【破文标题】Savage's Slayer's Crackme # 1算法分析
【破文作者】XXNB
【作者邮箱】支持PYG
【作者主页】http://free.ys168.com/?binbinbin7456
【破解工具】OD
【破解平台】xpsp2
【软件名称】Savage's Slayer's Crackme # 1
【软件大小】
【原版下载】
【保护方式】
【软件简介】Easy for first :)
Difficulty: 1 - Very easy, for newbies
Platform: Windows
Language: Assembler
Published: 25. Nov, 2006
Downloads: 7
【破解声明】向大侠们学习！！！只为学习！请尊重作者的劳动成功！
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【破解过程】
可以去 http://www.crackmes.de 这个网站下载
1、这个crackme非常有趣。我早上下的然后看啊看啊，就是不知道在哪里下断好。不过我在跟踪的过程发现，要用到计算机名，还要用到一个“reg.key”文件，于是我就建立了个“reg.key”文件。写入：123456789。运行，还是没能找到关键点，因为它根本就没有读取文件，怎么回事？？。然后我就复制我的用户名BINBIN到“reg.key”文件里。跟踪的时候，发现居然能读取BINBIN这个字符串，我当时奇怪得很。
最终，在我的试验下原来发现下面的00401292这句开始是对计算机名运算的，先是得到计算机名Ascii码累加值，然后一个循环减。最终结果要等于0，于是我知道当然是减用户名的Ascii码累加值才能等于0。原来，程序是读取你的剪切板的数据的。
所以，如果要使得register按钮有效，就必须复制自己的计算机名，然后按下check，就会出现“Step 1 ok -> now Register it!”了、。
注册按钮有效后就是读取文件了。首先，读取“reg.key”文件的字节数，一定要大于等于8位字节才有效。具体看下面代码注释。
0040108B |. 51 push ecx ; /pBufferSize => KeyMe1.0040333A
0040108C |. 68 3B324000 push 0040323B ; |BINBIN
00401091 |. E8 86020000 call <jmp.&KERNEL32.GetComputerNameA> ; \GetComputerNameA
00401096 |. 68 3B324000 push 0040323B ; /BINBIN
0040109B |. E8 A0020000 call <jmp.&KERNEL32.lstrlenA> ; \lstrlenA
004010A0 |. A3 46334000 mov dword ptr [403346], eax ; 计算机名长度
004010A5 |. 59 pop ecx
004010A6 |. 58 pop eax
004010A7 |. 60 pushad
004010A8 |. 9C pushfd
004010A9 |. BE 3B324000 mov esi, 0040323B ; 计算机名
004010AE |. C705 3E334000>mov dword ptr [40333E], 0
004010B8 |> AC /lods byte ptr [esi] ; 这里这个循环就是得到计算机名的累加值了
004010B9 |. 0FB6C0 |movzx eax, al
004010BC |. 0105 3E334000 |add dword ptr [40333E], eax
004010C2 |. 3C 00 |cmp al, 0
004010C4 |.^ 75 F2 \jnz short 004010B8
004010C6 |. 9D popfd
004010C7 |. 61 popad
004010C8 |. 33C0 xor eax, eax
004010CA |. C9 leave
004010CB |. C2 1000 retn 10
004010CE |> 837D 0C 10 cmp dword ptr [ebp+C], 10
004010D2 |. 75 0A jnz short 004010DE
004010D4 |. 6A 00 push 0 ; /Result = 0
004010D6 |. FF75 08 push dword ptr [ebp+8] ; |hWnd
004010D9 |. E8 7A020000 call <jmp.&USER32.EndDialog> ; \EndDialog
004010DE |> 817D 0C 11010>cmp dword ptr [ebp+C], 111
004010E5 |. 0F85 19020000 jnz 00401304
004010EB |. 8B45 10 mov eax, dword ptr [ebp+10]
004010EE |. 66:83F8 68 cmp ax, 68 ; (Initial CPU selection)
004010F2 |. 75 0A jnz short 004010FE
