
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):
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=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考高分网。
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