Meisell-Lehmer算法(求1...n范围内的素数个数)

求1...n范围内的素数个数有多少了，如果数据不大的话素筛肯定就可以了，但像HDU-5901中n的范围为10^11肯定就不能素筛了，所以这里介绍Meisell-Lehmer算法，代码颇是复杂，摘的网上的模板然后改成自己的代码习惯。

//Meisell-Lehmer算法

#include <cstdio>
#include <cmath>
#define LL long long
using namespace std;
const int N = 5e6 + 5;
const int M = 7;
const int PM = 2 * 3 * 5 * 7 * 11 * 13 * 17;
int np[N];
int prime[N], pi[N];
int phi[PM+1][M+1], sz[M+1];
int getprime()
{
    int cnt = 0;
    np[0] = np[1] = 1;
    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]] = 1;
            if(i % prime[j] == 0) break;
        }
    }
    return cnt;
}
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);	//cbrt(x): x的立方根
    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()
{
    LL n; init();
    while(~scanf("%lld",&n))
    {
        printf("%lld\n", lehmer_pi(n));
    }
    return 0;
}

还有一种简单的，但似乎慢一些.

#include <iostream>
#include <cstdio>
#include <cmath>
#define LL long long
using namespace std;
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;
}

还有一个方法来自一个官方题解

#include <cstdio>
#include <cmath>

#define MAXN 100  
#define MAXM 10001  
#define MAXP 40000  
#define MAX 400000 
#define LL long long
#define chkbit(ar, i) (((ar[(i) >> 6]) & (1 << (((i) >> 1) & 31))))  
#define setbit(ar, i) (((ar[(i) >> 6]) |= (1 << (((i) >> 1) & 31))))  
#define isprime(x) (( (x) && ((x)&1) && (!chkbit(ar, (x)))) || ((x) == 2))   
using namespace std; 

namespace pcf{  
    LL dp[MAXN][MAXM];  
    unsigned int ar[(MAX >> 6) + 5] = {0};  
    int len = 0, primes[MAXP], counter[MAX];  
    void Sieve()
    {  
        setbit(ar, 0), setbit(ar, 1);  
        for(int i = 3; (i*i) < MAX; i++, i++)
        {  
            if(!chkbit(ar, i))
            {  
                int k = i << 1;  
                for(int j = (i*i); j < MAX; j += k) setbit(ar, j);  
            }  
        }
        for (int i = 1; i < MAX; i++)
        {  
            counter[i] = counter[i-1];  
            if(isprime(i)) primes[len++] = i, counter[i]++;  
        }  
    }  
   
    void init()
    {  
        Sieve();  
        for(int n = 0; n < MAXN; n++)
        {  
            for(int m = 0; m < MAXM; m++)
            {  
                if(!n) dp[n][m] = m;  
                else dp[n][m] = dp[n-1][m] - dp[n-1][m/primes[n-1]];  
            }  
        }  
    }  
   
    LL phi(LL m, int n)
    {  
        if (n == 0) return m;  
        if (primes[n-1] >= m) return 1;  
        if (m < MAXM && n < MAXN) return dp[n][m];  
        return phi(m, n-1) - phi(m/primes[n-1], n-1);  
    }  
   
    LL Lehmer(LL m)
    {  
        if (m < MAX) return counter[m];  
        LL w, res = 0;  
        int i, a, s, c, x, y;  
        s = sqrt(0.9 + m), y = c = cbrt(0.9 + m);  
        a = counter[y], res = phi(m, a) + a - 1;  
        for (i = a; primes[i] <= s; i++) res = res - Lehmer(m/primes[i]) + Lehmer(primes[i]) - 1;  
        return res;  
    }  
}  
   
LL solve(LL n)
{  
    int i, j, k, l;  
    LL x, y, res = 0;  
    for (i = 0; i < pcf::len; i++)
    {  
        x = pcf::primes[i], y = n/x;  
        if ((x*x) > n) break;  
        res += (pcf::Lehmer(y) - pcf::Lehmer(x));  
    }  
    for (i = 0; i < pcf::len; i++)
    {  
        x = pcf::primes[i];  
        if ((x*x*x) > n) break;  
        res++;  
    }
    return res;  
}

int main()  
{  
    LL n;
    pcf::init();
    while(~scanf("%lld", &n))  
    {
        printf("%lld\n", pcf::Lehmer(n));  
    }  
    return 0;  
}

继续加油~~~