BUUCTF [NCTF2019]Sore

很有意思的一道题
题目源码:

from string import ascii_letters
from flag import flag


ctoi = lambda x: ascii_letters.index(x)
itoc = lambda x: ascii_letters[x]

key = flag.strip('NCTF{}')
len_key = len(key)

plaintext = open('plaintext.txt', 'r').read()

plain = ''.join(p for p in plaintext if p in ascii_letters)
cipher = ''.join( itoc( ( ctoi(p) + ctoi( key[i % len_key] ) ) % 52 ) \
                  for i,p in enumerate(plain) )

open('ciphertext.txt', 'w').write(cipher)

我们得到的只有ciphertext的信息 然而想要获得作为key的flag肯定是需要plaintext
这里可以采用 词频分析 其实也就是类似维吉尼亚加密的方法

这里还顺带给了我们key
只是要注意 这里大小写是不符合实际的
应该改为
Shewouldntwalkrightnext
所以重新计算key:

from string import ascii_letters

ctoi = lambda x: ascii_letters.index(x)
itoc = lambda x: ascii_letters[x]

with open(r'\ciphertext.txt','r+') as f:
    cipher = f.read()
with open(r'\plaintext.txt','r+') as f:
    plaintext = f.read()
plaintext = "Shewouldntwalkrightnext"
cipher    = "nsfAIHFrMuLynuCApeEstxJ"
plain = ''.join(p for p in plaintext if p in ascii_letters)
flag = ""
for i in range(len(cipher)):
    c = cipher[i]
    p = plaintext[i]
    for key in ascii_letters:
        cc = itoc( ( ctoi(p) +  ctoi(key)  ) % 52 )
        if(cc == c):
            flag += key
            break
    print(flag)
# VlbEUNUoZbPyckLSjXLfpaQ

最终flag:vlbeunuozbpycklsjXlfpaq