小编Jon*_*ung的帖子

我正在学习使用Python的Multiprocessing软件包来解决令人尴尬的并行问题,因此我编写了串行和并行版本来确定小于或等于自然数n的素数的数量.根据我从博客文章和Stack Overflow问题中读到的内容,我想出了以下代码:

串行

import math
import time

def is_prime(start, end):
    """determine how many primes within given range"""
    numPrime = 0
    for n in range(start, end+1):
        isPrime = True
        for i in range(2, math.floor(math.sqrt(n))+1):
            if n % i == 0:
                isPrime = False
                break
        if isPrime:
            numPrime += 1
    if start == 1:
        numPrime -= 1  # since 1 is not prime
    return numPrime

if __name__ == "__main__":
    natNum = 0
    while natNum < 2: …