2020 SCTF 部分WriteUp

signin

准备

signin.exe：https://wwa.lanzous.com/inIQdec11zi

 

程序分析

可以判断出，这个程序实际上是由Python打包成的可执行文件，且在运行这个程序时，在同目录下产生了一个tmp.dll文件，猜测是程序调用某些函数的接口。

 

反编译

使用archive_viewer.py反编译为字节码文件

python archive_viewer.py signin.exe

 

修补文件

55 0D 0D 0A 00 00 00 00 70 79 69 30 10 01 00 00

 

程序是在Python3.8环境下打包，因此我们需要在Python3.8下使用uncompyle6

uncompyle6 main.pyc > main.py

得到py文件

 1 # uncompyle6 version 3.7.2
 2 # Python bytecode 3.8 (3413)
 3 # Decompiled from: Python 3.8.0 (tags/v3.8.0:fa919fd, Oct 14 2019, 19:37:50) [MSC v.1916 64 bit (AMD64)]
 4 # Embedded file name: main.py
 5 # Compiled at: 1995-09-28 00:18:56
 6 # Size of source mod 2**32: 272 bytes
 7 import sys
 8 from PyQt5.QtCore import *
 9 from PyQt5.QtWidgets import *
10 from signin import *
11 from mydata import strBase64
12 from ctypes import *
13 import _ctypes
14 from base64 import b64decode
15 import os
16 
17 class AccountChecker:
18 
19     def __init__(self):
20         self.dllname = './tmp.dll'
21         self.dll = self._AccountChecker__release_dll()
22         self.enc = self.dll.enc
23         self.enc.argtypes = (c_char_p, c_char_p, c_char_p, c_int)
24         self.enc.restype = c_int
25         self.accounts = {b'SCTFer': b64decode(b'PLHCu+fujfZmMOMLGHCyWWOq5H5HDN2R5nHnlV30Q0EA')}
26         self.try_times = 0
27 
28     def __release_dll(self):
29         with open(self.dllname, 'wb') as (f):
30             f.write(b64decode(strBase64.encode('ascii')))
31         return WinDLL(self.dllname)
32 
33     def clean(self):
34         _ctypes.FreeLibrary(self.dll._handle)
35         if os.path.exists(self.dllname):
36             os.remove(self.dllname)
37 
38     def _error(self, error_code):
39         errormsg = {0:'Unknown Error', 
40          1:'Memory Error'}
41         QMessageBox.information(None, 'Error', errormsg[error_code], QMessageBox.Abort, QMessageBox.Abort)
42         sys.exit(1)
43 
44     def __safe(self, username: bytes, password: bytes):
45         pwd_safe = b'\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00'
46         status = self.enc(username, password, pwd_safe, len(pwd_safe))
47         return (pwd_safe, status)
48 
49     def check(self, username, password):
50         self.try_times += 1
51         if username not in self.accounts:
52             return False
53         encrypted_pwd, status = self._AccountChecker__safe(username, password)
54         if status == 1:
55             self._AccountChecker__error(1)
56         if encrypted_pwd != self.accounts[username]:
57             return False
58         self.try_times -= 1
59         return True
60 
61 
62 class SignInWnd(QMainWindow, Ui_QWidget):
63 
64     def __init__(self, checker, parent=None):
65         super().__init__(parent)
66         self.checker = checker
67         self.setupUi(self)
68         self.PB_signin.clicked.connect(self.on_confirm_button_clicked)
69 
70     @pyqtSlot()
71     def on_confirm_button_clicked(self):
72         username = bytes((self.LE_usrname.text()), encoding='ascii')
73         password = bytes((self.LE_pwd.text()), encoding='ascii')
74         if username == b'' or password == b'':
75             self.check_input_msgbox()
76         else:
77             self.msgbox(self.checker.check(username, password))
78 
79     def check_input_msgbox(self):
80         QMessageBox.information(None, 'Error', 'Check Your Input!', QMessageBox.Ok, QMessageBox.Ok)
81 
82     def msgbox(self, status):
83         msg_ex = {0:'', 
84          1:'', 
85          2:"It's no big deal, try again!", 
86          3:'Useful information is in the binary, guess what?'}
87         msg = 'Succeeded! Flag is your password' if status else 'Failed to sign in\n' + msg_ex[(self.checker.try_times % 4)]
88         QMessageBox.information(None, 'SCTF2020', msg, QMessageBox.Ok, QMessageBox.Ok)
89 
90 
91 if __name__ == '__main__':
92     app = QApplication(sys.argv)
93     checker = AccountChecker()
94     sign_in_wnd = SignInWnd(checker)
95     sign_in_wnd.show()
96     app.exec()
97     checker.clean()
98     sys.exit()
99 # okay decompiling main.pyc

 

代码分析

通过代码我们能够了解到这些信息

1.

