正方加密解密算法及获取密钥

作者: BccSafe 分类: Code[Delphi] 发布时间: 2014-07-18 01:22 ? 6没有评论

 这学期挂了2门，很是不爽

于是乎重操旧业，话说我已经很久没碰渗透测试这方面的东西了

各种坎坷…

具体细节以后再写

拿到系统管理员的密文后，百度他的算法资料 

也不用反编译那个zjdx.dll了，网上资料一片一片的

 

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Public Function jiam(ByVal PlainStr As String, ByVal key As String) As String         Dim strChar, KeyChar, NewStr As String         Dim Pos As Integer         Dim i, intLen As Integer         Dim Side1, Side2 As String         Pos = 1         For i = 1 To Len(PlainStr)             strChar = Mid(PlainStr, i, 1)             KeyChar = Mid(key, Pos, 1)             If ((Asc(strChar) Xor Asc(KeyChar)) < 32) Or ((Asc(strChar) Xor Asc(KeyChar)) > 126) Or (Asc(strChar) < 0) Or (Asc(strChar) > 255) Then                 NewStr = NewStr & strChar             Else                 NewStr = NewStr & Chr(Asc(strChar) Xor Asc(KeyChar))             End If             If Pos = Len(key) Then Pos = 0             Pos = Pos + 1         Next         If Len(NewStr) Mod 2 = 0 Then             Side1 = StrReverse(Left(NewStr, CInt((Len(NewStr) / 2))))             Side2 = StrReverse(Right(NewStr, CInt((Len(NewStr) / 2))))             NewStr = Side1 & Side2         End If         jiam = NewStr     End Function public string jiemi(string PlainStr, string key) {       string text3;       int num2 = 1;       if ((Strings.Len(PlainStr) % 2) == 0)       {             string text4 = Strings.StrReverse(Strings.Left(PlainStr, (int) Math.Round(((double) Strings.Len(PlainStr)) / 2)));             string text5 = Strings.StrReverse(Strings.Right(PlainStr, (int) Math.Round(((double) Strings.Len(PlainStr)) / 2)));             PlainStr = text4 + text5;       }       int num3 = Strings.Len(PlainStr);       for (int num1 = 1; num1 <= num3; num1++)       {             string text6 = Strings.Mid(PlainStr, num1, 1);             string text2 = Strings.Mid(key, num2, 1);             if (((((Strings.Asc(text6) ^ Strings.Asc(text2)) < 0x20) | ((Strings.Asc(text6) ^ Strings.Asc(text2)) > 0x7e)) | (Strings.Asc(text6) < 0)) | (Strings.Asc(text6) > 0xff))             {                   text3 = text3 + text6;             }             else             {                   text3 = text3 + StringType.FromChar(Strings.Chr(Strings.Asc(text6) ^ Strings.Asc(text2)));             }             if (num2 == Strings.Len(key))             {                   num2 = 0;             }             num2++;       }       return text3; }

 

奉上delphi版的方正加密解密算法(他的算法很渣，看看就好了)

 

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//uses StrUtils; function RightCopy(S: string; Index, count: Integer): string; begin   Result := Copy(S,Length(S)-count+1-(Index-1),count); end; function jiami(PlainStr, key: string): string; var   StrChar, Keychar: Char;   Pos: Integer;   i : Integer;   StrLeft, StrRight: String; begin   Pos := 1;   for i := 1 to Length(PlainStr) do   begin     StrChar := copy(PlainStr, i, 1)[1];     KeyChar := copy(key, Pos, 1)[1];     if ((ord(StrChar) Xor ord(KeyChar)) < 32) Or       ((ord(StrChar) Xor ord(KeyChar)) > 126) Or       (ord(StrChar) > 255) Then   //ord(StrChar) < 0 永远为false 去掉了       Result := Result + StrChar     Else       Result := Result + Char(ord(StrChar) Xor ord(KeyChar));     if Pos = Length(key) then Pos := 0;     Inc(Pos);   end;   If Length(Result) Mod 2 = 0 Then   begin     StrLeft := ReverseString(copy(Result, 0, Round(Length(Result) / 2)));     StrRight := ReverseString(RightCopy(Result, 1, Round(Length(Result) / 2)));     Result := StrLeft + StrRight;   end; end; function jiemi(PlainStr, key: string): string; var   StrLeft, StrRight: string;   num2, i : Integer;   StrChar, Keychar: Char; begin   num2 := 1;   if Length(PlainStr) mod 2 = 0 then   begin     StrLeft := ReverseString(copy(PlainStr, 0, Round(Length(PlainStr) / 2)));     StrRight := ReverseString(RightCopy(PlainStr, 1,       Round(Length(PlainStr) / 2)));     PlainStr := StrLeft + StrRight;   end;   for i := 1 to Length(PlainStr) do   begin     StrChar := copy(PlainStr, i, 1)[1];     Keychar := copy(key, num2, 1)[1];     if (((((ord(StrChar) Xor ord(Keychar)) < 32) Or ((ord(StrChar) Xor ord(Keychar)) >       126)) ) Or (ord(StrChar) > 255)) then  //ord(StrChar) < 0 永远为false 去掉了       Result := Result + StrChar     else       Result := Result + Char(ord(StrChar) xor ord(Keychar));     if num2 = Length(key) then num2 := 0;     Inc(num2);   end; end;

 

 

然后问题出现了，我们学校的密钥并不是网上流传的Acxylf365jw

渣渣表示只能继续百度- -

发现核攻击写了一片关于这个的算法

GET!

根据他写的，改的delphi版的GetKey

 

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function GetKey(PlainStr: String; Pass: String): String; var   StrChar, Keychar: Char;   Pos: Integer;   i: Integer;   StrLeft, StrRight: String; begin   Pos := 0;   If Length(PlainStr) Mod 2 = 0 Then   begin     StrLeft := ReverseString(copy(PlainStr, 0, Round(Length(PlainStr) / 2)));     StrRight := ReverseString(RightCopy(PlainStr, 1, Round(Length(PlainStr) / 2)));     PlainStr := StrLeft + StrRight;   end;   for i := 0 to Length(PlainStr) - 1 do   begin     StrChar := Copy(PlainStr, i, 1)[1];     KeyChar := Copy(Pass, Pos, 1)[1];     If StrChar = KeyChar Then       Result := Result + '*'     else       Result := Result + Char(ord(StrChar) Xor ord(KeyChar));     Pos := Pos + 1;   end; end;

 

从数据库里找了点数据出来(有些账号密码就是账号名)

成功拿到了密钥Encrypt01

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