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使用IDA破解TraceMe.exe

时间:2017-07-03 19:11:43      阅读:279      评论:0      收藏:0      [点我收藏+]

标签:ext   跳转   data-   net   打开   gravity   username   ==   log   

我发现用IDA破解TraceMe.exe比ODeasy多了。

打开IDA 后。直接搜索“序列号”。得到

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双击跳转到反汇编窗体,按F5转换为类C++代码

 signed int __stdcall DialogFunc(HWND hWnd, int a2, unsigned __int16 a3, int a4)
{
  signed int v5; // ebx@9
  int v6; // eax@11
  HWND v7; // eax@12
  HWND v8; // eax@12
  HWND v9; // eax@12
  HICON v10; // eax@20
  const CHAR String2[4]; // [sp+8h] [bp-F4h]@1
  int v12; // [sp+Ch] [bp-F0h]@1
  int v13; // [sp+10h] [bp-ECh]@1
  __int16 v14; // [sp+14h] [bp-E8h]@1
  char v15; // [sp+16h] [bp-E6h]@1
  char v16; // [sp+18h] [bp-E4h]@1
  __int16 v17; // [sp+2Ch] [bp-D0h]@1
  char v18; // [sp+2Eh] [bp-CEh]@1
  char v19; // [sp+30h] [bp-CCh]@1
  __int16 v20; // [sp+44h] [bp-B8h]@1
  char v21; // [sp+46h] [bp-B6h]@1
  CHAR String; // [sp+48h] [bp-B4h]@9
  const CHAR String1; // [sp+98h] [bp-64h]@9


  memcpy(&v16, "你输入字符要大于四个。", 0x14u);
  v17 = *(_WORD *)"!";
  v12 = dword_405054;
  v18 = aUFIKGb[22];
  v15 = byte_40505E;
  memcpy(&v19, "序列号错误。再来一次!", 0x14u);
  *(_DWORD *)String2 = dword_405050;
  v20 = *(_WORD *)"!";
  v14 = word_40505C;
  v13 = dword_405058;
  v21 = aAGmGb[22];
  if ( a2 == 16 )
  {
    DestroyWindow(hWnd);
    return 1;
  }
  if ( a2 == 272 )
  {
    v10 = LoadIconA(hInstance, (LPCSTR)0x70);
    SendMessageA(hWnd, 0x80u, 1u, (LPARAM)v10);
    SendDlgItemMessageA(hWnd, 110, 0xC5u, 0x50u, 0);
    return 1;
  }
  if ( a2 != 273 )
    return 0;
  if ( (signed int)a3 > 1013 )
  {
    if ( a3 == 1014 || a3 == 40002 )
      DialogBoxParamA(hInstance, (LPCSTR)0x67, hWnd, sub_401020, 0);
    return 0;
  }
  if ( a3 != 1013 )
  {
    if ( a3 == 2 || a3 == 1002 )
    {
      SendMessageA(hWnd, 0x10u, 0, 0);
      return 0;
    }
    return 0;
  }
  v5 = GetDlgItemTextA(hWnd, 110, &String, 81);//username
  GetDlgItemTextA(hWnd, 1000, (LPSTR)&String1, 101);//序列号
  if ( String && v5 >= 5 )//推断输入的username长度是不是大于5
  {
    LOBYTE(v6) = sub_401340(&String1, &String, v5);//此函数依据username生成序列号。并与输入的序列号进行比較
    if ( v6 )//v6==1,表示username和序列号匹配
    {
      lstrcpyA(::String1, String2);
      v7 = GetDlgItem(hWnd, 110);
      EnableWindow(v7, 0);
      v8 = GetDlgItem(hWnd, 1000);
      EnableWindow(v8, 0);
      v9 = GetDlgItem(hWnd, 1000);
    }
    else
    {
      lstrcpyA(::String1, &v19);
      v9 = GetDlgItem(hWnd, 1000);
    }
  }
  else
  {
    lstrcpyA(::String1, &v16);//username长度小于5。提示“你输入字符要大于四个”
    v9 = GetDlgItem(hWnd, 110);
  }
  SetFocus(v9);
  MessageBeep(0);
  DialogBoxParamA(hInstance, (LPCSTR)0x79, hWnd, sub_401060, 0);
  return 0;
}


bool __cdecl sub_401340(LPCSTR lpString1, LPSTR lpString2, signed int a3)
{
  signed int v3; // ecx@1
  int v4; // esi@1
  signed int i; // eax@1


  v3 = 3;
  v4 = 0;
  for ( i = 0; v3 < a3; ++i )
  {
    if ( i > 7 )
      i = 0;
    v4 += (unsigned __int8)byte_405030[i] * (unsigned __int8)lpString2[v3++];
  }
  wsprintfA(lpString2, "%ld", v4);
  return lstrcmpA(lpString1, lpString2) == 0;
}

byte_405030[]数组在.data区:

char byte_405030[8]={0x0C,0x0A,0x13,0x09,0x0C,0x0B,0x0A,0x08}

假如我输入username为:tanyo

v4=121*12+111*10=2562


输入usernametanyo,序列号2562进行验证:

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使用IDA破解TraceMe.exe

标签:ext   跳转   data-   net   打开   gravity   username   ==   log   

原文地址:http://www.cnblogs.com/cxchanpin/p/7112523.html

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