crackme采用多线程，分步骤解码方式，最终需要解密一段代码再运行并弹出提示消息框。反调试为CRC校验，具体逆向步骤如下：
IDA 打开crackerme，发现 GetDlgItemTextA，查找参考并来到 sub_4012E7

  v3 = 0;
  v4 = 8;
  v5 = &sn;
  v6 = (unsigned __int8)sn;
  while ( v4 )
  {
    v3 += v6;
    LOBYTE(v6) = *++v5;
    --v4;
  }
  if ( (_BYTE)v6 )
    result = (char *)MessageBoxA(hWnd, err, err, 0);
  else
    result = sub_401B0A(v3);

char *__stdcall sub_401B0A(int a1)
{
  char *result; // eax@2

  if ( dword_403037 )
  {
    ++byte_403C99;
    switch ( a1 )
    {
      case 0x566:
        result = (char *)sub_401DA0();
        break;
      case 0x79A:
        result = (char *)sub_401DC1();
        break;
      case 0x86B:
        result = sub_401E16();
        break;
      case 0x5D5:
        result = (char *)sub_401DE2();
        break;
      case 0x325:
        result = (char *)sub_401DE2();
        break;
      default:
        byte_403C99 = 0;
        result = (char *)MessageBoxA(0, err, err, 0);
        break;
    }
  }
  else
  {
    result = (char *)MessageBoxA(0, err, err, 0);
  }
  return result;
}

      a3 = 8;
      while ( a3 )
      {
        v8 += v11;
        *v10++ = v11 ^ 0x66;
        ++v9;
        --a3;
        LOBYTE(v11) = *v9;
      }

    if ( v8 == 0x353 )
    {
      v12 = 12;
      goto LABEL_18;
    }
    if ( v8 == 0x325 )
    {
      v12 = 4;
      goto LABEL_18;
    }
    if ( v8 == 0x29B )
      break;
    if ( v8 == 0x363 )
    {
      v12 = 20;
LABEL_18:
      v13 = (char *)&dword_4032F1 + v12;
      LOWORD(v12) = *(_WORD *)((char *)&unk_403BEA + 7);
      a3 = v12 << 16;
      LOWORD(a3) = *(_WORD *)((char *)&unk_403BEA + 9) + a3;
      ((void (__stdcall *)(_DWORD))a3)(*(_DWORD *)v13);
    }