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
