春节红包第三题解题记录

本文来源吾爱破解论坛

分析在前辈的帖子里https://www.52pojie.cn/thread-873265-1-1.html，我用python解了一下，需安装pycrypto库

[Python] 纯文本查看 复制代码

#!/usr/bin/python
#-*- coding:utf-8-*-
from    Crypto.Cipher   import AES
#from    binascii    import b2a_hex,a2b_hex
import base64,binascii,re,hashlib
class AESCBC:
    def __init__(self, key):
        self.key = key
        self.mode = AES.MODE_CBC
        self.bs = 16  # block size
        self.PADDING = lambda s: s + (self.bs - len(s) % self.bs) * chr(self.bs - len(s) % self.bs)
    
    def encrypt(self, text):
        generator = AES.new(self.key, self.mode,self.key) 
        try:
            crypt = generator.encrypt(self.PADDING(text))
            crypted_str = base64.b64encode(crypt)
            result = crypted_str.decode()
        except Exception:
            result = '加密失败，请确认输入是否有误!'
        return result
    
    def decrypt(self, text):
        generator = AES.new(self.key, self.mode,self.key) 
        text += (len(text) % 4) * '='
        try:
            decrpyt_bytes = base64.b64decode(text)
            meg = generator.decrypt(decrpyt_bytes)
            # 去除解码后的非法字符
            result = re.compile('[\\x00-\\x08\\x0b-\\x0c\\x0e-\\x1f\n\r\t]').sub('', meg.decode())
        except Exception:
            result = '解密失败，请确认输入是否有误!'
        return result
def     main():

        name = input("input uid:")
        sha1 = hashlib.sha1()
        sha1.update(name.encode('utf8'))
        #key = binascii.a2b_hex('8380cf291d73c51efb5351d7df0268cb89be530e'[:32])#秘钥的长度必须为16
        key = binascii.a2b_hex(sha1.hexdigest()[:32])
        result = 'HappyNewYearFrom52PoJie.Cn'
        cryptor = AESCBC(key)#这里选用的是CBC模式

        plainText = cryptor.encrypt(result)#解密函数
        print((plainText))



if      __name__=="__main__":
        main()