crackme.py（好像是python的逆向）

无聊的时候翻电脑玩发现一个脚本，就试着玩一下

解题时发现还可以使用暴力破解来做这个题
上网简单一搜就搜到了，最近在看汇编想学一下逆向，就想着能不能把这个程序反过输出
有其他人

#Embedded file name: G:/2/build/scripts-2.7/crackme.py

def encrypt(key, seed, string):
    rst = []
    for v in string:
        rst.append((ord(v) + seed ^ ord(key[seed])) % 255)
        seed = (seed + 1) % len(key)
    return rst


if __name__ == '__main__':
    print('Welcome to idf\'s python crackme')
    flag = input('Enter the Flag: ')
    KEY1 = 'Maybe you are good at decryptint Byte Code, have a try!'
    KEY2 = [124, 48, 52, 59, 164, 50, 37, 62, 67, 52, 48, 6, 1, 122, 3, 22, 72, 1, 1, 14, 46, 27, 232]
    en_out = encrypt(KEY1, 5, flag)
    if KEY2 == en_out:
        print('You Win')
    else:
        print('Try Again !')

是一个简单的逆向，很简单但是不会，搞了好久才做出来

def de(keys,seeds,strings):
    seeds = 5
    for i in range(23):
        v = (strings[i] ^ ord(keys[seeds])) - seeds
        print(chr(v),end="")
        seeds = (seeds + 1) % len(keys)

 这行代码搞死个人啊
rst.append((ord(v) + seed ^ ord(key[seed])) % 255)

学艺不精弄了好长时间才发现是这样的
rst.append(((ord(v) + seed) ^ ord(key[seed])) % 255)
后面这个%255就没用到搞得我试了半天