PC-Washer 2.0最新版算法分析及Keygen源码
标 题: 【原创】PC-Washer 2.0最新版 算法分析作 者: **sen
时 间: 2008-07-26,12:40
链 接:https://www.chinapyg.com/redirect.php?tid=36091&goto=lastpost#lastpost
【破文标题】PC-Washer 2.0最新版 算法分析
【破文作者】**sen
【作者邮箱】**[email protected]
【作者主页】**sen2.photo.163.com
【破解工具】PEiD,OD
【破解平台】Windows XP
【软件名称】PC Washer - ACleaner 2.0 build 07-24-08
【软件大小】1.01 MB
【软件类别】国外软件/系统辅助
【软件授权】共享版
【软件语言】英文
【更新时间】2008-7-24
【原版下载】http://qymm.onlinedown.net/down/pcwashersetup.zip
【保护方式】注册码
【软件简介】PC Washer是一款系统辅助软件，可以清理广告软体、注册表、快捷方式、程序垃圾等等，可以使你的系统更快速更健康！
【破解声明】我是一只小菜鸟，偶得一点心得，愿与大家分享：）
初学破解与编程，只是感兴趣，没有其它目的。失误之处敬请诸位大侠赐教！
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【破解内容】
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每天都要找一些软件来练手，从上次发表文章到现在已经一个月了，到时间遛遛了！这破解就和做数学题一样，几天不摸就手生，应该勤加练习啊o(∩_∩)o...
试运行程序发现是单一注册码保护，有错误提示“Invalid Regisered...”，PEID查壳得知是VC++的程序，无壳省了不少劲。经过两遍初步跟踪，我不断调整假码，最终我们选择输入这个假码“**sen8419876-54321-25814"（具体原因下面会详细分析）。
OD加载查找字符串“invalid”找到下面的关键点：0041AA40/$6A FF push -1
0041AA42|.68 182C4300 push 00432C18 ;SE 处理程序安装
0041AA47|.64:A1 0000000>mov eax, dword ptr fs:
0041AA4D|.50 push eax
0041AA4E|.64:8925 00000>mov dword ptr fs:, esp
0041AA55|.83EC 3C sub esp, 3C
0041AA58|.53 push ebx
0041AA59|.55 push ebp
0041AA5A|.8D4424 54 lea eax, dword ptr
0041AA5E|.8D4C24 08 lea ecx, dword ptr
0041AA62|.50 push eax
0041AA63|.C74424 50 000>mov dword ptr , 0
0041AA6B|.E8 A2480100 call <jmp.&MFC42.#535_CString::CString>
0041AA70|.8D4C24 0C lea ecx, dword ptr
0041AA74|.C64424 4C 01mov byte ptr , 1
0041AA79|.E8 1A470100 call <jmp.&MFC42.#540_CString::CString>
0041AA7E|.8B4C24 08 mov ecx, dword ptr ;取假码放入ECX中
0041AA82|.C64424 4C 02mov byte ptr , 2
0041AA87|.8379 F8 19 cmp dword ptr , 19 ;注册码大于等于25位，否则跳向错误处
0041AA8B|.0F8C C4020000 jl 0041AD55
0041AA91|.8D5424 1C lea edx, dword ptr
0041AA95|.6A 0C push 0C
0041AA97|.52 push edx
0041AA98|.8D4C24 10 lea ecx, dword ptr ;假码内存地址放入ECX
0041AA9C|.E8 2B490100 call <jmp.&MFC42.#4129_CString::Left>
0041AAA1|.50 push eax
0041AAA2|.8D4C24 10 lea ecx, dword ptr
0041AAA6|.C64424 50 03mov byte ptr , 3
0041AAAB|.E8 DC460100 call <jmp.&MFC42.#858_CString::operator=>
0041AAB0|.8D4C24 1C lea ecx, dword ptr
0041AAB4|.C64424 4C 02mov byte ptr , 2
0041AAB9|.E8 C8460100 call <jmp.&MFC42.#800_CString::~CString>
0041AABE|.8D4424 1C lea eax, dword ptr
0041AAC2|.6A 0C push 0C
0041AAC4|.50 push eax
0041AAC5|.8D4C24 10 lea ecx, dword ptr
0041AAC9|.E8 0A490100 call <jmp.&MFC42.#4277_CString::Mid>
0041AACE|.50 push eax
