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

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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

2 
3 
10

 

Sample Output

1 
2 
4

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: ITAKMotto:今日的我要超越昨日的我，明日的我要胜过今日的我，以创作出更好的代码为目标，不断地超越自己。**/#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 = 1e-7;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;}