No, a+b-1 would return values outside of the desired range.
Don

On Aug 29, 12:26 pm, nishaanth <[email protected]> wrote:
> @Don...Nice solution.. But the return statement should be a+b-1
>
>
>
> On Mon, Aug 29, 2011 at 10:33 PM, Don <[email protected]> wrote:
> > If you draw the nxn grid and assign a value to each diagonal:
>
> > (For n = 5)
>
> >  ----------b
> > |  12345
> > |  2345
> > |  345
> > |  45
> > |  5
> > a
>
> > You want the result to be the orthogonal distance from the diagonal.
> > That is what the formula computes.
> > Don
>
> > On Aug 29, 11:28 am, Piyush Grover <[email protected]> wrote:
> > > I understand what you are doing in the loop but return statement is not
> > > clear to me. Can you explain.
>
> > > On Mon, Aug 29, 2011 at 9:48 PM, Don <[email protected]> wrote:
> > > > int custRand(int n)
> > > > {
> > > >        int a,b;
> > > >        do
> > > >        {
> > > >                a = rand(n);
> > > >                b = rand(n);
> > > >        } while(a < b);
> > > >        return n - a + b;
> > > > }
>
> > > > On Aug 29, 10:48 am, Piyush Grover <[email protected]> wrote:
> > > > > Given a function rand(n) which returns a random value between 1...n
> > > > assuming
> > > > > equal probability.
> > > > > Write a function CustRand(n) using rand(n) which returns a value
> > between
> > > > > 1...n such that
> > > > > the probability of occurrence of each number is proportional to its
> > > > value.
>
> > > > > I have a solution but wondering if I can get better than this or some
> > > > other
> > > > > approaches:
>
> > > > > CustRand(n){
>
> > > > >     sum  = n*(n+1)/2;
>
> > > > >     a = rand(sum);
> > > > >     for(i = 1; i <= n; i++){
> > > > >          h = i*(i+1)/2;
> > > > >          l = i*(i-1)/2;
> > > > >          if(a <= h && a > l)
> > > > >              return i;
> > > > >     }
>
> > > > > }
>
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> --
> S.Nishaanth,
> Computer Science and engineering,
> IIT Madras.

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