No...if a = b = 5 then return 9 doesn't make any sense.
n - (a-b) is very much fine.
for n = 5
Total # of cases when a - b = 0; 5 return 5
Total # of cases when a-b = 1; 4 return 4
Total # of cases when a-b = 2; 3 return 3
and so on.
On Mon, Aug 29, 2011 at 10:56 PM, nishaanth <nishaant...@gmail.com> wrote:
> @Don...Nice solution.. But the return statement should be a+b-1
>
>
> On Mon, Aug 29, 2011 at 10:33 PM, Don <dondod...@gmail.com> 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 <piyush4u.iit...@gmail.com> 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 <dondod...@gmail.com> 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 <piyush4u.iit...@gmail.com> 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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