elf.dllname = './tmp.dll'

调用了tmp.dll文件作为接口。

 

2.

    def __safe(self, username: bytes, password: bytes):
        pwd_safe = b'\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00'
        status = self.enc(username, password, pwd_safe, len(pwd_safe))
        return (pwd_safe, status)

    def check(self, username, password):
        self.try_times += 1
        if username not in self.accounts:
            return False
        encrypted_pwd, status = self._AccountChecker__safe(username, password)
        if status == 1:
            self._AccountChecker__error(1)
        if encrypted_pwd != self.accounts[username]:
            return False
        self.try_times -= 1
        return True
self.accounts = {b'SCTFer': b64decode(b'PLHCu+fujfZmMOMLGHCyWWOq5H5HDN2R5nHnlV30Q0EA')}

调用tmp.dll文件中的enc函数，传入username, password, pwd_safe, len(pwd_safe)，实际就是将password加密后存储到pwd_safe字节码中。最后用pwd_safe与b64decode(b'PLHCu+fujfZmMOMLGHCyWWOq5H5HDN2R5nHnlV30Q0EA')比较，且我们能够了解到用户名应该是SCTFer，且最后返回的status一个为非1。

 

打开tmp.dll文件，找到enc函数

观察代码，实际操作可以分为两部分，逆向分析

 

异或操作

第47~54行代码实际上就是将Dst与用户名循环异或，最后得到b64decode(b'PLHCu+fujfZmMOMLGHCyWWOq5H5HDN2R5nHnlV30Q0EA')，因此我们只需要逆向异或就能得到加密后的Dst

from base64 import *

username = "SCTFer"
pwd_safe = b64decode('PLHCu+fujfZmMOMLGHCyWWOq5H5HDN2R5nHnlV30Q0EA')
# print (len(pwd_safe))
num = ["%02x" % x for x in pwd_safe]
hex_num = [int(x, 16) for x in num]

print(num)
# print (len(num))
for i in range(32):
    hex_num[i] ^= ord(username[i % len(username)])
# print (hex_num)
hex_nums = bytes.fromhex(''.join([hex(x)[2:].rjust(2, '0') for x in hex_num]))

print (hex_nums)

得到

b'o\xf2\x96\xfd\x82\x9c\xde\xb52v\x86yK3\xe6\x1f\x06\xd8\xb7=\x13J\xb8\xe3\xb52\xb3\xd38\x86\x10\x02\x00'

 

加密操作

每次传入了8字节数据进行加密（总共64字节），打开sub_180011311函数

仔细观察代码，实际上这部分代码是使用CRC32的查表法，对数据进行加密。

加密原理实际上就是CRC32算法---输入一组长度32的字符串，每8个字节分为1组，共4组。对每一组取首位，判断正负。正值，左移一位；负值，左移一位，再异或0xB0004B7679FA26B3。重复判断操作64次，得到查表法所用的表。

因此我们只需要将整个加密过程逆向操作得到查表法的表，再进行CRC32计算，就能得到输入。

secret = []

# for i in range(4):
#     secret.append(int(hex_nums[i*8:(i + 1) * 8][::-1].hex(),16))

for i in range(4):
    secret.append(int.from_bytes(hex_nums[i*8:(i + 1) * 8], byteorder="little"))

print (secret)

key = 0xB0004B7679FA26B3

flag = ""

for s in secret:
    for i in range(64):
        sign = s & 1
        if sign == 1:
            s ^= key
        s //= 2
        if sign == 1:
            s |= 0x8000000000000000
    print(hex(s))
    j = 0
    while j < 8:
        flag += chr(s&0xFF)
        s >>= 8
        j += 1
print(flag)

因为计算机中采用小端排序，因此需要注意分组倒序。得到

 

脚本

from base64 import *

username = "SCTFer"
pwd_safe = b64decode('PLHCu+fujfZmMOMLGHCyWWOq5H5HDN2R5nHnlV30Q0EA')
# print (len(pwd_safe))
num = ["%02x" % x for x in pwd_safe]
hex_num = [int(x, 16) for x in num]

print(num)
# print (len(num))
for i in range(32):
    hex_num[i] ^= ord(username[i % len(username)])
# print (hex_num)
hex_nums = bytes.fromhex(''.join([hex(x)[2:].rjust(2, '0') for x in hex_num]))

print (hex_nums)

secret = []

# for i in range(4):
#     secret.append(int(hex_nums[i*8:(i + 1) * 8][::-1].hex(),16))

for i in range(4):
    secret.append(int.from_bytes(hex_nums[i*8:(i + 1) * 8], byteorder="little"))

print (secret)

key = 0xB0004B7679FA26B3

flag = ""

for s in secret:
    for i in range(64):
        sign = s & 1
        if sign == 1:
            s ^= key
        s //= 2
        if sign == 1:
            s |= 0x8000000000000000
    print(hex(s))
    j = 0
    while j < 8:
        flag += chr(s&0xFF)
        s >>= 8
        j += 1
print(flag)

