我发现您的问题有点不清楚，但我相信下面的宏可以满足您的要求。
如果你有一个变体 Result，你可以将 Result 设置为一个数组。然后，您可以依次将 Result(1)、Result(1)(1)、Result(1)(1)(1) 等设置为嵌套数组。使用合适的递归例程，我相信您可以在 Excel 限制内创建任何大小的数组。但是，我认为这种方法很难理解。
我不相信有更简单的方法来创建具有可变维数的数组。然而，改变尺寸的大小不是问题。
由于您最多有五个维度，我决定使用固定数量的维度，尾随未使用的维度宽度为 1。对于您的示例（第 1 步 1 到 10、第 10 步 10 到 50、第 5 步 5 到 25），这将需要：
Dim Result(1 To 10, 1 To 5, 1 To 5, 1 To 1, 1 To 1)
前三个维度有 10、5 和 5 个元素，可以容纳一系列值。最后两个维度只是占位符。
您正在让您的用户输入维度详细信息。我已经从工作表“Dyn Dims”中加载了详细信息。对于与您的示例匹配的测试，我将此工作表设置为：
Min Max Step
1 10 1
10 50 10
5 25 5
我将此信息加载到长数组要求（1 到 3、1 到 5）。列是最小值、最大值和步长。这些行最多允许五个维度。如果第 3 列（步骤）为零，则不使用维度。我不允许负步长值，但如果有必要，请指出需要更改的位置。
您需要根据用户输入的数据初始化此数组。
根据数组需求，宏计算每个维度的元素个数。我已经用值测试了这个计算，例如 1 step 2 to 10，其中 N 没有值，例如 Min + N * Step = Max。
宏然后根据需要维度数组结果。
你没有说你想要数组中的什么值，所以我将它们设置为“N:N:N”形式的值，其中 Ns 是来自 Min-To-Max-Step 计算的值。这个我已经在宏里解释过了，这里不再赘述。
最后，我将数组的内容输出到以日期和时间命名的文件中。在您的示例中，输出为：
Dimensions
1 2 3 Value
1 1 1 1:10:5
2 1 1 2:10:5
3 1 1 3:10:5
4 1 1 4:10:5
5 1 1 5:10:5
6 1 1 6:10:5
7 1 1 7:10:5
8 1 1 8:10:5
9 1 1 9:10:5
10 1 1 10:10:5
1 2 1 1:20:5
: : : :
5 5 5 5:50:25
6 5 5 6:50:25
7 5 5 7:50:25
8 5 5 8:50:25
9 5 5 9:50:25
10 5 5 10:50:25
我相信我已经包含了足够多的 cmets 来解释宏，但如果有必要，我会提出问题。
Option Explicit
Sub DD()
Const ColReqMin As Long = 1
Const ColReqMax As Long = 2
Const ColReqStep As Long = 3
Dim DimCrnt As Long
Dim Entry(1 To 5) As Long
Dim EntryStepped As Boolean
Dim FileOutNum As Long
Dim Index(1 To 5) As Long
Dim IndexStepped As Boolean
Dim NumEntries(1 To 5) As Long
Dim Requirements(1 To 3, 1 To 5) As Long
Dim Result() As String
Dim RowDDCrnt As Long
Dim Stg As String
Dim Value As String
' Load Requirements with the required ranges
With Worksheets("Dyn Dims")
RowDDCrnt = 2 ' First data row of worksheet Dyn Dims
' Note this macro does not check for blank lines in the middle
' of the table.
For DimCrnt = 1 To 5
If IsEmpty(.Cells(RowDDCrnt, ColReqStep)) Then
' No step value so this dimension not required for this run
Requirements(ColReqStep, DimCrnt) = 0
Else
Requirements(ColReqMin, DimCrnt) = .Cells(RowDDCrnt, ColReqMin)
Requirements(ColReqMax, DimCrnt) = .Cells(RowDDCrnt, ColReqMax)
Requirements(ColReqStep, DimCrnt) = .Cells(RowDDCrnt, ColReqStep)
End If
RowDDCrnt = RowDDCrnt + 1
Next
End With
' Calculate number of entries for each dimension
For DimCrnt = 1 To 5
If Requirements(ColReqStep, DimCrnt) = 0 Then
' Dummy dimension
NumEntries(DimCrnt) = 1
Else
NumEntries(DimCrnt) = (Requirements(ColReqMax, DimCrnt) - _
Requirements(ColReqMin, DimCrnt) + _
Requirements(ColReqStep, DimCrnt)) \ _
Requirements(ColReqStep, DimCrnt)
End If
Next
' Size array
ReDim Result(1 To NumEntries(1), _
1 To NumEntries(2), _
1 To NumEntries(3), _
1 To NumEntries(4), _
1 To NumEntries(5))
' Initialise entry for each dimension to minimum value, if any,
' and index for each dimension to 1
For DimCrnt = 1 To 5
Index(DimCrnt) = 1
If Requirements(ColReqStep, DimCrnt) <> 0 Then
Entry(DimCrnt) = Requirements(ColReqMin, DimCrnt)
End If
Next
' Starting with Entry(1), this loop steps the entry if the dimension is used.
