@above
Correction.. its 2m C m
@saurabh..
Sorry i missed your post.. seems that u are doing exactly the same..
Just a suggestion.. to make the code run faster, we can store f(n) in
a variable and then use it in the expression to avoid recalculation
of !(n)
i,e t = f(m);
printf("%d", f(2*m) / pow(t,2) );
//In the original code we are making 4m multiplications.. in the above
given one we are doing 3m multiplcations..
We can further optimize (2m muls and m adds)..
Further optimization could be:
int a=2*m; b=m, k=m;
while(--k)
{ b*=k; a*=(k+m);}
printf("%d", a/b);
On Dec 16, 6:14 pm, Lucifer <sourabhd2...@gmail.com> wrote:
> @anubhav..
>
> Its 2n C n...
>
> Basically they are asking u to finds no. of paths that can be formed
> using m Tops and m rights..which is nothing but no. of unique ways of
> arranging m T's and m R's .. i,e 2n C n
>
> On Dec 16, 5:38 pm, sourabh singh <singhsourab...@gmail.com> wrote:
>
>
>
>
>
>
>
> > limit is just v.smll can ny1 help me with this code: [ c code ] get it
> > under 120 bytes .
> > is there any way to take inputs. without using scanf ??? in c . m
> > thinking about inputs getting into
> > argc array directly.???
>
> > #define s(n) scanf("%d",&n);
> > f(int n){return n==1?1:(n*f(n-1));}
> > main()
> > {
> > int t,n;
> > s(t)
> > while(t--)
> > { s(n)
> > printf("%d",(f(2*n)/f(n))/f(n));
> > }
>
> > }
> > On Thu, Dec 15, 2011 at 10:08 PM, anubhav raj <anubhaw....@gmail.com> wrote:
> > > #include<stdio.h>
> > > #define s(n) scanf("%d",&n)
> > > #define I int
> > > I a[14][14];
> > > I d(I m,I n)
> > > {
> > > I s=a[m][n];
> > > I k;
> > > if(!m||!n)
> > > k=1;
> > > else if(s)
> > > k=s;
> > > else
> > > {
> > > k=d(m-1,n)+d(m,n-1);
> > > s=k;
> > > }
> > > return k;
> > > }
> > > main()
> > > {
> > > I t,n,k;s(t);
> > > while(t--)
> > > {
> > > s(n);
> > > k=d(n,n);
> > > printf("%d\n",k);
> > > }
> > > }
>
> > > @saurav:this is the actual and very simple code...........may be u cn
> > > reduce its no of bytes........its simple dp approach...tnx....
>
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