63

整数を書かれた数値に変換する効率的な方法はありますか。たとえば、次のようになります。

string Written = IntegerToWritten(21);

"Twenty One" を返します。

大規模なルックアップ テーブルを使用せずにこれを行う方法はありますか?

4

11 に答える 11

68

これはかなりうまくいくはずです：

public static class HumanFriendlyInteger
{
    static string[] ones = new string[] { "", "One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine" };
    static string[] teens = new string[] { "Ten", "Eleven", "Twelve", "Thirteen", "Fourteen", "Fifteen", "Sixteen", "Seventeen", "Eighteen", "Nineteen" };
    static string[] tens = new string[] { "Twenty", "Thirty", "Forty", "Fifty", "Sixty", "Seventy", "Eighty", "Ninety" };
    static string[] thousandsGroups = { "", " Thousand", " Million", " Billion" };

    private static string FriendlyInteger(int n, string leftDigits, int thousands)
    {
        if (n == 0)
        {
            return leftDigits;
        }

        string friendlyInt = leftDigits;

        if (friendlyInt.Length > 0)
        {
            friendlyInt += " ";
        }

        if (n < 10)
        {
            friendlyInt += ones[n];
        }
        else if (n < 20)
        {
            friendlyInt += teens[n - 10];
        }
        else if (n < 100)
        {
            friendlyInt += FriendlyInteger(n % 10, tens[n / 10 - 2], 0);
        }
        else if (n < 1000)
        {
            friendlyInt += FriendlyInteger(n % 100, (ones[n / 100] + " Hundred"), 0);
        }
        else
        {
            friendlyInt += FriendlyInteger(n % 1000, FriendlyInteger(n / 1000, "", thousands+1), 0);
            if (n % 1000 == 0)
            {
                return friendlyInt;
            }
        }

        return friendlyInt + thousandsGroups[thousands];
    }

    public static string IntegerToWritten(int n)
    {
        if (n == 0)
        {
            return "Zero";
        }
        else if (n < 0)
        {
            return "Negative " + IntegerToWritten(-n);
        }

        return FriendlyInteger(n, "", 0);
    }
}

(100 万、10 億などのバグを修正するために編集)

于 2008-08-06T10:31:24.170 に答える
21

Humanizer という便利なライブラリを使用しています。

https://github.com/Humanizr/Humanizer

複数の文化をサポートし、数字だけでなく日付も変換します。使い方は非常に簡単です。

使用方法は次のとおりです。

int someNumber = 543;
var culture = System.Globalization.CultureInfo("en-US");
var result = someNumber.ToWords(culture); // 543 -> five hundred forty-three

そして出来上がり！

于 2016-10-07T12:46:09.923 に答える
9

このコードを使用します。これは VB コードですが、簡単に C# に変換できます。できます

Function NumberToText(ByVal n As Integer) As String

   Select Case n
Case 0
  Return ""

Case 1 To 19
  Dim arr() As String = {"One","Two","Three","Four","Five","Six","Seven", _
    "Eight","Nine","Ten","Eleven","Twelve","Thirteen","Fourteen", _
      "Fifteen","Sixteen","Seventeen","Eighteen","Nineteen"}
  Return arr(n-1) & " "

Case 20 to 99
  Dim arr() as String = {"Twenty","Thirty","Forty","Fifty","Sixty","Seventy","Eighty","Ninety"}
  Return arr(n\10 -2) & " " & NumberToText(n Mod 10)

Case 100 to 199
  Return "One Hundred " & NumberToText(n Mod 100)

Case 200 to 999
  Return NumberToText(n\100) & "Hundreds " & NumberToText(n mod 100)

Case 1000 to 1999
  Return "One Thousand " & NumberToText(n Mod 1000)

Case 2000 to 999999
  Return NumberToText(n\1000) & "Thousands " & NumberToText(n Mod 1000)

Case 1000000 to 1999999
  Return "One Million " & NumberToText(n Mod 1000000)

Case 1000000 to 999999999
  Return NumberToText(n\1000000) & "Millions " & NumberToText(n Mod 1000000)

Case 1000000000 to 1999999999
  Return "One Billion " & NumberTotext(n Mod 1000000000)

Case Else
  Return NumberToText(n\1000000000) & "Billion " _
    & NumberToText(n mod 1000000000)
End Select
End Function