 

get flag!

username:SCTFer

password:SCTF{We1c0m3_To_Sctf_2020_re_!!}

 

 

get_up

下载：https://wwa.lanzous.com/i6fDyedivgh

 

代码分析

这是实际就是在说明输入字符串长度不大于6，且通过sub_401DF0返回1

返回1，实际就是满足加密后的Dst与32c1d123c193aecc4280a5d7925a2504相同，实际是MD5加密，得到输入为sycsyc

 

代码解析

接下来，程序查找.reioc段，并将.reioc段数据与sycsyc循环异或，写出IDAPython脚本

beg_adr = 0x405000
dst = "sycsyc"
for i in range(0,0x200,16):
    for j in range(16):
        PatchByte(beg_adr+i+j,Byte(beg_adr+i+j)^ord(dst[j%6]))

然后跟去混淆方法类似，先转换为Data，再强制分析数据，转换为函数。

 

打开sub_4027F0函数

 

这里实际和上面差不多，输入长度为30的flag，取前5个字符，与.ebata段的部分循环异或。（实际就是得到下面sub_404000函数的代码）。这里可以合理猜测flag的前五个字符为SCTF{,写出脚本

beg_adr = 0x404000
dst = "SCTF{"
for i in range(16,96):
    PatchByte(beg_adr+i,Byte(beg_adr+i)^ord(dst[i%5]))

跟上面一样将.ebata段转换为data，再分析代码，转换为函数，开头如果提示栈不平衡，重新单独分析一下就行。

 

解密

这里一个函数实际就是RC4生成S-BOX，另一个函数就是加密函数，key值为syclover

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
typedef unsigned longULONG;

#pragma warning(disable:4996)

/*初始化函数*/
void rc4_init(unsigned char* s, unsigned char* key, unsigned long Len)
{
    int i = 0, j = 0;
    char k[256] = { 0 };
    unsigned char tmp = 0;
    for (i = 0; i < 256; i++)
    {
        s[i] = i;
        k[i] = key[i % Len];
    }
    for (i = 0; i < 256; i++)
    {
        j = (j + s[i] + k[i]) % 256;
        tmp = s[i];
        s[i] = s[j]; // 交换s[i]和s[j]
        s[j] = tmp;
    }
}

/*加解密*/
void rc4_crypt(unsigned char* s, unsigned char* Data, unsigned long Len)
{
    int i = 0, j = 0, t = 0;
    unsigned long k = 0;
    unsigned char tmp;
    for (k = 0; k < Len; k++)
    {
        i = (i + 1) % 256;
        j = (j + s[i]) % 256;
        tmp = s[i];
        s[i] = s[j]; // 交换s[x]和s[y]
        s[j] = tmp;
        t = (s[i] + s[j]) % 256;
        Data[k] ^= s[t];
    }
}

int main()
{
    unsigned char s[256] = { 0 }, s2[256] = { 0 }; // S-box
    char key[256] = { "syclover" };
    char pData[512] = { 0 };
    int v12[] = { 128,85,126,45,209,9,37,171,60,86,149,196,54,19,237,114,36,147,178,200,69,236,22,107,103,29,249,163,150,217 };
    int i;
    //char m[32] = { 0 };

    for (int i = 0; i < 30; ++i) {
        pData[i] = (char)v12[i];
    }

    unsigned long len = strlen(pData);

    printf("pData=%s\n", pData);
    printf("key=%s,length=%d\n\n", key, strlen(key));
    rc4_init(s, (unsigned char*)key, strlen(key)); // 已经完成了初始化
    printf("完成对S[i]的初始化，如下：\n\n");
    for (i = 0; i < 256; i++)
    {
        printf("%02X", s[i]);
        if (i && (i + 1) % 16 == 0)putchar('\n');
    }
    printf("\n\n");
    for (i = 0; i < 256; i++) // 用s2[i]暂时保留经过初始化的s[i]，很重要的！！！
    {
        s2[i] = s[i];
    }

    printf("已经初始化，现在解密:\n\n");
    printf("len = %d\n", len);
    rc4_crypt(s, (unsigned char*)pData, len); 
    printf("pData=%s\n\n", pData);

    system("PAUSE");
    return 0;
}

 

get flag！

SCTF{zzz~(|3[___]_rc4_5o_e4sy}

 

Can you hear

robot36 app：https://wwa.lanzous.com/it7Zsedix5i

 

音频是SSTV，用软件Robot36接受音频中的数据

 

doudizhu

 只要将手里牌出完，就能获得flag

 

posted @ 2020-07-06 09:31  Hk_Mayfly  阅读(1094)  评论(5)  编辑  收藏  举报