buuctf MISC - key不在这里

最新推荐文章于 2025-05-26 08:45:00 发布

cr@c3r

最新推荐文章于 2025-05-26 08:45:00 发布

阅读量256

点赞数

CC 4.0 BY-SA版权

分类专栏： buuctf - MISC 文章标签： python 开发语言网络安全

本文链接：https://blog.youkuaiyun.com/c868954104/article/details/134598226

buuctf - MISC 专栏收录该内容

5 篇文章

订阅专栏

得到一张图片，放到ps 反向得到

扫码得到

https://cn.bing.com/search?q=key%E4%B8%8D%E5%9C%A8%E8%BF%99%E9%87%8C&m=10210897103375566531005253102975053545155505050521025256555254995410298561015151985150375568&qs=n&form=QBRE&sp=-1&sc=0-38&sk=&cvid=2CE15329C18147CBA4C1CA97C8E1BB8C

可以发现其中m参数像是一段ascii码连在一起

构造python脚本解密得到flag

def extract_numbers(input_string):
    # 初始化指针
    pointer = 0
    result = []

    while pointer < len(input_string):
        # 如果开头是1，向后取3位
        if input_string[pointer] == '1':
            extracted_number = input_string[pointer:pointer + 3]
            pointer += 3
        else:
            # 否则向后取2位
            extracted_number = input_string[pointer:pointer + 2]
            pointer += 2

        # 将提取的数字添加到结果列表中
        result.append(int(extracted_number))

    return result

input_string = '10210897103375566531005253102975053545155505050521025256555254995410298561015151985150375568'
numbers = extract_numbers(input_string)
flag = ''
for i in numbers:
    flag += chr(i)
flag = flag.replace('%7B','{')
flag = flag.replace('%7D','}')
print(flag)

flag{5d45fa256372224f48746c6fb8e33b32}