记录一下刷题&比赛中遇到的一些古典密码题
最近更新:2024-12-03
 
[FSCTF 2023]兔帽的奇妙冒险 附件:
1 2 有一只名叫贝斯64 先生的兔子。它戴着一顶古老的兔帽子,帽子上绣着奇异的符文,传说这是一种能与神秘力量连接的魔法帽子。 OFR3 c05 RTEFKM2 Q3 QkxqRGlTbWthWExxSXhrZFo3 SU5 DVWRkOTVFQ2 JUSE1 VcXpRQk9 hc3 hqMzRrbHRJMD0 = 
 
base64 解密一下
8TwsNQLAJ3d7BLjDiSmkaXLqIxkdZ7INCUdd95ECbTHMUqzQBOasxj34kltI0=
再加上 Rabbit 特征的密文头 U2FsdGVkX1  再Rabbit 解密
FSCTF{Bunny_is_so_cute_Why_e@t_bunny} 
[MoeCTF 2021]OldButPopular 
附件
 
1 2 qj3r_y31s{vxk7_hig__0_a}g4ax_x0x540slv 你能解密出这段是什么意思吗? 观察一下这段密文格式有什么不一样 
 
一看就是移了顺序的,而且 {} 也移了,栅栏密码:所有字母变换位置但字母不变
这题的flag头 不是 flag就是 moectf,密文里没有,那就是字母变换了,大概是凯撒加密了
因为凯撒只对26个字母加密,所以{}是不动的,所以我们可以先进行凯撒解密通过 {} 来筛选正确的密文
1 qsigxj {g453v_a4rx_x0_k0_sy7_xl3_a0v1h} 
 
在线网址枚举或者随波逐流一把梭
1 moectf{c453 r_w4 nt_t0 _g0 _ou7 _th3 _w0 r1 d} 
 
 
加密原理:
凯撒密码在线加密解密 - 千千秀字 (qqxiuzi.cn) 
栅栏密码在线加密解密 - 千千秀字 (qqxiuzi.cn) 
户外天堂 
湘赣边武功山论剑 
 
1 这里是修炼之人的梦想之地,风景如画、宁静安详,少侠可以联系尚宝司少卿“张程”大人,在这里稍作休息,恢复体力,继续冒险。 
 
1 张程大人:或许维吉尼亚密码在比武中能派上用场。这种加密方法通过变换字母顺序和使用不同的密钥,能有效保护通信内容,对于保密传输至关重要,值得深入研究和应用。 
 
1 2 kr0 gMDY1 NTY5 MDghNTU5 MsozMhU5 MzUgNiM3 NTI0 ODI5 MsA2 MrE3 CmU9 NsU1 MhcKYz01 NTkyOTk4 MDM2 Mio3 Mrg0 OTYfODE0 OTcfNjUhMiA0 MhUiODM= key:rsa 
 
小改了一下字典,加密字典变成了小写字母,除小写外其他字符不变
因为字典的不同,所以在线网站和 工具都解不对,只能自己敲代码
所以我们需要理解原理来进行编写
1,$len(key) \le len(message)$ ,当密钥的长度小于明文长度时,需要重复拼接key到相同长度
2,
注:模数 26 是 字典的长度,即 p = len(table) ,字典长度为几,模数为几
Mi 就是 M这个明文在字母表中的索引值,比如表是 abcdefghijklmnopqrstuvwxyz, 假如 M 为 b, 那 Mi 就表示 1   (表是a-z,下标也就是0-25)
同样,Ki 就是 Mi 所对应的 K的索引值,假设明文 M 对应的是密钥 r,在表中的索引值就是 17,于是 Ki=17
具体可以结合下面的exp进行理解:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 import  base64def  pad_key (key, length ):    Key = key * length     return  Key[:length] def  encrypt (message, table, key ):    enc = ''      Key = pad_key(key, length)     for  i in  range (length):         if  message[i].islower():             enc += table[(table.index(message[i]) + table.index(Key[i])) % len (table)]         else :             enc += message[i]     return  enc message = 'kr0gMDY1NTY5MDghNTU5MsozMhU5MzUgNiM3NTI0ODI5MsA2MrE3CmU9NsU1MhcKYz01NTkyOTk4MDM2Mio3Mrg0OTYfODE0OTcfNjUhMiA0MhUiODM='  table = 'abcdefghijklmnopqrstuvwxyz'  key = 'rsa'  length = len (message) print (base64.b64decode(encrypt(message,table,key)).decode())
 
rsa 就不解释了
1 2 3 4 5 6 7 8 9 10 11 12 13 import  libnumfrom  xenny.ctf.crypto.modern.asymmetric.rsa import  factorn=1065569082559283359352737524829206217  e=65537  c=557299803638728496181497065330435383  p,q = factor.attack(n) d=libnum.invmod(e,(p-1 )*(q-1 )) m = pow (c,d,n) print (libnum.n2s(pow (c,d,n)))
 
easy_crypto 
2024-09-05
 
1 2 3 4 key.txt 1091091153210977773210977109457732774646324677831153277464546324611511545838377321098377 flag.rar 
 
