碰到过类似的原题，但是还是记录一下，这样的思路很有趣。
这道题的原理就是将base64原本的替换顺序打乱，改成相应的字符顺序，然后使用新的字符顺序去进行base64的加解密。
class base64:
def __init__(self,alphabet = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/"):
self.alphabet = alphabet
def _EnInsideManage(self,strlist):
strflag = ""
temp = ord(strlist[0]) >> 2
strflag += self.alphabet[temp]
temp = ((ord(strlist[0])&3)<<4)|(ord(strlist[1])>>4)
strflag += self.alphabet[temp]
temp = ((ord(strlist[1])&15)<<2)|(ord(strlist[2])>>6)
strflag += self.alphabet[temp]
temp = (ord(strlist[2])&63)
strflag += self.alphabet[temp]
return strflag
def enbase64(self,charString):
encode = ""
for i in range(len(charString)//3):
encode += self._EnInsideManage(charString[i*3:i*3+3])
if len(charString)%3!=0:
if len(charString)%3 == 1:
encode += self._EnInsideManage(charString[-1:]+chr(0)+chr(0))[:2]+"=="
if len(charString)%3 == 2:
encode += self._EnInsideManage(charString[-2:]+chr(0))[:3]+'='
return encode
def TenToBin(self,tenum):
binstr = ""
for i in range(5,-1,-1):
if 1 == (tenum//(2**i)):
binstr += '1'
tenum = tenum%(2**i)
else:
binstr += '0'
return binstr
def BinToStr(self,strbin):
"Turn the binary string to a ASCII string"
strten = ""
for i in range(len(strbin)//8):
num = 0
test = strbin[i*8:i*8+8]
for j in range(8):
num += int(test[j])*(2**(7-j))
strten += chr(num)
return strten
def debase64(self,base64string):
binstr = ""
for i in base64string:
binstr += self.TenToBin(self.alphabet.find(i))
return self.BinToStr(binstr)
from itertools import combinations, permutations
for i in list(permutations(['3','4','j','u'], 4)):
try:
password = 'JASGBWcQPRXEFLbCDIlmnHUVKTYZdMovwipatNOefghq56rs{}kxyz012789+/'.format(''.join(i))
print password
newobj = base64(alphabet=password)
print(newobj.debase64("MyLkTaP3FaA7KOWjTmKkVjWjVzKjdeNvTnAjoH9iZOIvTeHbvD=="))
except:
pass