@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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