(int j=1; i*p[j]<N; j++){
      vis[i*p[j]] = 1;
      if(i%p[j]==0){phi[i*p[j]] = p[j]*phi[i];break;}
      phi[i*p[j]]=(p[j]-1)*phi[i];
    }
  }
  for(int i=1;i<N;i++)
    sum[i]=(sum[i-1]+1LL*i*i%P*phi[i]%P)%P;
}
LL S2(LL x){x%=P;return x*(x+1)%P*(2*x+1)%P*inv%P;}
LL S3(LL x){x%=P;return(x*(x+1)/2)%P*((x*(x+1)/2)%P)%P;}
LL Sn(LL x){
  if(x<N) return sum[x];
  if(mp[x]) return mp[x];
  LL res=S3(x);
  for(LL l=2,r;l<=x;l=r+1){
    r=x/(x/l);
    res=(res-(S2(r)-S2(l-1))%P*Sn(x/l)%P+P)%P;
  }
  return mp[x]=res;
}
LL calc(){
  LL ans=0;
  for(LL l=1,r;l<=n;l=r+1){
    r=n/(n/l);
    ans=(ans+(Sn(r)-Sn(l-1))%P*S3(n/l)%P+P)%P;
  }
  return ans;
}
int main(){
  scanf("%lld%lld",&P,&n);
  init();
  printf("%lld\n",calc());
  return 0;
}