How could I check if a number is a perfect square? Speed is of no concern, for now, just working. See also: Integer square root in python.

I created a function that will test to see if a given parameter is a square number. Read about square numbers here: https://en.wikipedia.org/?title=Square_number If the number is a square number...

I would like to square every value in data, and I am thinking about using a for loop like this: data = rnorm(100, mean=0, sd=1) Newdata = {L = NULL; for (i in data) {i = i*i} L = i return (L)}

How could I write an if-then statement that checks if an inputted integer is a perfect square or not (i.e. if I took the square root, it would be an integer as well: 4, 9, 16, 25, 36, etc.) i...

This question is a follow-up on the post here: Fastest way to determine if an integer's square root is an integer, What's a good algorithm to determine if an input is a perfect square?. On...

Part of my code is as follows: class Array def square! self.map {|num| num ** 2} self end end When I call: [1,2,3].square! I expect to get [1,4,9], but instead I get [1,2,3]. Why is this ...

I've just started learning Python and have started doing some problems just to help buid my skills however I am pretty stuck on this question. Make a list containing all positive integers up to 1...

This is a code to check if a number is perfect square or not. Why does it work ? static bool IsSquare(int n) { int i = 1; for (; ; ) { if (n < 0) return false; if (n == 0) return true; ...

Can I rely on sqrt((float)a)*sqrt((float)a)==a or (int)sqrt((float)a)*(int)sqrt((float)a)==a to check whether a number is a perfect square? Why or why not? int a is the number to be judged....

Possible Duplicate: Fastest way to determine if an integer's square root is an integer What's a way to see if a number is a perfect square? bool IsPerfectSquare(long input) { /...

I'm looking for the fastest way to determine if a long value is a perfect square (i.e. its square root is another integer): I've done it the easy way, by using the built-in Math.sqrt() function,...