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[休闲灌水] 文本加密求破解哈

文本加密求破解哈

我自己改的文本加密,谁破解出来奖励人民币100以下为加密后的文本代码

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


用户系统信息:Mozilla/5.0 (Windows; U; Windows NT 6.1; en-US) AppleWebKit/533.1 (KHTML, like Gecko) Maxthon/3.0.11.8 Safari/533.1
最后编辑QYCity 最后编辑于 2010-06-12 01:02:43
分享到:
gototop
 

回复:文本加密求破解哈

鬼知道你这是什么算法,我猜是RSA,因为资料比较多。

RSA的话,即使你放n和e别人也是搞不出的
gototop
 

回复:文本加密求破解哈

看不懂,转成16进制了?
gototop
 

回复:文本加密求破解哈

该用户帖子内容已被屏蔽
gototop
 

回复:文本加密求破解哈

使用用asp加密函数 XOR_DES
不过使用的key是用中文的哈
gototop
 

回复:文本加密求破解哈

看起来有点像MD5,不过没解过这么长的
如果是ESCAPE或JS/VB的简单加密,我可以解
ASP中可以实现RSA加密和解密,:根据RSA
这个好像功力不够,惭愧

根据前辈资料整理出在asp中加密与解密函数
rsa.asp
<%
rem 在ASP中实现加密与解密,加密方法:根据RSA
rem 联系:admin#boaichina.com (请将#换成@)
Class clsRSA
   Public PrivateKey
   Public PublicKey
   Public Modulus
  
  
  
   Public Function Crypt(pLngMessage, pLngKey)
     On Error Resume Next
     Dim lLngMod
     Dim lLngResult
     Dim lLngIndex
     If pLngKey Mod 2 = 0 Then
       lLngResult = 1
       For lLngIndex = 1 To pLngKey / 2
         lLngMod = (pLngMessage ^ 2) Mod Modulus
         ' Mod may error on key generation
         lLngResult = (lLngMod * lLngResult) Mod Modulus
         If Err Then Exit Function
       Next
     Else
       lLngResult = pLngMessage
       For lLngIndex = 1 To pLngKey / 2
         lLngMod = (pLngMessage ^ 2) Mod Modulus
         On Error Resume Next
         ' Mod may error on key generation
         lLngResult = (lLngMod * lLngResult) Mod Modulus
         If Err Then Exit Function
       Next
     End If
     Crypt = lLngResult
   End Function

   Public Function Encode(ByVal pStrMessage)
     Dim lLngIndex
     Dim lLngMaxIndex
     Dim lBytAscii
     Dim lLngEncrypted
     lLngMaxIndex = Len(pStrMessage)
     If lLngMaxIndex = 0 Then Exit Function
     For lLngIndex = 1 To lLngMaxIndex
       lBytAscii = Asc(Mid(pStrMessage, lLngIndex, 1))
       lLngEncrypted = Crypt(lBytAscii, PublicKey)
       Encode = Encode & NumberToHex(lLngEncrypted, 4)
     Next
   End Function
  
   Public Function Decode(ByVal pStrMessage)
     Dim lBytAscii
     Dim lLngIndex
     Dim lLngMaxIndex
     Dim lLngEncryptedData
     Decode = ""
     lLngMaxIndex = Len(pStrMessage)
     For lLngIndex = 1 To lLngMaxIndex Step 4
       lLngEncryptedData = HexToNumber(Mid(pStrMessage, lLngIndex, 4))
       lBytAscii = Crypt(lLngEncryptedData, PrivateKey)
       Decode = Decode & Chr(lBytAscii)
     Next
   End Function
  
   Private Function NumberToHex(ByRef pLngNumber, ByRef pLngLength)
     NumberToHex = Right(String(pLngLength, "0") & Hex(pLngNumber), pLngLength)
   End Function
   Private Function HexToNumber(ByRef pStrHex)
     HexToNumber = CLng("&h" & pStrHex)
   End Function
End Class
%>

test.asp
<!--#INCLUDE FILE="RSA.asp"-->
<%
function Encryptstr(Message)
Dim LngKeyE
Dim LngKeyD
Dim LngKeyN
Dim StrMessage
Dim ObjRSA

   LngKeyE = "32823"
   LngKeyD = "20643"
   LngKeyN = "29893"
   StrMessage = Message
  
   Set ObjRSA = New clsRSA
  
 
       ObjRSA.PublicKey = LngKeyE
       ObjRSA.Modulus = LngKeyN
       Encryptstr = ObjRSA.Encode(StrMessage)
   Set ObjRSA = Nothing
end function


function decryptstr(Message)
Dim LngKeyE
Dim LngKeyD
Dim LngKeyN
Dim StrMessage
Dim ObjRSA

   LngKeyE = "32823"
   LngKeyD = "20643"
   LngKeyN = "29893"
   StrMessage = Message
  
   Set ObjRSA = New clsRSA
       ObjRSA.PrivateKey =LngKeyD
       ObjRSA.Modulus=LngKeyN
       decryptstr=ObjRSA.Decode(StrMessage)
   Set ObjRSA = Nothing
end function

dim last,first
first="sohu"
Response.Write "加密前为:"&first
last=Encryptstr(first)
Response.Write "加密后为"&last
Response.Write "解密后为" &decryptstr(last)
%>
最后编辑tianxunmycool 最后编辑于 2010-06-12 01:52:30


