@ Special Nos..
Well the actual logic would be :
int count = 0;for ( int i = 2; i <= LOGbase2(N) ; i+=2) count+=
[ (i-1) C (i/2) ] ; // here xCy is nothing but the no. of ways y items
can be selected from a collection of x items.
Hence, the working code:
int totalCount = 0;
int interCnt = 1;
if ( LOGbase2(N) > 1)
{
totalCount = 1; // for LOGbase2(N) = 2...
for ( int i = 4; i <= LOGbase2(N) ; i+=2)
{
interCnt = (i-1)*(i-2) * interCnt / ( i/2 * (i/2 -1));
totalCount += interCnt;
}
printf("%d", totalCount);
}
else
printf("%d", 0);
On Dec 27, 7:38 pm, Tasvinder Singh <[email protected]> wrote:
> I think the first problem involves some mathematics...
> In this we fix the first bit and if the remaining no. of bits are odd then
> we calculate the no. as follows
>
> If we have 2^4=16 then total bits 5 so we do not include this.
> Total no. of bits in one less than the given no. (in this eg. 15) is 4.
> Fix first bit, no. of bits remaining = 3
> Now let 2 bits are 0 and one bit 1. We have total 3!/(2!*1!) = 3
> combinations.
>
> Now go for next even no which is 2 in this case again do the same
> Fix first bit, no. of bits remaining = 1
> Now let 1 bit is 0. We have total 1!/(0!*1!) = 1 combinations.
>
> Next even 0. stop here.
> You can go for this by starting from 2 till no. is less than given N
>
>
>
>
>
>
>
>
>
> On Tue, Dec 27, 2011 at 7:28 PM, kumar rajat <[email protected]> wrote:
> > Hi
> > I have jus started to learn DP and I have coded the standard
> > algorithms(LCS,etc).
> > I have been trying these problems in SPOJ:
>
> >http://www.spoj.pl/problems/NOVICE63/
> >http://www.spoj.pl/problems/NY10E/
>
> > I understand these may be solved elegantly using DP,but I dont get to
> > code the same.
> > Can any1 help me how to solve these types of problems using DP?
> > Any help regarding the same will be greatly appreciated.
>
> > --
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>
> --
> Tasvinder Singh
> B.Tech Final Year,
> Department of Computer Engineering,
> Malaviya National Institute Of Technology, Jaipur.
> Contact: +91-9460627331
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