关于中文编码的问题

[tantao1982]

小弟我做了一个rsa加密解密算法的class，用vb写的，现在发现他对于英文解密是好的，但是做到中文解密的时候就出了问题。请高手大哥们给我看看，小弟我感激不尽。
Class clsRSA

Public PrivateKey
Public PublicKey
Public Modulus

Public Sub GenKey()
Dim lLngPhi
Dim q
Dim p

Randomize

Do
Do

' 2 random primary numbers (0 to 1000)
Do
p = Rnd * 1000 \ 1
Loop While Not IsPrime(p)

Do
q = Rnd * 1000 \ 1
Loop While Not IsPrime(q)


' n = product of 2 primes
Modulus = p * q \ 1

' random decryptor (2 to n)
PrivateKey = Rnd * (Modulus - 2) \ 1 + 2

lLngPhi = (p - 1) * (q - 1) \ 1
PublicKey = Euler(lLngPhi, PrivateKey)

Loop While PublicKey = 0 Or PublicKey = 1

' Loop if we can't crypt/decrypt a byte
Loop While Not TestCrypt(255)

End Sub

Private Function TestCrypt(ByRef pBytData)
Dim lStrCrypted
lStrCrypted = Crypt(pBytData, PublicKey)
TestCrypt = Crypt(lStrCrypted, PrivateKey) = pBytData
End Function

Private Function Euler(ByRef pLngPHI, ByRef pLngKey)

Dim lLngR(3)
Dim lLngP(3)
Dim lLngQ(3)

Dim lLngCounter
Dim lLngResult

Euler = 0

lLngR(1) = pLngPHI: lLngR(0) = pLngKey
lLngP(1) = 0: lLngP(0) = 1
lLngQ(1) = 2: lLngQ(0) = 0

lLngCounter = -1

Do Until lLngR(0) = 0

lLngR(2) = lLngR(1): lLngR(1) = lLngR(0)
lLngP(2) = lLngP(1): lLngP(1) = lLngP(0)
lLngQ(2) = lLngQ(1): lLngQ(1) = lLngQ(0)

lLngCounter = lLngCounter + 1

lLngR(0) = lLngR(2) Mod lLngR(1)
lLngP(0) = ((lLngR(2)\lLngR(1)) * lLngP(1)) + lLngP(2)
lLngQ(0) = ((lLngR(2)\lLngR(1)) * lLngQ(1)) + lLngQ(2)

Loop

lLngResult = (pLngKey * lLngP(1)) - (pLngPHI * lLngQ(1))

If lLngResult > 0 Then
Euler = lLngP(1)
Else
Euler = Abs(lLngP(1)) + pLngPHI
End If

End Function

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

Private Function IsPrime(ByRef pLngNumber)
Dim lLngSquare
Dim lLngIndex
IsPrime = False
If pLngNumber < 2 Then Exit Function
If pLngNumber Mod 2 = 0 Then Exit Function
lLngSquare = Sqr(pLngNumber)
For lLngIndex = 3 To lLngSquare Step 2
If pLngNumber Mod lLngIndex = 0 Then Exit Function
Next
IsPrime = True
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