@above
Correction:
int a=2*m-1; b=m/2, k=m/2, r=a;
while(--k)
{ b*=k; r-=2; a*=r;}
printf("%d", a*pow(2, (m+1)/2) / b);
On Dec 16, 7:09 pm, Lucifer <sourabhd2...@gmail.com> wrote:
> @All
>
> As i see this question is more about the size of code than runtime..
> Hence, my next post might not help a lot on the front of size but may
> help in case u look at runtime.. If not of interest plz do skip this
> post..
>
> // M muls...
>
> int a=2*m-1; b=m/2, k=m/2; while(--k) { b*=k; a*=(a-2); a-
> =2;} printf("%d", a*pow(2, (m+1)/2) / b);
>
> On Dec 16, 6:39 pm, Lucifer <sourabhd2...@gmail.com> wrote:
>
>
>
>
>
>
>
> > @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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