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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