004010F4 |. 6A 00 push 0 ; /Result = 0
004010F6 |. FF75 08 push dword ptr [ebp+8] ; |hWnd
004010F9 |. E8 5A020000 call <jmp.&USER32.EndDialog> ; \EndDialog
004010FE |> 66:83F8 69 cmp ax, 69
00401102 |. 75 13 jnz short 00401117
00401104 |. 6A 40 push 40 ; /Style = MB_OK|MB_ICONASTERISK|MB_APPLMODAL
00401106 |. 68 00304000 push 00403000 ; |[ info ]
0040110B |. 68 09304000 push 00403009 ; |[ KeyMe # 1 coded by Slayer ]\r\n\r\nRules : No patching at all. >> Keygen it <<\r\n\r\nSolutions : [email]slayeronthefly@yahoo.com[/email]\r\n\r\nAll accepted solutions will be awarded by source code :)
00401110 |. 6A 00 push 0 ; |hOwner = NULL
00401112 |. E8 59020000 call <jmp.&USER32.MessageBoxA> ; \MessageBoxA
00401117 |> 66:83F8 6A cmp ax, 6A
0040111B |. 0F85 41010000 jnz 00401262
00401121 |. 6A 00 push 0 ; /hTemplateFile = NULL
00401123 |. 68 80000000 push 80 ; |Attributes = NORMAL
00401128 |. 6A 03 push 3 ; |Mode = OPEN_EXISTING
0040112A |. 6A 00 push 0 ; |pSecurity = NULL
0040112C |. 6A 01 push 1 ; |ShareMode = FILE_SHARE_READ
0040112E |. 68 00000080 push 80000000 ; |Access = GENERIC_READ
00401133 |. 68 26314000 push 00403126 ; |reg.key
00401138 |. E8 D3010000 call <jmp.&KERNEL32.CreateFileA> ; \CreateFileA
0040113D |. 83F8 FF cmp eax, -1
00401140 |. 75 05 jnz short 00401147
00401142 |. E9 E7000000 jmp 0040122E
00401147 |> A3 4A334000 mov dword ptr [40334A], eax
0040114C |. 6A 00 push 0 ; /pFileSizeHigh = NULL
0040114E |. FF35 4A334000 push dword ptr [40334A] ; |hFile = 00000088 (window)
00401154 |. E8 C9010000 call <jmp.&KERNEL32.GetFileSize> ; \GetFileSize
00401159 |. 83F8 08 cmp eax, 8 ; 文件的字节一定不能小于8
0040115C |. 73 10 jnb short 0040116E
0040115E |. FF35 4A334000 push dword ptr [40334A] ; /hObject = 00000088 (window)
00401164 |. E8 A1010000 call <jmp.&KERNEL32.CloseHandle> ; \CloseHandle
00401169 |. E9 C0000000 jmp 0040122E
0040116E |> A3 52334000 mov dword ptr [403352], eax ; 存储计算机名的位数
00401173 |. FF35 52334000 push dword ptr [403352] ; /MemSize = 8
00401179 |. 6A 40 push 40 ; |Flags = GPTR
0040117B |. E8 AE010000 call <jmp.&KERNEL32.GlobalAlloc> ; \GlobalAlloc
00401180 |. 0BC0 or eax, eax
00401182 |. 75 1F jnz short 004011A3
00401184 |. 68 0D314000 push 0040310D ; /Memory allocation error!
00401189 |. 6A 66 push 66 ; |ControlID = 66 (102.)
