Sum multiples of 3 and 5: Difference between revisions

=={{header|VBSA}}==

{{trans|VBScript}}

<lang vb>Private Function SumMult3and5VBScript(n As Double) As Double

Dim i As Double

For i = 1 To n - 1

If i Mod 3 = 0 Or i Mod 5 = 0 Then

SumMult3and5VBScript = SumMult3and5VBScript + i

End If

Next

End Function</lang>

Other way :

<lang vb>Private Function SumMult3and5(n As Double) As Double

Dim i As Double

For i = 3 To n - 1 Step 3

SumMult3and5 = SumMult3and5 + i

Next

For i = 5 To n - 1 Step 5

If i Mod 15 <> 0 Then SumMult3and5 = SumMult3and5 + i

Next

End Function</lang>

Better way :

<lang vb>Private Function SumMult3and5BETTER(n As Double) As Double

Dim i As Double

For i = 3 To n - 1 Step 3

SumMult3and5BETTER = SumMult3and5BETTER + i

Next

For i = 5 To n - 1 Step 5

SumMult3and5BETTER = SumMult3and5BETTER + i

Next

For i = 15 To n - 1 Step 15

SumMult3and5BETTER = SumMult3and5BETTER - i

Next

End Function</lang>

Call :

<lang vb>Option Explicit

Sub Main()

Dim T#

T = Timer

Debug.Print SumMult3and5VBScript(100000000) & " " & Format(Timer - T, "0.000 sec.")

T = Timer

Debug.Print SumMult3and5(100000000) & " " & Format(Timer - T, "0.000 sec.")

T = Timer

Debug.Print SumMult3and5BETTER(100000000) & " " & Format(Timer - T, "0.000 sec.")

Debug.Print "-------------------------"

Debug.Print SumMult3and5BETTER(1000)

End Sub</lang>

{{Out}}

<pre>2,33333331666667E+15 9,059 sec.

2,33333331666667E+15 2,107 sec.

2,33333331666667E+15 1,799 sec.

-------------------------

233168

</pre>