2023中国工业互联网安全大赛总决赛部分题解

来旅游的，头次场景题0解（全场），运气太差抽的都是0解题。。。。

部分环境很糟糕（狗头保命），搞了个相册站，欢迎参观：https://photo.iloli.moe，头次打线下CTF（坐牢坐牢）

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，逆推一下即可解得flag（实际上就是某密码管理软件的加密过程，只不过脚本要上网才能搜到，结合比赛时选手开的热点很难不怀疑有人偷偷上网，小声说）

# 脚本给的，套，就嗯套
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}

断网试一下午是真难绷（），只能说自己知识储备量不太全（）

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")

…等待更新