n≥1的斐波那契数列. 46爪哇 [英] Fibonacci sequence for n > 46 Java

问题描述

我有以下代码，可为n<提供正确的值； 47.

I have the following code which provides the correct values for n < 47.

public static int fib(int n) {
    int nthTerm = 0;
    if (n == 2)
        nthTerm = 1;
    else {
        double goldenRatio = (1 + Math.sqrt(5)) / 2;
        nthTerm = (int) (Math.round(Math.pow(goldenRatio, n)
                - Math.pow(1 - goldenRatio, n)) / Math.sqrt(5));

        if (n % 2 == 1 && n < 45)
            nthTerm++;
    }
    return nthTerm;
}

n的任何值> 46都超出int范围.在n> 46的情况下，我该如何适应这种方法?

Any value for n > 46 is out of int range. How could I adapt this approach to work for n > 46?

P.S.我知道BigInteger，但不怎么擅长，所以我也很感谢使用BigInteger的示例.

P.S. I know of BigInteger but am not very adept at it so I would appreciate an example using BigInteger, too.

推荐答案

您可以将其用于将代码转换成BigInteger.

You can use this for transformated code into BigInteger.

package your.pack

import java.math.BigDecimal;
import java.math.BigInteger;

/**
 * Created on 3/6/16.
 */
public class Fibonacci {

    private static BigDecimal goldenRatio = new BigDecimal((1 + Math.sqrt(5)) / 2);
    private static BigDecimal goldenRatioMin1 = goldenRatio.subtract(BigDecimal.ONE);
    private static BigDecimal sqrt5 = new BigDecimal(Math.sqrt(5));

    private static BigInteger fib(int n) {
        BigInteger nthTerm = new BigInteger("0");
        if (n == 2)
            nthTerm = BigInteger.ONE;
        else {
            BigDecimal minResult = goldenRatio.pow(n).subtract(goldenRatioMin1.pow(n));
            nthTerm = minResult.divide(sqrt5,0).toBigInteger();

            if (n % 2 == 1 && n < 45){
                nthTerm = nthTerm.add(BigInteger.ONE);
            }

        }
        return nthTerm;
    }

    private static int fib2(int n) {
        int nthTerm = 0;
        if (n == 2)
            nthTerm = 1;
        else {
            double goldenRatio = (1 + Math.sqrt(5)) / 2;
            nthTerm = (int) (Math.round(Math.pow(goldenRatio, n)
                    - Math.pow(1 - goldenRatio, n)) / Math.sqrt(5));

            if (n % 2 == 1 && n < 45)
                nthTerm++;
        }
        return nthTerm;
    }

    public static void main(String []args){
        System.out.println(
                fib(47)
        );
    }

}

方法fib2是您的代码，fib被转换为BigInteger.干杯

Method fib2 is your code, fib is the transformed into BigInteger. Cheers

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