刚才好像看不到文章的内容啊！呵呵
【文章标题】: 对happytown的“一个适合初学者的crackme ”的简单分析
【文章作者】: 水中花
【下载地址】: 自己搜索下载
【作者声明】: 只是感兴趣，没有其他目的。失误之处敬请诸位大侠赐教!
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【详细过程】
可以用字符串下断，断在此处
0040112C . 50 push eax
0040112D . 53 push ebx
0040112E . 55 push ebp
0040112F . 68 00020000 push 200 ; /Count = 200 (512.)
00401134 . 68 49634000 push crackme.00406349 ; |Buffer = crackme.00406349
00401139 . 68 EA030000 push 3EA ; |ControlID = 3EA (1002.)
0040113E . FF75 08 push dword ptr [ebp+8] ; |hWnd
00401141 . E8 4A020000 call <jmp.&user32.GetDlgItemTextA> ; \GetDlgItemTextA 获取用户名的长度
00401146 . 83F8 03 cmp eax, 3 必须大于3位
00401149 . 77 18 ja short crackme.00401163
0040114B . 6A 10 push 10 ; /Style = MB_OK|MB_ICONHAND|MB_APPLMODAL
0040114D . 68 06634000 push crackme.00406306 ; |bad boy...
00401152 . 68 0A624000 push crackme.0040620A ; |Text = "Username must have at least 4 chars..."
00401157 . FF75 08 push dword ptr [ebp+8] ; |hOwner
0040115A . E8 3D020000 call <jmp.&user32.MessageBoxA> ; \MessageBoxA
0040115F . C9 leave
00401160 . C2 1000 retn 10
00401163 > 8D15 49634000 lea edx, dword ptr [406349]
00401169 . 52 push edx ; /String => "j与zi>"
0040116A . E8 8D020000 call <jmp.&kernel32.lstrlenA> ; \lstrlenA 去除用户名的第一位字符，该串设了A
0040116F . 8BE8 mov ebp, eax
00401171 . B9 05000000 mov ecx, 5 ; 循环计数器
00401176 . 33F6 xor esi, esi
00401178 . 33C0 xor eax, eax
0040117A > 8A0C16 mov cl, byte ptr [esi+edx] ; 以下是将A的各个字符按顺序与AA,89,C4,FE,46（超过5位的，分别A串的前面字符）相异或，所得串设为B
0040117D . 8AD9 mov bl, cl
0040117F . 3298 28634000 xor bl, byte ptr [eax+406328] 相异或
00401185 . 40 inc eax
00401186 . 83F8 05 cmp eax, 5
00401189 . 881C32 mov byte ptr [edx+esi], bl 存入[edx+esi]中
0040118C . 8888 27634000 mov byte ptr [eax+406327], cl
00401192 . 75 02 jnz short crackme.00401196
00401194 . 33C0 xor eax, eax
00401196 > 46 inc esi
00401197 . 3BF5 cmp esi, ebp
00401199 .^ 72 DF jb short crackme.0040117A
0040119B . 33FF xor edi, edi
0040119D . 33C9 xor ecx, ecx
0040119F . 85ED test ebp, ebp
004011A1 . 76 26 jbe short crackme.004011C9
004011A3 > 8A9F 2D634000 mov bl, byte ptr [edi+40632D] ; 以下是将B串反取按顺序与78 F0 D0 03 E7（超过5位的与B串的反取字符）相异或，所得串设以C
004011A9 . 8BF5 mov esi, ebp
004011AB . 2BF1 sub esi, ecx
004011AD . 4E dec esi
004011AE . 8A0432 mov al, byte ptr [edx+esi]
004011B1 . 32D8 xor bl, al 异或
004011B3 . 47 inc edi
004011B4 . 881C32 mov byte ptr [edx+esi], bl
004011B7 . 8887 2C634000 mov byte ptr [edi+40632C], al
004011BD . 83FF 05 cmp edi, 5
004011C0 . 75 02 jnz short crackme.004011C4
004011C2 . 33FF xor edi, edi
004011C4 > 41 inc ecx
004011C5 . 3BCD cmp ecx, ebp
004011C7 .^ 72 DA jb short crackme.004011A3
004011C9 > 33F6 xor esi, esi
004011CB . 33FF xor edi, edi
004011CD . 85ED test ebp, ebp
004011CF . 76 21 jbe short crackme.004011F2
004011D1 > 8A043A mov al, byte ptr [edx+edi] 以下是将C串反取按顺序与F7 FD F4 E7 B9（超过5位的与C串的反取字符）相异或，所得串设以D
