问题
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)
{
// TODO
}
I\'m using C# but this is language agnostic.
Bonus points for clarity and simplicity (this isn\'t meant to be code-golf).
Edit: This got much more complex than I expected! It turns out the problems with double precision manifest themselves a couple ways. First, Math.Sqrt takes a double which can\'t precisely hold a long (thanks Jon).
Second, a double\'s precision will lose small values ( .000...00001) when you have a huge, near perfect square. e.g., my implementation failed this test for Math.Pow(10,18)+1 (mine reported true).
回答1:
bool IsPerfectSquare(long input)
{
long closestRoot = (long) Math.Sqrt(input);
return input == closestRoot * closestRoot;
}
This may get away from some of the problems of just checking "is the square root an integer" but possibly not all. You potentially need to get a little bit funkier:
bool IsPerfectSquare(long input)
{
double root = Math.Sqrt(input);
long rootBits = BitConverter.DoubleToInt64Bits(root);
long lowerBound = (long) BitConverter.Int64BitsToDouble(rootBits-1);
long upperBound = (long) BitConverter.Int64BitsToDouble(rootBits+1);
for (long candidate = lowerBound; candidate <= upperBound; candidate++)
{
if (candidate * candidate == input)
{
return true;
}
}
return false;
}
Icky, and unnecessary for anything other than really large values, but I think it should work...
回答2:
bool IsPerfectSquare(long input)
{
long SquareRoot = (long) Math.Sqrt(input);
return ((SquareRoot * SquareRoot) == input);
}
回答3:
In Common Lisp, I use the following:
(defun perfect-square-p (n)
(= (expt (isqrt n) 2)
n))
来源:https://stackoverflow.com/questions/343852/whats-a-good-algorithm-to-determine-if-an-input-is-a-perfect-square