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CTF对抗- KCTF 2023 第十二题 wp - 98k

推荐 原创

周杰伦

文章分类 CTF对抗 阅读数 : 390 阅读时长 : 9分钟


【CTF对抗- KCTF 2023 第十二题 wp - 98k】此文章归类为：CTF对抗。

程序中的一些关键函数加了混淆，直接把 0x49A0E8 处 patch 为 double 类型的 100.0 ，再声明其类型为 const double 这样 ida F5 就能直接去除无关的逻辑。

题目实现了一个虚拟机，程序启动后会先解密得到虚拟机执行的命令，之后来到函数 0x460D0E 执行：

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char __thiscall global_vtable_2_func_18(C * this )
{
   C *v1; // ebx
   int _pc; // esi
   B *ins_start; // edi MAPDST
   char *v6; // esi
   _DWORD *v7; // eax
   const char *v8; // ecx
   unsigned int v9; // edx
   unsigned int v10; // edi
   _BYTE *v11; // ebx
   char v12; // al
   int v14; // [esp+8h] [ebp-6Ch]
   char v16[88]; // [esp+1Ch] [ebp-58h] BYREF
 
   v1 = this ;
   if ( v1->instructions.start == v1->instructions.finish )
     return 0;
   this ->pc = 0;
   ins_start = this ->instructions.start;
   if ( v1->instructions.finish - ins_start > 0 )
   {
     _pc = 0;
     ins_start = v1->instructions.start;
     do
     {
       if ( _pc == -1 )
       {
         v1->pc = 0;
         _pc = 0;
       }
       if ( !(*((unsigned __int8 (__thiscall **)(C *, B *))v1->vtable + 15))(v1, &ins_start[_pc]) )
       {
         // show error message
       }
       _pc = v1->pc + 1;
       v1->pc = _pc;
       ins_start = v1->instructions.start;
     }
     while ( _pc < v1->instructions.finish - ins_start );
   }
   return 1;
}

v1->vtable + 15 就是虚表中的第 15 项，地址为 0x4525EB 。

相关数据结构：

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struct A
{
   int field_0;
   std::string s0;
   int i1;
   int i2;
   int i3;
   std::string s4;
   std::string s5;
   std::string s6;
};
 
struct std::vector$A$
{
   A *start;
   A *finish;
   A *end_of_storage;
};
 
struct B
{
   int is_label;
   int field_4;
   std::string s1;
   std::string s2;
   std::vector$A$ vector;
};
 
struct std::vector$B$
{
   B *start;
   B *finish;
   B *end_of_storage;
};
 
struct RBTree // std::map<std::string, std::string>
{
   struct RBTree *left;
   struct RBTree *parent;
   struct RBTree *right;
   char color;
   char isnil;
   char padding[2];
   std::string key;
   std::string value;
};
 
struct C
{
   int **vtable;
   RBTree *tree;
   int node_count;
   std::vector$B$ instructions;
   int pc;
   char valid_data;
};

