Count primes

Count primes

Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 521    Accepted Submission(s): 221

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
  

 

Source

2016 ACM/ICPC Asia Regional Shenyang Online

题意不说了，很简单。

下面是网上大神的代码，表示现在自己也是看不懂啊。（以后弄懂了再补充）

#include    <stdio.h>
#include    <string.h>
#include    <stdlib.h>
#include    <time.h>
#include    <math.h>
#define MAXN 100    // pre-calc max n for phi(m, n)
#define MAXM 10010 // pre-calc max m for phi(m, n)
#define MAXP 40000 // max primes counter
#define MAX 400010    // max prime
#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))

long long n;
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  res = 0;
    int i, a, s, c,  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()
{
    while(scanf("%I64d",&n)!=EOF)
    {
        init();
        printf("%I64d\n",Lehmer(n));
    }
    return 0;
}