Question

现在要进行更改，以便每个数字都是最后2个数字的总和，因此请使其每个数字都是最后3个数字的总和并且该系列的前3个数字是0、1、2

    static void PrintFibonocciNumbers(int count)
    {
        int[] result = GetFibonocciNumbers(count);
        PrintArray(result);
    }

    static int[] GetFibonocciNumbers(int count)
    {
        int[] result = new int[count];

        for (int i = 0; i< count; ++i)
        {
            //need a special case if i == 0...we know for 0, the number is 0...so
            //also need a special case if i == 1...we know that will be 1
            // so manually set the first 2 values to the numbers that we know...0 and 1
            // then the rest can be calculated automaticlly
            if(i <= 1)
            {
                result[i] = i;
            }
            else
            {
                result[i] = result[i - 3] + result[i - 1];
            }

            //lets run it through the debugger
        }

        return result;
    }

    static int AddInts(int a, int b)
    {
        int c = a + b;

        return c;
    }
}

}

Answer 1

我实际上会使用IEnumerable来生成序列。在我看来，表达起来更容易：

public static IEnumerable<int> GetSequence()
{
        //define the initial values for the iteration
        int first = 0;
        int second = 1;
        int next = 2;

        //return the first 3 items
        yield return first;
        yield return second;
        yield return next;

        //loop forever (in reality, we might overflow) you can also limit this to
        //a specific n
        while (true)
        {
            //shift the 3 values, losing the first one (we no longer need it)
            (first, second, next) = (second, next, first + second + next);
            //return our result and yield the thread
            yield return next;
        }
}

//take the first 15 values of the IEnumerable
foreach (var val in GetSequence().Take(15))
{
  Console.WriteLine(val);
}

视觉工作室中的斐波那契阵列序列

1 个答案: