MISC刷题9

[MoeCTF 2022]cccrrc

题目提示crc爆破

下载得到cccccrc.zip，有密码

出现报错

使用末初大佬脚本：https://blog.csdn.net/mochu7777777/article/details/110206427

import binascii
import string

def crack_crc():
    print('-------------Start Crack CRC-------------')
    crc_list = [0xc0a3a573, 0x3cb6ab1c, 0x85bb0ad4, 0xf4fde00b]#文件的CRC32值列表，注意顺序
    comment = ''
    chars = string.printable
    for crc_value in crc_list:
        for char1 in chars:
            for char2 in chars:
                for char3 in chars:
                    for char4 in chars:
                        res_char = char1 + char2 + char3 + char4#获取遍历的任意4Byte字符
                        char_crc = binascii.crc32(res_char.encode())#获取遍历字符的CRC32值
                        calc_crc = char_crc & 0xffffffff#将遍历的字符的CRC32值与0xffffffff进行与运算
                        if calc_crc == crc_value:#将获取字符的CRC32值与每个文件的CRC32值进行匹配
                            print('[+] {}: {}'.format(hex(crc_value),res_char))
                            comment += res_char
    print('-----------CRC Crack Completed-----------')
    print('Result: {}'.format(comment))

if __name__ == '__main__':
    crack_crc()

最终可得：

另一个脚本：https://github.com/AabyssZG/CRC32-Tools

[GKCTF 2021]你知道apng吗

下载得到：girl.apng

用浏览器将其打开，打开后的效果有点像gif，是动态的

使用工具将其分为静态的几张图片：http://www.downcc.com/soft/40902.html

有几张图有明显二维码：

还有个需要用ps扭曲一下二维码：

调整一下，然后QR识别即可

还有一个要用stegsolve:

QR扫描

最后拼接为 flag{a3c7e4e5-9b9d-ad20-0327-288a235370ea}

[SCTF 2014]misc400

下载得到附件--sctf.png

思路：

• 宽高改写
• lsb隐写
• 010
• exif信息查看
• binwalk

binwalk sctf.png

发现存在两个数据块

应该是数据块出现了人为的修改

然后使用pngcheck

https://github.com/wkliu19/pngcheck

make -f Makefile.unx
cp pngcheck /usr/local/bin/pngcheck

然后就可以在任意目录下执行pngcheck命令了

pngcheck -v sctf.png

详细查看每个数据块的情况

发现最后的数据块没有填满(45027没到65524)，之后又出现了138字节，证明是人为添加了数据块

接下来分离sctf.png，查看IDAT的数据块：

binwalk -e sctf.png --run-as=root

查看较小的文件，发现15AFFB文件中有二进制：

1111111000100001101111111100000101110010110100000110111010100000000010111011011101001000000001011101101110101110110100101110110000010101011011010000011111111010101010101111111000000001011101110000000011010011000001010011101101111010101001000011100000000000101000000001001001101000100111001111011100111100001110111110001100101000110011100001010100011010001111010110000010100010110000011011101100100001110011100100001011111110100000000110101001000111101111111011100001101011011100000100001100110001111010111010001101001111100001011101011000111010011100101110100100111011011000110000010110001101000110001111111011010110111011011

猜测有可能是01转二维码

脚本：

from PIL import Image
MAX = 25
pic = Image.new("RGB", (MAX, MAX))
str = "1111111000100001101111111100000101110010110100000110111010100000000010111011011101001000000001011101101110101110110100101110110000010101011011010000011111111010101010101111111000000001011101110000000011010011000001010011101101111010101001000011100000000000101000000001001001101000100111001111011100111100001110111110001100101000110011100001010100011010001111010110000010100010110000011011101100100001110011100100001011111110100000000110101001000111101111111011100001101011011100000100001100110001111010111010001101001111100001011101011000111010011100101110100100111011011000110000010110001101000110001111111011010110111011011"
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")

ToolsFx: