@varun
I think u want to write

while (k % m == 0)

On Sat, Jul 2, 2011 at 1:56 PM, varun pahwa <varunpahwa2...@gmail.com>wrote:

> k -> rope of desired length.
> l -> rope of given length
> m = 2;
> while(k % m)
> m *= 2;
> ans :: (log2(l) - log2(m) + 1).
> ex.
> k = 6,l = 8
> so initially m = 2;
> after 1st iteration m = 4;
> then break;
> so min = log2(8) - log2(4) + 1  = 3 -2 + 1 = 2.
>
>
>
> On Sat, Jul 2, 2011 at 1:16 AM, cegprakash <cegprak...@gmail.com> wrote:
>
>> nope
>>
>> On Jul 2, 1:14 pm, keyan karthi <keyankarthi1...@gmail.com> wrote:
>> > yup :)
>> >
>> > On Sat, Jul 2, 2011 at 1:38 PM, Shalini Sah <
>> shalinisah.luv4cod...@gmail.com
>> >
>> > > wrote:
>> > > i guess the no. of 1s in the binary representation of the number is
>> the
>> > > answer..for 6 its 2...
>> >
>> > > On Sat, Jul 2, 2011 at 1:32 PM, cegprakash <cegprak...@gmail.com>
>> wrote:
>> >
>> > >> the length of the rope is l units.
>> > >> I can only cut any rope into two halves.
>> >
>> > >> for example if the length of the rope is 8 and we need a length of
>> > >> rope 6
>> >
>> > >> we first cut into two halves and we get 4, 4
>> > >> now we cut any of the half again and we get 4,2,2
>> >
>> > >> now we can merge 4 and 2 and form a rope of length 6.
>> >
>> > >> in this example we need a minimum of 2 cuts to get the length of rope
>> > >> 6 from 8
>> >
>> > >> assume that l is always a power of 2 and we need always a even length
>> > >> of rope from it how to find the number of minimum cuts needed to get
>> > >> the new rope?.
>> >
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>
>
> --
> Varun Pahwa
> B.Tech (IT)
> 7th Sem.
> Indian Institute of Information Technology Allahabad.
> Ph : 09793899112 ,08011820777
> Official Email :: rit2008...@iiita.ac.in
> Another Email :: varunpahwa.ii...@gmail.com
>
> People who fail to plan are those who plan to fail.
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-- 
Sunny Aggrawal
B-Tech IV year,CSI
Indian Institute Of Technology,Roorkee

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