0040118B |. FF75 08 push dword ptr [ebp+8] ; |hWnd
0040118E |. E8 EF010000 call <jmp.&USER32.SetDlgItemTextA> ; \SetDlgItemTextA
00401193 |. FF35 4A334000 push dword ptr [40334A] ; /hObject = 00000088 (window)
00401199 |. E8 6C010000 call <jmp.&KERNEL32.CloseHandle> ; \CloseHandle
0040119E |. E9 B4000000 jmp 00401257
004011A3 |> A3 4E334000 mov dword ptr [40334E], eax
004011A8 |. 6A 00 push 0 ; /pOverlapped = NULL
004011AA |. 68 2E314000 push 0040312E ; |A
004011AF |. FF35 52334000 push dword ptr [403352] ; |BytesToRead = 8
004011B5 |. FF35 4E334000 push dword ptr [40334E] ; |Buffer = 00149A20
004011BB |. FF35 4A334000 push dword ptr [40334A] ; |hFile = 00000088 (window)
004011C1 |. E8 74010000 call <jmp.&KERNEL32.ReadFile> ; \ReadFile
004011C6 |. 0BC0 or eax, eax
004011C8 |. 75 18 jnz short 004011E2
004011CA |. FF35 4A334000 push dword ptr [40334A] ; /hObject = 00000088 (window)
004011D0 |. E8 35010000 call <jmp.&KERNEL32.CloseHandle> ; \CloseHandle
004011D5 |. FF35 4E334000 push dword ptr [40334E] ; /hMem = 00149A20
004011DB |. E8 54010000 call <jmp.&KERNEL32.GlobalFree> ; \GlobalFree
004011E0 |. EB 4C jmp short 0040122E
004011E2 |> FF35 4A334000 push dword ptr [40334A] ; /hObject = 00000088 (window)
004011E8 |. E8 1D010000 call <jmp.&KERNEL32.CloseHandle> ; \CloseHandle
004011ED |. 56 push esi
004011EE |. 52 push edx
004011EF |. 8B35 4E334000 mov esi, dword ptr [40334E]
004011F5 |. 8B06 mov eax, dword ptr [esi] ; 第一个值到eax。这个值是key文件中前4位倒过来的ascii码值
004011F7 |. 83C6 04 add esi, 4
004011FA |. 8B16 mov edx, dword ptr [esi] ; 第二个值到edx。这个当然就是5～8位倒过来的ascii码值了
004011FC |. 33C2 xor eax, edx ; 前面两个值异或。结果在eax
004011FE |. 8B15 42334000 mov edx, dword ptr [403342] ; 这里好像总是0的
00401204 |. 03C2 add eax, edx ; 加0相当于没有加
00401206 |. 5A pop edx
00401207 |. 5E pop esi
00401208 |. 3B05 3E334000 cmp eax, dword ptr [40333E] ; 计算机名的ascii累加值和上面运算的结果比较。相等就成功
0040120E 75 1E jnz short 0040122E ; 这里就是关键跳了。
00401210 |. 68 D9304000 push 004030D9 ; /Good work. You have done it!
00401215 |. 6A 66 push 66 ; |ControlID = 66 (102.)
00401217 |. FF75 08 push dword ptr [ebp+8] ; |hWnd
0040121A |. E8 63010000 call <jmp.&USER32.SetDlgItemTextA> ; \SetDlgItemTextA
0040121F |. 6A 00 push 0 ; /Enable = FALSE
00401221 |. FF35 5A334000 push dword ptr [40335A] ; |hWnd = 002A03C2 ('-> Register <-',class='Button',parent=00280320)
00401227 |. E8 26010000 call <jmp.&USER32.EnableWindow> ; \EnableWindow
0040122C |. EB 29 jmp short 00401257
0040122E |> 68 F6304000 push 004030F6 ; /Registration failed!!!
00401233 |. 6A 66 push 66 ; |ControlID = 66 (102.)
00401235 |. FF75 08 push dword ptr [ebp+8] ; |hWnd
00401238 |. E8 45010000 call <jmp.&USER32.SetDlgItemTextA> ; \SetDlgItemTextA
0040123D |. 6A 00 push 0 ; /Enable = FALSE
0040123F |. FF35 5A334000 push dword ptr [40335A] ; |hWnd = 002A03C2 ('-> Register <-',class='Button',parent=00280320)
00401245 |. E8 08010000 call <jmp.&USER32.EnableWindow> ; \EnableWindow
0040124A |. 6A 01 push 1 ; /Enable = TRUE
0040124C |. FF35 56334000 push dword ptr [403356] ; |hWnd = 001B0330 ('Check',class='Button',parent=00280320)
00401252 |. E8 FB000000 call <jmp.&USER32.EnableWindow> ; \EnableWindow
