素数打表

O(n)
#include    <stdio.h>  
#include    <string.h>  
#include    <stdlib.h>  
#include    <time.h>  
#include    <math.h>  
   
__int64 *primarr, *v;  
__int64 q = 1, p = 1;  
   
//π(n)  
__int64 pi(__int64 n, __int64 primarr[], __int64 len)  
{  
    __int64 i = 0, mark = 0;  
    for (i = len - 1; i > 0; i--) {  
        if (primarr[i] < n) {  
            mark = 1;  
            break;  
        }  
    }  
    if (mark)  
        return i + 1;  
    return 0;  
}  
   
//Φ(x,a)  
__int64 phi(__int64 x, __int64 a, __int64 m)  
{  
    if (a == m)  
        return (x / q) * p + v[x % q];  
    if (x < primarr[a - 1])  
        return 1;  
    return phi(x, a - 1, m) - phi(x / primarr[a - 1], a - 1, m);  
}  
   
__int64 prime(__int64 n)  
{  
    char *mark;  
    __int64 mark_len;  
    __int64 count = 0;  
    __int64 i, j, m = 7;  
    __int64 sum = 0, s = 0;  
    __int64 len, len2, len3;  
   
    mark_len = (n < 10000) ? 10002 : ((__int64)exp(2.0 / 3 * log(n)) + 1);  
   
    //筛选n^(2/3)或n内的素数  
    mark = (char *)malloc(sizeof(char) * mark_len);  
    memset(mark, 0, sizeof(char) * mark_len);  
    for (i = 2; i < (__int64)sqrt(mark_len); i++) {  
        if (mark[i])  
            continue;  
        for (j = i + i; j < mark_len; j += i)  
            mark[j] = 1;  
    }  
    mark[0] = mark[1] = 1;  
   
    //统计素数数目  
    for (i = 0; i < mark_len; i++)  
        if (!mark[i])  
            count++;  
   
    //保存素数  
    primarr = (__int64 *)malloc(sizeof(__int64) * count);  
    j = 0;  
    for (i = 0; i < mark_len; i++)  
        if (!mark[i])  
            primarr[j++] = i;  
   
    if (n < 10000)  
        return pi(n, primarr, count);  
   
    //n^(1/3)内的素数数目  
    len = pi((__int64)exp(1.0 / 3 * log(n)), primarr, count);  
    //n^(1/2)内的素数数目  
    len2 = pi((__int64)sqrt(n), primarr, count);  
    //n^(2/3)内的素数数目  
    len3 = pi(mark_len - 1, primarr, count);  
   
    //乘积个数  
    j = mark_len - 2;  
    for (i = (__int64)exp(1.0 / 3 * log(n)); i <= (__int64)sqrt(n); i++) {  
        if (!mark[i]) {  
            while (i * j > n) {  
                if (!mark[j])  
                    s++;  
                j--;  
            }  
            sum += s;  
        }  
    }  
    free(mark);  
    sum = (len2 - len) * len3 - sum;  
    sum += (len * (len - 1) - len2 * (len2 - 1)) / 2;  
   
    //欧拉函数  
    if (m > len)  
        m = len;  
    for (i = 0; i < m; i++) {  
        q *= primarr[i];  
        p *= primarr[i] - 1;  
    }  
    v = (__int64 *)malloc(sizeof(__int64) * q);  
    for (i = 0; i < q; i++)  
        v[i] = i;  
    for (i = 0; i < m; i++)  
        for (j = q - 1; j >= 0; j--)  
            v[j] -= v[j / primarr[i]];  
   
    sum = phi(n, len, m) - sum + len - 1;  
    free(primarr);  
    free(v);  
    return sum;  
}  
   
int main()  
{  
    __int64 n;  
    __int64 count;  
    int h;  
    clock_t start, end;  
    while(scanf("%I64d", &n)!=EOF)  
    {  
  
        p=1;  
        q=1;  
        start = clock();  
        count = prime(n);  
        end = clock() - start;  
        printf("%I64d(%d亿)内的素数个数为%I64d\n",n,n/100000000,count);  
        printf("用时%lf毫秒\n",(double)end/1000);  
    }  
    return 0;  
}
Meissel-Lehmer
/*
遇到素数需要打表时,先估算素数的个数:
num = n / lnx;
num为大概数字,越大误差越小(只是估计,用于估算素数表数组大小)
这个打表法效率貌似很高,网上说几乎达到了线性时间(不知道是真是假=。=)
*/
#include<iostream>
#include<cstring>
#include<cstdio>
#include<ctime>
#include<algorithm>
using namespace std;
int n;
bool visit[10100000];
int prime[10000000];


void init_prim()
{
    memset(visit, true, sizeof(visit));
    int num = 0;
    for (int i = 2; i <= n; ++i)
    {
        if (visit[i] == true)
        {
            num++;
            prime[num] = i;
        }
        for (int j = 1; ((j <= num) && (i * prime[j] <= n));  ++j)
        {
            visit[i * prime[j]] = false;
            if (i % prime[j] == 0) break; //点睛之笔
        }
    }
}

int main()
{
    memset(prime, 0, sizeof(prime));
    int count = 0;
    cin>>n;
    init_prim();
    for(int i = 0; i <= n; ++i)
        if(prime[i])
        {
            cout<<prime[i]<<" ";
            count++;
        }
        cout<<endl;
        cout<<"素数个数为:"<<count<<endl;
}   
usual

 

posted @ 2017-08-02 22:10  Aragaki  阅读(151)  评论(0编辑  收藏  举报