HDU 5901 - Count Primes

Problem Description
Easy question! Calculate how many primes between [1...n]!

Input
Each line contain one integer n(1 <= n <= 1e11).Process to end of file.

Output
For each case, output the number of primes in interval [1...n]

Sample Input

2
3
10

Sample Output

1
2

4

题意：给出一个数，求出 1 到这个数之间总共有多少个素数。

题目无比简单，数据范围无比恶心。后来得知这是模板题，在这提供两个模板。

一、快速求范围内素数

#include <cstdio>#include <algorithm>using namespace std;long long f[340000 + 5], g[340000 + 5];long long n;void Init(){    long long m;    for (m = 1; m*m <= n; ++m)        f[m] = n/m - 1;    for (long long i = 1; i <= m; ++i)        g[i] = i - 1;    for (long long i = 2; i <= m; ++i)    {        if (g[i] == g[i-1])            continue;        for (long long 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 (long long j = m; j >= i*i; --j)            g[j] -= (g[j/i] - g[i-1]);    }}int main(){    while (scanf("%lld", &n) != EOF)    {        Init();        printf("%lld\n", f[1]);    }    return 0;}

二、传说中的π函数

<span style="color:#000000;">#include<cstdio>#include<cmath>using namespace std;#define LL long longconst 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;}int main(){    init();    LL n;    while(~scanf("%lld",&n))    {        printf("%lld\n",lehmer_pi(n));    }    return 0;}</span>