Checking if a number is perfect square?

Question

I think there is a precision problem with the following code:

bool isPerfectSquare(long long n){
    long long squareRootN=(long long)(sqrt(n)+0.5);

    return          


        

Answer 1

mmmm don't use float/double for your true/false outcome (you'll end up having approximation problems) Much better the integer apporach:

   boolean is_square(long long n)
   {
      long long a=n;
      while (a*a>n)
      {
         a=div(a+div(n,a),2);
      }
      return a*a==n;
   }

div() is the integer division without remainder (you can use GCD() in some ways)

I know, I know... care must be taken for overflow issues