概述
解除 CTF 也有很多年了, 但是真正的将网上的题目通关刷题还是没有过的, 同时感觉水平下降的太厉害, 这两个月准备把网上目前公开有的 CTF 环境全部刷一遍, 同时收集题目做为素材, 为后面的培训及靶场搭建做好准备. 本文是 2018 年 7 月 8 日前所有密码类的题目通关 Writeup.
Writeup
变异凯撒
普通凯撒密码参考, https://en.wikipedia.org/wiki/Caesar_cipher .
加密密文: afZ_r9VYfScOeO_UL^RWUc
格式: flag{ }
解题过程分这几部分, 首先 afZ_r9VYfScOeO_UL^RWUc 的 ascii 码为
97 102 90 95 114 57 86 89 102 83 99 79 101 79 95 85 76 94 82 87 85 99
flag{ } 的前几位 ascii 码为
102 108 97 103 123
按位做一个比较就可以发现 102-97=5,108-102=6,97-90=7, 所以此题目凯撒的规律为第一个字符 ascii 加 5, 其他每个字符按位 ascii 自增 1, 所以解密代码如下. 解密代码如下
- #!/usr/bin/env python
- #-*- coding: utf-8 -*-
- """
- @Author : darkN0te
- @Create date : 2018-07-07
- @description : 凯撒轮转密码解密
- @Update date :
- """
- INIT_ADD = 5
- input = raw_input()
- output = ""
- for char in input:
- output += chr(ord(char) + INIT_ADD)
- INIT_ADD += 1
- print output
输出
- afZ_r9VYfScOeO_UL^RWUc
- flag{
- Caesar_variation
- }
结束.
传统知识 + 古典密码
题目描述:
小明某一天收到一封密信, 信中写了几个不同的年份
辛卯, 癸巳, 丙戌, 辛未, 庚辰, 癸酉, 己卯, 癸巳.
信的背面还写有 "+ 甲子", 请解出这段密文.
key 值: CTF{XXX}
根据天干地支纪年法
1. 甲子 2. 乙丑 3. 丙寅 4. 丁卯 5. 戊辰 6. 己巳 7. 庚午 8. 辛未 9. 壬申 10. 癸酉
11. 甲戌 12. 乙亥 13. 丙子 14. 丁丑 15. 戊寅 16. 己卯 17. 庚辰 18. 辛巳 19. 壬午 20. 癸未
21. 甲申 22. 乙酉 23. 丙戌 24. 丁亥 25. 戊子 26. 己丑 27. 庚寅 28. 辛卯 29. 壬辰 30. 癸巳
31. 甲午 32. 乙未 33. 丙申 34. 丁酉 35. 戊戌 36. 己亥 37. 庚子 38. 辛丑 39. 壬寅 40. 癸卯
41. 甲辰 42. 乙巳 43. 丙午 44. 丁未 45. 戊申 46. 己酉 47. 庚戌 48. 辛亥 49. 壬子 50. 癸丑
51. 甲寅 52. 乙卯 53. 丙辰 54. 丁巳 55. 戊午 56. 己未 57. 庚申 58. 辛酉 59. 壬戌 60. 癸亥
写出题中所给组合的数字编码
28 30 23 8 17 10 16 30
加上一个甲子 (60)
88 90 93 68 77 70 76 90
转换成 ASCII 字母:
XZSDMFLZ
栅栏密码 (两栏):
XMZFSLDZ
凯撒:
SHUANGYU
最后按格式提交 CTF{SHUANGYU} 即可.
what's wrong with this
题目描述:
We managed to get this package of the robots servers. We managed to determine that it is some kind of compiled bytecode. But something is wrong with it. Our usual analysis failed - so we have to hand this over to you pros. We only know this: The program takes one parameter and it responds with "Yup" if you have found the secret code, with "Nope" else. We expect it should be obvious how to execute it.
