I was asked to find out the no of rows given n . i.e. how we say 4 rows for
10 ???
On 14 Aug 2011 10:54, "sagar pareek" <sagarpar...@gmail.com> wrote:
> here is sudo code
>
> take two variables
>
> int max=1;
> int count=1;
> int i,total_no_of_rows;
>
> while(total_no_of_rows--)
> {
> for(i=0;i<count;i++)
> {
> printf("%d ",max-i);
> }
> printf("\n");
> count++;
> max+=count;
> }
>
> On Sun, Aug 14, 2011 at 10:25 AM, Rahul <raikra...@gmail.com> wrote:
>
>> @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 <ankur.kkhur...@gmail.com
>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 <
>>> aashish.barn...@gmail.com> wrote:
>>>
>>>> @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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>>>>>
>>>>
>>>>
>>>>
>>>> --
>>>> *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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>
>
>
> --
> **Regards
> SAGAR PAREEK
> COMPUTER SCIENCE AND ENGINEERING
> NIT ALLAHABAD
>
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