HDU 5901 Count primes(1e11以内的素数个数)

#include <map>
#include <set>
#include <stack>
#include <queue>
#include <cmath>
#include <string>
#include <vector>
#include <cstdio>
#include <cctype>
#include <cstring>
#include <sstream>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#pragma comment(linker, "/STACK:102400000,102400000")

using namespace std;
#define   MAX           400005
#define   maxnode       10
#define   sigma_size    30
#define   lson          l,m,rt<<1
#define   rson          m+1,r,rt<<1|1
#define   lrt           rt<<1
#define   rrt           rt<<1|1
#define   middle        int m=(r+l)>>1
#define   LL            long long
#define   ull           unsigned long long
#define   mem(x,v)      memset(x,v,sizeof(x))
#define   lowbit(x)     (x&-x)
#define   pii           pair<int,int>
#define   bits(a)       __builtin_popcount(a)
#define   mk            make_pair
#define   limit         10000

//const int    prime = 999983;
const int    INF   = 0x3f3f3f3f;
const LL     INFF  = 0x3f3f;
const double pi    = acos(-1.0);
//const double inf   = 1e18;
const double eps   = 1e-8;
const LL     mod   = 1e9+7;
const ull    mx    = 133333331;

/*****************************************************/
inline void RI(int &x) {
char c;
while((c=getchar())<'0' || c>'9');
x=c-'0';
while((c=getchar())>='0' && c<='9') x=(x<<3)+(x<<1)+c-'0';
}
/*****************************************************/

const int MAXN = 105;
const int MAXM = 10005;
const int MAXP = 40005;
//const int MAX  = 400005;
#define chkbit(ar, i) (((ar[(i) >> 6]) & (1 << (((i) >> 1) & 31))))
#define setbit(ar, i) (((ar[(i) >> 6]) |= (1 << (((i) >> 1) & 31))))
#define isprime(x) (( (x) && ((x)&1) && (!chkbit(ar, (x)))) || ((x) == 2))
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
[m] = m;
else dp
[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
[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;
}
}
using namespace pcf;

int main(){
init();
LL n;
while(cin>>n){
cout<<Lehmer(n)<<endl;
}
return 0;
}