python素数筛选法浅析

#! prime.py 
import time 
 
def primes(n): 
 P = [] 
 f = [] 
 for i in range(n+1): 
  if i > 2 and i%2 == 0: 
   f.append(1) 
  else: 
   f.append(0) 
 
 i = 3 
 while i*i <= n: 
  if f[i] == 0: 
   j = i*i 
   while j <= n: 
    f[j] = 1 
    j += i+i 
  i += 2 
 
 P.append(2) 
 for i in range(3,n,2): 
  if f[i] == 0: 
   P.append(i) 
 
 return P 
 
def isPrime(n): 
 if n > 2 and n%2 == 0: 
  return 0 
 
 i = 3 
 while i*i <= n: 
  if n%i == 0: 
   return 0 
  i += 2 
 
 return 1 
 
def primeCnt(n): 
 cnt = 0 
 for i in range(2,n): 
  if isPrime(i): 
   cnt += 1 
 return cnt 
 
if __name__ == '__main__': 
 start = time.clock() 
 n = 10000000 
 P = primes(n); 
 print("There are %d primes less than %d"%(len(P),n)) 
 #for i in range(10): 
 # print(P[i]) 
 print("Time: %f"%(time.clock()-start)) 
 #for n in range(2,100000): 
 # if isPrime(n): 
 #  print("%d is prime"%n) 
  #print("%d is "%n + ("prime" if isPrime(n) else "not prime")) 
 
 start = time.clock() 
 n = 1000000 
 print("There are %d primes less than %d"%(primeCnt(n),n)) 
 print("Time: %f"%(time.clock()-start) 

#include <iostream> 
#include <cstdio> 
#include <algorithm> 
using namespace std; 
#define N 100000001 
 
unsigned f[(N5)+5]; 
int p[5761456],m; 
void init() 
{ 
  int i,j; 
  for(i=4;i<N;i+=2) 
    f[i5]|=1<<(i&0x1F); 
  p[m++]=2; 
  for(i=3;i*i<N;i+=2) 
    if(!(f[i5]&(1<<(i&0x1F)))) 
    { 
      p[m++]=i; 
      for(j=i*i;j<N;j+=i+i) 
        f[j5]|=1<<(j&0x1F); 
    } 
  for(;i<N;i+=2) 
    if(!(f[i5]&(1<<(i&0x1F)))) 
      p[m++]=i; 
} 
int is_prime(int n) 
{ 
  int i; 
  for(i=0;p[i]*p[i]<=n;i++) 
    if(n%p[i]==0) 
      return 0; 
  return 1; 
} 
int isPrime(int n) 
{ 
  if(n>2 && n%2==0) 
    return 0; 
  int i=3; 
  while(i*i<=n) 
  { 
    if(n%i==0) 
      return 0; 
    i+=2; 
  } 
  return 1; 
} 
int main() 
{ 
  int n=0,i; 
  clock_t st=clock(); 
  init(); 
  /*for(i=2;i<10000000;i++) 
    if(isPrime(i)) 
      n++;*/ 
  printf("%d %dms\n",m,clock()-st); 
  /*while(~scanf("%d",&n),n) 
  { 
    i=lower_bound(p,p+m,n+1)-p; 
    printf("%d\n",i); 
  }*/ 
  return 0; 
}