Fibonacci numbers

Mathematica
Note, that this implementation exists. Algorithm is:

Fibonacci[n_] := Round[GoldenRatio^n/Sqrt[5]]

Be aware, that this formula:

can not be trivially applied in other languages.

Java
In Java, we have arbitrarily large integers and decimals, but prebuilt classes lack the functionality (it's a shame). For example we can not easily take root of 5. To do this, we would need to use a library like JScience or write Newton approximation method (what can be quite error prone process).

	/**
	* Gives the Fibonacci number.
	*
	* @param n
	* element to return.
	* @return Fibonacci and Negafibonacci numbers.
	*/
	public static BigInteger fibonacci(final int n) {
	if (n == 0)
	return BigInteger.ZERO;
	int an = Math.abs(n);
	BigInteger[] result = { BigInteger.ONE, BigInteger.ONE };
	BigInteger[] tmp = new BigInteger[2];
	for (int i = 1; i < an; i++) {
	tmp[0] = result[0].add(result[1]);
	tmp[1] = result[0];
	result = Arrays.copyOf(tmp, tmp.length);
	}
	return ((n < 0) && (n % 2 == 0)) ? result[1].negate() : result[1];
	}

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C#
In C#, there is no arbitrarily large decimals. It does have class decimal (with 28-29 significant digits), but assuming that Math class supports it, is just asking too much (it's a shame). But thumbs up, for it at least has BigInteger.

	/// <summary>
	/// Gives the Fibonacci number.
	/// </summary>
	/// <remarks>using System.Numerics;</remarks>
	/// <param name="n">element to return.</param>
	/// <returns>Fibonacci and Negafibonacci numbers.</returns>
	public static BigInteger fibonacci(int n)
	{
	if (n == 0)
	return BigInteger.Zero;
	int an = Math.Abs(n);
	int nr = 2;
	BigInteger[] result = { BigInteger.One, BigInteger.One };
	BigInteger[] tmp = new BigInteger[nr];
	for (int i = 1; i < an; i++)
	{
	tmp[0] = BigInteger.Add(result[0], result[1]);
	tmp[1] = result[0];
	Array.Copy(tmp, 0, result, 0, 2);
	}
	return ((n < 0) && (n % 2 == 0)) ? -result[1] : result[1];
	}

view raw gistfile1.cs hosted with ❤ by GitHub