产生随即数问题，小妹请教各位大虾！！！！

nslhao 2003-03-21 12:10:55

产生52个随机数，不能重复，请教算法？？不胜感激。

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nik_Amis 2003-03-21

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手误，呵呵

for n=1 to 80
a=int(rnd*51):b=int(rnd*51)
s=y(a):y(a)=y(b):y(b)=s
next
考虑改为
for n=0 to 50
a=int(rnd*51):s=y(n):y(n)=y(a):y(a)=s
next
这样会好一点？讨论讨论

Cooly 2003-03-21

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:) nik的算法很不错的。
小改一下。:)

private y(51) as long

sub gen()
dim n as long,s as long,a as long,b as long
for n=0 to 51
'y(51)=n+1 <==这里
y(n)=n+1
next

for n=1 to 80
a=int(rnd*51):b=int(rnd*51)
s=y(a):y(a)=y(b):y(b)=s
next

end sub

nik_Amis 2003-03-21

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private y(51) as long

sub gen()
dim n as long,s as long,a as long,b as long
for n=0 to 51
y(51)=n+1
next

for n=1 to 80
a=int(rnd*51):b=int(rnd*51)
s=y(a):y(a)=y(b):y(b)=s
next

end sub
这个算法速度快得多

AechoJohn 2003-03-21

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楼上的

Cooly 2003-03-21

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'生成52个1到100的不重复随机数
Dim i, j As Long
Dim ran(1 To 52) As Long
Dim tNum As Long
Dim isExist As Boolean
Randomize
For i = 1 To 52
isExist = False

tNum = Int(100 * Rnd + 1)
Debug.Print tNum
For j = 1 To i
If ran(j) = tNum Then
isExist = True
i = i - 1
Exit For
End If
Next

If isExist = False Then
ran(i) = tNum
End If

Next

For i = 1 To 52
msg = msg & ran(i) & Space(1)
Next
MsgBox msg

northwolves 2003-03-21

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Private Sub Command1_Click()
Dim x As New Collection, a(0 To 51) As String
For i = 0 To 99
x.Add i
Next
For i = 0 To 51
Randomize
num = Int(Rnd * x.Count + 1)
a(i) = Str(x.Item(num))
x.Remove num
Next
MsgBox Join(a(), ","), 64, "52 random number between 0 and 99"
End Sub

Cooly 2003-03-21