' If the stepped entry is not greater than the maximum, then this repeat of
' the loop has finished. If the stepped entry is greater than the maximum,
' it is reset to its minimum and the next entry stepped and checked in the
' same way. If no entry is found that can be stepped, the loop is finished.
' If the dimensions after all 1 to 3 step 1, the values created by this loop
' are:
' 1 1 1 1 1
' 2 1 1 1 1
' 3 1 1 1 1
' 1 2 1 1 1
' 2 2 1 1 1
' 3 2 1 1 1
' 1 3 1 1 1
' 2 3 1 1 1
' 3 3 1 1 1
' 1 1 2 1 1
' 2 1 2 1 1
' 3 1 2 1 1
' : : : : :
' 3 3 3 3 3
Do While True
' Concatenate entries to create value for initial element
' or for element identified by last loop
Value = Entry(1)
For DimCrnt = 2 To 5
If Requirements(ColReqStep, DimCrnt) = 0 Then
Exit For
End If
Value = Value & ":" & Entry(DimCrnt)
Next
Result(Index(1), Index(2), Index(3), Index(4), Index(5)) = Value
' Find an entry to step
EntryStepped = False
For DimCrnt = 1 To 5
If Requirements(ColReqStep, DimCrnt) = 0 Then
Exit For
End If
Index(DimCrnt) = Index(DimCrnt) + 1
Entry(DimCrnt) = Entry(DimCrnt) + _
Requirements(ColReqStep, DimCrnt)
' ### Changes required her if a negative step value is allow
If Entry(DimCrnt) <= Requirements(ColReqMax, DimCrnt) Then
' This stepped entry is within permitted range
EntryStepped = True
Exit For
End If
' This entry past its maximum so reset to minimum
' and let for loop step entry for next dimension
Index(DimCrnt) = 1
Entry(DimCrnt) = Requirements(ColReqMin, DimCrnt)
Next
If Not EntryStepped Then
' All elements of Result initialised
Exit Do
End If
Loop
' All elements of Result initialised
' Output values as test.
FileOutNum = FreeFile
Open ActiveWorkbook.Path & "\" & Format(Now(), "yymmdd hhmmss") & ".txt" _
For Output As #FileOutNum
' Initialise Index
For DimCrnt = 1 To 5
Index(DimCrnt) = 1
Next
' Create header line for table
Print #FileOutNum, "Dimensions"
Stg = ""
For DimCrnt = 1 To 5
If Requirements(ColReqStep, DimCrnt) = 0 Then
Exit For
End If
Stg = Stg & Right(" " & DimCrnt, 4)
Next
Stg = Stg & " Value"
Print #FileOutNum, Stg
' Similar logic to loop that intialised Result but using Index and UBound.
Do While True
' Output initial element or element identified by previous loop
Stg = ""
For DimCrnt = 1 To 5
If Requirements(ColReqStep, DimCrnt) = 0 Then
Exit For
End If
Stg = Stg & Right(" " & Index(DimCrnt), 4)
Next
Stg = Stg & " " & Result(Index(1), Index(2), Index(3), Index(4), Index(5))
Print #FileOutNum, Stg
' Identify next element, if any
IndexStepped = False
For DimCrnt = 1 To 5
If Requirements(ColReqStep, DimCrnt) = 0 Then
Exit For
End If
Index(DimCrnt) = Index(DimCrnt) + 1
If Index(DimCrnt) <= UBound(Result, DimCrnt) Then
IndexStepped = True
Exit For
Else
Index(DimCrnt) = 1
End If
Next
If Not IndexStepped Then
' All entries output
Exit Do
End If
Loop
Close #FileOutNum
End Sub