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HDU 5901 Count Primes （模板 + 数论知识）——2016 ACM/ICPC Asia Regional Shenyang Online

来源：程序员人生发布时间：2016-09-25 09:13:22 阅读次数：2600次

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Count primes

Time Limit: 12000/6000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 635 Accepted Submission(s): 299

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

Sample Output

题目大意：

求 1−n(n≤1011) 之间有多少个素数，

解题思路：

板子题目，可以参考 Codeforces 655 F

My Code：

/**
2016 - 09 - 19 晚上
Author: ITAK

Motto:

本日的我要超出昨日的我，明日的我要胜过本日的我，
以创作出更好的代码为目标，不断地超出自己。
**/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <algorithm>
#include <set>
#include <time.h>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
const LL INF = 1e9+5;
const LL MAXN = 1e6+5;
const LL MOD = 1e9+7;
const double eps = 1e⑺;
const double PI = acos(-1);
using namespace std;
LL Scan_LL()///输入外挂
{
    LL res=0,ch,flag=0;
    if((ch=getchar())=='-')
        flag=1;
    else if(ch>='0'&&ch<='9')
        res=ch-'0';
    while((ch=getchar())>='0'&&ch<='9')
        res=res*10+ch-'0';
    return flag?-res:res;
}
void Out(LL a)///输出外挂
{
    if(a>9)
        Out(a/10);
    putchar(a%10+'0');
}

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 get_phi(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 get_phi(x, s - 1) - get_phi(x / prime[s], s - 1);
}
LL getpi(LL x)
{
    if(x < N)   return pi[x];
    LL ans = get_phi(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 = get_phi(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("%I64d",&n))
    {
        Out(lehmer_pi(n));
        puts("");
    }
    return 0;
}

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