0041AACF|.8D4C24 0C lea ecx, dword ptr
0041AAD3|.C64424 50 04mov byte ptr , 4
0041AAD8|.E8 AF460100 call <jmp.&MFC42.#858_CString::operator=>
0041AADD|.8D4C24 1C lea ecx, dword ptr
0041AAE1|.C64424 4C 02mov byte ptr , 2
0041AAE6|.E8 9B460100 call <jmp.&MFC42.#800_CString::~CString>
0041AAEB|.8B4C24 08 mov ecx, dword ptr
0041AAEF|.8379 F8 11 cmp dword ptr , 11 ;上面一堆操作是去除假码的前12位；剩余的假码设为K2，其实K2还要大于等于17位
0041AAF3|.0F8C 5C020000 jl 0041AD55
0041AAF9|.6A 05 push 5 ;参数“5”
0041AAFB|.6A 2D push 2D ;参数“-”
0041AAFD|.8D4C24 10 lea ecx, dword ptr
0041AB01|.E8 544B0100 call <jmp.&MFC42.#6662_CString::Find> ;此处在验证假码K2；进入可知K2中每隔5位必须有个连字符“-”
0041AB06|.83CD FF or ebp, FFFFFFFF
0041AB09|.3BC5 cmp eax, ebp
0041AB0B|.0F84 6A020000 je 0041AD7B从0041AB01的call进入至此：73D46911 >8BFF mov edi, edi
73D46913 56 push esi
73D46914 8BF1 mov esi, ecx
73D46916 8B06 mov eax, dword ptr
73D46918 8B4C24 0C mov ecx, dword ptr
73D4691C 3B48 F8 cmp ecx, dword ptr
73D4691F 7D 15 jge short 73D46936
73D46921 03C1 add eax, ecx
73D46923 0FB64C24 08 movzx ecx, byte ptr
73D46928 51 push ecx
73D46929 50 push eax
73D4692A FF15 3467DD73 call dword ptr [<&msvcrt._mbsc>; 此处看右下角堆栈处已经显示了相关参数，要进入看看
73D46930 85C0 test eax, eax ; 若ECX的第一位是“-”，则此处EAX的值就是“-54321-25814”
73D46932 59 pop ecx
73D46933 59 pop ecx
73D46934 75 05 jnz short 73D4693B
73D46936 83C8 FF or eax, FFFFFFFF
73D46939 EB 02 jmp short 73D4693D
73D4693B 2B06 sub eax, dword ptr ; ESI的指针是指向假码K2的
73D4693D 5E pop esi
73D4693E C2 0800 retn 8返回到0041AB06。
从73D4692A处的call进来至此：77C017FF >8BFF mov edi, edi
77C01801 55 push ebp
77C01802 8BEC mov ebp, esp
77C01804 E8 1C870000 call 77C09F25
77C01809 8B40 60 mov eax, dword ptr
77C0180C 3B05 4425C377 cmp eax, dword ptr
77C01812 74 05 je short 77C01819
77C01814 E8 14EEFFFF call 77C0062D
77C01819 8378 08 00 cmp dword ptr , 0
77C0181D 75 06 jnz short 77C01825
77C0181F 5D pop ebp
77C01820 E9 3B5E0100 jmp strchr
77C01825 8B4D 08 mov ecx, dword ptr ; 把K2的后半部分"-54321-25814"放入ECX
77C01828 53 push ebx
77C01829 56 push esi
77C0182A EB 2F jmp short 77C0185B
77C0182C 0FB6F2 movzx esi, dl
77C0182F F64406 1D 04 test byte ptr , 4
77C01834 74 1C je short 77C01852
77C01836 41 inc ecx
77C01837 8A19 mov bl, byte ptr
77C01839 84DB test bl, bl
77C0183B 74 3B je short 77C01878
77C0183D 0FB7D2 movzx edx, dx
77C01840 0FB6F3 movzx esi, bl
77C01843 C1E2 08 shl edx, 8
77C01846 0BD6 or edx, esi
77C01848 3955 0C cmp dword ptr , edx
77C0184B 75 0D jnz short 77C0185A
77C0184D 8D41 FF lea eax, dword ptr