题目内容:谐音一点都不好玩,但好在差别不大。
尝试了很多,最后可能是拼凑的ascii码,
1 2 3 4 5 6 7 8 9 10 str  = '1091091153210977773210977109457732774646324677831153277464546324611511545838377' num = 0  while  num <= len (str ):    for  j in  range (2 , 4 ):         if  32  <= int (str [num:num + j]) <= 126 :             print (chr (int (str [num:num + j])), end='' )             num += j 
 
ms,摩斯
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 str  = 'mms mMM mMm-M M.. .MSs M.-. .ss-SSM mSM' M = ['M' , 'm' , '.' ] S = ['S' , 's' , '.' ] for  i in  str :    if  i in  M:         print ('.' , end='' )     elif  i == ' ' :         print ('/' , end='' )     else :         print ('-' , end='' ) print ()for  i in  str :    if  i in  S:         print ('.' , end='' )     elif  i == ' ' :         print ('/' , end='' )     else :         print ('-' , end='' ) 
 
得到key:GO0DLC$K
打开flag压缩包是一大堆base
由结果可知进行了50次base64加密
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 import  re, base64s = open ('flag.txt' , 'rb' ).read() base16_dic = r'^[A-F0-9=]*$'  base32_dic = r'^[A-Z2-7=]*$'  base64_dic = r'^[A-Za-z0-9/+=]*$'  n= 0  while  True :    n += 1      t = s.decode()     if  '{'  in  t:         print (t)         break      elif  re.match (base16_dic, t):         s = base64.b16decode(s)         print (str (n) + ' base16' )     elif  re.match (base32_dic, t):         s = base64.b32decode(s)         print (str (n) + ' base32' )     elif  re.match (base64_dic, t):         s = base64.b64decode(s)         print (str (n) + ' base64' ) 
 
Affine 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 import  stringimport  libnumfrom  tqdm import  trangeqwer= 'flag{'     sercet='aipx{823j56p37ap92p93pd4g7ad6a0p01p21}'  dic = string.ascii_lowercase def  affine_decrypt (ciphertext, a, b, dic ):    m = len (dic)     a_inv = libnum.invmod(a, m)     plaintext = ''      for  char in  ciphertext:         if  char in  dic:             y = dic.index(char)             x = (a_inv * (y - b)) % m             plaintext += dic[x]         else :             plaintext += char     return  plaintext n = 26  for  a in  trange(100 ):    for  b in  range (100 ):         c=''          for  j in  qwer:             if  ord (j) >= ord ("a" ) and  ord (j) <= ord ("z" ):                 cha=ord (j)-ord ('a' )                 c+=chr ((a * cha+ b) % n+ord ('a' ))             else :                 c+=j             if  'aipx'  in  c:                 print (c)                 print (a,b)                 print ((affine_decrypt(sercet,a,b,dic)))                 break  
 
cs 
古剑山2024
 
1 jhjnjcjikmkfjjjkkejkkekdjgjcjnjhjnjcjiko 
 
https://en.wiktionary.org/wiki/Appendix:Two-letter_combinations 
容易知道是j和k开头的两个字母来表示一个字母
 
猜测凯撒密码
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 hncimfjkekedgcnhncio gmbhleijdjdcfbmgmbhn flagkdhicicbealflagm ekzfjcghbhbadzkekzfl djyeibfgagazcyjdjyek cixdhaefzfzybxicixdj bhwcgzdeyeyxawhbhwci agvbfycdxdxwzvgagvbh zfuaexbcwcwvyufzfuag yetzdwabvbvuxteyetzf xdsycvzauautwsdxdsye wcrxbuyztztsvrcwcrxd vbqwatxysysruqbvbqwc uapvzswxrxrqtpauapvb tzouyrvwqwqpsoztzoua syntxquvpvpornysyntz rxmswptuouonqmxrxmsy qwlrvostntnmplwqwlrx pvkqunrsmsmlokvpvkqw oujptmqrlrlknjuoujpv ntioslpqkqkjmitntiou mshnrkopjpjilhsmshnt lrgmqjnoioihkgrlrgms kqflpimnhnhgjfqkqflr jpekohlmgmgfiepjpekq iodjngklflfehdoiodjp 
 
像也不想像,是k那部分不对,应该是 j和k 分成两部分位移了
1 2 3 4 5 6 7 8 9 10 11 12 a = 'jhjnjcjikmkfjjjkkejkkekdjgjcjnjhjnjcjiko'  for  j in  range (-26 ,26 ):    flag = ''      for  i in  range (0 ,len (a),2 ):         if  a[i] == 'j' :             flag += chr (ord (a[i+1 ])-2 )         else :             flag+=chr (ord (a[i+1 ])+j)     print (flag) 
 
apple 
古剑山
0 sloved
 
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