将解密之后的指令所在的内存 dump 出来，写脚本恢复原始指令：

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def vm(ins):
     opcode = int (ins[ 2 ])
     operands = ins[ 3 ]
     if opcode = = 0x21344D4938CE0640 :
         print ( 'system({0})' . format ( repr (operands[ 0 ][ 0 ].decode())))
     elif opcode = = 0x3820EA1739C3E154 :
         print ( 'map[{0}] = hex(vigenere(map[{0}], "恭喜发财".encode("gbk")))' . format ( repr (operands[ 0 ][ 4 ])))
     elif opcode = = 0x4B0134D06B40680 :
         print ( 'if (len(map[{0}]) {1} {2}) goto {3}' . format ( repr (operands[ 0 ][ 4 ]), operands[ 1 ][ 5 ].decode(), operands[ 2 ][ 1 ], operands[ 3 ][ 6 ].decode()))
     elif opcode = = 0x5304FD305CA8C22A :
         print ( 'sleep({})' . format (operands[ 0 ][ 1 ]))
     elif opcode = = 0x5AC009C0F14B76E8 :
         print ( 'map[{0}] = input()' . format ( repr (operands[ 0 ][ 4 ])))
     elif opcode = = 0x6975C7A3C07CD226 :
         print ( 'map[{0}] += {1}' . format ( repr (operands[ 0 ][ 4 ]), repr (operands[ 1 ][ 0 ])))
     elif opcode = = 0x7929CBF0A1496FB0 :
         print ( 'map[{0}] = base64encode(map[{0}])' . format ( repr (operands[ 0 ][ 4 ])))
     elif opcode = = 0x885F75A1461ECEBB :
         print ( 'if (map[{0}] == map[{1}]) goto {2}' . format ( repr (operands[ 0 ][ 4 ]), repr (operands[ 1 ][ 4 ]), operands[ 2 ][ 6 ].decode()))
     elif opcode = = 0x8DB9D83D80004137 :
         print ( 'print({})' . format ( repr (operands[ 0 ][ 0 ].replace(b '\\n' , b '\n' ).decode( 'gbk' ))))
     elif opcode = = 0xA43CBF9D015186F1 :
         if len (operands): code = operands[ 0 ][ 1 ]
         else : code = 0
         print ( 'exit({0})' . format (code))
     elif opcode = = 0xAA9C8E70F01F8D61 :
         print ( 'map[{0}] = map[{1}]' . format ( repr (operands[ 0 ][ 4 ]), repr (operands[ 1 ][ 4 ])))
     elif opcode = = 0xDDEEFF2200112233 :
         print ( 'nop' )
     elif opcode = = 0xE56D33B21C50A892 :
         print ( 'goto {0}' . format (operands[ 0 ][ 6 ].decode()))
     elif opcode = = 0xF4CC06C2E34200F0 :
         print ( 'nop 0' )
     elif opcode = = 0xF4CC06F2E3420459 :
         print ( 'nop 1' )
     elif opcode = = 0xFD1D1DFB19850CA1 :
         print ( 'map[{0}] = md5(map[{0}])' . format ( repr (operands[ 0 ][ 4 ])))
     else :
         assert False , hex (opcode)
 
mem_offset = 0xea0000
mem_size = 0x20000
mem_dump = open ( 'MEM_00EA0000_00020000.mem' , 'rb' ).read()
 
ins_start = 0xeb4b00
ins_end = 0xeb62a4
 
def read_dword(addr):
     assert mem_offset < = addr < mem_offset + mem_size
     return int .from_bytes(mem_dump[addr - mem_offset: addr - mem_offset + 4 ], 'little' )
 
def read_data(addr, size):
     assert mem_offset < = addr < mem_offset + mem_size - size
     return mem_dump[addr - mem_offset: addr - mem_offset + size]
 
def read_str(addr):
     size = read_dword(addr + 0x10 )
     cap = read_dword(addr + 0x14 )
     if cap > 0x10 : addr = read_dword(addr)
     return read_data(addr, size)
 
def read_A(addr):
     return (read_str(addr + 4 ), read_dword(addr + 0x1c ), read_dword(addr + 0x20 ), read_dword(addr + 0x24 ), read_str(addr + 0x28 ), read_str(addr + 0x40 ), read_str(addr + 0x58 ))
 
def read_B(addr):
     B = (read_dword(addr), read_dword(addr + 4 ), read_str(addr + 0x20 ), [])
     A_start = read_dword(addr + 0x38 )
     A_end = read_dword(addr + 0x3c )
     for addr in range (A_start, A_end, 0x70 ):
         B[ - 1 ].append(read_A(addr))
     return B
 
ins = []
for addr in range (ins_start, ins_end, 0x44 ):
     ins.append(read_B(addr))
 
