首先这个算法没什么特殊之处，只是怕以后找不到，所以放到了这上面         每个字节加密后有6种结果（占两个字节，如果需要大于6种的话，就要多用1个字节，即占3 个字节），也就是说如果字串占n个字节的话，可能产生的结果为6的n次方个，这个算法破解的强度不大，大家可以完善一下： '窗体上一个按钮，两个listboxOption Explicit Private Sub Command1_Click()    Dim i As Long    Dim s As String    For i = 1 To 100        s = encode("这是一个测试 hello world")        List1.AddItem s        s = decode(s)        List2.AddItem s    NextEnd SubPrivate Function encode(ByVal s As String) As String '加密    If Len(s) = 0 Then Exit Function    Dim buff() As Byte    buff = StrConv(s, vbFromUnicode)    Dim i As Long    Dim j As Byte    Dim k As Byte, m As Byte    Dim mstr As String    mstr = "ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789abcdefghijklmnopqrstuvwxyz"    Dim outs As String    i = UBound(buff) + 1    outs = Space(2 * i)    Dim temps As String    For i = 0 To UBound(buff)        Randomize Time        j = CByte(5 * (Math.Rnd()) + 0) '最大产生的随机数只能是5，不能再大了,再大的话，就要多用一个字节        buff(i) = buff(i) Xor j        k = buff(i) Mod Len(mstr)        m = buff(i) \ Len(mstr)        m = m * 2 ^ 3 + j        temps = Mid(mstr, k + 1, 1) + Mid(mstr, m + 1, 1)        Mid(outs, 2 * i + 1, 2) = temps     Next     encode = outsEnd Function Private Function decode(ByVal s As String) As String '解密    On Error GoTo myERR    Dim i As Long    Dim j As Byte    Dim k As Byte    Dim m As Byte    Dim mstr As String    mstr = "ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789abcdefghijklmnopqrstuvwxyz"    Dim t1 As String, t2 As String    Dim buff() As Byte    Dim n As Long    n = 0    For i = 1 To Len(s) Step 2        t1 = Mid(s, i, 1)        t2 = Mid(s, i + 1, 1)        k = InStr(1, mstr, t1) - 1        m = InStr(1, mstr, t2) - 1        j = m \ 2 ^ 3        m = m - j * 2 ^ 3        ReDim Preserve buff(n)        buff(n) = j * Len(mstr) + k        buff(n) = buff(n) Xor m        n = n + 1     Next     decode = StrConv(buff, vbUnicode)     Exit FunctionmyERR:     decode = ""End Function 作者Blog：http://blog.csdn.net/rainstormmaster/

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