解题链接: http://ctf5.shiyanbar.com/crypto/What-s-wrong-with-this/hello.tar.gz
中文题干: 我们设法获得了这个机器人服务器包. 我们设法确定它是某种编译的字节码. 但它有些问题. 我们通常的分析失败了 - 所以我们不得不把它交给你们. 我们只知道这个: 程序接受一个参数, 如果你找到了密码, 它会以 "Yup" 响应, 而 "Nope" 则是. 我们期望它应该是如何执行它的.
网站给出了一个答案 PDF, 请查看
http://hebin.me/wp-content/uploads/2017/09/2017090715264378.pdf
解压题目给出的文件 hello.tar.gz, 我们知道了一些程序特征会打印 Yup 和 Nope, 然后使用命令
grep -R 'Yup\|Nope'
进行搜索, 找到匹配文件.
使用 uncompyle 进行反编译不可以, 使用 Decompyle++ 进行反编译. 安装过程是这样.
- Git clone https://github.com/zrax/pycdc.git
- cd pycdc
- cmake .
- make
- make install
- pycdc __main__hello__.pyc
我们用过 pycdc 反编译出__main__hello__.pyc 的源码
- # Source Generated with Decompyle++
- # File: __main__hello__.pyc (Python 2.7)
- import sys
- import dis
- import multiprocessing
- import UserList
- def encrypt_string(s):
- Unsupported opcode: <255>
- new_str = []
- # WARNING: Decompyle incomplete
- def rot_chr(c, amount):
- None = chr(((ord(c) + 33) % amount) / 94 % 33)
- SECRET = 'w*0;CNU[\\gwPWk}3:PWk"#&:ABu/:Hi,M'
- if encrypt_string(sys.argv - 1) == SECRET:
- print>>'Yup'
- else:
- print>>'Nope'
- None = None
可以看到, 有一部分字节码没有被反编译出来, 这是由于一部分字节码没有被识别造成的, 使用 pycdas 查看一下信息.
- ? what's wrong with this ~/Safe/03_tools/pycdc/pycdas __main__hello__.pyc
- __main__hello__.pyc (Python 2.7)
File Name: chall.py
- Object Name:
- Arg Count: 0
- Locals: 0
- Stack Size: 3
- Flags: 0x00000040 (CO_NOFREE)
- [Names]
- 'sys'
- 'hashlib'
- 'sha256'
- 'dis'
- 'multiprocessing'
- 'UserList'
- 'encrypt_string'
- 'rot_chr'
- 'SECRET'
- 'argv'
- [Var Names]
- [Free Vars]
- [Cell Vars]
- [Constants]
- -1
- None
- (
- 'sha256'
- )
File Name: chall.py
- Object Name: encrypt_string
- Arg Count: 1
- Locals: 4
- Stack Size: 8
- Flags: 0x00000043 (CO_OPTIMIZED | CO_NEWLOCALS | CO_NOFREE)
- [Names]
- 'enumerate'
- 'append'
- 'rot_chr'
- 'ord'
- 'join'
- [Var Names]
- 's'
- 'new_str'
- 'index'
- 'c'
- [Free Vars]
- [Cell Vars]
- [Constants]
- None
- 0
- 10
- 1
- ''
- [Disassembly]
- 0 BUILD_LIST 0
- 3 STORE_FAST 1: new_str
- 6 SETUP_LOOP 99 (to 108)
- 9 LOAD_GLOBAL 0: enumerate
- 12 LOAD_FAST 0: s
- 15 CALL_FUNCTION 1
- 18
- 19 FOR_ITER 85 (to 107)
- 22 UNPACK_SEQUENCE 2
- 25 STORE_FAST 2: index
- 28 STORE_FAST 3: c
- 31 LOAD_FAST 2: index
- 34 LOAD_CONST 1: 0
- 37 COMPARE_OP 2 (==)
- 40 POP_JUMP_IF_FALSE 68
- 43 LOAD_FAST 1: new_str
- 46 LOAD_ATTR 1: append
- 49 LOAD_GLOBAL 2: rot_chr
