Question

我有一个双输入“ a”。我的目标是找到一个整数“ b”，这样“ a * b”将产生一个整数，加上一些允许的误差。例如“ 100.227273 * 22 = 2205（+ 0.000006错误）”，我要查找的答案是“ 22”。

我已经研究过this post，但我仅部分理解了最佳答案。我真的可以使用一些帮助来创建完成该任务的算法。我下面有一些适用于某些情况的代码，但不是全部。

    private int FindSmallestMultiplier(double input)
    {
        int numerator = 1;
        int temp;
        double output = input;
        double invert = input;
        int denominator = 1;
        List<int> whole = new List<int>();
        double dec = input;
        int i = -1;
        while (Math.Abs(Math.Round(numerator * input)-numerator*input) > 0.001)
        {
            i = i + 1;
            //seperate out the whole and decimal portions of the input
            whole.Add(Convert.ToInt32(Math.Floor(invert)));
            dec = output - whole[i];
            //get the decimal portion of the input, and invert
            invert =  1 / dec;

            //create nested fraction to determine how far off from a whole integer we are
            denominator = 1;
            numerator = 1;
            for(int j = whole.Count-1; j >= 0; j--)
            {
                temp = whole[j] * denominator + numerator;
                numerator = denominator;
                denominator = temp;
            }
        }
        return numerator;
    }

上面的代码适用于许多输入情况，例如0.3333、0.5。一个不起作用的示例是0.75或0.101，只是为了将对命名为无穷大。请帮助我弄清楚我的代码有什么问题，或者提供一个可以产生预期结果的代码示例。谢谢！

Answer 1

这是链接的问题中描述的方法的示例实现。如果迭代计算连续分数的新系数。这样做，它检查重构的数字是否达到所需的精度。如果是这样，它将返回重构分数的分母作为结果

// Reconstructs a fraction from a continued fraction with the given coefficients
static Tuple<int, int> ReconstructContinuedFraction(List<int> coefficients)
{
    int numerator = coefficients.Last();
    int denominator = 1;

    for(int i = coefficients.Count - 2; i >= 0; --i)
    {
        //swap numerator and denominator (= invert number)
        var temp = numerator;
        numerator = denominator;
        denominator = temp;

        numerator += denominator * coefficients[i];
    }
    return new Tuple<int, int>(numerator, denominator);
}

static int FindSmallestMultiplier(double input, double error)
{
    double remainingToRepresent = input;
    List<int> coefficients = new List<int>();
    while (true)
    {
        //calculate the next coefficient
        var integer = (int)Math.Floor(remainingToRepresent);                
        remainingToRepresent -= integer;
        remainingToRepresent = 1 / remainingToRepresent;
        coefficients.Add(integer);

        //check if we reached the desired accuracy
        var reconstructed = ReconstructContinuedFraction(coefficients);

        var multipliedInput = input * reconstructed.Item2;
        var multipliedInputRounded = Math.Round(multipliedInput);
        if (Math.Abs(multipliedInput - multipliedInputRounded) < error)
            return reconstructed.Item2;
    }
}

一个示例程序，尝试找到Pi的乘数...

public static void Main()
{
    var number = Math.PI;
    var multiplier = FindSmallestMultiplier(number, 0.001);
    Console.WriteLine(number + " * " + multiplier + " = " + number * multiplier);

}

...给出以下输出：

3.14159265358979 * 113 = 354.999969855647

找出最小的整数，该整数乘以已知的双精度数将产生一个整数

1 个答案: