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listbox
Option 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
    Next
End Sub
Private 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 = outs
End 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 Function
myERR:
     decode = ""
End Function