- 52 LOAD_FAST 3: c
- 55 LOAD_CONST 2: 10
- 58 CALL_FUNCTION 2
- 61 CALL_FUNCTION 1
- 64 ROT_TWO
- 65 JUMP_ABSOLUTE 19
- 68 LOAD_FAST 1: new_str
- 71 LOAD_ATTR 1: append
- 74 LOAD_GLOBAL 2: rot_chr
- 77 LOAD_FAST 3: c
- 80 LOAD_GLOBAL 3: ord
- 83 LOAD_FAST 1: new_str
- 86 LOAD_FAST 2: index
- 89 LOAD_CONST 3: 1
- 92 BINARY_ADD
- 93 BINARY_SUBTRACT
- 94 CALL_FUNCTION 1
- 97 CALL_FUNCTION 2
- 100 CALL_FUNCTION 1
- 103 ROT_TWO
- 104 JUMP_ABSOLUTE 19
- 107 END_FINALLY
- 108 LOAD_CONST 4: ''
- 111 LOAD_ATTR 4: join
- 114 LOAD_FAST 1: new_str
- 117 CALL_FUNCTION 1
- 120 IMPORT_STAR
File Name: chall.py
- Object Name: rot_chr
- Arg Count: 2
- Locals: 2
- Stack Size: 3
- Flags: 0x00000043 (CO_OPTIMIZED | CO_NEWLOCALS | CO_NOFREE)
- [Names]
- 'chr'
- 'ord'
- [Var Names]
- 'c'
- 'amount'
- [Free Vars]
- [Cell Vars]
- [Constants]
- None
- 33
- 94
- [Disassembly]
- 0 LOAD_GLOBAL 0: chr
- 3 LOAD_GLOBAL 1: ord
- 6 LOAD_FAST 0: c
- 9 CALL_FUNCTION 1
- 12 LOAD_CONST 1: 33
- 15 BINARY_ADD
- 16 LOAD_FAST 1: amount
- 19 BINARY_MODULO
- 20 LOAD_CONST 2: 94
- 23 BINARY_DIVIDE
- 24 LOAD_CONST 1: 33
- 27 BINARY_MODULO
- 28 CALL_FUNCTION 1
- 31 IMPORT_STAR
- 'w*0;CNU[\\gwPWk}3:PWk"#&:ABu/:Hi,M'
- 1
- 'Yup'
- 'Nope'
- [Disassembly]
- 0 LOAD_CONST 0: -1
- 3 LOAD_CONST 1: None
- 6 IMPORT_NAME 0: sys
- 9 STORE_NAME 0: sys
- 12 LOAD_CONST 0: -1
- 15 LOAD_CONST 2: ('sha256',)
- 18 IMPORT_NAME 1: hashlib
- 21 IMPORT_FROM 2: sha256
- 24 STORE_NAME 2: sha256
- 27 ROT_TWO
- 28 LOAD_CONST 0: -1
- 31 LOAD_CONST 1: None
- 34 IMPORT_NAME 3: dis
- 37 STORE_NAME 3: dis
- 40 LOAD_CONST 0: -1
- 43 LOAD_CONST 1: None
- 46 IMPORT_NAME 4: multiprocessing
- 49 STORE_NAME 4: multiprocessing
- 52 LOAD_CONST 0: -1
- 55 LOAD_CONST 1: None
- 58 IMPORT_NAME 5: UserList
- 61 STORE_NAME 5: UserList
- 64 LOAD_CONST 3: <CODE> encrypt_string
- 67 MAKE_FUNCTION 0
- 70 STORE_NAME 6: encrypt_string
- 73 LOAD_CONST 4: <CODE> rot_chr
- 76 MAKE_FUNCTION 0
- 79 STORE_NAME 7: rot_chr
- 82 LOAD_CONST 5: 'w*0;CNU[\\gwPWk}3:PWk"#&:ABu/:Hi,M'
- 85 STORE_NAME 8: SECRET
- 88 LOAD_NAME 6: encrypt_string
- 91 LOAD_NAME 0: sys
- 94 LOAD_ATTR 9: argv
- 97 LOAD_CONST 6: 1
- 100 BINARY_SUBTRACT
- 101 CALL_FUNCTION 1
- 104 LOAD_NAME 8: SECRET
- 107 COMPARE_OP 2 (==)
- 110 POP_JUMP_IF_FALSE 121
- 113 LOAD_CONST 7: 'Yup'
- 116 PRINT_NEWLINE
- 117 PRINT_ITEM_TO
- 118 JUMP_FORWARD 5 (to 126)
