我想从一个范围内的0-9产生10个不同的号码。 所期望的输出可能看起来像这样,9 0 8 6 5 3 2 4 1 7
Dim arraynum(9) As Integer
Dim crmd As Boolean
Dim rmd as integer
For i = 0 To 9
arraynum(i) = -1
Next i
crmd = True
Randomize Timer
For i = 0 To 9
rmd = Int(Rnd * 10)
For j = 0 To 9
If arraynum(j) = rmd Then
j = 9
If crmd = False Then
i = i - 1
End If
crmd = True
Else
crmd = False
End If
Next j
If crmd = False Then
arraynum(i) = rmd
QuestionILabel.Caption = QuestionILabel.Caption + Str(arraynum(i))
End If
Next i
选择随机值,然后扔掉那些你已经使用是一个坏主意。 它使较长的运行时间可用号码池变少,因为你把自己的越来越多。
你想要的是我会用下面的代码(因为它的功课伪代码)实现一个洗牌列表:
dim n[10] // gives n[0] through n[9]
for each i in 0..9:
n[i] = i // initialize them to their indexes
nsize = 10 // starting pool size
do 10 times:
i = rnd(nsize) // give a number between 0 and nsize-1
print n[i]
nsize = nsize - 1 // these two lines effectively remove the used number
n[i] = n[nsize]
通过简单地选择从池中随机数,然后用从池上面的数字替换它,减少池的大小,你会得到一个洗牌,而不必担心大量掉了前面的。 这一点很重要,如果数字是,它不引入不必要的启动延迟高。
例如,检查下面的板凳检查:
<--------- n[x] ---------->
for x = 0 1 2 3 4 5 6 7 8 9 nsize rnd(nsize) output
--------------------------- ----- ---------- ------
0 1 2 3 4 5 6 7 8 9 10 4 4
0 1 2 3 9 5 6 7 8 9 7 7
0 1 2 3 9 5 6 8 8 2 2
0 1 8 3 9 5 6 7 6 6
0 1 8 3 9 5 6 0 0
5 1 8 3 9 5 2 8
5 1 9 3 4 1 1
5 3 9 3 0 5
9 3 2 1 3
9 1 0 9
你可以看到在游泳池降低,当您去,因为你总是有一个未使用的一个替代使用一个,你永远不会有重复。
现在你的功课,包括转向到这一点VB的:-)
而且,由于该作业是现在几乎可以肯定逾期(约一年前),我会后展示如何做到这一点,对于一个完整的解决方案VBA。
Option Explicit
Option Base 0
Sub Macro1()
Randomize
Dim list(10) As Integer
Dim i As Integer
Dim size As Integer
Dim pos As Integer
Dim result As String
For i = 0 To 9
list(i) = i
Next
size = 10
result = ":"
For i = 1 To 10
pos = Int(Rnd() * size)
result = result & list(pos) & ":"
size = size - 1
list(pos) = list(size)
Next
MsgBox result
End Sub
这产生的,在三个不同的运行:
:5:7:4:2:9:1:0:8:3:6:
:3:9:6:0:7:8:5:4:2:1:
:7:6:3:5:1:8:9:0:4:2:
你需要一个随机排列在0数组9。
我忘了怎么写的基本..类似:
dim a(10)
for i=0 to 9 do a(i) = i
rem do random permute over a:
for i=0 to 9 do
j = rand() mod (i+1)
tmp = a(j)
a(i) = a(j)
a(j) = tmp
next i
Option Explicit 'Force variable declaration
Private Sub Form_Load()
Dim Ar(1 To 100) As Integer 'The array to store it in
Dim i, j As Integer 'Counters for loops
Dim X As Integer 'Variable to store the random generated number
Dim bFound As Boolean 'Boolean to check if the value has been generated before
Randomize 'Just once to ensure that we get random values
For i = 1 To 100
Do 'Start the loop that generates a random number and checks if it has already been generated
X = RandomInteger(1, 100) 'Generate a random number
bFound = False 'Set the boolean to false, if we find the number while searching the array, we'll set it to true which means that we already have that number
For j = 1 To i 'We only need to check up to i (since we haven't put any values in the rest of the array)
If Ar(j) = X Then 'If an item of the arrray is the same as the last generated number
bFound = True 'Set the boolean to true (it already exists)
DoEvents 'To not freeze until the looping is done
Exit For 'Since we found it there is no need to check the rest
End If
Next
Loop Until bFound = False 'If it wasn't found then we'll add it, if it was found then we go back to generating a new number and comparing it with all the items of the array
Ar(i) = X 'Add it to the array
Next
ShowInTextBox Text1, Ar 'Just to print the data and see it
End Sub
Private Function RandomInteger(Lowerbound As Integer, Upperbound As Integer) As Integer 'The random number generator code
RandomInteger = Int((Upperbound - Lowerbound + 1) * Rnd + Lowerbound)
End Function
Private Sub ShowInTextBox(TB As TextBox, A() As Integer) 'Just a sub to show the data in a textbox
Dim i As Integer
TB.Text = ""
For i = 1 To UBound(A)
TB.Text = TB.Text & CStr(A(i)) & vbCrLf
Next
TB.Text = Left$(TB.Text, Len(TB.Text) - 2)
End Sub
这是最简单不过的工作代码。 也有没有的API!
Dim a(1 To 10), tmp, count, isRepeated
count = 1
Randomize
While count <= 10
isRepeated = False
tmp = Left(Rnd * 10, 1)
For i = 1 To 10
If tmp = a(i) Then isRepeated = True: Exit For
Next
If isRepeated = False Then a(count) = tmp: count = count + 1
Wend
YourLine = Join(a, "")
而这一切,乡亲!