CODEWITHSUNDEEP
vb.net
Zack
Zack

Reputation: 117

Display the sum of the even numbers between two integers

I have a Visual Basic program that expects two integers from the user.

I need tobe able to display the sum of the even numbers between the two integers entered by the user. If the user's entry is even, it should be included in the sum.

I am having trouble figuring out the algorithm.

Here is my code:

Public Class frmMain

Private Sub btnExit_Click(sender As Object, e As EventArgs) Handles btnExit.Click
    Me.Close()
End Sub

Private Sub txtFirstNum_KeyPress(sender As Object, e As KeyPressEventArgs) Handles txtFirstNum.KeyPress
    ' allows the text box to accept only numbers, and the Backspace key

    If (e.KeyChar < "0" OrElse e.KeyChar > "9") AndAlso
        e.KeyChar <> ControlChars.Back Then
        e.Handled = True
    End If
End Sub

Private Sub txtSecondNum_KeyPress(sender As Object, e As KeyPressEventArgs) Handles txtSecondNum.KeyPress
    ' allows numbers and Backspace only

    If (e.KeyChar < "0" OrElse e.KeyChar > "9") AndAlso
        e.KeyChar <> ControlChars.Back Then
        e.Handled = True
    End If
End Sub

Private Sub btnDisplay_Click(sender As Object, e As EventArgs) Handles btnDisplay.Click
    ' display the sum of even numbers

    ' declarations
    Dim intFirstNum As Integer
    Dim intSecondNum As Integer

    ' convert string to number with TryParse method
    Integer.TryParse(txtFirstNum.Text, intFirstNum)
    Integer.TryParse(txtSecondNum.Text, intSecondNum)

    ' find the even numbers
    If intFirstNum Mod 2 = 0 Then

    ElseIf intFirstNum Mod 2 = 1 Then

    End If

    ' add the even numbers

    ' display the result

End Sub
End Class

How can I fix this?

Upvotes: 1

Views: 7058

Answers (3)

Joel Coehoorn
Joel Coehoorn

Reputation: 415725

If you know the first number is even by checking that at the start, you don't have to check the modulus of every digit along the way, and can just increment by 2s:

Public Function SumEvenNumbers(ByVal Start As Integer, ByVal End As Integer) As Integer
    If Start Mod 2 = 1 Then Start += 1

    Dim Result As Integer = 0
    While Start <= End
       Result += Start
       Start += 2
    End While

    Return Result
End Function

And just for fun, here's an example that uses linq and Enumerable.Range:

Public Function SumEvenNumbers(ByVal Start As Integer, ByVal End As Integer) As Integer
     If Start Mod 2 = 1 Then Start += 1
     Return Enumerable.Range(Start, End - Start).Sum(Function(i) i Mod 2 == 0)
End Function

But the best answer is the one that's already posted, and just uses Math to reduce it all to a fixed average ( O(1), rather than O(n) ).

Upvotes: 1

Guffa
Guffa

Reputation: 700302

You can calculate how many even numbers there are, and from that calculate the sum:

' adjust the numbers to the closest even number
If (intFirstNum mod 2) = 1 Then intFirstNum += 1
If (intSecondNum mod 2) = 1 Then intSecondNum -= 1

' determine how many even numbers there are in the range
Dim count As Integer = (intSecondNum - intFirstNum) \ 2 + 1

' the sum is the average times the count
Dim sum As Integer = (intFirstNum + intSecondNum) * count \ 2

The TryParse methods returns a boolean, which tells you if it was possible to parse the string. You shouldn't ignore the return value, so use it as:

If Integer.TryParse(txtFirstNum.Text, intFirstNum) Then
  ' yes, it was a number
End If

Upvotes: 2

David Khuu
David Khuu

Reputation: 967

Public Function GetEvenNumberSum(min As Integer, max As Integer) As Integer
    Dim sum As Integer = 0
    For i As Integer = min To max
        If i Mod 2 = 0 Then
            sum += i
        End If
    Next
    Return sum
End Function

Upvotes: -1

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