系统 : Windows xp
程序 : cztria~1
程序下载地址 :http://pan.baidu.com/s/1slUwmVr
要求 : 爆破
使用工具 : OD
可在看雪论坛中查找关于此程序的破文:传送门
废话不多说,直接查询到字符串:“ you did it!”,双击定位:
0040137B |. 6A 40 push 40 ; /Count = 40 (64.) 0040137D |. 68 20334000 push 00403320 ; |pediy 00401382 |. FF35 66324000 push dword ptr [403266] ; |hWnd = 000405D8 (class=‘Edit‘,parent=000505C0) 00401388 |. E8 A3080000 call <jmp.&USER32.GetWindowTextA> ; \GetWindowTextA 0040138D |. 83F8 04 cmp eax, 4 ; 小于等于4? 00401390 |. 0F8E 9F000000 jle 00401435 00401396 |. 6A 40 push 40 ; /Count = 40 (64.) 00401398 |. 68 60334000 push 00403360 ; |12345 0040139D |. 68 B90B0000 push 0BB9 ; |ControlID = BB9 (3001.) 004013A2 |. FF75 08 push dword ptr [ebp+8] ; |hWnd 004013A5 |. E8 6E080000 call <jmp.&USER32.GetDlgItemTextA> ; \GetDlgItemTextA 004013AA |. 83F8 04 cmp eax, 4 ; 小于等于4? 004013AD |. 0F8E 82000000 jle 00401435 004013B3 |. A3 62324000 mov dword ptr [403262], eax 004013B8 |. FF35 66324000 push dword ptr [403266] ; /hWnd = 000405D8 (class=‘Edit‘,parent=000505C0) 004013BE |. E8 AF080000 call <jmp.&USER32.SetFocus> ; \SetFocus 004013C3 |. BF 20334000 mov edi, 00403320 ; pediy 004013C8 |. BE 20334000 mov esi, 00403320 ; pediy 004013CD |> AC /lods byte ptr [esi] ; 循环迭代用户名字符串 004013CE |. 0C 00 |or al, 0 004013D0 |. 74 05 |je short 004013D7 004013D2 |. 0C 20 |or al, 20 004013D4 |. AA |stos byte ptr es:[edi] 004013D5 |.^ EB F6 \jmp short 004013CD 004013D7 |> BF A0324000 mov edi, 004032A0 004013DC |. BE 60334000 mov esi, 00403360 ; 12345 004013E1 |. 8D1D 20334000 lea ebx, dword ptr [403320] 004013E7 |. 33C9 xor ecx, ecx 004013E9 |> AC /lods byte ptr [esi] ; 循环迭代 密码 004013EA |. 0C 00 |or al, 0 004013EC |. 74 17 |je short 00401405 004013EE |. 8A13 |mov dl, byte ptr [ebx] ; 循环迭代 用户名 004013F0 |. 2AD0 |sub dl, al ; 用户名字符 - 密码字符 004013F2 |. 80CA 00 |or dl, 0 ; 如果相同,则跳转出错 004013F5 |. 74 3E |je short 00401435 004013F7 |. 8AC2 |mov al, dl 004013F9 |. 24 0F |and al, 0F 004013FB |. 0C 00 |or al, 0 ; al为0? 004013FD |. 74 36 |je short 00401435 ; 为0则跳转出错 004013FF |. AA |stos byte ptr es:[edi] ; 保存al成表 00401400 |. 02C8 |add cl, al ; 结果累加 00401402 |. 43 |inc ebx 00401403 |.^ EB E4 \jmp short 004013E9 00401405 |> 890D 6A324000 mov dword ptr [40326A], ecx ; 保存累加结果 0040140B |. E8 27020000 call 00401637 ; 关键call 00401410 |. BE A0324000 mov esi, 004032A0 00401415 |. 8B15 62324000 mov edx, dword ptr [403262] ; 取密码长度 0040141B |. C1EA 02 shr edx, 2 ; 逻辑右移 0040141E |. 03F2 add esi, edx 00401420 |. 8A06 mov al, byte ptr [esi] ; 表中取值 00401422 |. 33D2 xor edx, edx 00401424 |. 8B15 6E324000 mov edx, dword ptr [40326E] 0040142A |. 2BD0 sub edx, eax 0040142C |. A1 6A324000 mov eax, dword ptr [40326A] 00401431 |. 3BC2 cmp eax, edx 00401433 75 31 jz short 00401466 00401435 |> 68 00200000 push 2000 ; /Style = MB_OK|MB_TASKMODAL 0040143A |. 68 D1314000 push 004031D1 ; | error 0040143F |. 68 F9314000 push 004031F9 ; | sorry cracker, wrong. 00401444 |. FF75 08 push dword ptr [ebp+8] ; |hOwner 00401447 |. E8 02080000 call <jmp.&USER32.MessageBoxA> ; \MessageBoxA 0040144C |. 6A 40 push 40 ; /Length = 40 (64.) 0040144E |. 68 E0324000 push 004032E0 ; |Destination = cztria~1.004032E0 00401453 |. E8 56080000 call <jmp.