我喜欢培根分值：20
- 来源： Ph0enix
- 难度：中
- 参与人数：5331人
- Get Flag：2533人
- 答题人数：2876人
- 解题通过率：88%
key: CTF{}
解题链接： http://ctf5.shiyanbar.com/crypto/enc1.txt
肯定摩斯密码
解码后：
m o r s e _ i s _ c o o l _ b u t _ b a c o n _ i s _ c o o l e r _ d c c d c c c d d d c d c c c d d c c c c c c c c c d d c d c c c c d c c c c c c d c c c d c c d c c c c d c c d d d c c d d d c c d c d d
转化为：
baabaaabbbabaaabbaaaaaaaaabbabaaaabaaaaa/abaaabaaba/aaabaabbbaabbbaababb
image.png
实在没想到，真的是败给了实验吧的格式。
CTF{SHIYANBA IS COOL}
附上python解密培根：
#!/usr/bin/python
# -*- coding: utf-8 -*-
import re
alphabet=['a','b','c','d','e','f','g','h','i','j','k',
'l','m','n','o','p','q','r','s','t','u','v','w','x','y','z']
one_biao=["aaaaa","aaaab","aaaba","aaabb","aabaa","aabab",
"aabba","aabbb","abaaa","abaab","ababa","ababb","abbaa",
"abbab","abbba","abbbb","baaaa","baaab","baaba","baabb",
"babaa","babab","babba","babbb","bbaaa","bbaab"]
def encode():
#python3.0版本后用input替换了raw_input
string=input('请输入字符串加密')#明文
e_string=""
for index in string:
for i in range(0,26):
if(index==alphabet[i]):#字母匹配
e_string+=one_biao
break
print('编码'+e_string)
return
def decode():
e_string=input('请输入暗文解密')
e_array=re.findall(".{5}",e_string)
d_string=""
for index in e_array:
for i in range(0,26):
if index==one_biao[i]:
d_string+=alphabet[i]
print("解码为："+d_string)
return
if __name__=="__main__":
number=input('输入数1或2，1加密，2解密: ')
if number=="1":
encode()
elif number=='2':
decode()