Maths question

[ChrisL]

A CD can hold upto 200 images

I need to know what formula to use to work out the quantity of CD's

If $images = 0 then there would be 0 CD

If $images = 12 then there would be 1 CD

If $images =201 then there would be 2 CD's

I have this

$images = 12
$maxCD = 200
Msgbox(0,"",Int($images/$maxCD +1))

Which works fine except for $images = 0 because you still get 1 cd.

Am I better changing the code so that if $images = 0 then $CD = 0 or is there a better way

$images = 12
$maxCD = 200

If $images >  0 then 

       $CD = Int($images/$maxCD +1)
    Else
    $CD = 0
EndIf

Just need someone to confirm I'm right or an Arse

(Note If replying Arse then the reply must be justified )

[PsaltyDS]

That was fun...

$maxCD = 200
While 1
    $images = InputBox("CD Count", "How many images?")
    If @error Then Exit

    $CDs = Int($images / $maxCD)
    If Mod($images, $maxCD) Then $CDs += 1
    MsgBox(64, "Results", "At " & $maxCD & " images per CD, it will take " & $CDs & " CDs to hold " & $images & " images.")
WEnd

[the_lord_mephy]

Take a look at the Ceiling function

Edit: well if you're give him code i'll do it too

$images = 12
$maxCD = 200
Msgbox(0,"",Ceiling($images/$maxCD))

Edited August 16, 2007 by the_lord_mephy

[PsaltyDS]

Take a look at the Ceiling function

Edit: well if you're give him code i'll do it too

$images = 12
$maxCD = 200
Msgbox(0,"",Ceiling($images/$maxCD))

Ooh... I like! Never noticed that function before and had to go look it up!

[ChrisL]

Thanks both

Never heard of Ceiling() before.