&KERNEL32.RtlZeroMemory> ; \RtlZeroMemory 00401458 |. 6A 40 push 40 ; /Length = 40 (64.) 0040145A |. 68 A0334000 push 004033A0 ; |Destination = cztria~1.004033A0 0040145F |. E8 4A080000 call <jmp.&KERNEL32.RtlZeroMemory> ; \RtlZeroMemory 00401464 |. EB 2F jmp short 00401495 00401466 |> 68 00200000 push 2000 ; /Style = MB_OK|MB_TASKMODAL 0040146B |. 68 E5314000 push 004031E5 ; | <registered> 00401470 |. 68 10324000 push 00403210 ; | you did it! 00401475 |. FF75 08 push dword ptr [ebp+8] ; |hOwner 00401478 |. E8 D1070000 call <jmp.&USER32.MessageBoxA> ; \MessageBoxA 0040147D |. 6A 40 push 40 ; /Length = 40 (64.) 0040147F |. 68 E0324000 push 004032E0 ; |Destination = cztria~1.004032E0 00401484 |. E8 25080000 call <jmp.&KERNEL32.RtlZeroMemory> ; \RtlZeroMemory 00401489 |. 6A 40 push 40 ; /Length = 40 (64.) 0040148B |. 68 A0334000 push 004033A0 ; |Destination = cztria~1.004033A0 00401490 |. E8 19080000 call <jmp.&KERNEL32.RtlZeroMemory> ; \RtlZeroMemory
跟入 0040140B |. E8 27020000 call 00401637 ; 关键call
00401637 /$ BE A0324000 mov esi, 004032A0 0040163C |. 8B15 62324000 mov edx, dword ptr [403262] ; 取密码长度 00401642 |. 52 push edx 00401643 |. 33C0 xor eax, eax 00401645 |. 83EA 01 sub edx, 1 00401648 |. 03F2 add esi, edx 0040164A |. 8A06 mov al, byte ptr [esi] ; 表中取值 0040164C |. F7E0 mul eax 0040164E |. 5A pop edx 0040164F |. 83EA 01 sub edx, 1 00401652 |. F7E2 mul edx 00401654 |. B9 01000000 mov ecx, 1 00401659 |> 2BC1 /sub eax, ecx 0040165B |. 83F8 00 |cmp eax, 0 ; eax为0? 0040165E |. 7E 08 |jle short 00401668 00401660 |. 83C2 01 |add edx, 1 00401663 |. 83C1 02 |add ecx, 2 00401666 |.^ EB F1 \jmp short 00401659 00401668 |> 52 push edx ; 保存edx 00401669 |. BE A0324000 mov esi, 004032A0 0040166E |. 8BFE mov edi, esi 00401670 |. 8B15 62324000 mov edx, dword ptr [403262] ; 取密码长度 00401676 |. 33C0 xor eax, eax 00401678 |. 83EA 01 sub edx, 1 0040167B |. 03F2 add esi, edx 0040167D |. 8A06 mov al, byte ptr [esi] ; 表中取值 0040167F |. 83C0 01 add eax, 1 00401682 |. 5A pop edx 00401683 |. 03C2 add eax, edx 00401685 |. D1E8 shr eax, 1 00401687 |. 8B15 62324000 mov edx, dword ptr [403262] ; 取密码长度 0040168D |. 03FA add edi, edx 0040168F |. AA stos byte ptr es:[edi] 00401690 |. F7E0 mul eax 00401692 |. 8B15 62324000 mov edx, dword ptr [403262] ; 取密码长度 00401698 |. 83EA 01 sub edx, 1 0040169B |. F7E2 mul edx 0040169D |. B9 01000000 mov ecx, 1 004016A2 |> 2BC1 /sub eax, ecx 004016A4 |. 83F8 00 |cmp eax, 0 ; eax为0? 004016A7 |. 7E 08 |jle short 004016B1 004016A9 |. 83C2 01 |add edx, 1 004016AC |. 83C1 02 |add ecx, 2 004016AF |.^ EB F1 \jmp short 004016A2 004016B1 |> 52 push edx 004016B2 |. BE A0324000 mov esi, 004032A0 004016B7 |. 8B15 62324000 mov edx, dword ptr [403262] ; 取密码长度 004016BD |. 33C0 xor eax, eax 004016BF |. 03F2 add esi, edx 004016C1 |. 8A06 mov al, byte ptr [esi] ; 取表中末位 004016C3 |. 83C0 01 add eax, 1 004016C6 |. 5A pop edx 004016C7 |. 03C2 add eax, edx 004016C9 |. D1E8 shr eax, 1 004016CB |. A3 6E324000 mov dword ptr [40326E], eax ; 保存结果 004016D0 \. C3 retn
这是一个典型的二元函数加密,将用户名与密码的差值生成一个表 和 累加值。再根据表生成两个特殊值。
输入的结果差值要符合 特殊值1 - 特殊值2 == 累加结果
我们可以直接将判断的条件修改成:
00401433 /75 31 jnz short 00401466
就可以完成爆破了。
时间: 2024-11-16 19:54:31