題目連結

https://learn.cylabacademy.org/library/245

題目解法

將 level1.py 以及 level1.flag.txt.enc 兩個檔案放在同個目錄下。 可以看到 level1.py 第 19 行有寫出密碼

user_pw == "1e1a"

執行腳本後輸入密碼 1e1a 即可得到 flag。

$ python level1.py            
Please enter correct password for flag: 1e1a
Welcome back... your flag, user:
picoCTF{545---------------060}

題目原始碼

level1.py

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### THIS FUNCTION WILL NOT HELP YOU FIND THE FLAG --LT ########################
def str_xor(secret, key):
    #extend key to secret length
    new_key = key
    i = 0
    while len(new_key) < len(secret):
        new_key = new_key + key[i]
        i = (i + 1) % len(key)        
    return "".join([chr(ord(secret_c) ^ ord(new_key_c)) for (secret_c,new_key_c) in zip(secret,new_key)])
###############################################################################


flag_enc = open('level1.flag.txt.enc', 'rb').read()



def level_1_pw_check():
    user_pw = input("Please enter correct password for flag: ")
    if( user_pw == "1e1a"):
        print("Welcome back... your flag, user:")
        decryption = str_xor(flag_enc.decode(), user_pw)
        print(decryption)
        return
    print("That password is incorrect")



level_1_pw_check()