Using pow in php

Question

Hi can I use pow instead of writing it like this?

                    $round1 = $max / 2;
                    $round2 = $round1 + ($max / 2 / 2);
                    $round3 = $round2 + ($max / 2 / 2 / 2);
                    $round4 = $round3 + ($max / 2 / 2 / 2 / 2);
                    $round5 ...
                    $final = $max - 1;
                    $third = $max;

Thanks for help!

Maybe add some kind of explanation of your expected output etc? — Amal Murali, Aug 12 '13 at 12:14
Dividing by 2 is equivalent to shifting right by one: `$max >> 1` is identical to `$max / 2`. — nickb, Aug 12 '13 at 12:24

score 2 · Answer 1 · answered Aug 12 '13 at 12:24

2

You wouldn't happen to be looking for an easier way to do this, would you:

$round = array($max/2);
for ($i=1;i<$max;$i++)
{
    $round[] = $round[$i -1] + $max/pow(2, $i + 1);
}

Of course, array_sum might be worth a peek, too.

answered Aug 12 '13 at 12:24

Elias Van Ootegem

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score -1 · Answer 2 · answered Aug 12 '13 at 12:27

-1

try this,

 $rount[0]=0;
 for($i=1;$1<upto n rounds;$i++)
 {
   $rount[i]=$round[i-1]+($max/(2*i));
 }

answered Aug 12 '13 at 12:27

Magendran V

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3

this is correct for `i` = 1 or 2 but `($max/2/2/2) != ($max/(2*3))` and will not work following this. – Kami Aug 12 '13 at 12:41

Using pow in php

2 Answers2