简单的python逆向分析

52lxw

拿到一个findkey.txt文件，但不上纯文本的，于是放进010 editor里面看


findkey.txt (2.6 KB, 下载次数: 59)

2018-3-30 23:21 上传

点击文件名下载附件

看到是03 F3 0D 0A开头，可以确定是python的可执行文件

改后缀名为findkey.pyc放到https://tool.lu/pyc/里反编译:

[Python] 纯文本查看 复制代码

#!/usr/bin/env python
# visit http://tool.lu/pyc/ for more information
import sys
lookup = [196,153,149,206,17,221,10,217,167,18,36,135,103,61,111,31,92,152,21,228,105,191,173,41,2,245,23,144,1,246,89,178,182,119,38,85,48,226,165,241,166,214,71,90,151,3,109,169,150,224,69,156,158,57,181,29,200,37,51,252,227,93,65,82,66,80,170,77,49,177,81,94,202,107,25,73,148,98,129,231,212,14,84,121,174,171,64,180,233,74,140,242,75,104,253,44,39,87,86,27,68,22,55,76,35,248,96,5,56,20,161,213,238,220,72,100,247,8,63,249,145,243,155,222,12,32,43,186,0,102,216,126,15,42,115,138,240,147,229,204,117,223,141,159,131,232,124,254,60,116,46,113,79,16,128,6,251,40,205,137,199,83,54,188,19,
    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 = raw_input('Input your Key:').strip()
if len(flag) != 17:
    print 'Wrong Key!!'
    sys.exit(1)
flag = flag[::-1]
for i in range(0, len(flag)):
    if ord(flag[i]) + pwda[i] & 255 != lookup[i + pwdb[i]]:
        print 'Wrong Key!!'
        sys.exit(1)

print 'Congratulations!!'

看到flag必须为17位了，得到以下破解脚本:

[Python] 纯文本查看 复制代码

#!/usr/bin/env python
# visit http://tool.lu/pyc/ for more information
import sys,string
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 = raw_input('Input your Key:').strip()
if len(flag) != 17:
    print 'Wrong Key!!'
    sys.exit(1)
flag = flag[::-1]
'''
result = []
for i in range(17):
    for flag in string.printable:
        if ord(flag) + pwda[i] & 255 == lookup[i + pwdb[i]]:
            result.append(flag)
            break
result.reverse()
print len(result),''.join(result)
print 'Congratulations!!'

jhsunnyshine

强啊懒猪

hcl909

本以为pyhtone 编译好了 基本上很难破解了 没有想到和当年的VB一样 稍稍复杂了点，不过也是楼主比较厉害

ss8651twtw

感谢简单的 python 逆向分析 赞!

丑到变态

感谢分享

chen4321

666，很给力

kernel

不是很明白。。。。

Heln

学习了，赞

sitiger

赞一个，不错不错

Caesar93

逆向出直接有加密算法

sh1ner

666学习了