Analysis of python prime number embedding method and python Prime Number Analysis

Analysis of python prime number embedding method and python Prime Number Analysis

Principle:

Prime Number refers to the number in a natural number greater than 1 that cannot be divisible by any other natural number except 1 and itself. It plays an important role in encryption applications. For example, in the well-known RSA algorithm, it is a type-based Decomposition Problem Based on big integers, finding two super-large prime numbers and then multiplying them as keys. A common method for prime numbers isThe Sieve of Eratosthenes)To put it simply, draw a table and delete the table ,:

Start from 2 and start from 2. If the current number is a prime number, delete or mark all its multiples from the table, and finally obtain all the prime numbers.

There is an optimization:

When the multiples of 2 and 3 are marked, 6 is marked twice. Therefore, starting from the square of I, we can reduce the time.

For example, 3 is marked from 9, instead of 6, and 6 is added each time.

All prime numbers except 2 are odd. The square of an odd number is still an odd number. If an odd number is added, the even number is not a prime number. Therefore, an even number is added (2 times the prime number ).

All the even numbers are pre-processed.

Note: 1 is neither a prime number nor a union number, and 1 is not processed here.

#! 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) 

5.767 s is used to calculate a prime number of less than 10 million using the prime number embedding method,

The ordinary prime number embedding method uses 9.642 s to calculate a prime number within 1 million,

The prime number less than 0.1 billion is obtained by using the prime number embedding method of C ++, and 0.948 s is used,

3.965 s is used to calculate a prime number of less than 10 million using the C ++ common prime number embedding method,

The explanation language is indeed much slower than the compilation language.

Attaches a C ++ program and uses bit compression to optimize the space.

#include <iostream> #include <cstdio> #include <algorithm> using namespace std; #define N 100000001  unsigned f[(N>>5)+5]; int p[5761456],m; void init() {   int i,j;   for(i=4;i<N;i+=2)     f[i>>5]|=1<<(i&0x1F);   p[m++]=2;   for(i=3;i*i<N;i+=2)     if(!(f[i>>5]&(1<<(i&0x1F))))     {       p[m++]=i;       for(j=i*i;j<N;j+=i+i)         f[j>>5]|=1<<(j&0x1F);     }   for(;i<N;i+=2)     if(!(f[i>>5]&(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; } 

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