返回第 N 个斐波那契数列？

Question

我对课堂作业有疑问，我需要知道如何使用迭代返回斐波那契数列的第 n 个数（不允许递归）。

我需要一些关于如何执行此操作的提示，以便我可以更好地理解我做错了什么。我在我的program.cs中输出到控制台，因此它在下面的代码中不存在。

    // Q1)
    //
    // Return the Nth Fibonacci number in the sequence
    //
    // Input: uint n (which number to get)
    // Output: The nth fibonacci number
    //

    public static UInt64 GetNthFibonacciNumber(uint n)
    {

    // Return the nth fibonacci number based on n.


    if (n == 0 || n == 1)
        {
            return 1;
        }

        // The basic Fibonacci sequence is 
        // 1, 1, 2, 3, 5, 8, 13, 21, 34...
        // f(0) = 1
        // f(1) = 1
        // f(n) = f(n-1) + f(n-2)
        ///////////////
        //my code is below this comment

        uint a = 0;
        uint b = 1;

        for (uint i = 0; i < n; i++)
        {
            n = b + a;
            a = b;
            b = n;
        }
        return n;

Answer 1

:)

static ulong Fib(int n) 
{
    double sqrt5 = Math.Sqrt(5);
    double p1 = (1 + sqrt5) / 2;
    double p2 = -1 * (p1 - 1);


    double n1 = Math.Pow(p1, n + 1);
    double n2 = Math.Pow(p2, n + 1);
    return (ulong)((n1 - n2) / sqrt5);
}

Answer 2

只是为了一点乐趣，你可以使用无限的斐波那契列表和一些 IEnumerable 扩展来做到这一点

public IEnumerable<int> Fibonacci(){
   var current = 1;
   var b = 0;
   while(true){
       var next = current + b;
       yield return next;
       b = current;
       current = next;
   }
}

public T Nth<T>(this IEnumerable<T> seq, int n){
    return seq.Skip.(n-1).First();
}

得到第n个数字就是

Fibonacci().Nth(n);

Answer 3

        public static int GetNthFibonacci(int n)
    {
        var previous = -1;
        var current = 1;
        int index = 1;
        int element = 0;

        while (index++ <= n)
        {
            element = previous + current;
            previous = current;
            current = element;
        }

        return element;
    }

Answer 4

我认为这应该可以解决问题：

    uint a = 0;
    uint b = 1;
    uint c = 1;

    for (uint i = 0; i < n; i++)
    {
        c = b + a;
        a = b;
        b = c;
    }
    return c;

Answer 5

    public IEnumerable<BigInteger> FibonacciBig(int maxn)
    {
        BigInteger Fn=1;
        BigInteger Fn_1=1;
        BigInteger Fn_2=1;

        yield return Fn;
        yield return Fn;

        for (int i = 3; i < maxn; i++)
        {
            Fn = Fn_1 + Fn_2;

            yield return Fn;

            Fn_2 = Fn_1;
            Fn_1 = Fn;
        }


    }

你可以通过

得到第n个数字

   FibonacciBig(100000).Skip(n).First();

Answer 6

这是你作业的答案，你应该从 3 开始，因为你已经有了 f1 和 f2 的数字（前两个数字）。请注意，获取第 0 个斐波那契数是没有意义的。

public static UInt64 GetNthFibonacciNumber(uint n)
    {

    // Return the nth fibonacci number based on n.


if (n == 1 || n == 2)
    {
        return 1;
    }


    uint a = 1;
    uint b = 1;
    uint c;

    for (uint i = 3; i <= n; i++)
    {
        c = b + a;
        a = b;
        b = c;
    }
    return c;

}

Answer 7

    public static UInt64 GetNthFibonacciNumber(uint n)
    {
        if (n == 0 || n == 1)
        {
            return 1;
        }
        UInt64 a = 1, b = 1;
        uint i = 2;
        while (i <= n)
        {
            if (a > b) b += a;
            else a += b;
            ++i;
        }
        return (a > b) ? a : b;
    }

Answer 8

 public static List<int> PrintFibonacci(int number)
        {
            List<int> result = new List<int>();
            if (number == 0)
            {
                result.Add(0);
                return result;
            }
            else if (number == 1)
            {
                result.Add(0);
                return result;
            }
            else if (number == 2)
            {
                result.AddRange(new List<int>() { 0, 1 });
                return result;
            }
            else
            {
                //if we got thus far,we should have f1,f2 and f3 as fibonacci numbers
                int f1 = 0,
                    f2 = 1;

                result.AddRange(new List<int>() { f1, f2 });
                for (int i = 2; i < number; i++)
                {
                    result.Add(result[i - 1] + result[i - 2]);
                }
            }
            return result;

        }

Answer 9

只需要 2 个变量（在 for 循环中声明一个也算）。

public int NthFib(int n)
{
    int curFib = 0;
    int nextFib = 1;
    while (--n > 0)
    {
        nextFib += curFib;
        curFib = nextFib - curFib;
    }
    return curFib;
}

如果您想查看 n 的序列，请将其更改为：

public IEnumerable<int> NthFib(int n)
{
    int curFib = 0;
    int nextFib = 1;
    while (n-- > 0)
    {
        yield return curFib;
        nextFib += curFib;
        curFib = nextFib - curFib;
    }
}

Answer 10

    public long Fibonacci(int n)
    {
        int a = 0;

        int b = 1;

        int c = 0;

        for (int i = 0; i < n; i++)
        {
            c = a + b;
            a = b;
            b = c;
        }

        return a;
    }