On Thu, 22 Apr 2010 10:17:10 -0400
Dan Joseph <dmjoseph@xxxxxxxxx> wrote:
> On Thu, Apr 22, 2010 at 10:12 AM, Stephen <stephen-d@xxxxxxxxxx>
> wrote:
>
> > 1,252,398 DIV 30 = 41,746 groups of 30.
> >
> > 1,252,398 MOD 30 = 18 items in last group
> >
> Well, the only problem with going that route, is the one group is not
> equally sized to the others. 18 is ok for a group in this instance,
> but if it was a remainder of only 1 or 2, there would be an issue.
> Which is where I come to looking for a the right method to break it
> equally.
>
My take on it:
$Items=1252398;
$MaxInGroup=30;
for ($x=$MaxInGroup; $x>1;$x--) {
$remainder=$Items % $x;
// Change 17 to the max amount allowed in the last group
if ($remainder == 0 || $remainder >= 17) { //
$groups = (int) ($Items /$x)+1;
echo $groups."\n";
echo $remainder;
break;
}
}
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Peter van der Does
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