77C01850 EB 22 jmp short 77C01874
77C01852 0FB7F2 movzx esi, dx
77C01855 3975 0C cmp dword ptr , esi ; 比较ECX的第一位是不是“-”
77C01858 74 0A je short 77C01864
77C0185A 41 inc ecx
77C0185B 66:0FB611 movzx dx, byte ptr ; 取ECX的第一位
77C0185F 66:85D2 test dx, dx
77C01862^ 75 C8 jnz short 77C0182C
77C01864 8B45 0C mov eax, dword ptr
77C01867 0FB7D2 movzx edx, dx
77C0186A 2BC2 sub eax, edx ; 如果ECX第一位是“-”，则此处EAX位0
77C0186C F7D8 neg eax ; NEG ( NEGate ) 求补指令
77C0186E 1BC0 sbb eax, eax ; SBB ( SuBtract with Borrow) 带借位减法指令
77C01870 F7D0 not eax ; EAX取反
77C01872 23C1 and eax, ecx
77C01874 5E pop esi
77C01875 5B pop ebx
77C01876 5D pop ebp
77C01877 C3 retn返回到上面的73D46930处。0041AB11|.6A 0B push 0B
0041AB13|.6A 2D push 2D
0041AB15|.8D4C24 10 lea ecx, dword ptr
0041AB19|.E8 3C4B0100 call <jmp.&MFC42.#6662_CString>;同上，比较K2的第十二位是不是“-”
0041AB1E|.3BC5 cmp eax, ebp
0041AB20|.0F84 55020000 je 0041AD7B
0041AB26|.33DB xor ebx, ebx
0041AB28|>8AD3 /mov dl, bl ;这一段在比较K2有没有哪一位包含小写的26个字母，若包含则错
0041AB2A|.6A 01 |push 1
0041AB2C|.80C2 61 |add dl, 61
0041AB2F|.8D4C24 0C |lea ecx, dword ptr
0041AB33|.885424 20 |mov byte ptr , dl
0041AB37|.8B4424 20 |mov eax, dword ptr
0041AB3B|.50 |push eax
0041AB3C|.E8 194B0100 |call <jmp.&MFC42.#6662_CStrin>
0041AB41|.3BC5 |cmp eax, ebp
0041AB43|.0F85 32020000 |jnz 0041AD7B
0041AB49|.43 |inc ebx
0041AB4A|.83FB 1A |cmp ebx, 1A
0041AB4D|.^ 7C D9 \jl short 0041AB28
0041AB4F|.33DB xor ebx, ebx
0041AB51|>8ACB /mov cl, bl ;这一段在比较K2有没有哪一位包含大写的26个字母，若包含则错
0041AB53|.6A 01 |push 1
0041AB55|.80C1 41 |add cl, 41
0041AB58|.884C24 20 |mov byte ptr , cl
0041AB5C|.8B5424 20 |mov edx, dword ptr
0041AB60|.52 |push edx
0041AB61|.8D4C24 10 |lea ecx, dword ptr
0041AB65|.E8 F04A0100 |call <jmp.&MFC42.#6662_CStrin>
0041AB6A|.3BC5 |cmp eax, ebp
0041AB6C|.0F85 09020000 |jnz 0041AD7B
0041AB72|.43 |inc ebx
0041AB73|.83FB 1A |cmp ebx, 1A
0041AB76|.^ 7C D9 \jl short 0041AB51
0041AB78|.56 push esi ;也就是说K2必须全部是数字
0041AB79|.57 push edi
0041AB7A|.8D4C24 20 lea ecx, dword ptr
0041AB7E|.E8 15460100 call <jmp.&MFC42.#540_CString:>
0041AB83|.8D4C24 1C lea ecx, dword ptr
0041AB87|.C64424 54 05mov byte ptr , 5
0041AB8C|.E8 07460100 call <jmp.&MFC42.#540_CString:>
0041AB91|.8D4C24 18 lea ecx, dword ptr
0041AB95|.C64424 54 06mov byte ptr , 6
0041AB9A|.E8 F9450100 call <jmp.&MFC42.#540_CString:>
0041AB9F|.8D4424 24 lea eax, dword ptr
0041ABA3|.B3 07 mov bl, 7
0041ABA5|.6A 05 push 5
0041ABA7|.50 push eax
0041ABA8|.8D4C24 18 lea ecx, dword ptr
0041ABAC|.885C24 5C mov byte ptr , bl
0041ABB0|.E8 17480100 call <jmp.&MFC42.#4129_CString>
0041ABB5|.50 push eax
0041ABB6|.8D4C24 24 lea ecx, dword ptr
0041ABBA|.C64424 58 08mov byte ptr , 8
0041ABBF|.E8 C8450100 call <jmp.&MFC42.#858_CString:>
0041ABC4|.8D4C24 24 lea ecx, dword ptr
0041ABC8|.885C24 54 mov byte ptr , bl
0041ABCC|.E8 B5450100 call <jmp.&MFC42.#800_CString:>
0041ABD1|.8D4C24 24 lea ecx, dword ptr
0041ABD5|.6A 05 push 5
0041ABD7|.51 push ecx
0041ABD8|.8D4C24 18 lea ecx, dword ptr
0041ABDC|.E8 2B4A0100 call <jmp.&MFC42.#5710_CString>
0041ABE1|.50 push eax
0041ABE2|.8D4C24 1C lea ecx, dword ptr
0041ABE6|.C64424 58 09mov byte ptr , 9
0041ABEB|.E8 9C450100 call <jmp.&MFC42.#858_CString:>
0041ABF0|.8D4C24 24 lea ecx, dword ptr
0041ABF4|.885C24 54 mov byte ptr , bl
0041ABF8|.E8 89450100 call <jmp.&MFC42.#800_CString:>
0041ABFD|.6A 05 push 5
0041ABFF|.8D5424 28 lea edx, dword ptr
0041AC03|.6A 06 push 6
0041AC05|.52 push edx
0041AC06|.8D4C24 1C lea ecx, dword ptr
0041AC0A|.E8 C1490100 call <jmp.&MFC42.#4278_CString>
0041AC0F|.50 push eax
0041AC10|.8D4C24 20 lea ecx, dword ptr
0041AC14|.C64424 58 0Amov byte ptr , 0A
0041AC19|.E8 6E450100 call <jmp.&MFC42.#858_CString:>
0041AC1E|.8D4C24 24 lea ecx, dword ptr
0041AC22|.885C24 54 mov byte ptr , bl
0041AC26|.E8 5B450100 call <jmp.&MFC42.#800_CString:>
0041AC2B|.8B7C24 20 mov edi, dword ptr ;把K2中的三组数字分别设为N1、N2、N3，则此处取N1进行运算
0041AC2F|.8BCD mov ecx, ebp
0041AC31|.33C0 xor eax, eax
0041AC33|.8D5424 28 lea edx, dword ptr
0041AC37|.F2:AE repne scas byte ptr es: ;相当于计算N1长度，不明白请参看http://bbs.pediy.com/showthread.php?p=404309
0041AC39|.F7D1 not ecx ;经过此次NOT运算其实ECX=6
0041AC3B|.2BF9 sub edi, ecx
0041AC3D|.8BC1 mov eax, ecx ;把ECX的值放入EAX保存
0041AC3F|.8BF7 mov esi, edi ;把N1放入ESI
0041AC41|.8BFA mov edi, edx
0041AC43|.8D5424 34 lea edx, dword ptr
0041AC47|.C1E9 02 shr ecx, 2 ;ECX右移2，结果放入ECX
0041AC4A|.F3:A5 rep movs dword ptr es:, >;同上
0041AC4C|.8BC8 mov ecx, eax
0041AC4E|.33C0 xor eax, eax
0041AC50|.83E1 03 and ecx, 3 ;ECX和3进行AND运算
0041AC53|.F3:A4 rep movs byte ptr es:, b>
0041AC55|.8B7C24 1C mov edi, dword ptr ;把N2放入EDI进行运算
0041AC59|.8BCD mov ecx, ebp
0041AC5B|.F2:AE repne scas byte ptr es:
0041AC5D|.F7D1 not ecx
0041AC5F|.2BF9 sub edi, ecx
0041AC61|.8BC1 mov eax, ecx ;把ECX的值放入EAX保存
0041AC63|.8BF7 mov esi, edi ;N2放入ESI
0041AC65|.8BFA mov edi, edx
0041AC67|.C1E9 02 shr ecx, 2
0041AC6A|.F3:A5 rep movs dword ptr es:, >;同上
0041AC6C|.8BC8 mov ecx, eax
0041AC6E|.33C0 xor eax, eax
0041AC70|.83E1 03 and ecx, 3
0041AC73|.8D5424 40 lea edx, dword ptr
0041AC77|.F3:A4 rep movs byte ptr es:, b>
0041AC79|.8B7C24 18 mov edi, dword ptr ;N3放入EDI
0041AC7D|.8BCD mov ecx, ebp
0041AC7F|.F2:AE repne scas byte ptr es: ;同上
0041AC81|.F7D1 not ecx
0041AC83|.2BF9 sub edi, ecx
0041AC85|.8BC1 mov eax, ecx
0041AC87|.8BF7 mov esi, edi ;把N3放入ESI保存
0041AC89|.8BFA mov edi, edx
0041AC8B|.C1E9 02 shr ecx, 2
0041AC8E|.F3:A5 rep movs dword ptr es:, >
0041AC90|.8BC8 mov ecx, eax
0041AC92|.83E1 03 and ecx, 3
0041AC95|.F3:A4 rep movs byte ptr es:, b>
0041AC97|.8B3D 10574300 mov edi, dword ptr [<&MSVCRT.>;C函数atoi(char *str) 将字符串str转换成整数并返回这个整数到EAX中
0041AC9D|.8D4C24 28 lea ecx, dword ptr ;N1的内存地址放入ECX
0041ACA1|.51 push ecx ; /s
0041ACA2|.FFD7 call edi ; \函数atoi
0041ACA4|.8D5424 38 lea edx, dword ptr ;此处看EAX的值为4DA4h，十进制正好是19876（我们假码中N1的值）
0041ACA8|.8BE8 mov ebp, eax ;这句指令是把N1的十六进制值储存到EBP，下面关键点要用到
0041ACAA|.52 push edx
0041ACAB|.FFD7 call edi
0041ACAD|.8BF0 mov esi, eax ;此处看EAX的值为D431h，十进制正好是54321（我们假码中N2的值）；这句指令是把N2的十六进制值储存到ESI，下面关键点要用到
0041ACAF|.8D4424 48 lea eax, dword ptr
0041ACB3|.50 push eax
0041ACB4|.FFD7 call edi ;此处看EAX的值为64D6h，十进制正好是25814（我们假码中N3的值）
0041ACB6|.33F5 xor esi, ebp ;XOR N2,N1
0041ACB8|.83C4 0C add esp, 0C
0041ACBB|.81F6 B0410000 xor esi, 41B0 ;ESI与41B0h（十进制就是16816）异或，值放入ESI
0041ACC1|.C64424 54 06mov byte ptr , 6
0041ACC6|.5F pop edi
0041ACC7|.3BF0 cmp esi, eax ;还记得吗？EAX此时的值是假码中N3的十六进制哦！
0041ACC9|.5E pop esi
0041ACCA|.8D4C24 10 lea ecx, dword ptr
0041ACCE|.75 64 jnz short 0041AD34 ;此处如果没有跳，则注册成功了！！！
0041ACD0|.E8 B1440100 call <jmp.&MFC42.#800_CString:>近来正好在自学Delphi编程，就顺手自己写个注册机，关键源码公开如下：procedure TForm1.btn1Click(Sender: TObject);
var
b,c,Temp:integer;
begin
randomize; //初始化随机数
repeat
b:=10000+Random(90000);//算出一个随机5位数，即N1
c:=10000+Random(90000);//算出一个N2
Temp:=16816 xor (b xor c);//计算出N3
until Length(IntToStr(Temp))=5;//经测试发现N3必须是5位，否则会提示注册错误
Form1.Edit1.Text:='Senhuanpediy'+ inttostr(b) + '-' + inttostr(c) + '-' + inttostr(Temp);
end;
end.
[ 本帖最后由 playboysen 于 2008-9-4 20:05 编辑 ] 版主，不好意思，我又来了/:017 喜欢算法分析的文章，学习 坛主好像有点“偷懒”了哦/:018 好久没看到了 8月1日，更新注册机源代码，主要是界面更新，截图如下
加入xm音乐播放，水花效果，半透明效果等
Delphi 7编译通过 支持楼主
俺们菜鸟有福了 汗，这篇文章难道就写那么X吗？发出来了这么多天，只有我们坛子里的弟兄们好像有点“不屑一顾”，坛主也没来看一眼，写了一上午，没有功劳也有苦劳吧/:L 写的挺好的，向兄弟学习算法。。 原帖由 qifeon 于 2008-8-3 20:04 发表 https://www.chinapyg.com/images/common/back.gif
写的挺好的，向兄弟学习算法。。
谢谢/:013
俺又有了动力啦，再接再厉~~ 不错,再鼓励你一下哈,加油/:good
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