2023 China Industrial Internet Security Competition Finals Retrospective

免责声明：本文为个人技术学习与工程实践笔记，所涉操作仅应在获得授权的环境中进行。因不当使用造成的后果由使用者自行承担。

首次遇到场景题全场 0 解，抽到的都是 0 解题，运气确实不太好。

部分环境很糟糕，另外做了一个相册站，欢迎参观：https://photo.iloli.moe，这也是首次参加线下赛事

gigached

下载题目，010 打开发现是一个 mp4 文件，加个后缀发现是 gigached（Can you Feel My Heart~）

结合题目猜测是要让我们获取视频的 LSB，ffmpeg 简单分离一下视频帧

ffmpeg -i gigached.mp4 -o frames_%4b.png

然后写个脚本挨张提取 PNG LSB 通道数据

# encoding=utf-8
from PIL import Image
def getlsb(f1, f2):
    tmp = Image.open(f1)
    w, h = tmp.size
    cnt = 0
    data = ""
    for height in range(0, h):
        for weight in range(0, w):
            pixel = tmp.getpixel((weight, height))
            # 处理三通道（RGB）
            if cnt % 3 == 0:
                cnt += 1
                data += str((int(pixel[0]) % 2))
            if cnt % 3 == 1:
                cnt += 1
                data += str((int(pixel[1]) % 2))
            if cnt % 3 == 2:
                cnt += 1
                data += str((int(pixel[2]) % 2))
    with open(f2, "wb") as file:
        for i in range(0, len(data), 8):
            t = int(data[i: i + 8], 2)
            file.write(chr(t))
            t = ""
    file.close()
for i in range(1, 423):
    filename = str("frames_" + str(i).zfill(5) + ".png")
    new = filename
    output = str("output_" + str(i).zfill(5) + ".txt")
    out = output
    getlsb(new, out)

结果得到

然后挨个查看 txt 文本数据，看了一圈后发现只有前几个文件有 flag

./output_00002.txt: binary file matches
./output_00003.txt: binary file matches
./output_00004.txt: binary file matches
./output_00005.txt: binary file matches
./output_00006.txt: binary file matches
./output_00007.txt: binary file matches
./output_00008.txt: binary file matches
./output_00009.txt: binary file matches
./output_00010.txt: binary file matches
./output_00011.txt: binary file matches
./output_00012.txt: binary file matches
./output_00013.txt: binary file matches
./output_00014.txt: binary file matches
./output_00015.txt: binary file matches
./output_00016.txt: binary file matches

整理得到

8:dp6UYHfO
8:NZhyAEYx
8:Js8o3024
8:l+je3k5k
8:aJmsL+Hx
8:w43oQ3oz
8:XjTT3/LQ
8:qF4fuu/w
8:MNqZVqiA
8:Q3vzixxY
8:QYbnuQ==
8: the key
8: isgYpt3
8:GvM7X1A1
3:CoT

根据上文，结合压缩包注释给出的代码逆向推导，即可解得 flag（本题实际上是一套密码管理软件的加密过程，只是相关脚本需要联网检索才能找到）

# 题目提供的加密脚本
def encrypt(key, source):
        source = source.encode()
	key = key.encode()
	key = SHA256.new(key).digest()
	IV = Random.new().read(AES.block_size)
	encryptor = AES.new(key, AES.MODE_CBC, IV)
	padding = AES.block_size - len(source) % AES.block_size
	source += bytes([padding]) * padding
	data = IV + encryptor.encrypt(source)
	return data

