攻防世界 reverse Windows_Reverse1

Windows_Reverse1  2019_DDCTF

查壳

脱壳

 脱壳后运行闪退，(或许需要修复下IAT??)，先IDA 静态分析一下

int __cdecl main(int argc, const char **argv, const char **envp)
{
  char v4; // [esp+4h] [ebp-804h]
  char v5; // [esp+5h] [ebp-803h]
  char v6; // [esp+404h] [ebp-404h]
  char Dst; // [esp+405h] [ebp-403h]

  v6 = 0;
  memset(&Dst, 0, 0x3FFu);
  v4 = 0;
  memset(&v5, 0, 0x3FFu);
  printf("please input code:");
  scanf("%s", &v6);
  sub_401000(&v6);
  if ( !strcmp(&v4, "DDCTF{reverseME}") )       
    printf("You've got it!!%s\n", &v4);
  else
    printf("Try again later.\n");
  return 0;
}

关键函数sub_401000

unsigned int __cdecl sub_401000(const char *a1)
{
  _BYTE *v1; // ecx
  unsigned int v2; // edi
  unsigned int result; // eax
  int v4; // ebx

  v2 = 0;
  result = strlen(a1);
  if ( result )
  {
    v4 = a1 - v1;
    do
    {
      *v1 = byte_402FF8[(char)v1[v4]];
      ++v2;
      ++v1;
      result = strlen(a1);
    }
    while ( v2 < result );
  }
  return result;
}

View Code

应该就是一个简单的替换算法，然后程序将结果与“DDCTF{reverseME}“进行比较。

接下来进行动态调试dump出置换数组

wp:

# -*- coding: UTF-8 -*-
hexData=[
    0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0xF8, 0xAE, 0x1D, 0x3E, 0x07, 0x51, 0xE2, 0xC1,
    0xFF, 0xFF, 0xFF, 0xFF, 0xFF, 0xFF, 0xFF, 0xFF, 0xFE, 0xFF, 0xFF, 0xFF, 0x01, 0x00, 0x00, 0x00,
    0x7E, 0x7D, 0x7C, 0x7B, 0x7A, 0x79, 0x78, 0x77, 0x76, 0x75, 0x74, 0x73, 0x72, 0x71, 0x70, 0x6F,
    0x6E, 0x6D, 0x6C, 0x6B, 0x6A, 0x69, 0x68, 0x67, 0x66, 0x65, 0x64, 0x63, 0x62, 0x61, 0x60, 0x5F,
    0x5E, 0x5D, 0x5C, 0x5B, 0x5A, 0x59, 0x58, 0x57, 0x56, 0x55, 0x54, 0x53, 0x52, 0x51, 0x50, 0x4F,
    0x4E, 0x4D, 0x4C, 0x4B, 0x4A, 0x49, 0x48, 0x47, 0x46, 0x45, 0x44, 0x43, 0x42, 0x41, 0x40, 0x3F,
    0x3E, 0x3D, 0x3C, 0x3B, 0x3A, 0x39, 0x38, 0x37, 0x36, 0x35, 0x34, 0x33, 0x32, 0x31, 0x30, 0x2F,
    0x2E, 0x2D, 0x2C, 0x2B, 0x2A, 0x29, 0x28, 0x27, 0x26, 0x25, 0x24, 0x23, 0x22, 0x21, 0x20, 0x00,
    0x01, 0x00, 0x00, 0x00, 0x90, 0x19, 0x9F, 0x00, 0xA8, 0x2C, 0x9F, 0x00, 0x00, 0x00, 0x00, 0x00,
    0x00, 0x00, 0x00, 0x00 ]
str='DDCTF{reverseME}'
flag=''
for i in range(len(str)):
    flag+=chr(hexData[ord(str[i])])
print('flag{'+flag+'}')

flag{ZZ[JX#,9(9,+9QY!}