これがc#のコードです

public static string AmountInWords(double amount)
{
        var n = (int)amount;

        if (n == 0)
            return "";
        else if (n > 0 && n <= 19)
        {
            var arr = new string[] { "One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine", "Ten", "Eleven", "Twelve", "Thirteen", "Fourteen", "Fifteen", "Sixteen", "Seventeen", "Eighteen", "Nineteen" };
            return arr[n - 1] + " ";
        }
        else if (n >= 20 && n <= 99)
        {
            var arr = new string[] { "Twenty", "Thirty", "Forty", "Fifty", "Sixty", "Seventy", "Eighty", "Ninety" };
            return arr[n / 10 - 2] + " " + AmountInWords(n % 10);
        }
        else if (n >= 100 && n <= 199)
        {
            return "One Hundred " + AmountInWords(n % 100);
        }
        else if (n >= 200 && n <= 999)
        {
            return AmountInWords(n / 100) + "Hundred " + AmountInWords(n % 100);
        }
        else if (n >= 1000 && n <= 1999)
        {
            return "One Thousand " + AmountInWords(n % 1000);
        }
        else if (n >= 2000 && n <= 999999)
        {
            return AmountInWords(n / 1000) + "Thousand " + AmountInWords(n % 1000);
        }
        else if (n >= 1000000 && n <= 1999999)
        {
            return "One Million " + AmountInWords(n % 1000000);
        }
        else if (n >= 1000000 && n <= 999999999)
        {
            return AmountInWords(n / 1000000) + "Million " + AmountInWords(n % 1000000);
        }
        else if (n >= 1000000000 && n <= 1999999999)
        {
            return "One Billion " + AmountInWords(n % 1000000000);
        }
        else
        {
            return AmountInWords(n / 1000000000) + "Billion " + AmountInWords(n % 1000000000);
        }
    }
于 2008-08-06T09:46:18.863 に答える
3

なぜ大規模なルックアップテーブルなのですか?

string GetWrittenInteger(int n)
{
  string[] a = new string[] {"One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine" }
  string[] b = new string[] { "Ten", "Eleven", "Twelve", "Thirteen", "Fourteen", "Fifteen", "Sixteen", "Seventeen", "Eighteen", "Nineteen" }
  string[] c = new string[] {"Twenty", "Thirty", "Forty", "Sixty", "Seventy", "Eighty", "Ninety"};
  string[] d = new string[] {"Hundred", "Thousand", "Million"}
  string s = n.ToString();

  for (int i = 0; i < s.Length; i++)
  {
    // logic (too lazy but you get the idea)
  }
}
于 2008-08-06T09:22:35.993 に答える
2

受け入れられた答えは完全に機能していないようです。これは、21 のような数字のダッシュを処理しません。また、「100 と 1」のような数字に「and」という単語を入れません。また、再帰的です。

これが私の答えです。「and」という単語をインテリジェントに追加し、数字を適切にハイフネーションします。変更が必要な場合はお知らせください。

これを呼び出す方法は次のとおりです（明らかに、これをどこかのクラスに入れたいと思うでしょう）：

for (int i = int.MinValue+1; i < int.MaxValue; i++)
{
    Console.WriteLine(ToWords(i));
}

コードは次のとおりです。

private static readonly string[] Ones = {"", "One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine"};

private static readonly string[] Teens =
{
    "Ten", "Eleven", "Twelve", "Thirteen", "Fourteen", "Fifteen", "Sixteen",
    "Seventeen", "Eighteen", "Nineteen"
};

private static readonly string[] Tens =
{
    "", "", "Twenty", "Thirty", "Forty", "Fifty", "Sixty", "Seventy", "Eighty",
    "Ninety"
};

public static string ToWords(int number)
{
    if (number == 0)
        return "Zero";

    var wordsList = new List<string>();

    if (number < 0)
    {
        wordsList.Add("Negative");
        number = Math.Abs(number);
    }

    if (number >= 1000000000 && number <= int.MaxValue) //billions
    {
        int billionsValue = number / 1000000000;
        GetValuesUnder1000(billionsValue, wordsList);
        wordsList.Add("Billion");
        number -= billionsValue * 1000000000;

        if (number > 0 && number < 10)
            wordsList.Add("and");
    }

    if (number >= 1000000 && number < 1000000000) //millions
    {
        int millionsValue = number / 1000000;
        GetValuesUnder1000(millionsValue, wordsList);
        wordsList.Add("Million");
        number -= millionsValue * 1000000;

