爆炸快求1~n有多少素数

这个求一千亿以内的素数大约用6780ms

#include <stdio.h>
#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <time.h>
#include <math.h>
using namespace std;
__int64  *primarr, *v;
__int64  q = 1, p = 1;

void Out(__int64 a)    //输出外挂
{
    if(a>9)
        Out(a/10);
    putchar(a%10+'0');
}

__int64 Scan()     //输入外挂
{
    __int64 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;
}
//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;
    //int h;
    ///clock_t start, end;
    //freopen("C:\Users\acer\Desktop\in.txt","r",stdin);
    //std::ios::sync_with_stdio(false);
    while(~scanf("%lld",&n))
    {
        if(n==2)
        {
            printf("1
");
            continue;
        }
        //p=1;
        //q=1;
        //start = clock();
        Out(prime(n+1));
        printf("
");
        //end = clock() - start;
        //printf("用时%lfms
",(double)end);
    }
    return 0;
}

这个完全就是爆炸了

#include <bits/stdtr1c++.h>

#define MAXN 100    // pre-calc max n for phi(m, n)
#define MAXM 100010 // pre-calc max m for phi(m, n)
#define MAXP 666666 // max primes counter
#define MAX 10000010    // max prime
#define clr(ar) memset(ar, 0, sizeof(ar))
#define read() freopen("lol.txt", "r", stdin)
#define dbg(x) cout << #x << " = " << x << endl
// compressed bool flag for sieve prime. (i >> 1) because even numbers are omitted.
#define setbit(ar, i) (((ar[(i) >> 6]) |= (1 << (((i) >> 1) & 31))))
#define chkbit(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{
    long long 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]];
            }
        }
    }

    long long phi(long long 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);
    }

    long long Lehmer(long long m){
        if (m < MAX) return counter[m];

        long long 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;
    }
}



int main(){
    pcf::init();
    long long n, res;

    while (scanf("%lld", &n) != EOF){
        printf("%lld
", pcf::Lehmer(n));
    }
    return 0;
}