ISC2016训练赛 phrackCTF--findkey

测试文件：https://static2.ichunqiu.com/icq/resources/fileupload/phrackCTF/REVERSE/findkey

 

1.准备

获得信息

• Python文件

 

2.文件分析

分析得到是Python文件，首先我把文件后缀改为了.py但显示为乱码，因此猜测是.pyc文件。可以到https://tool.lu/pyc/将文件反编译为py文件。得到代码

 1 # uncompyle6 version 3.4.0
 2 # Python bytecode 2.7 (62211)
 3 # Decompiled from: Python 3.7.1 (default, Dec 10 2018, 22:54:23) [MSC v.1915 64 bit (AMD64)]
 4 # Embedded file name: findkey
 5 # Compiled at: 2016-04-30 17:54:18
 6 import sys
 7 lookup = [
 8  196,
 9  153, 149,
10  206, 17,
11  221, 10, 217, 167, 18, 36, 135, 103, 61, 111, 31, 92, 152, 21, 228, 105, 191, 173, 41, 2, 245, 23, 144, 1, 246, 89, 178, 182, 119, 38, 85, 48, 226, 165, 241, 166, 214, 71, 90, 151, 3, 109, 169, 150, 224, 69, 156, 158, 57, 181, 29, 200, 37, 51, 252, 227, 93, 65, 82, 66, 80, 170, 77, 49, 177, 81, 94, 202, 107, 25, 73, 148, 98, 129, 231, 212, 14, 84, 121, 174, 171, 64, 180, 233, 74, 140, 242, 75, 104, 253, 44, 39, 87, 86, 27, 68, 22, 55, 76, 35, 248, 96, 5, 56, 20, 161, 213, 238, 220, 72, 100, 247, 8, 63, 249, 145, 243, 155, 222, 122, 32, 43, 186, 0, 102, 216, 126, 15, 42, 115, 138, 240, 147, 229, 204, 117, 223, 141, 159, 131, 232, 124, 254, 60, 116, 46, 113, 79, 16, 128, 6, 251, 40, 205, 137, 199, 83, 54, 188, 19, 184, 201, 110, 255, 26, 91, 211, 132, 160, 168, 154, 185, 183, 244, 78, 33, 123, 28, 59, 12, 210, 218, 47, 163, 215, 209, 108, 235, 237, 118, 101, 24, 234, 106, 143, 88, 9, 136, 95, 30, 193, 176, 225, 198, 197, 194, 239, 134, 162, 192, 11, 70, 58, 187, 50, 67, 236, 230, 13, 99, 190, 208, 207, 7, 53, 219, 203, 62, 114, 127, 125, 164, 179, 175, 112, 172, 250, 133, 130, 52, 189, 97, 146, 34, 157, 120, 195, 45, 4, 142, 139]
12 pwda = [188, 155, 11, 58, 251, 208, 204, 202, 150, 120, 206, 237, 114, 92, 126, 6, 42]
13 pwdb = [53, 222, 230, 35, 67, 248, 226, 216, 17, 209, 32, 2, 181, 200, 171, 60, 108]
14 flag = raw_input('Input your Key:').strip()
15 if len(flag) != 17:
16     print 'Wrong Key!!'
17     sys.exit(1)
18 flag = flag[::-1]
19 for i in range(0, len(flag)):
20     if ord(flag[i]) + pwda[i] & 255 != lookup[(i + pwdb[i])]:
21         print 'Wrong Key!!'
22         sys.exit(1)
23 
24 print 'Congratulations!!'
25 # okay decompiling findkey.pyc

 

3.代码分析

 

1.从第15行代码得知，flag长度为17

 

2.都18~22行代码，首先对flag进行reverse字符，再进行ord(flag[i]) + pwda[i] & 255 != lookup[(i + pwdb[i])]操作。

 

因此我们可以逆向操作，获取flag

 

4.脚本获取

lookup = [
 196,
 153, 149,
 206, 17,
 221, 10, 217, 167, 18, 36, 135, 103, 61, 111, 31, 92, 152, 21, 228, 105, 191, 173, 41, 2, 245, 23, 144, 1, 246, 89, 178, 182, 119, 38, 85, 48, 226, 165, 241, 166, 214, 71, 90, 151, 3, 109, 169, 150, 224, 69, 156, 158, 57, 181, 29, 200, 37, 51, 252, 227, 93, 65, 82, 66, 80, 170, 77, 49, 177, 81, 94, 202, 107, 25, 73, 148, 98, 129, 231, 212, 14, 84, 121, 174, 171, 64, 180, 233, 74, 140, 242, 75, 104, 253, 44, 39, 87, 86, 27, 68, 22, 55, 76, 35, 248, 96, 5, 56, 20, 161, 213, 238, 220, 72, 100, 247, 8, 63, 249, 145, 243, 155, 222, 122, 32, 43, 186, 0, 102, 216, 126, 15, 42, 115, 138, 240, 147, 229, 204, 117, 223, 141, 159, 131, 232, 124, 254, 60, 116, 46, 113, 79, 16, 128, 6, 251, 40, 205, 137, 199, 83, 54, 188, 19, 184, 201, 110, 255, 26, 91, 211, 132, 160, 168, 154, 185, 183, 244, 78, 33, 123, 28, 59, 12, 210, 218, 47, 163, 215, 209, 108, 235, 237, 118, 101, 24, 234, 106, 143, 88, 9, 136, 95, 30, 193, 176, 225, 198, 197, 194, 239, 134, 162, 192, 11, 70, 58, 187, 50, 67, 236, 230, 13, 99, 190, 208, 207, 7, 53, 219, 203, 62, 114, 127, 125, 164, 179, 175, 112, 172, 250, 133, 130, 52, 189, 97, 146, 34, 157, 120, 195, 45, 4, 142, 139]
pwda = [188, 155, 11, 58, 251, 208, 204, 202, 150, 120, 206, 237, 114, 92, 126, 6, 42]
pwdb = [53, 222, 230, 35, 67, 248, 226, 216, 17, 209, 32, 2, 181, 200, 171, 60, 108]

flag = ""

for i in range(0,17):
    flag += chr(lookup[(i + pwdb[i])] - pwda[i] & 255)
                

flag = flag[::-1]
print(flag)

 

5.get flag！

PCTF{PyC_Cr4ck3r}