@Beginner
If You Know How to implement a series whose difference in terms form An
Arithmetic Progression

Then Is quite easy
to print the (n)TH Line
Add  n to the highest number seen so far and then subtract the 1 for n times
(p = n-- , check for p )

@Everyone Else
Can anyone one think of a more elegant method , to find #rows


Rahul


On Sun, Aug 14, 2011 at 10:09 AM, Ankur Khurana <[email protected]>wrote:

> to print from 10 to 7 in the row number 4. here prev will be 6.
>
>
> On Sun, Aug 14, 2011 at 10:05 AM, AASHISH SUMAN <[email protected]
> > wrote:
>
>> @ankur
>>
>> what is the need of
>> prev=((i-)*i)/2;
>>
>>
>>
>>
>> On Sun, Aug 14, 2011 at 9:51 AM, Ankur Khurana 
>> <[email protected]>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 <[email protected]> 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 <[email protected]> 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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>>
>>
>>
>> --
>> *WITH BEST REGARDS :
>>
>> AASHISH SUMAN
>> MCA FINAL YEAR
>> *
>> *NIT DURGAPUR*
>> *+91-9547969906*
>>
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>
>
>
> --
> Ankur Khurana
> Computer Science
> Netaji Subhas Institute Of Technology
> Delhi.
>
>  --
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>

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