004011D4 . 8A8E 32634000 mov cl, byte ptr [esi+406332]
004011DA . 32C8 xor cl, al 异或
004011DC . 46 inc esi
004011DD . 880C3A mov byte ptr [edx+edi], cl
004011E0 . 8886 31634000 mov byte ptr [esi+406331], al
004011E6 . 83FE 05 cmp esi, 5
004011E9 . 75 02 jnz short crackme.004011ED
004011EB . 33F6 xor esi, esi
004011ED > 47 inc edi
004011EE . 3BFD cmp edi, ebp
004011F0 .^ 72 DF jb short crackme.004011D1
004011F2 > 33FF xor edi, edi
004011F4 . 33C9 xor ecx, ecx
004011F6 . 85ED test ebp, ebp
004011F8 . 76 26 jbe short crackme.00401220
004011FA > 8A9F 37634000 mov bl, byte ptr [edi+406337] 以下是将D串反取按顺序与B5 1B C9 50 73（超过5位的与D串的反取字符）相异或，所得串设为E
00401200 . 8BF5 mov esi, ebp
00401202 . 2BF1 sub esi, ecx
00401204 . 4E dec esi
00401205 . 8A0432 mov al, byte ptr [edx+esi]
00401208 . 32D8 xor bl, al 异或
0040120A . 47 inc edi
0040120B . 881C32 mov byte ptr [edx+esi], bl
0040120E . 8887 36634000 mov byte ptr [edi+406336], al
00401214 . 83FF 05 cmp edi, 5
00401217 . 75 02 jnz short crackme.0040121B
00401219 . 33FF xor edi, edi
0040121B > 41 inc ecx
0040121C . 3BCD cmp ecx, ebp
0040121E .^ 72 DA jb short crackme.004011FA
00401220 > 8D3D 45634000 lea edi, dword ptr [406345]
00401226 . 33C0 xor eax, eax
00401228 . 85ED test ebp, ebp
0040122A . C705 45634000>mov dword ptr [406345], 0
00401234 . 76 17 jbe short crackme.0040124D
00401236 > 8BC8 mov ecx, eax 以下是将E串反取按顺序与00，00，00，00，（超过5位的与E串的反取字符）相add,超过4位有E串的反取相add
00401238 . 83E1 03 and ecx, 3
0040123B . 8A1C0F mov bl, byte ptr [edi+ecx]
0040123E . 8D340F lea esi, dword ptr [edi+ecx]
00401241 . 8A0C02 mov cl, byte ptr [edx+eax]
00401244 . 02D9 add bl, cl 相add
00401246 . 40 inc eax
00401247 . 3BC5 cmp eax, ebp
00401249 . 881E mov byte ptr [esi], bl 存入[esi]中
0040124B .^ 72 E9 jb short crackme.00401236
0040124D > 5D pop ebp
0040124E . B9 0A000000 mov ecx, 0A ecx赋值为A，十进制10
00401253 . A1 45634000 mov eax, dword ptr [406345] 将[406345](为[esi])中的内容取出
00401258 . 33DB xor ebx, ebx
0040125A > 33D2 xor edx, edx
0040125C . F7F1 div ecx 除A
0040125E . 80C2 30 add dl, 30 低位与30相add
00401261 . 8893 49654000 mov byte ptr [ebx+406549], dl 存入[ebx+406549]中
00401267 . 43 inc ebx
00401268 . 85C0 test eax, eax
0040126A .^ 75 EE jnz short crackme.0040125A
0040126C . 68 49654000 push crackme.00406549 ; /String = "" 此为上面计算所得的值，设为F
00401271 . E8 86010000 call <jmp.&kernel32.lstrlenA> ; \lstrlenA
00401276 . 33DB xor ebx, ebx
00401278 > 8A88 48654000 mov cl, byte ptr [eax+406548] 将F串反取，即为注册码
0040127E . 888B 49674000 mov byte ptr [ebx+406749], cl
00401284 . 43 inc ebx
00401285 . 48 dec eax
00401286 .^ 75 F0 jnz short crackme.00401278
00401288 . 68 49674000 push crackme.00406749 ; /String2 = "" 此处出现真码
0040128D . 68 49654000 push crackme.00406549 ; |String1 = crackme.00406549
00401292 . E8 5F010000 call <jmp.&kernel32.lstrcpyA> ; \lstrcpyA
00401297 . 68 00020000 push 200 ; /Count = 200 (512.)
