c# - 我的 Pandigital To Roman Number 不工作

Question

我必须编写一个 C# 程序，在该程序中，我根据输入的数字生成罗马数字。出于某种原因，我无法在输入时获取代码来给我一个值。

namespace RomaiSzam

    class ToRomanNumber
    {
        string s = "";

        public string ToRoman(int number)
        {

            if ((number < 0) || (number > 3999))
            {
                s = s + "Invalid input";
            }
            if (number < 1) return s;
            if (number >= 1000) { s = s + "M"; ToRoman(number - 1000); }
            if (number >= 900) { s = s + "CM"; ToRoman(number - 900); }
            if (number >= 500) { s = s + "D"; ToRoman(number - 500); }
            if (number >= 400) { s = s + "CD"; ToRoman(number - 400); }
            if (number >= 100) { s = s + "C"; ToRoman(number - 100); }
            if (number >= 90) { s = s + "XC"; ToRoman(number - 90); }
            if (number >= 50) { s = s + "L"; ToRoman(number - 50); }
            if (number >= 40) { s = s + "XL"; ToRoman(number - 40); }
            if (number >= 10) { s = s + "X"; ToRoman(number - 10); }
            if (number >= 9) { s = s + "IX"; ToRoman(number - 9); }
            if (number >= 5) { s = s + "V"; ToRoman(number - 5); }
            if (number >= 4) { s = s + "IV"; ToRoman(number - 4); }
            if (number >= 1) { s = s + "I"; ToRoman(number - 1); }
            return s;

        }
  }

命名空间 RomaiSzam

class Program
{
    static void Main(string[] args)
    {
        Console.WriteLine("Write the number");
        int num = int.Parse(Console.ReadLine());
        ToRomanNumber tr = new ToRomanNumber();
        string opt = tr.ToRoman(num);
        Console.WriteLine(opt);

    }
}

这是代码。有什么问题？

Answer 1

您正在遍历所有各种if...您不应该...

这可能是使其正常工作所需的最小逻辑更改。

class ToRomanNumber
{
    string s = "";

    public string ToRoman(int number)
    {

        if ((number < 0) || (number > 3999))
        {
            s = s + "Invalid input";
        }
        if (number < 1) return s;
        if (number >= 1000) { s = s + "M"; ToRoman(number - 1000); return s; }
        if (number >= 900) { s = s + "CM"; ToRoman(number - 900); return s; }
        if (number >= 500) { s = s + "D"; ToRoman(number - 500); return s; }
        if (number >= 400) { s = s + "CD"; ToRoman(number - 400); return s; }
        if (number >= 100) { s = s + "C"; ToRoman(number - 100); return s; }
        if (number >= 90) { s = s + "XC"; ToRoman(number - 90); return s; }
        if (number >= 50) { s = s + "L"; ToRoman(number - 50); return s; }
        if (number >= 40) { s = s + "XL"; ToRoman(number - 40); return s; }
        if (number >= 10) { s = s + "X"; ToRoman(number - 10); return s; }
        if (number >= 9) { s = s + "IX"; ToRoman(number - 9); return s; }
        if (number >= 5) { s = s + "V"; ToRoman(number - 5); return s; }
        if (number >= 4) { s = s + "IV"; ToRoman(number - 4); return s; }
        if (number >= 1) { s = s + "I"; ToRoman(number - 1); return s; }
        return s;

    }
}

请注意，您的写法非常难看。像这样使用实例变量 s 让我起鸡皮疙瘩。

至少你应该这样做static，像这样:

class ToRomanNumber
{
    public static string ToRoman(int number)
    {
        if ((number < 0) || (number > 3999))
        {
            return "Invalid input";
        }
        if (number >= 1000) { return "M" + ToRoman(number - 1000); }
        if (number >= 900) { return "CM" + ToRoman(number - 900); }
        if (number >= 500) { return "D" + ToRoman(number - 500); }
        if (number >= 400) { return "CD" + ToRoman(number - 400); }
        if (number >= 100) { return "C" + ToRoman(number - 100); }
        if (number >= 90) { return "XC" + ToRoman(number - 90); }
        if (number >= 50) { return "L" + ToRoman(number - 50); }
        if (number >= 40) { return "XL" + ToRoman(number - 40); }
        if (number >= 10) { return "X" + ToRoman(number - 10); }
        if (number >= 9) { return "IX" + ToRoman(number - 9); }
        if (number >= 5) { return "V" + ToRoman(number - 5); }
        if (number >= 4) { return "IV" + ToRoman(number - 4); }
        if (number >= 1) { return "I" + ToRoman(number - 1); }
        return string.Empty;
    }
}

像这样使用它

string str = ToRomanNumber.ToRoman(5);

而且我认为它仍然有点难看，因为它是递归的...为此使用递归是.... bleah :-)

我通常会这样写:

class ToRomanNumber
{
    public static string ToRoman(int number)
    {
        if ((number < 0) || (number > 3999))
        {
            return "Invalid input";
        }

        var sb = new StringBuilder();

        while (number > 0)
        {
            if (number >= 1000) { sb.Append("M"); number -= 1000; }
            else if (number >= 900) { sb.Append("CM"); number -= 900; }
            else if (number >= 500) { sb.Append("D"); number -= 500; }
            else if (number >= 400) { sb.Append("CD"); number -= 400; }
            else if (number >= 100) { sb.Append("C"); number -= 100; }
            else if (number >= 90) { sb.Append("XC"); number -= 90; }
            else if (number >= 50) { sb.Append("L"); number -= 50; }
            else if (number >= 40) { sb.Append("XL"); number -= 40; }
            else if (number >= 10) { sb.Append("X"); number -= 10; }
            else if (number >= 9) { sb.Append("IX"); number -= 9; }
            else if (number >= 5) { sb.Append("V"); number -= 5; }
            else if (number >= 4) { sb.Append("IV"); number -= 4; }
            else if (number >= 1) { sb.Append("I"); number -= 1; }
        }

        return sb.ToString();
    }
}

所以有一个循环。