如何限制2十进制变量转换为第三个变量，反之亦然？ [英] How to convert 2 restricted decimal variables to a third variable and vice versa?

问题描述

我有如下2转换器的方法：

I have 2 convertor methods as below:

private const decimal MaxValidValue = 99.99m;
public decimal ConvertABToC(decimal a, decimal b)
{
    return a * b;
}

public void ConvertCtoAB(decimal c, ref decimal a, ref decimal b)
{
    if (c > MaxValidValue*MaxValidValue)
    {
        throw new ApplicationException();
    }

    if (c <= MaxValidValue)
    {
        a = 1.00m;
        b = c;
    }
    else 
    {
        // need to introduce some logic or assumptions here
    }
}

有知道的3个重要的事情：

There are 3 important things to know:

1）的a和b的变量是在0.00到99.99的范围内，因此C不能有一个大于99.99 * 99.99

1) The a and b variables are in the range of 0.00 to 99.99 therefore c can't have a value greater than 99.99*99.99

2）的a，b和c不能有多于2小数precession如A = 99.123将是无效的。

2) the a, b and c must not have more than 2 decimal precession e.g. a = 99.123 would be invalid.

3），可以使用四舍五入，如果你需要，只要decimal.Round（A * B，2）==℃。

3) you can use rounding if you'd need to as long as decimal.Round(a * b, 2) == c.

4）的组合等（1,3），（3,1），（2,2），（1,4），（0.5，8）或甚至（0.25，16）都是有效的;不要紧，只要为c。将a和b的产物。

4) combinations like (1, 3), (3, 1), (2, 2), (1, 4), (0.5, 8) or even (0.25, 16) are all valid; it doesn't matter as long as c would be the product of a and b.

你将如何完成ConvertCtoAB执行？

How would you complete the implementation of ConvertCtoAB?

非常感谢，

推荐答案

乘 C 10000。然后因素这个数字到它的首要因素。然后找到的质因子的分隔成两个集合，使得在每一组数的乘积小于10,000。如果这样的分区可以发现，然后返回这两款产品由100分为 A 和 B 。否则，添加一个数字，然后重试。

Multiply C by 10,000. Then factor this number into its prime factors. Then find a partition of the prime factors into two sets such that the product of the numbers in each set is less than 10,000. If such a partition can be found, then return these two products divided by 100 as A and B. Otherwise, add one to the number and try again.

例如，如果 C = 100.07 ，然后因素是 2，2，5，5，10007 。由于产品之一必须包括系数 10007 ，这是一个素数，第一个条件就决不会满意。因此，我们尝试再次 1000701 = 3 * 3 * 3 * 13 * 2851 。这个时候，我们可以分区的数量，我们有 A = 3.51 和 B = 28.51 作为一个可能的解决方案。

For example, if C=100.07, then the factors are 2, 2, 5, 5, 10007. Because one of the products must include the factor 10007, which is a prime number, the first condition can never be satisfied. So we try again with 1000701 = 3*3*3*13*2851. This time, we can partition the number, and we have A=3.51 and B=28.51 as a possible solution.

您可以最多99次做到这一点。如果你需要100以上，比不已经产生，从 ConvertABToC 的输入值。

You can do this at most 99 times. If you need 100 or more, than the input value cannot have been generated from ConvertABToC.

这不仅确保了 ConvertCtoAB 的结果，反馈到 ConvertABtoC时将产生相同的 C ，而不是其他的方式。这似乎违反规则＃3，但随后在其他地方的问题是关于四舍五入。

This only guarantees that the result of ConvertCtoAB, when fed back into ConvertABtoC will produce the same C, not the other way around. It appears to violate rule #3, but then elsewhere the question is about rounding.

如果没有舍入都被允许，那么应该停止，并试图在原有的10000 * C后报告不可行。

If no rounding at all is allowed, then one should stop and report infeasibility after trying the original 10000*C.

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