Question

我已经看过很多教程，但似乎对我没什么用。我需要生成随机数而不重复。

这是我的代码：

Dim intNumber As Integer
    Dim arrNumber(0 To 0) As Integer
    Dim i, x, y As Integer
    'Make sure the Label is clear
    Label1.Text = ""



    For x = 0 To 0
Start:
        intNumber = Int((25 * Rnd()) + 1) ' Random number 1 to 25
        For y = 0 To 0
            ' Check arrNumber (y)
            'If intnumber has already been selected,
            'Then go and select another one.
            If intNumber = arrNumber(y) Then
                GoTo Start
            End If
        Next y

        'Place the next non-repeated number in the arrNumber(x).

        arrNumber(x) = intNumber


    Next x

    '----------------------------------------------------
    For i = 0 To 0

        Label1.Text = Label1.Text & (arrNumber(i))
        broj1.random.Text = Label1.Text
    Next

End Sub

Answer 1

你的问题是，特别是在vb中，难以解决。您正在寻找随机交易而不是随机滚动。也就是说，您正在寻找赌场经销商的模拟一张接一张地拖着25张卡片，而不是经销商转动一个25格的轮盘赌。

让事情变得艰难的一件事就是第26张牌。改组？处理相同的订单？

这是一篇关于这个主题的体面论文。 http://www.4guysfromrolla.com/articles/070208-1.aspx

这里有一些C＃代码要处理。

    private static List<int> _deck = null;
    private static readonly Random NumberGenerator  = new Random();
    public static int Deal()
    {
        if (_deck == null || _deck.Count == 0)
        {
            /* new deck */
            _deck = new List<int>();
            for (var i = 0; i <= 24; i++) _deck.Add(i);
        }

        /* get a random card from the remaining deck */
        var next = NumberGenerator.Next(0, _deck.Count);
        /* retrieve the card's number */
        var q = _deck[next];
        /* and remove the card from the deck */
        _deck.RemoveAt(next);
        /* return in range 1-25 */
        return q + 1;
    }

Answer 2

有时您需要学习如何构建时钟，有时您只需知道时间：

    Const HowMany As Integer = 25     ' how many numbers do you want?

    Dim Used As New List(Of Integer)
    Used.Add(0)                       ' position zero has 0, all other positions have a generated counting number

    For i = 1 To HowMany
        Dim OK As Boolean = False     ' OK becomes true when a unique number has been generated
        Do
            Dim num As Integer = Int((HowMany * Rnd()) + 1)
            If Used.Contains(num) Then
                OK = False         ' we'll try again shortly
            Else
                OK = True          ' this round is complete -- found a unique number
                Used.Add(num)      ' add the generated number to the list
            End If
        Loop Until OK     ' in other words, loop until we've found a unique number
    Next

    For i = 1 To HowMany
        MsgBox(Used(i))   ' or just use the list in however you need
    Next

生成随机数而不重复

2 个答案: