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find if a number is perfect square

How to find if a number is a perfect square?

I am NOT looking for the solution which says check if sqrt is integer or not
1 Solution
Rog DManager Inforamtion SystemsCommented:
Could you give us more information?  How about a sample of what you are looking for?  When is this due? ;)
bsarvanikumarAuthor Commented:
for those who need clarification

I don't want this solution

int n;
cin >> n;
if (sqrt (N) == Whole Number)
  cout << "A perfect square"
  cout << "Not a perfect square"
I was hoping more in terms of bit wise operators (>>,<<,&,|) etc

 I actually read the solution in terms of bit wise operators many weeks ago. I could not recall the solution.

This is not due as in for assignment or hw. Hence I alotted only few points. Take your time and hopefully provide a solution i am looking for.

Perhaps then you'd be interested in researching assembly square root solutions and convert them manually?  Here is one for z80 which has a similar feel to x86:

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I came across this website which might be closer to what you are looking for.


Please disregard my question if it is irrelevant, but are you sure you saw the biwise solution for N^2?
It is possible that you remember seeing the bitwise solution for checking whether the number is power of 2 - it would be indeed simple in bitwise terms:
  M is 2^N  if and only if  (M & (M-1)) equals zero (assuming N>0).
For N^2 I cannot remember anything simple (if I do, I'll let you know, of course).

there is a very easy to do it using mod... unfortunatly i cant quite remember it but if i do ill tell u

l8rz... suma
No comment has been added to this question in more than 21 days, so it is now classified as abandoned.

I will leave the following recommendation for this question in the Cleanup topic area:
   PAQ - no points refunded (of 50)

Any objections should be posted here in the next 4 days. After that time, the question will be closed.

EE Cleanup Volunteer
Question finalized per recommendation.

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