解密

def decrypt(key, source):
    source = base64.b64decode(source)
    key = key.encode()  # 转换为字节类型
    key = SHA256.new(key).digest()
    IV = source[:AES.block_size]
    decryptor = AES.new(key, AES.MODE_CBC, IV)
    data = decryptor.decrypt(source[AES.block_size:])
    padding = data[-1]
    if data[-padding:] != bytes([padding]) * padding:
        raise ValueError("Invalid padding...")
    return data[:-padding].decode()
key = "gYpt3GvM7X1A1CoT"
flag = "dp6UYHfONZhyAEYxJs8o3024l+je3k5kaJmsL+Hxw43oQ3ozXjTT3/LQqF4fuu/wMNqZVqiAQ3vzixxYQYbnuQ=="
print(decrypt(key, flag))

flag{d53dbf36-444a-471d-a663-91a9e6f400e0}

断网环境下尝试了一下午，难度确实不小，也说明自身知识储备仍有欠缺

DNS_Query

给了流量包，tshark 提取流量，写脚本提取 dns 记录

11111110100111100100101111111
10000010011001100001101000001
10111010001011110011101011101
10111010000011000110101011101
10111010010100100011001011101
10000010001011000100101000001
11111110101010101010101111111
00000000101101101001000000000
11011010011100010101101000001
11011101110010011011100111100
01100111001101110110110011000
10101001011110111100110011011
11110011110110110011011100011
10001101010001010000101111001
01101011111111110001100100101
11001101111100111010011110101
10001110100001111001110001001
11100000111110111111010110000
11011011100101110111011110001
11011101101011100000000000101
11110011000010100101111111101
00000000101001011011100010100
11111110001010010101101011100
10000010010000100100100010000
10111010111001110011111110010
10111010111010110111100101001
10111010010000110001000010011
10000010101011111001011111101
11111110110110010000100010000

然后写脚本还原

from PIL import Image
from zlib import *
MAX = 29
pic = Image.new("RGB",(MAX,MAX))
str ="1111111010011110010010111111110000010011001100001101000001101110100010111100111010111011011101000001100011010101110110111010010100100011001011101100000100010110001001010000011111111010101010101010111111100000000101101101001000000000110110100111000101011010000011101110111001001101110011110001100111001101110110110011000101010010111101111001100110111111001111011011001101110001110001101010001010000101111001011010111111111100011001001011100110111110011101001111010110001110100001111001110001001111000001111101111110101100001101101110010111011101111000111011101101011100000000000101111100110000101001011111111010000000010100101101110001010011111110001010010101101011100100000100100001001001000100001011101011100111001111111001010111010111010110111100101001101110100100001100010000100111000001010101111100101111110111111110110110010000100010000"
i=0
for y in range(0,MAX):
    for x in range(0,MAX):
        if(str[i] == '1'):
            pic.putpixel([x,y],(0,0,0))
        else:pic.putpixel([x,y],(255,255,255))
        i = i+1
pic.show()
pic.save("flag.png")

其余题目待后续补充。

防御启示

1. 媒体文件是隐写的常见载体：本题把信息藏在视频帧的像素最低有效位中。业务侧对上传的图片、视频应做重编码与元数据清洗，并对 LSB 平面做统计分析，避免隐写内容随业务数据对外发布。
2. 口令派生不应只做一次哈希：给出的加密实现直接把 SHA256(口令) 当作 AES 密钥，没有盐值与迭代次数，低熵口令可被离线快速爆破。应改用 PBKDF2、bcrypt 或 Argon2 派生密钥，并为每个用户使用独立盐值。
3. 解密失败必须统一响应：给出的实现会对 padding 校验失败抛出区分性异常，若出现在服务端，这类可区分的错误响应与耗时差异会形成类似 padding oracle 的侧信道。错误提示、状态码与处理耗时都应统一。
4. CBC 模式需要额外的完整性保护：AES-CBC 只提供机密性，不防篡改。新系统应优先采用带认证的 AES-GCM，或对密文附加 HMAC 校验，并保证 IV 随机且不复用。
5. DNS 是常见的隐蔽外传通道：对边界 DNS 应做审计与限制——关注异常长度的子域、短时间内同一父域下大量随机标签、非常规记录类型，并禁止终端直连外部递归解析。