【模板】快速区间素数计数

【Process returned 0 (0x0)   execution time : 0.365 s】

<pre name="code" class="cpp">/*
*******************
*******************
HDU 5901 Count primes
*******************
*******************
*/
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<string>
#include<climits>
#include<string>
#include<queue>
#include<algorithm>
using namespace std;
#define rep(i,j,k)for(i=j;i<k;i++)
#define per(i,j,k)for(i=j;i>k;i--)
#define MS(x,y)memset(x,y,sizeof(x))
typedef long long LL;
const int INF=0x7ffffff;

const int N = 5e6 + 2;
bool np
;
int prime
, pi
;
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;
}
int main()
{
init();
LL n;
while(~scanf("%lld",&n))
{
printf("%lld\n",lehmer_pi(n));
}
return 0;
}

不能运行但是能AC

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

const ll maxn=1e11;

const ll maxp=sqrt(maxn)+10;

ll f[maxp],g[maxp];

ll solve(ll n)

{

    ll i,j,m;

    for(m=1;m*m<=n;m++)

    f[m]=n/m-1;

    for(i=1;i<=m;i++)

    g[i]=i-1;

    for(i=2;i<=m;i++)

    {

        if(g[i]==g[i-1]) continue;

        for(j=1;j<=min(m-1,n/i/i);j++)

        {

            if(i*j<m)

            f[j]-=f[i*j]-g[i-1];

            else

            f[j]-=g[n/i/j]-g[i-1];

        }

        for(j=m;j>=i*i;j--)

        g[j]-=g[j/i]-g[i-1];

    }

    return f[1];

}

int main()

{

    ll n;

    while(scanf("%lld",&n)!=EOF)

    printf("%lld\n",solve(n));

    return 0;

}