00401257 |> FF35 4E334000 push dword ptr [40334E] ; /hMem = 00149A20
0040125D |. E8 D2000000 call <jmp.&KERNEL32.GlobalFree> ; \GlobalFree
00401262 |> 66:83F8 67 cmp ax, 67
00401266 |. 0F85 98000000 jnz 00401304
0040126C |. 6A 00 push 0 ; /hWnd = NULL
0040126E |. E8 03010000 call <jmp.&USER32.OpenClipboard> ; \OpenClipboard
00401273 |. 6A 01 push 1 ; /Format = CF_TEXT
00401275 |. E8 E4000000 call <jmp.&USER32.GetClipboardData> ; \GetClipboardData
0040127A |. 0BC0 or eax, eax
0040127C |. 74 72 je short 004012F0
0040127E |. A3 33324000 mov dword ptr [403233], eax
00401283 |. 60 pushad
00401284 |. 9C pushfd
00401285 |. BE 33324000 mov esi, 00403233
0040128A |. 8B36 mov esi, dword ptr [esi]
0040128C |. FF35 3E334000 push dword ptr [40333E]
00401292 |. 8F05 42334000 pop dword ptr [403342] ; 计算机名累加值出栈
00401298 |. 8B0D 46334000 mov ecx, dword ptr [403346] ; 用户名位数
0040129E |> 8A5431 FF /mov dl, byte ptr [ecx+esi-1] ; 从固定字符串的第六位开始倒着取
004012A2 |. 0FB6D2 |movzx edx, dl
004012A5 |. 2915 42334000 |sub dword ptr [403342], edx ; 计算机名累加值－取得的ascii码值
004012AB |. 49 |dec ecx
004012AC |.^ 75 F0 \jnz short 0040129E
004012AE |. 9D popfd
004012AF |. 61 popad
004012B0 |. A1 42334000 mov eax, dword ptr [403342] ; 这里就是上面循环减后的值了
004012B5 |. 6BC0 FF imul eax, eax, -1 ; eax×(-1)放到eax
004012B8 |. 83C0 01 add eax, 1 ; eax+1
004012BB |. 6BC0 FF imul eax, eax, -1 ; eax×(-1)放到eax。又乘以-1干嘛
004012BE |. 83C0 01 add eax, 1 ; 又+1。相当于没事在这里乱弄
004012C1 |. 0BC0 or eax, eax ; 跟到这里，发现计算机名的累加值减甚么东西能得到0能。当然就是减自己了
004012C3 |. 75 2B jnz short 004012F0 ; 所以，一定要复制计算机名到剪切板才能成功。
004012C5 |. 68 BB304000 push 004030BB ; /Step 1 ok -> now Register it!
004012CA |. 6A 66 push 66 ; |ControlID = 66 (102.)
004012CC |. FF75 08 push dword ptr [ebp+8] ; |hWnd
004012CF |. E8 AE000000 call <jmp.&USER32.SetDlgItemTextA> ; \SetDlgItemTextA
004012D4 |. 6A 00 push 0 ; /Enable = FALSE
004012D6 |. FF35 56334000 push dword ptr [403356] ; |hWnd = 001B0330 ('Check',class='Button',parent=00280320)
004012DC |. E8 71000000 call <jmp.&USER32.EnableWindow> ; \EnableWindow
004012E1 |. 6A 01 push 1 ; /Enable = TRUE
004012E3 |. FF35 5A334000 push dword ptr [40335A] ; |hWnd = 002A03C2 ('-> Register <-',class='Button',parent=00280320)
004012E9 |. E8 64000000 call <jmp.&USER32.EnableWindow> ; \EnableWindow
004012EE |. EB 0F jmp short 004012FF
004012F0 |> 68 F6304000 push 004030F6 ; /Registration failed!!!
004012F5 |. 6A 66 push 66 ; |ControlID = 66 (102.)
004012F7 |. FF75 08 push dword ptr [ebp+8] ; |hWnd
004012FA |. E8 83000000 call <jmp.&USER32.SetDlgItemTextA> ; \SetDlgItemTextA
004012FF |> E8 42000000 call <jmp.&USER32.CloseClipboard> ; [CloseClipboard
00401304 |> 33C0 xor eax, eax
00401306 |. C9 leave
00401307 \. C2 1000 retn 10
1、先假设我的计算机名是：BINBIN。
我在“reg.key”文件里输入的是：12345678
读取“reg.key”文件的算法是：读取前4位，倒过来取ascii码值的16进制字符串，得到：34333231；
再读取后5～8位，倒过来取ascii码值的16进制字符串，得到：38373635
两个数异或，得到0C040404。这个数和计算机名的ascii码累加值比较，如果相等就注册成功。“Good work. You have done it!”
2、所以，相对于我的计算机名：BINBIN。 Ascii码累加值是01B2。甚么东西和甚么东西异或能得到01B2呢？？？
由于我的计算机名比较特殊经过我的测试，在“reg.key”文件里输入“2210?10（16进制ascii码是：3232313080333130）”就能注册成功。
[注意]传递专业知识、拓宽行业人脉——看雪讲师团队等你加入！
