如何用Python编写穷举特定长度密码的示例代码？

本文共计576个文字，预计阅读时间需要3分钟。

程序根据给定的字符集生成指定长度的所有可能字符串：pythondef get_pwd(str, num): if num==1: for x in str: yield x else: for x in str: for y in get_pwd(str, num-1): yield x + y

strKey=abcfor x in get_pwd(strKey, 3): print(x)

本程序可根据给定的字符字典，穷举指定长度的所有字符串：

def get_pwd(str, num): if(num == 1): for x in str: yield x else: for x in str: for y in get_pwd(str, num-1): yield x+y strKey="abc" for x in get_pwd(strKey,3): print x

结果：

aaa aab aac aba abb abc aca acb acc baa bab bac bba bbb bbc bca bcb bcc caa cab cac cba cbb cbc cca ccb ccc

本程序占用内存小，生成速度快，欢迎尝试！！！

补充知识：Python 穷举法, 二分法 与牛顿-拉夫逊方法求解平方根的性能对比

穷举法, 二分法 与牛顿-拉夫逊方法求解平方根的优劣，从左到右依次递优。

经过测试，穷举法基本超过 1 分钟，还没有出数据；

二分法只要区区1秒不到就出结果了。

牛顿-拉夫逊是秒出，没有任何的停顿。

numberTarget =int(input("Please enter a number:")) numberSqureRoot = 0 while(numberSqureRoot<abs(numberTarget)): if numberSqureRoot**2 >= abs(numberTarget): break numberSqureRoot = numberSqureRoot + 1 if numberSqureRoot**2 != numberTarget: print("Your number %s is not a perfect squre, the square root is %s " % ( numberTarget,numberSqureRoot) ) else: if numberTarget < 0 : numberSqureRoot = -numberSqureRoot print("Your number %s is a perfect squre, the square root is %s " % ( numberTarget, numberSqureRoot)) print("now we begin to calculate the binary search...") numberTarget=int(input("Please enter the number for binary search...")) numberSqureRoot = 0 lowValue = 0.0 highValue=numberTarget*1.0 epsilon = 0.01 numberSqureRoot = (highValue + lowValue)/2 while abs(numberSqureRoot**2 - numberTarget) >=epsilon: print("lowValue:%s, highValue:%s, currentValue:%s"%(lowValue,highValue,numberSqureRoot)) if numberSqureRoot**2<numberTarget: lowValue=numberSqureRoot else: highValue=numberSqureRoot numberSqureRoot = (lowValue+highValue) /2 print("The number %s has the squre root as %s " %(numberTarget,numberSqureRoot)) print("now we begin to calculate the newTon search...") numberTarget=int(input("Please enter the number for newTon search...")) numberSqureRoot = 0 epsilon = 0.01 k=numberTarget numberSqureRoot = k/2.0 while( abs(numberSqureRoot*numberSqureRoot - k)>=epsilon): numberSqureRoot=numberSqureRoot-(((numberSqureRoot**2) - k)/(2*numberSqureRoot)) print("squre root of %s is %s " %(numberTarget,numberSqureRoot))

以上这篇python 穷举指定长度的密码例子就是小编分享给大家的全部内容了，希望能给大家一个参考，也希望大家多多支持易盾网络。