Question

我正在尝试创建一个函数，将一个数字舍入到给定数字的最接近的倍数作为整数。

所以，如果数字是15，我们有

14,4轮至15
-14,4轮到-15
14,5轮到15
28轮到30
-28轮到-30

等等。我已经有了一些代码，但似乎没有按预期工作：

public static int RoundToFactor(float number, float Factor)
{
        int returnNumber;

        if((number%Factor) == 0) {
            returnNumber = (int)number;
        }

        returnNumber = (int) (Mathf.Round(number/Factor)*Factor);

        return returnNumber;
}

Answer 1

这是我制作的方法。它应该适合您的需求。我添加了一个额外的参数，要求在它同样接近时向上或向下舍入。（如果你注意到这些数字看起来像是错误的，例如199舍入到最接近因子2舍入，如果必要，则为200.将199改为-199，结果变为-198，这不是它只是一个小问题。）

public static double RoundToFactor(double Number, int Factor, bool RoundDirection = true)
    {/*round direction: in the event that the distance is 
      * equal from both next factors round up (true) down (false)*/

        double multiplyBy;
        if ((Number % Factor).Equals(0f))
        {
            return Number;
        }
        else
        {
            multiplyBy = Math.Round(Number / Factor);
            int Low = (int)multiplyBy - 1;
            int Mid = (int)multiplyBy;
            int High = (int)multiplyBy + 1;

            List<double> li = new List<double>() { Low, Mid, High };
            double minDelta = double.MaxValue;
            double Closest = 0d;

            foreach (double dbl in li)
            {
                double db = dbl * Factor;
                double Delta = (db < Number) ? (Number - db) : (db - Number);
                if (RoundDirection ? Delta <= minDelta : Delta < minDelta)
                {
                    minDelta = Delta;
                    Closest = db;
                }
            }
            return Closest;

        }

        throw new Exception("Math has broken!!!");
    }

舍入到数字的倍数

1 个答案: