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。