pyc re
这是一道pyc题目
*反编译
反编译如下1
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data = "M,\x1d-\x18}E'\x1ezN~\x1b*\x19+\x12%\x1d-" + 'I\x7fM(I{I\x7fJ.\x16wWcRj\x0e6\x0fn' + 'Zo\nn\x0fk\t1R7\x03g\x067\x00eUb\x043' + '\x014\x071Rr\x14x\x19~D?q"a5s,A%' + "\x10'\x11uLyA%\x1d|DrFv\x12t\x11#B&" + 'GsKzK*O)\x1c%GuC>\x1e\x7f\x1b+\x19*' + '\x1e&\x14-\x1f/\x1axAqBq@yO-LtE}' + '\x1b,MuBp\x12'
import os
import sys
import struct
import cStringIO
import string
import dis
import marshal
import types
import random
count = 0
def reverse(string):
return string[::-1]
data_list = list(reverse(data)[1:])
def decrpyt(c, key2):
global count
data_list[count] = c ^ key2
count += 1
def GetFlag1():
key = struct.unpack('B', data[len(data) - 8])[0]
for c in data_list:
if count == 0:
decrpyt(struct.unpack('B', c)[0], key)
continue
key = struct.unpack('B', data[len(data) - 3])[0]
decrpyt(struct.unpack('B', c)[0], key)
for c in data_list[::-1]:
print chr(c),
def GetFlag2():
key = struct.unpack('B', data[len(data) - 11])[0]
for c in data_list:
if count == 0:
decrpyt(struct.unpack('B', c)[0], key)
continue
key = struct.unpack('B', data[len(data) - 4 - count])[0]
decrpyt(struct.unpack('B', c)[0], key)
for c in data_list[::-1]:
print chr(c),
def GetFlag3():
key = struct.unpack('B', data[len(data) - 5])[0]
for c in data_list:
if count == 0:
decrpyt(struct.unpack('B', c)[0], key)
continue
key = struct.unpack('B', data[len(data) - 2 - count])[0]
decrpyt(struct.unpack('B', c)[0], key)
for c in data_list[::-1]:
print chr(c),
def GetFlag4():
global count
key = struct.unpack('B', data[len(data) - 1])[0]
for c in data_list:
if count == 0:
decrpyt(struct.unpack('B', c)[0], key)
continue
key = struct.unpack('B', data[len(data) - 1 - count])[0]
decrpyt(struct.unpack('B', c)[0], key)
count = 0
for c in data_list[::-1]:
print chr(c),
def GetFlag5():
key = struct.unpack('B', data[len(data) - 9])[0]
for c in data_list:
if count == 0:
decrpyt(struct.unpack('B', c)[0], key)
continue
key = struct.unpack('B', data[len(data) - 3 - count])[0]
decrpyt(struct.unpack('B', c)[0], key)
for c in data_list[::-12]:
print chr(c),
GetFlag1()
*结果
直接把几个GetFlag运行一遍就行了
最终结果在GetFlag4里面