for i in range ( len (ins)):
     t = ins[i]
     if t[ 0 ]:
         print (t[ 3 ][ 0 ][ 6 ].decode() + ':' )
     else :
         # print(i, end=': ')
         print ( '    ' , end = '')
         vm(ins[i])
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     sleep( 25 )
     print ( '                                           \n' )
     sleep( 25 )
     print ( '                  _oo0oo_                  \n' )
     sleep( 25 )
     print ( '                 o8888888o                 \n' )
     sleep( 25 )
     print ( '                 88" . "88                 \n' )
     sleep( 25 )
     print ( '                 (| -_- |)                 \n' )
     sleep( 25 )
     print ( '                 0\\  =  /0                 \n' )
     sleep( 25 )
     print ( "               ___/`---'\\___               \n" )
     sleep( 25 )
     print ( "             .'\\.\\|     |/./'.             \n" )
     sleep( 25 )
     print ( '            / \\.\\||  :  ||/./ \\            \n' )
     sleep( 25 )
     print ( '           / _||||| -:- |||||- \\           \n' )
     sleep( 25 )
     print ( '          | | \\.\\.\\  -  /././ | |          \n' )
     sleep( 25 )
     print ( "          | \\_|  ''\\---/''  |_/ |          \n" )
     sleep( 25 )
     print ( "          \\  .-\\__  '-'  ___/-. /          \n" )
     sleep( 25 )
     print ( "        ___'. .'  /--.--\\  `. .'___        \n" )
     sleep( 25 )
     print ( '     ."" \'<  `.___\\_<|>_/___.\' >\' "".      \n' )
     sleep( 25 )
     print ( '    | | :  `- \\`.;`\\ _ /`;.`/ - ` : | |    \n' )
     sleep( 25 )
     print ( '    \\  \\ `_.   \\_ __\\ /__ _/   .-` /  /    \n' )
     sleep( 25 )
     print ( "=====`-.____`.___ \\_____/___.-`___.-'===== \n" )
     sleep( 25 )
     print ( "                  `=---='                  \n" )
     sleep( 25 )
     print ( '                                           \n' )
     sleep( 25 )
     print ( '             佛祖保佑 永无BUG              \n' )
     sleep( 25 )
     print ( '            论坛 KCTF2023年度赛            \n' )
     sleep( 25 )
     print ( '            出题战队:中午吃什么            \n' )
     sleep( 25 )
     print ( '                                           \n' )
     sleep( 25 )
     print ( '                                           \n' )
     nop
     sleep( 25 )
     print ( '请输入用户名：\n--> ' )
     map [b '2569430338759937617' ] = input ()
     nop
     sleep( 25 )
     print ( '请输入序列号：\n--> ' )
     map [b '1509181994979340817' ] = input ()
     nop
     nop 0
     nop 1
     if ( len ( map [b '1509181994979340817' ]) ! = 32 ) goto 10547232137042693405
     nop
     map [b '18097274335226857185' ] = map [b '2569430338759937617' ]
     map [b '18097274335226857185' ] + = b 'KCTF2023'
     map [b '18097274335226857185' ] = hex (vigenere( map [b '18097274335226857185' ], "恭喜发财" .encode( "gbk" )))
     map [b '18097274335226857185' ] = base64encode( map [b '18097274335226857185' ])
     map [b '18097274335226857185' ] = md5( map [b '18097274335226857185' ])
     nop
     if ( map [b '18097274335226857185' ] = = map [b '1509181994979340817' ]) goto 15317636321340901566
     goto 15315557398280875957
     nop
10547232137042693405 :
     sleep( 25 )
     print ( 'error\n\n\n' )
     system( 'pause' )
     exit( 0 )
     nop
15317636321340901566 :
     sleep( 25 )
     print ( 'Success GoodJob!\n\n\n' )
     system( 'pause' )
     exit( 0 )
     nop
15315557398280875957 :
     sleep( 25 )
     print ( 'error\n\n\n' )
     system( 'pause' )
     exit( 0 )

求解：

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#!/usr/bin/env python3
 
from base64 import b64encode
from hashlib import md5
 
def vigenere(x, y):
     return bytes(x[i] ^ y[i % len (y)] for i in range ( len (x)))
 
def solve(username):
     return md5(b64encode(vigenere(username + b 'KCTF2023' , "恭喜发财" .encode( "gbk" )). hex ().upper().encode())).hexdigest()
 
 
'''
User-Name : 4070382B95A4F0ED
Serial-Number : 47a62c6eb8a72031a27b89abc3d976f7
'''
 
assert solve(b '4070382B95A4F0ED' ) = = '47a62c6eb8a72031a27b89abc3d976f7'
print (solve(b 'KCTF' )) # 35090e1336f1d1e872ba798256db1bfb


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