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
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;
}