@sukhmeet: plz explain ur method if the no is 50

On Sun, Mar 13, 2011 at 8:24 PM, sukhmeet singh <[email protected]>wrote:

> I think it can be done like this that assume that the number is represented
> as 1.. like 4=1111 , 3= 111 ans so on
> now if u have to fint the total number of ways then it is similar to
> placing a '+' sign in the mid places
> Like 3= 1 1 1 so possible are 1+11 (1+2) or 11+1 (2+1) or 1+1+1(1+1+1)
> hence 3 ways...!! similar approach can be taken ..
> hence the problem reduces to placing + sign in b/w .. placing 1 '+' , 2
> '+'  uptill the number it can be formed...!! Correct me if i am wrong..!!
>
>
> On Fri, Mar 11, 2011 at 10:02 PM, saurabh agrawal <[email protected]>wrote:
>
>> Do it assuming as two cases.
>>
>> On Fri, Mar 11, 2011 at 9:46 PM, Dave <[email protected]> wrote:
>>
>>> This is a good place to use recursion. One thing you have to know is
>>> whether order matters. E.g., are 2 + 1 and 1 + 2 the same or different
>>> ways to represent 3 as a sum of positive integers. This choice will
>>> affect the way you write the recursive function.
>>>
>>> Dave
>>>
>>> On Mar 11, 9:33 am, saurabh agrawal <[email protected]> wrote:
>>> > Given an integer n , you have to print all the ways in which n can be
>>> > represented as sum of positive integers
>>>
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