2020.10.16 打比赛抄板子记录
作者:
徐州煤炭工人
,
2020-10-17 18:32:27
,
所有人可见
,
阅读 691
const int N=5e6+2;
bool np[N];
int prime[N],pi[N];
int getprime()
{
int cnt=0;
np[0]=np[1]=true;
pi[0]=pi[1]=0;
for(int i=2;i<N;++i)
{
if(!np[i])
prime[++cnt]=i;
pi[i]=cnt;
for(int j=1;j<=cnt&&i*prime[j]<N;++j)
{
np[i*prime[j]]=true;
if(i%prime[j]==0) break;
}
}
return cnt;
}
const int M=7;
const int PM=2*3*5*7*11*13*17;
int phi[PM+1][M+1],sz[M+1];
void init()
{
getprime();
sz[0]=1;
for(int i=0;i<=PM;++i)
phi[i][0]=i;
for(int i=1;i<=M;++i)
{
sz[i]=prime[i]*sz[i-1];
for(int j=1;j<=PM;++j)
phi[j][i]=phi[j][i-1]-phi[j/prime[i]][i-1];
}
}
int sqrt2(ll x)
{
ll r=(ll)sqrt(x-0.1);
while(r*r<=x) ++r;
return int(r-1);
}
int sqrt3(ll x)
{
ll r=(ll)cbrt(x-0.1);
while(r*r*r<=x) ++r;
return int(r-1);
}
ll getphi(ll x,int s){
if(s==0)
return x;
if(s<=M)
return phi[x%sz[s]][s]+(x/sz[s])*phi[sz[s]][s];
if(x<=prime[s]*prime[s])
return pi[x]-s+1;
if(x<=prime[s]*prime[s]*prime[s]&&x<N)
{
int s2x=pi[sqrt2(x)];
ll ans=pi[x]-(s2x+s-2)*(s2x-s+1)/2;
for(int i=s+1;i<=s2x;++i)
ans+=pi[x/prime[i]];
return ans;
}
return getphi(x,s-1)-getphi(x/prime[s],s-1);
}
ll getpi(ll x)
{
if(x<N) return pi[x];
ll ans=getphi(x,pi[sqrt3(x)])+pi[sqrt3(x)]-1;
for(int i=pi[sqrt3(x)]+1,ed=pi[sqrt2(x)];i<=ed;++i)
ans-=getpi(x/prime[i])-i+1;
return ans;
}
ll lehmer_pi(ll x)
{
if(x<N) return pi[x];
int a=(int)lehmer_pi(sqrt2(sqrt2(x)));
int b=(int)lehmer_pi(sqrt2(x));
int c=(int)lehmer_pi(sqrt3(x));
ll sum=getphi(x,a)+ll(b+a-2)*(b-a+1)/2;
for(int i=a+1;i<=b;i++)
{
ll w=x/prime[i];
sum-=lehmer_pi(w);
if(i>c)
continue;
ll lim=lehmer_pi(sqrt2(w));
for(int j=i;j<=lim;j++)
sum-=lehmer_pi(w/prime[j])-(j-1);
}
return sum;
}