@all
plz point out wat's wrong with this code. works for small numbers but fails
for large number's
#include<conio.h>
#include<iostream>
using namespace std;
unsigned int find_nCp(unsigned int n,unsigned int p)
{ unsigned int j=p;
unsigned int num=1,den=1;
for(;j;j--)
num*=n--;
for(;p;p--)
den*=p;
return (num/den);
}
unsigned int num(unsigned int n,unsigned int k)
{
if(n==1 || k==0)
return 1;
return find_nCp(n,k) + num(n-2,(n-1)/2);
}
main()
{ uint64_t n,p,test=1;
unsigned int t,i,sum,j;
scanf("%u",&t);
while(t--)
{ n=test<<t;
scanf("%llu",&n);
if(n<4)
cout<<0<<endl;
else
{ for(p=1,i=0;p<=n;p=p<<1,i++);
i=i-2;
if(i%2==0)
{
i--;
}
cout<<num(i,i/2)<<endl;
getch();
}
}
return 0;
}
On Tue, Dec 27, 2011 at 10:33 AM, kumar rajat <thebossku...@gmail.com>wrote:
> *I am not able to follow..
>
> On Tue, Dec 27, 2011 at 11:47 PM, kumar rajat <thebossku...@gmail.com>
> wrote:
> > @Lucifier:
> > Thanks a lot..
> > But,I am able to follow the code that u posted for Non-Decreasing Digits.
> > Can u jus explain ur algo instead of givin the code directly?
> > Thanks once again.
> >
> > On Tue, Dec 27, 2011 at 10:44 PM, Lucifer <sourabhd2...@gmail.com>
> wrote:
> >> @ 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 <tasvinde...@gmail.com> 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 <thebossku...@gmail.com>
> 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.
> >>>
> >>> > --
> >>> > You received this message because you are subscribed to the Google
> Groups
> >>> > "Algorithm Geeks" group.
> >>> > To post to this group, send email to algogeeks@googlegroups.com.
> >>> > To unsubscribe from this group, send email to
> >>> > algogeeks+unsubscr...@googlegroups.com.
> >>> > For more options, visit this group at
> >>> >http://groups.google.com/group/algogeeks?hl=en.
> >>>
> >>> --
> >>> Tasvinder Singh
> >>> B.Tech Final Year,
> >>> Department of Computer Engineering,
> >>> Malaviya National Institute Of Technology, Jaipur.
> >>> Contact: +91-9460627331
> >>
> >> --
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