@ankur
what is the need of
prev=((i-)*i)/2;
On Sun, Aug 14, 2011 at 9:51 AM, Ankur Khurana <ankur.kkhur...@gmail.com>wrote:
> see , we can see that
> first lline have 1 number, second have 2 and third have 3 .....and so on.
> we can observe that first number of every row is sum of first k narutal
> number, where k is the row number. so for k=4, n=((4+1)*4)/2=10;
>
> so run a for loop,
>
>
> for(int i=1;i<num_rows;i++)
> {
> prev=((i-)*i)/2;
> k=((i+1)*i)/2;
> for(int j=k;i;k>prev;j--)
> {
> cout<<prev<<" ";
> }
> cout<<endl;
> }
>
>
> On Sun, Aug 14, 2011 at 9:43 AM, Dave <dave_and_da...@juno.com> wrote:
>
>> @Beginner: The largest number n in row r satisfies n = (r^2 + r) / 2.
>> So using the Quadratic Formula gives
>>
>> r = ( sqrt( 8*n + 1 ) - 1 ) / 2.
>>
>> For the row number r for any n (not necessarily the largest one in a
>> row),
>>
>> r = ceil( sqrt( 8*n + 1 ) - 1 ) / 2 )
>>
>> where ceil( x ) is the smallest integer not exceeding x.
>>
>> Dave
>>
>> On Aug 13, 10:22 pm, Beginner <murugavidya1...@gmail.com> wrote:
>> > How to print this triangle?
>> > 1
>> > 3 2
>> > 6 5 4
>> > 10 9 8 7
>> > and how to find the number of rows if n is given?
>> > For ex if n=10 how to find num of rows=4??
>> > Is it Log 10 to the base of 2!!!!!!
>>
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>>
>
>
> --
> Ankur Khurana
> Computer Science
> Netaji Subhas Institute Of Technology
> Delhi.
>
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