使用二维数组的第 N 个斐波那契数

Question

我怎样才能做这样的事情？我需要创建一个斐波那契 class 和一个 next () 方法 return 下一个斐波那契字符串值。后续调用应return：0、1、1、2、3、5、8等

程序从用户那里获取一个整数，returns 指定数量的字符串值。应使用数组进行计算。

public class Fibonacci {
    final long[][] A = {{1, 1}, {1, 0}};

    public static void  main(String[] args) {
        System.out.print("Enter the n th word of the sequence: ");
        long n = initialStatement(readValue());

        Fibonacci f = new Fibonacci();
        for (int i = 0; i < n; i++) {
            long[][] b = f.next();
            System.out.print(b[0][1]);
        }

    }


    public static long readValue() {
        Scanner scanner = new Scanner(System.in);
        return scanner.nextLong();

    }

    public static long initialStatement(long n) {
        do {
            if (n == 0 || n == 1) {
                return n;
            } else if (n < 0) {
                System.out.print("Wrong value, please enter correct value: ");
                n = Fibonacci.readValue();
            }
        } while (n < 0);
        return n;
    }

    public long[][] next() {

        long[][] a =new long[2][2];


        for (int i = 0; i < 2; i++) {
            for (int j = 0; j < 2; j++) {
                    a[i][j] += a[i][j] * A[j][i];
            }
        }
        return a ;
    }
}

Answer 1

斐波那契数列可以描述为：

fibo(0) = 0
fibo(1) = 1
fibo(n) = fibo(n - 1) + fibo(n - 2)

在代码中，最简单的实现不使用矩阵。您可以在 class:

中执行类似的操作

private int a = 0;
private int b = 0;

public void next() {
    
    // Special case: Return 0 as first number in the sequence.
    if(a == 0 && b == 0) {
        b = 1;
        return 0;
    }

    // Return the sum of the last two numbers, and store the new last two numbers in a and b.
    int result = a + b;
    a = b;
    b = result;
    return result;
}

这将 return 从 0 开始的斐波那契数列。

Answer 2

你的矩阵很混乱。

这里是用两个 int 而不是 matrix 对代码进行的简单更改。

测试一下然后告诉我。

public class Fibonacci {
    long i = -1; // clone of the loop counter i
    long fibo1 = 1;
    long fibo2 = 1;

    public  void  main(String[] args) {
        System.out.print("Enter the n th word of the sequence: ");
        long n = initialStatement(readValue());

        Fibonacci f = new Fibonacci();
        for (int i = 0; i < n; i++) {
            System.out.print(f.next());
        }

    }


    public static long readValue() {
        Scanner scanner = new Scanner(System.in);
        return scanner.nextLong();

    }

    public static long initialStatement(long n) {
        do {
            if (n == 0 || n == 1) {
                return 1; // fibo(0) == fibo(1) == 1
            } else if (n < 0) {
                System.out.print("Wrong value, please enter correct value: ");
                n = Fibonacci.readValue();
            }
        } while (n < 0);
        return n;
    }

    public String next() {
        i++; // simulation of loop i because it is not a param of next()
        if(i >= 2) {
            long aux = fibo1 + fibo2;
            fibo1 = fibo2;
            fibo2 = aux;
        }
        return ""+ fibo2;
    }
}

使用二维数组的第 N 个斐波那契数

Nth Fibonacci number using 2D arrays

java

arrays

matrix

sequence

fibonacci