0040129C . 68 49694000 push crackme.00406949 ; |Buffer = crackme.00406949
004012A1 . 6A 64 push 64 ; |ControlID = 64 (100.)
004012A3 . FF75 08 push dword ptr [ebp+8] ; |hWnd
004012A6 . E8 E5000000 call <jmp.&user32.GetDlgItemTextA> ; \GetDlgItemTextA
004012AB . 68 49654000 push crackme.00406549 ; /String2 = ""
004012B0 . 68 49694000 push crackme.00406949 ; |String1 = "123456"
004012B5 . E8 36010000 call <jmp.&kernel32.lstrcmpA> ; \lstrcmpA 真假码比较
004012BA . 0BC0 or eax, eax
004012BC . 75 16 jnz short crackme.004012D4 跳向出错处
004012BE . 6A 40 push 40 ; /Style = MB_OK|MB_ICONASTERISK|MB_APPLMODAL
004012C0 . 68 DB624000 push crackme.004062DB ; |good boy...nope, thats not it!\n\ntry againbad boy...
004012C5 . 68 AC624000 push crackme.004062AC ; |yep, thats the right code!\n\ngo write a keygen!good boy...nope, thats not it!\n\ntry againbad boy...
004012CA . FF75 08 push dword ptr [ebp+8] ; |hOwner
004012CD . E8 CA000000 call <jmp.&user32.MessageBoxA> ; \MessageBoxA
004012D2 . EB 14 jmp short crackme.004012E8
004012D4 > 6A 10 push 10 ; /Style = MB_OK|MB_ICONHAND|MB_APPLMODAL
004012D6 . 68 06634000 push crackme.00406306 ; |bad boy...
004012DB . 68 E7624000 push crackme.004062E7 ; |nope, thats not it!\n\ntry againbad boy...
004012E0 . FF75 08 push dword ptr [ebp+8] ; |hOwner
004012E3 . E8 B4000000 call <jmp.&user32.MessageBoxA> ; \MessageBoxA
004012E8 > 68 00020000 push 200 ; /Length = 200 (512.)
004012ED . 68 49654000 push crackme.00406549 ; |Destination = crackme.00406549
004012F2 . E8 ED000000 call <jmp.&kernel32.RtlZeroMemory> ; \RtlZeroMemory
004012F7 . 68 00020000 push 200 ; /Length = 200 (512.)
004012FC . 68 49634000 push crackme.00406349 ; |Destination = crackme.00406349
00401301 . E8 DE000000 call <jmp.&kernel32.RtlZeroMemory> ; \RtlZeroMemory
00401306 . 68 00020000 push 200 ; /Length = 200 (512.)
0040130B . 68 49674000 push crackme.00406749 ; |Destination = crackme.00406749
00401310 . E8 CF000000 call <jmp.&kernel32.RtlZeroMemory> ; \RtlZeroMemory
00401315 . B9 16000000 mov ecx, 16
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【经验总结】
我是初手，也是第一次写这种破文，不好和不对之处，请大家多多指教！
算法总结：
1、去除用户名的第一位字符，设为A，将A按顺序分别与AA 89 C4 FE 46（超过5位的，分别取A串的前面字符）相异或，设
所得结果为B
2、将B串反取按顺序与78 F0 D0 03 E7（超过5位的与B串的反取字符）相异或，所得串设以C
3、将C串反取按顺序与F7 FD F4 E7 B9（超过5位的与C串的反取字符）相异或，所得串设以D
4、将D串反取按顺序与B5 1B C9 50 73（超过5位的与D串的反取字符）相异或，所得串设为E
5、将E串反取按顺序与00，00，00，00，（超过5位的与E串的反取字符）相add,超过4位有E串的反取相add ，所得设为F
6、将F div A 余数＋30，化为数字，所得值设为G
7、将G反取就为正确注册码
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【版权声明】: 本文原创于看雪技术论坛, 转载请注明作者并保持文章的完整, 谢谢!
2007年03月13日 22:12:48
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