- 121 LOAD_CONST 8: 'Nope'
- 124 PRINT_NEWLINE
- 125 PRINT_ITEM_TO
- 126 LOAD_CONST 1: None
- 129 IMPORT_STAR
在研究一下如何修复的
修复后反编译得到的结果为
- # Source Generated with Decompyle ++
- # File: __main__hello__.pyc (Python 2.7)
- import sys
- from hashlib import sha256
- import dis
- import multiprocessing
- import UserList
- def encrypt_string(s):
- new_str = []
- for (index , c) in enumerate(s):
- if index == 0:
- new_str.append(rot_chr(c, 10))
- continue
- new_str.append(rot_chr(c, ord(new_str[index - 1])))
- return ''.join(new_str)
- def rot_chr(c, amount):
- return chr(((ord(c) - 33) + amount) % 94 + 33)
- SECRET = 'w*0;CNU[\\gwPWk}3:PWk"#&:ABu/:Hi,M'
- if encrypt_string(sys.argv[1]) == SECRET:
- print 'Yup'
- else:
- print 'Nope'
写出解密代码:
- SECRET = 'w*0;CNU[\\gwPWk}3:PWk"#&:ABu/:Hi,M'
- def decrypt_string(s):
- new_str = []
- for (index , c) in enumerate(s):
- if index == 0:
- new_str.append(rot_chr(c, 10))
- continue
- new_str.append(rot_chr(c, ord(s[index - 1])))
- return ''.join(new_str)
- def rot_chr(c, amount):
- return chr(((ord(c) - 33) - amount) % 94 + 33)
- print decrypt_string(SECRET)
- # output : modified_in7erpreters_are_3vil!!!
- try them all
题目描述:
You have found a passwd file containing salted passwords. An unprotected configuration file has revealed a salt of 5948. The hashed password for the 'admin' user appears to be 81bdf501ef206ae7d3b92070196f7e98, try to brute force this password.
您找到了一个有盐的密码表. 已知密码的明文结尾为 5948, 密码表中哈希值为 81bdf501ef206ae7d3b92070196f7e98, 尝试暴力破解此密码.
原题干为英文, 但是感觉和题目本身的意思不一样, 写了一段和原题目意思一致的中文. 题目的意思就是一个简单的爆破 md5. 难点是不知道到底这个明文到底有多少位, 都包含什么字符.
这里直接使用在 cmd5 上查到的结果 sniper5948.
脚本
- #!/usr/bin/env python
- #-*- coding: utf-8 -*-
- """
- @Author : darkN0te
- @Create date : 2018-07-07
- @description : md5 爆破 单进程
- @Update date :
- """
- import string
- import hashlib
- endOutput = "5948"
- file=open("output.txt","a")
- md5input=raw_input("请输入 md5:\n")
- md5input=md5input.lower()
- # apt=string.printable[:-6]
- apt=string.letters
- def dfs(s,num):
- m=hashlib.md5()
- print s + endOutput
- m.update(s + endOutput)
- md5temp=m.hexdigest()
- if md5temp==md5input:
- file.write("md5 是:"+md5input+"明文是:"+s+"\n")
- file.close()
- exit(-1)
- if len(s)==num:
- return
- for i in apt:
- dfs(s+i,num)
- myinput=7 #生成字符的位数
- for j in range(1,myinput):
- dfs("",j)
来源: http://www.bubuko.com/infodetail-2958604.html