Question

VB 用随机数(20~80)生成一个4行4列的矩阵A，求A的转置矩阵

Private Sub Picture1_Click() Dim a(4, 4) As Integer, i As Integer, j As IntegerFor i = 1 To 4For j = 1 To 4a(i, j) = Int(Rnd() * 61 + 20)Picture1.Print a(i, j);NextPicture1.PrintNextPicture1.Print
End Sub

Answer 1

已经完成：

Private Sub Picture1_Click()

 Dim a(4, 4) As Integer, i As Integer, j As Integer, b(4, 4) As Integer

For i = 1 To 4

For j = 1 To 4

a(i, j) = Int(Rnd() * 61 + 20)

Picture1.Print a(i, j);

Next

Picture1.Print

Next

Picture1.Print

For i = 1 To 4

For j = 1 To 4

b(i, j) = a(j, i)

Picture1.Print b(i, j);

Next

Picture1.Print

Next

Picture1.Print

End Sub

追问

如果。。我想把转置的放在picture2呢？？单击“命令按钮转置”然后picture2出现picture1的矩阵的转置

追答

Dim a(4, 4) As Integer
Private Sub Picture1_Click()
Dim i As Integer, j As Integer
For i = 1 To 4
For j = 1 To 4
a(i, j) = Int(Rnd() * 61 + 20)
Picture1.Print a(i, j);
Next
Picture1.Print
Next
Picture1.Print

End Sub

Private Sub Command1_Click()
Dim b(4, 4) As Integer
For i = 1 To 4

For j = 1 To 4
b(i, j) = a(j, i)
Picture2.Print b(i, j);
Next
Picture2.Print
Next
Picture2.Print
End Sub

Answer 2

我懒得给你重写，直接给你一个转置的函数，调用形式：bt = MatTrn(B)，B为输入矩阵，bt为输出结果

Public Function MatTrn(x() As Double) As Double()

    Dim n As Long

    Dim m As Long

    Dim i As Long

    Dim j As Long, y() As Double

    n = UBound(x, 1)

    m = UBound(x, 2)

    'If Not IsDimed(Y) Then

        ReDim y(1 To m, 1 To n)

    'End If

    'ReDim Y(1 To m, 1 To n) As Double

    For i = 1 To n

        For j = 1 To m

            y(j, i) = x(i, j)

        Next j

    Next i

    MatTrn = y

End Function

Answer 3

要是只是显示的话，只需要一列一列输出就好！代码如下
For j= 1 To 4
For i= 1 To 4

Picture2.Print a(i, j);
Next
Picture2.Print
Next
Picture2.Print
要是想得到A的转置矩阵，代码如下：
dim b(4,4) as integer
For j= 1 To 4
For i= 1 To 4
b(j,i)=a(i, j) ;
Next

Next

追问

Private Sub Picture1_Click()
这是矩阵A出现的代码
End Sub
然后
Private Sub command1_Click()
这是转置的代码，但是这样的话，上面的a（i，j）就会因程序结束而回到初始值。。不能用在这里。。是不是会这样啊？我不太懂啊？
End Sub