质数码的优化

本文介绍了质数码的优化的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

这是我用python编写的代码，用于计算小于给定数的素数之和.我还能做些什么来优化它?

This is my code in python for calculation of sum of prime numbers less than a given number. What more can I do to optimize it?

import math primes = [2,] #primes store the prime numbers for i in xrange(3,20000,2): #i is the test number x = math.sqrt(i) isprime = True for j in primes: #j is the devider. only primes are used as deviders if j <= x: if i%j == 0: isprime = False break if isprime: primes.append(i,) print sum (primes,)

推荐答案

您可以使用一种不同的算法，称为 Eratosthenes 的筛网 这将更快但需要更多的内存.保留一组标志，表示每个数字是否为素数，并且对于每个新素数，将该素数的所有倍数设置为零.

You can use a different algorithm called the Sieve of Eratosthenes which will be faster but take more memory. Keep an array of flags, signifying whether each number is a prime or not, and for each new prime set it to zero for all multiples of that prime.

N = 10000 # initialize an array of flags is_prime = [1 for num in xrange(N)] is_prime[0] = 0 # this is because indexing starts at zero is_prime[1] = 0 # one is not a prime, but don't mark all of its multiples! def set_prime(num): "num is a prime; set all of its multiples in is_prime to zero" for x in xrange(num*2, N, num): is_prime[x] = 0 # iterate over all integers up to N and update the is_prime array accordingly for num in xrange(N): if is_prime[num] == 1: set_prime(num) primes = [num for num in xrange(N) if is_prime[num]]

如果您使用有效的位数组，您实际上可以为相当大的 N 执行此操作，例如在 这个示例中(向下滚动页面，您会找到一个埃拉托色尼筛法示例).

You can actually do this for pretty large N if you use an efficient bit array, such as in this example (scroll down on the page and you'll find a Sieve of Eratosthenes example).