Минимальный в столбце на обратнуую диагональ

Option Explicit Dim i, j As Integer Dim n, m As Integer Dim arr() As Integer Private Sub cancel_Click() Form1.Hide End Sub Private Sub ok_Click() If Option1.Value = True Then m = Int(Rnd * 10) n = Int(Rnd * 10) Else m = Val(InputBox("Aaaaeoa ?enei noieaoia", "Aaia aaiiuo")) - 1 n = Val(InputBox("Aaaaeoa ?enei no?ie", "Aaia aaiiuo")) - 1 End If ReDim arr(n, m) Call fillMatrix Label1.Caption = getStr(arr, n, m) Call findMin Label2.Caption = getStr(arr, n, m) End Sub Private Sub fillMatrix() For i = 0 To n For j = 0 To m arr(i, j) = Int(Rnd() * 10) Next j Next i End Sub Private Sub findMin() Dim min, temp, jk As Integer For j = 0 To m min = arr(0, m) For i = 0 To n If arr(i, j) < min Then min = arr(i, j) jk = i End If Next i arr(jk, j) = arr(n - j, j) arr(n - j, j) = min Next j End Sub Private Function getStr(ar, an, am) As String getStr = "" For i = 0 To an For j = 0 To am getStr = LTrim(getStr + Str(ar(i, j)) + " ") Next j getStr = getStr + vbNewLine Next i End Function