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攻防世界MISC练习区(SimpleRAR base64stego 功夫再高也怕菜刀)

nanyue 2024-08-30 21:00:35 技术文章 8 ℃

SimpleRAR

题目介绍

解题思路

1.打开附件

未发现有用数据

2.winhex打开RAR文件

winhex工具

链接:https://pan.baidu.com/s/1XCIKvAwewZ3o_l0EX_DFFA

提取码:qqzg

3.png部分块的7A改为74

4.解压文件

5.winhex打开图片

查看内容可知这是一个gif图

发现是gif格式,将其重命名并用PhotoShop打开,发现有两个空白的图层

将两个图层分别提取出来,用StegSolve打开,不断点击箭头直到显示出图像

将两幅二维码拼接到一起并补全定位点,扫描二维码得到flag

找到flag

flag{yanji4n_bu_we1shi}

base64stego

题目介绍

解题思路

1.打开附件

2.尝试解压缩

需要密码

3.winhex打开压缩文件

查找十六进制数值--504B--列出搜索结果--确定,发现zip为伪加密

将09 00改为00 00

4.解压获取

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5.运行python脚本解码

python脚本

import re
 
import base64
 
 
 
b64chars = 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/'
 
 
 
# stego.txt为待解密的base64隐写字符串所在的文件
 
f = open('F://stego.txt','r')
 
base64str = f.readline()
 
 
 
# pattern2用于匹配两个等号情况时,等号前的一个字符
 
# pattern2用于匹配一个等号情况时,等号前的一个字符
 
pattern2 = r'(\S)==#39;
 
pattern1 = r'(\S)=#39;
 
 
 
# 提取后的隐写二进制字符加入binstring中
 
binstring = ''
 
 
 
# 逐行读取待解密的base64隐写字符串,逐行处理
 
while(base64str):
 
    # 先匹配两个等号的情况,如果匹配不上,再配置一个等号的情况
 
    # 如果无等号,则没有隐藏,无需处理
 
    if re.compile(pattern2).findall(base64str):
 
        # mstr为等号前的一个字符,该字符为隐写二进制信息所在的字符
 
        mstr = re.compile(pattern2).findall(base64str)[0]
 
        # 确认mstr字符对应的base64二进制数,赋值给mbin
 
        mbin = bin(b64chars.find(mstr))
 
        # mbin格式如0b100,mbin[0:2]为0b
 
        # mbin[2:].zfill(6)为将0b后面的二进制数前面补0,使0b后面的长度为6
 
        mbin2 = mbin[0:2] + mbin[2:].zfill(6)
 
        # 两个等号情况隐写了4位二进制数,所以提取mbin2的后4bit
 
        # 赋值给stegobin,这就是隐藏的二进制信息
 
        stegobin = mbin2[-4:]
 
        binstring += stegobin
 
    elif re.compile(pattern1).findall(base64str):
 
        mstr = re.compile(pattern1).findall(base64str)[0]
 
        mbin = bin(b64chars.find(mstr))
 
        mbin2 = mbin[0:2] + mbin[2:].zfill(6)
 
        # 一个等号情况隐写了2位二进制数,所以提取mbin2的后2bit
 
        stegobin = mbin2[-2:]
 
        binstring += stegobin
 
    base64str = f.readline()
 
 
 
# stegobin将各行隐藏的二进制字符拼接在一起
 
# 从第0位开始,8bit、8bit处理,所以range的步进为8
 
for i in range(0,len(binstring),8):
 
    # int(xxx,2),将二进制字符串转换为10进制的整数,再用chr()转为字符
 
    print(chr(int(binstring[i:i+8],2)),end='')
 
print()

运行脚本

找到flag

flag{Base_sixty_four_point_five}

功夫再高也怕菜刀

题目介绍

解题思路

1.使用binwalk发现流量包里有个Zip压缩包

2.分离附件

得到flag,打开需要密码

3.Wireshark打开附件

4.搜索flag.txt

5.右键跟踪字节流

6.发现FFD8

FFD8FF是jpg文件的开头

找到DDF9

在末尾找到FFD9

7.以FF D8开头、FF D9结尾的这部分复制,并在winhex中新建文件并粘贴,注意粘贴格式选择为ASCII Hex

打开图片

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