2024鹏城杯-misc

网安第一课

找到key1

key2

6iMmn76ucYG9PDtsvu

解压之后

上脚本

from PIL import Image
 
images = [Image.open(f"{i}.png") for i in range(1, 38)]
qr_code = Image.new("RGB", (128, 128), (255, 255, 255))
for i in range(37):
    img1 = images[i]
    for j in range(128):
        print(i, j, img1.getpixel((j, 0))[1])
        pixel_value = img1.getpixel((j, 0))[1]
        if pixel_value & 1:
            qr_code.putpixel((i, j), (0, 0, 0))
        else:
            qr_code.putpixel((i, j), (255, 255, 255))
 
qr_code.save("qr_code.png")

需要改一下脚本

**Simple_steganography-pcb2024**

提取出来

之后得到一个二维码

假的

发现有隐藏图片，提取出来

这张图片

hint.txt，里面是a=7,b=35

用脚本

from PIL import Image

def arnold(infile: str, outfile: str = None, a: int = 1, b: int = 1, shuffle_times: int = 1, reverse: bool = False) -> None:
    """
    Arnold猫脸变换函数

    Parameters:
        infile - 输入图像路径
        outfile - 输出图像路径
        a - Anrold 变换参数
        b - Anrold 变换参数
        shuffle_times - 置乱次数
        reverse - 逆变换
    """
    inimg = Image.open(infile)
    width, height = inimg.size
    indata = inimg.load()
    outimg = Image.new(inimg.mode, inimg.size)
    outdata = outimg.load()

    for _ in range(shuffle_times):
        for x in range(width):
            for y in range(height):
                if reverse:
                    nx = ((a * b + 1) * x - a * y) % width
                    ny = (y - b * x) % height
                else:
                    nx = (x + a * y) % width
                    ny = (b * x + (a * b + 1) * y) % height
                outdata[ny, nx] = indata[y, x]
    
    outimg.save(outfile if outfile else "arnold_"+infile, inimg.format)

arnold("flag.jpg", "decode.jpg", 7, 35, 1, True)

解出来得到第二部分flag

再用bkcrack进行明文攻击

time ./bkcrack -C secret.zip -c flag.png -p png_header -o 0 > 1.log

跑出来了，解压得到第一部分的flag

We_l1k3_h4ck1ng