计算一个BigInteger（System.Numerics.BigInteger）的平方根 [英] Calculate square root of a BigInteger (System.Numerics.BigInteger)

问题描述

.NET 4.0提供了 System.Numerics.BigInteger 键入任意-大的整数。我需要计算平方根（或一个合理的近似 - 例如，整数的平方根）的的BigInteger 。所以，我没有重新实现轮子，没有任何人有一个很好的扩展方法为这个？

.NET 4.0 provides the System.Numerics.BigInteger type for arbitrarily-large integers. I need to compute the square root (or a reasonable approximation -- e.g., integer square root) of a BigInteger. So that I don't have to reimplement the wheel, does anyone have a nice extension method for this?

推荐答案

的检查的BigInteger不是有code来计算的Java BigInteger的整数的平方根一个完美的正方形。这被翻译成C＃，作为一个扩展的方法。

Check if BigInteger is not a perfect square has code to compute the integer square root of a Java BigInteger. Here it is translated into C#, as an extension method.

    public static BigInteger Sqrt(this BigInteger n)
    {
        if (n == 0) return 0;
        if (n > 0)
        {
            int bitLength = Convert.ToInt32(Math.Ceiling(BigInteger.Log(n, 2)));
            BigInteger root = BigInteger.One << (bitLength / 2);

            while (!isSqrt(n, root))
            {
                root += n / root;
                root /= 2;
            }

            return root;
        }

        throw new ArithmeticException("NaN");
    }

    private static Boolean isSqrt(BigInteger n, BigInteger root)
    {
        BigInteger lowerBound = root*root;
        BigInteger upperBound = (root + 1)*(root + 1);

        return (n >= lowerBound && n < upperBound);
    }

非正式测试表明，这是大约75X比的Math.sqrt慢，对于小的整数。该VS探查指向isSqrt乘法的热点。

Informal testing indicates that this is about 75X slower than Math.Sqrt, for small integers. The VS profiler points to the multiplications in isSqrt as the hotspots.

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