本程序可根據給定的字符字典，窮舉指定長度的所有字符串：

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

結果：

aaaaabaacabaabbabcacaacbaccbaababbacbbabbbbbcbcabcbbcccaacabcaccbacbbcbcccaccbccc

本程序占用內存小，生成速度快，歡迎嘗試?。。?/p>

補充知識：Python 窮舉法, 二分法 與牛頓-拉夫遜方法求解平方根的性能對比

窮舉法, 二分法 與牛頓-拉夫遜方法求解平方根的優劣，從左到右依次遞優。

經過測試，窮舉法基本超過 1 分鐘，還沒有出數據；

二分法只要區區1秒不到就出結果了。

牛頓-拉夫遜是秒出，沒有任何的停頓。

numberTarget =int(input('Please enter a number:'))numberSqureRoot = 0while(numberSqureRoot<abs(numberTarget)): if numberSqureRoot**2 >= abs(numberTarget): break numberSqureRoot = numberSqureRoot + 1if 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 = 0lowValue = 0.0highValue=numberTarget*1.0epsilon = 0.01numberSqureRoot = (highValue + lowValue)/2while 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) /2print('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 = 0epsilon = 0.01k=numberTargetnumberSqureRoot = k/2.0while( abs(numberSqureRoot*numberSqureRoot - k)>=epsilon): numberSqureRoot=numberSqureRoot-(((numberSqureRoot**2) - k)/(2*numberSqureRoot))print('squre root of %s is %s ' %(numberTarget,numberSqureRoot))

以上這篇python 窮舉指定長度的密碼例子就是小編分享給大家的全部內容了，希望能給大家一個參考，也希望大家多多支持好吧啦網。