Question

目前我有以下代码。它逐行读取文件db.lcz，并将每个结果作为项目放入数组中。

在db.lcz中，每行有一个10位数字，我想根据所述文件中的现有数字得到一个新的随机数。

-

任何想法都是受欢迎的，我已经尝试了一些选项，比如for循环等，但我一直迷路。我知道我必须针对概率方面检查每个字符值，但我不知道如何继续。感谢所有帮助。

谢谢！

 Private Sub Button1_Click(sender As Object, e As EventArgs) Handles Button1.Click

    Dim lines() As String = IO.File.ReadAllLines("db.lcz")
    Dim codeArray As New ArrayList()

    For x As Integer = 0 To lines.GetUpperBound(0)
        codeArray.Add(lines(x))
    Next
    For x = 0 To codeArray.Count - 1
        MsgBox(codeArray.Item(x))
    Next




    Dim rng As New Random
    Dim number As Integer = rng.Next(1, 1000000000)  'Test.GenerateDigits(rng, 9)
    Dim digits As String = number.ToString("0000000000")
    MsgBox(digits)
End Sub

Private Shared Function GenerateDigits(ByVal rng As Random, _
                 ByVal length As Integer) As String
    Dim chArray As Char() = New Char(length - 1) {}
    Dim i As Integer
    For i = 0 To length - 1
        chArray(i) = Convert.ToChar(rng.Next(10) + &H30)
    Next i
    Return New String(chArray)
End Function

Answer 1

您可以使用shuffle函数随机重新排序数字，并使用索引数组。

s = "1234567890"
s1 = ""
ix = shuffle(s.Length - 1)
For i As Integer = 0 To s.Length - 1
  s1 &= s(ix(i))
  Next i

shuffle功能可以类似于：

Function shuffle(ByVal ncards As Integer) As List(Of Integer)

Dim rand As New Random
Dim ix As New List(Of Integer)
Dim k, n, tmp As Integer

For i as integer = 0 To ncards
  ix.Add(i)
  Next i

n = ncards
For i as integer = ncards To 1 Step -1
  k = rand.Next(1, n + 1)
  tmp = ix(i)
  ix(i) = ix(k)
  ix(k) = tmp
  Next i

Return ix

End Function

Visual Basic - 基于现有数字列表的随机数

1 个答案: