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
print >>'Yup'
else:
print
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)
[Code]
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'
)
[Code]
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
[Code]
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:
")
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+"
")
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)