实验吧 trivial writeup

该部分为本科期间实验吧题解wp备份。

题目

An unlocked terminal is displaying the following:Encryption complete, ENC(???,T0pS3cre7key) = Bot kmws mikferuigmzf rmfrxrwqe abs perudsf! Nvm kda ut ab8bv_w4ue0_ab8v_DDUYou poke around and find this interesting file.

解题链接： http://ctf5.shiyanbar.com/crypto/trivial/encrypt.rar

解题

下载解题链接之后，发现是个encrypt的py文件，打开发现以下代码：

#!/usr/bin/env python
import sys

alphaL = "abcdefghijklnmopqrstuvqxyz"
alphaU = "ABCDEFGHIJKLMNOPQRSTUVQXYZ"
num    = "0123456789"
keychars = num+alphaL+alphaU

if len(sys.argv) != 3:
  print "Usage: %s SECRET_KEY PLAINTEXT"%(sys.argv[0])
  sys.exit()

key = sys.argv[1]
if not key.isalnum():
  print "Your key is invalid, it may only be alphanumeric characters"
  sys.exit()

plaintext = sys.argv[2]

ciphertext = ""
for i in range(len(plaintext)):
  rotate_amount = keychars.index(key[i%len(key)])
  if plaintext[i] in alphaL:
    enc_char = ord('a') + (ord(plaintext[i])-ord('a')+rotate_amount)%26
  elif plaintext[i] in alphaU:
    enc_char = ord('A') + (ord(plaintext[i])-ord('A')+rotate_amount)%26
  elif plaintext[i] in num:
    enc_char = ord('0') + (ord(plaintext[i])-ord('0')+rotate_amount)%10
  else:
    enc_char = ord(plaintext[i])
  ciphertext = ciphertext + chr(enc_char)

print "Encryption complete, ENC(%s,%s) = %s"%(plaintext,key,ciphertext)

其中for循环那块是真正加密的过程。显然，这是个对称加密的过程。

加密依赖一个一次同余式以及rotate_amount进行：

rotate_amount=keychars.index(key[i%len(key)]
enc_text[i]-C1 =(plaintext[i]+rotate_amount) mod C2 # ( 其中C1、C2为常数）

因此考虑写个python脚本进行解密，依赖以下两个公式：

rotate_amount=keychars.index(key[i%len(key)]
plaintext[i] =[(enc_text[i]-C1-rotate_amount) mod C2]+ C1 # ( 其中C1、C2为常数）

我的脚本代码：
https://github.com/FrancisQiu/-shiyanbar_Crypto/pull/1/commits/067673eeab7d238307804f8ef01c6cbe2f6b7edd