20220207-CTF-MISC-第11题--- base64隐写--附带脚本
攻防世界- MISC新手区–第11题–base64隐写
下载之后解压,是stego.txt
打开stego.txt
显然是base64编码之后的结果,base64解码,我还百度翻译了一下,也没什么发现。
于是想办法,因为没有MISC的基础,就去看了WP,知道是base64隐写。于是查阅资料了解了一下。
base64隐写的原理
(1)起因:
俄罗斯有个叫 Olympic_ctf 的 CTF, 在 2014 年有道 misc 题是关于 Base64 的隐写题.
大意就是给了你一段字符串, 让你找 flag.
这里放上字符串, 便于读者实验
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(2)复习base64编码
BASE64 是一种编码方式, 是一种可逆的编码方式.
编码后的数据是一个字符串, 包含的字符为: A-Za-z0-9+/
共 64 个字符:26 + 26 + 10 + 1 + 1 = 64
其实是 65 个字符, = 是填充字符.
64 个字符需要 6 位二进制来表示
2^5=32
2^6=64
表示成数值为 0~63.
比如, 字符串"Rascals"经过 Base64 编码后变为"UmFzY2Fscw=="
上面的 图,完整展示了为什么转化后面会多出两个=,原因是后面补了两个000000
概括来说,出现等号是因为:
需在原数据二进制值后面添加零, 使其字节数是 6 的倍数.
然后, 在编码后的字符串后面添加 1 个或 2 个等号"=", 表示所添加的零值字节数.
按照上面的逻辑,同理:
长度为 3 个字节的数据经过 Base64 编码后就变为 4 个字节 38=64
(3)base64隐写原理
从Base64编码的原理可得Base64解码的过程:
丢掉末尾的所有‘=’;
每个字符查表转换为对应的6位索引,得到一串二进制字符串;
从头开始,每次取8位转换为对应的ASCII字符,如果不足8位则丢弃。
以上面刚才用到的例子:
做解析:
UmFzY2Fscw== 首先丢掉末尾的== ,得到UmFzY2Fscw
然后 每个字符查表转换为对应的6位索引
后面的完整过程如下:
在解码的过程中中,会有部分数据被丢弃(即不会影响解码结果),这些数据正是我们在编码过程中补的0
也就是说,如果我们在编码过程中不全用0填充,而是用其他的数据填充,仍然可以正常编码解码,因此这些位置可以用于隐写。
解开隐写的方法就是将这些不影响解码结果的位提取出来组成二进制串,然后转换成ASCII字符串。
(4)base64隐写脚本(解密)
python脚本:
d='''U3RlZ2Fub2dyYXBoeSBpcyB0aGUgYXJ0IGFuZCBzY2llbmNlIG9m
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L3dpa2kvU3RlZ2Fub2dyYXBoeQ0K'''
e=d.splitlines()
binstr=""
base64="ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/"
for i in e :if i.find("==")>0:temp=bin((base64.find(i[-3])&15))[2:]#取倒数第3个字符,在base64找到对应的索引数(就是编码数),取低4位,再转换为二进制字符binstr=binstr + "0"*(4-len(temp))+temp #二进制字符补高位0后,连接字符到binstrelif i.find("=")>0:temp=bin((base64.find(i[-2])&3))[2:] #取倒数第2个字符,在base64找到对应的索引数(就是编码数),取低2位,再转换为二进制字符binstr=binstr + "0"*(2-len(temp))+temp #二进制字符补高位0后,连接字符到binstr
str=""
for i in range(0,len(binstr),8):str=str+chr(int(binstr[i:i+8],2)) #从左到右,每取8位转换为ascii字符,连接字符到字符串
print(str) #结果是 Base_sixty_four_point_five转换为
输出:
得到flag就是flag{Base_sixty_four_point_five}
总结
base64隐写解密,以后碰到直接用脚本,先成为脚本小子(小白没办法)
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