        if (number > 0 && number < 10)
            wordsList.Add("and");
    }

    if (number >= 1000 && number < 1000000) //thousands
    {
        int thousandsValue = number/1000;
        GetValuesUnder1000(thousandsValue, wordsList);
        wordsList.Add("Thousand");
        number -= thousandsValue * 1000;

        if (number > 0 && number < 10)
            wordsList.Add("and");
    }

    GetValuesUnder1000(number, wordsList);

    return string.Join(" ", wordsList);
}

private static void GetValuesUnder1000(int number, List<string> wordsList)
{
    while (number != 0)
    {
        if (number < 10)
        {
            wordsList.Add(Ones[number]);
            number -= number;
        }
        else if (number < 20)
        {
            wordsList.Add(Teens[number - 10]);
            number -= number;
        }
        else if (number < 100)
        {
            int tensValue = ((int) (number/10))*10;
            int onesValue = number - tensValue;

            if (onesValue == 0)
            {
                wordsList.Add(Tens[tensValue/10]);
            }
            else
            {
                wordsList.Add(Tens[tensValue/10] + "-" + Ones[onesValue]);
            }

            number -= tensValue;
            number -= onesValue;
        }
        else if (number < 1000)
        {
            int hundredsValue = ((int) (number/100))*100;
            wordsList.Add(Ones[hundredsValue/100]);
            wordsList.Add("Hundred");
            number -= hundredsValue;

            if (number > 0)
                wordsList.Add("and");
        }
    }
}
于 2015-05-26T20:18:11.677 に答える
2 
using System;
using System.Collections.Generic; 
using System.Linq; 
using System.Text; 

namespace tryingstartfror4digits 
{ 
    class Program 
    { 
        static void Main(string[] args)
        {
            Program pg = new Program();
            Console.WriteLine("Enter ur number");
            int num = Convert.ToInt32(Console.ReadLine());

            if (num <= 19)
            {
                string g = pg.first(num);
                Console.WriteLine("The number is " + g);
            }
            else if ((num >= 20) && (num <= 99))
            {
                    if (num % 10 == 0)
                    {
                        string g = pg.second(num / 10);
                        Console.WriteLine("The number is " + g);
                    }
                    else
                    {
                        string g = pg.second(num / 10) + pg.first(num % 10);
                        Console.WriteLine("The number is " + g);
                    }
            }
            else if ((num >= 100) && (num <= 999))
            {
                int k = num % 100;
                string g = pg.first(num / 100) +pg.third(0) + pg.second(k / 10)+pg.first(k%10);
                Console.WriteLine("The number is " + g);
            }
            else if ((num >= 1000) && (num <= 19999))
            {
                int h = num % 1000;
                int k = h % 100;
                string g = pg.first(num / 1000) + "Thousand " + pg.first(h/ 100) + pg.third(k) + pg.second(k / 10) + pg.first(k % 10);
                Console.WriteLine("The number is " + g);
            }

            Console.ReadLine();
        }

        public string first(int num)
        {
            string name;

            if (num == 0)
            {
                name = " ";
            }
            else
            {
                string[] arr1 = new string[] { "One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine" , "Ten", "Eleven", "Twelve", "Thirteen", "Fourteen", "Fifteen", "Sixteen", "Seventeen", "Eighteen", "Nineteen"};
                name = arr1[num - 1];
            }

            return name;
        }

        public string second(int num)
        {
            string name;

            if ((num == 0)||(num==1))
            {
                 name = " ";
            }
            else
            {
                string[] arr1 = new string[] { "Twenty", "Thirty", "Forty", "Fifty", "Sixty", "Seventy", "Eighty", "Ninety" };
                name = arr1[num - 2];
            }

            return name;
        }

        public string third(int num)
        {
            string name ;

            if (num == 0)
            {
                name = "";
            }
            else
            {
                string[] arr1 = new string[] { "Hundred" };
                name = arr1[0];
            }

            return name;
        }
    }
}

これは 1 から 19999 まで正常に動作します。完了後すぐに更新されます

于 2011-12-04T18:18:02.027 に答える
1

以下は、小数だけでなく整数も返すC# コンソール アプリケーションです。

于 2012-08-13T23:24:38.623 に答える
1

その文字列を取得して、string s=txtNumber.Text.Tostring(); のように変換するだけです。int i=Convert.ToInt32(s.Tostring()); 完全な整数値のみを書き込みます

于 2015-08-05T08:20:33.477 に答える