@Shady: I'd try something like this:
For each n:
Let m = the number of 1-bits in n.
If k >= 1<<m, then output -1.
For i = 1 to m do
Replace the ith 1-bit of n (counting from the right) with the ith
bit of k.
Output the updated n.
Example: n = 27 = binary 11011, k = 5 = binary 00101:
m = 4.
Replace the first 1-bit of n with the first bit of k: n = 11011.
Replace the 2nd 1-bit of n with the 2nd bit of k: n = 11001.
Replace the 3rd 1-bit of n with the 3rd bit of k: n = 11001.
Replace the 4th 1-bit of n with the 4th bit of k: n = 01001.
Output 5.
Dave
On Feb 11, 12:30 am, shady <sinv...@gmail.com> wrote:
> answer is always +ve,
>
> does anyone know how to code it ?
>
>
>
> On Sat, Feb 11, 2012 at 11:37 AM, Tushar <tushicom...@gmail.com> wrote:
> > for the given test case in the problem, answer could have been zero
>
> > what does it mean by "without even reading zero" in the problm
> > statement?
>
> > On Feb 9, 11:34 pm, shady <sinv...@gmail.com> wrote:
> > > Hi All,
> > > Anyway to implement this in a good mannerhttp://
> >www.spoj.pl/ARHN/problems/PRINCESS, solution is simple,
> > > check for ones and then generate numbers is an increasing number...
> > > so if a number n = 22(10110)
> > > then for k = 1, ans = 2(10)
> > > then for k = 2, ans = 4(100)
> > > then for k = 3, ans = 6(110)
>
> > > Shady
>
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