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python 穷举指定长度的密码例子

来源:互联网 收集:自由互联 发布时间:2021-04-09
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):

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))

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