使用Python判断质数(素数)的简单方法讲解
质数又称素数。指在一个大于1的自然数中，除了1和此整数自身外，不能被其他自然数整除的数。素数在数论中有着很重要的地位。比1大但不是素数的数称为合数。1和0既非素数也非合数。质数是与合数相对立的两个概念，二者构成了数论当中最基础的定义之一。基于质数定义的基础之上而建立的问题有很多世界级的难题，如哥德巴赫猜想等。算术基本定理证明每个大于1的正整数都可以写成素数的乘积，并且这种乘积的形式是唯一的。这个定理的重要一点是，将1排斥在素数集合以外。如果1被认为是素数，那么这些严格的阐述就不得不加上一些限制条件。 前几天偶尔的有朋友问python怎么判断素数的方法，走网上查了查，总结了python脚本判断一个数是否为素数的几种方法：
1.运用python的数学函数
import math
def isPrime(n):
if n <= 1:
return False
for i in range(2, int(math.sqrt(n)) + 1):
if n % i == 0:
return False
return True
2.单行程序扫描素数
from math import sqrt N = 100 [ p for p in range(2, N) if 0 not in [ p% d for d in range(2, int(sqrt(p))+1)] ]
运用python的itertools模块
from itertools import count
def isPrime(n): www.jb51.net
if n <= 1:
return False
for i in count(2):
if i * i > n:
return True
if n % i == 0:
return False
3.不使用模块的两种方法
方法1：
def isPrime(n):
if n <= 1:
return False
i = 2
while i*i <= n:
if n % i == 0:
return False
i += 1
return True
方法2：
def isPrime(n):
if n <= 1:
return False
if n == 2:
return True
if n % 2 == 0:
return False
i = 3
while i * i <= n:
if n % i == 0:
return False
i += 2
return True
eg：求出20001到40001之间的质数(素数)
既然只能被1或者自己整出,那说明只有2次余数为0的时候，代码如下:
#!/usr/bin/python L1=[] for x in xrange(20001,40001): n = 0 for y in xrange(1,x+1): if x % y == 0: n = n + 1 if n == 2 : print x L1.append(x) print L1
结果如下:
20011 20021 20023 20029 20047 20051 20063 20071 20089 20101 20107 20113 20117 20123 20129 20143 20147 20149 20161 20173 ….
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