☭ 说:不帅用刀砍,不爱用钱砸
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gototop
 

回复:文本加密求破解哈

ESCAPE或JS/VB的简单加密
这个根本算不上加密的
我用的加密函数送上哈

Function XOR_ENS(Source, Key) '加密函数
  Dim i, iKey, iKeyLen
  Dim SSA, SSB, SSS
  Dim XOR_STR_A
  Select Case Len(Source)
      Case 1
          Source = Source & Chr(32) & Chr(32) & Chr(32) & Chr(32)
      Case 2
          Source = Source & Chr(32) & Chr(32) & Chr(32)
      Case 3
          Source = Source & Chr(32) & Chr(32)
      Case 4
          Source = Source & Chr(32)
    End Select
    XOR_STR_A = ""
    iKeyLen = Len(Key)
    iKey = 1
    'Source = StrConv(Source, vbFromUnicode)
    For i = 1 To LenB(Source)
        SSA = CInt(AscB(MidB(Source, i, 1)))
        SSB = CInt(Asc(Mid(Key, iKey, 1)))
        iKey = iKey + 1
        If iKey > iKeyLen Then
            iKey = 1
        End If
        SSS = SSA Xor SSB
        XOR_STR_A = XOR_STR_A & Right("0" & Hex(SSS), 2)
    Next
    XOR_ENS = XOR_STR_A
    XOR_ENS = Replace(XOR_ENS, "1", "*")
    XOR_ENS = Replace(XOR_ENS, "9", "1")
    XOR_ENS = Replace(XOR_ENS, "*", "9")
    XOR_ENS = Replace(XOR_ENS, "8", "#")
    XOR_ENS = Replace(XOR_ENS, "2", "8")
    XOR_ENS = Replace(XOR_ENS, "#", "2")
    XOR_ENS = Replace(XOR_ENS, "4", "#")
    XOR_ENS = Replace(XOR_ENS, "7", "4")
    XOR_ENS = Replace(XOR_ENS, "#", "7")
    XOR_ENS = Replace(XOR_ENS, "3", "#")
    XOR_ENS = Replace(XOR_ENS, "6", "3")
    XOR_ENS = Replace(XOR_ENS, "#", "6")
    XOR_ENS = Replace(XOR_ENS, "D", "#")
    XOR_ENS = Replace(XOR_ENS, "F", "D")
    XOR_ENS = Replace(XOR_ENS, "#", "F")
    XOR_ENS = Replace(XOR_ENS, "A", "#")
    XOR_ENS = Replace(XOR_ENS, "B", "A")
    XOR_ENS = Replace(XOR_ENS, "#", "B")
End Function

Function XOR_DES(Source, Key)  ' 解密函数
    Dim i, iKey, iKeyLen
    Dim SSA, SSB, SSS
    Dim XOR_STR_A
    XOR_STR_A = ""
    iKeyLen = Len(Key)
    iKey = 1
    XOR_STR_A = ""
    Source = Replace(Source, "1", "*")
    Source = Replace(Source, "9", "1")
    Source = Replace(Source, "*", "9")
    Source = Replace(Source, "8", "#")
    Source = Replace(Source, "2", "8")
    Source = Replace(Source, "#", "2")
    Source = Replace(Source, "4", "#")
    Source = Replace(Source, "7", "4")
    Source = Replace(Source, "#", "7")
    Source = Replace(Source, "3", "#")
    Source = Replace(Source, "6", "3")
    Source = Replace(Source, "#", "6")
    Source = Replace(Source, "D", "#")
    Source = Replace(Source, "F", "D")
    Source = Replace(Source, "#", "F")
    Source = Replace(Source, "A", "#")
    Source = Replace(Source, "B", "A")
    Source = Replace(Source, "#", "B")
   
    For i = 1 To Len(Source) Step 2
        SSA = CInt("&H"&(Mid(Source, i, 2))&"")
        SSB = (Asc(Mid(Key, iKey, 1)))
        iKey = iKey + 1
          If iKey > iKeyLen Then
            iKey = 1
        End If
           SSS = SSA Xor SSB
        XOR_STR_A = XOR_STR_A & ChrB(SSS)
    Next
    'XOR_DES = StrConv(XOR_STR_A, vbUnicode)
    XOR_DES = replace(XOR_STR_A,vbcrlf,"",lenB(XOR_STR_A)-4)
    xor_des=trim(XOR_STR_A)
    xor_des=cstr(xor_des)
End Function
                                        '产生密钥算法
Function KeyGeN(iKeyLength)
    Dim k, iCount, strMyKey
    lowerbound = 35
    upperbound = 96
    Randomize
    For i = 1 To iKeyLength
        k = Int(((upperbound - lowerbound) + 1) * Rnd + lowerbound)
        strMyKey = strMyKey & Chr(k) & ""
    Next
    KeyGeN = strMyKey
End Function
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