Question

#include <string>
using namespace std;

int main()
{
    int num,
        leftDigit,
        rightDigit;
    string ones[] = { "Zero",
        "One",
        "Two",
        "Three",
        "Four",
        "Five",
        "Six",
        "Seven",
        "Eight",
        "Nine",
        "Ten",
        "Eleven",
        "Twelve",
        "Thirteen",
        "Fourteen",
        "Fifteen",
        "Sixteen",
        "Seventeen",
        "Eighteen",
        "Nineteen",
    };
    string tens[] = { "Twenty",
        "Thirty",
        "Forty",
        "Fifty",
        "Sixty",
        "Seventy",
        "Eighty",
        "Ninety",
    };
    cout << "Enter a Number " ;
    cin >> num; 

    if (num <=0 || num >=100 )
    {
        cout << "the Number is not Between 1-99" << endl;

    }

    else if (num >= 1 && num <= 19)
    {

        cout << "The Number you have entered is " << ones[num] << endl;

    }
    else if (num >= 20 && num <= 99)
    {

        leftDigit = num / 10;
        rightDigit = num % 10;

        cout << "the Number you have entered is " << tens[leftDigit - 2] << " " << ones[rightDigit] << endl;
    }

    system("PAUSE");
    return 0;
}

我一直在玩数字到单词，并找到了一个简单的方法来从1-99做任何2位数字。

有什么方法可以扩展我当前的代码以允许3位数字并允许数字1-999以及它有多容易？

Answer 1

您可以使用递归处理更大的数字。使用此程序已经适用于2位数的事实，使用num = num / 100解析数百位数并将余数r = num % 100传递给您创建的函数。通过一点点工作，您可以处理任何长度的数字。

Answer 2

这是SQL中的算法（对不起）。 dbo.NumberWord表格有两列：Number和WordEn。该表格包含数字1到20以及30,40,50,60,70,80和90的英文单词。（一，二，......二十，三十，四十，......，九十）。 / p>

这是一个递归T-SQL函数，用于将1到999,000之间的任何整数转换为英文单词：

CREATE FUNCTION [dbo].[fnNumberToEnglish] (@Number int)
    RETURNS nvarchar(1024)
AS
BEGIN   
    DECLARE
        @English nvarchar(1024) = 
            (SELECT CASE
                WHEN @Number = 0 THEN ''
                WHEN @Number BETWEEN 1 AND 19 THEN
                    (SELECT WordEn FROM dbo.NumberWord
                    WHERE Number = @Number)
                WHEN @Number BETWEEN 20 AND 99 THEN
                    (SELECT WordEn FROM dbo.NumberWord
                    WHERE (Number / 10) = @Number / 10) + N'-' + dbo.fnNumberToEnglish(@Number%10)
                WHEN @Number BETWEEN 100 AND 999 THEN
                    (dbo.fnNumberToEnglish(@Number / 100)) + N' Hundred ' + dbo.fnNumberToEnglish(@Number%100)
                WHEN @Number BETWEEN 1000 AND 999999 THEN
                    (dbo.fnNumberToEnglish(@Number / 1000)) + N' Thousand ' + dbo.fnNumberToEnglish(@Number%1000)
                ELSE
                    N' INVALID INPUT' 
            END);

    SET @English = RTRIM(@English);

    IF RIGHT(@English,1) = '-' BEGIN
        SET @English = RTRIM(LEFT(@English, LEN(@English) - 1));
    END

    RETURN @English;
END

Answer 3

这应该可以解决问题：

#include <string>
#include <iostream>
using namespace std;

string ones[] = { "",
"One",
"Two",
"Three",
"Four",
"Five",
"Six",
"Seven",
"Eight",
"Nine",
"Ten",
"Eleven",
"Twelve",
"Thirteen",
"Fourteen",
"Fifteen",
"Sixteen",
"Seventeen",
"Eighteen",
"Nineteen",
};
string tens[] = { "Twenty",
"Thirty",
"Forty",
"Fifty",
"Sixty",
"Seventy",
"Eighty",
"Ninety",
};
string hundred = "Hundred";
string thousand = "Thousand";
string million = "Million";
string intToWord(int num)
{
    if (num >= 1000000 && num < 1000000000)
    {
        int first = num / 1000000;
        return intToWord(first) + " " + million + " " + intToWord(num - first * 1000000);
    }
    else if (num >= 1000)
    {
        int first = num / 1000;
        return intToWord(first) + " " + thousand + " " + intToWord(num - first * 1000);
    }
    else if (num >= 100)
    {
        int first = num / 100;
        return ones[first] + " " + hundred + " " + intToWord(num - first * 100);
    }
    else if (num >= 20)
    {

        int leftDigit = num / 10;
        int rightDigit = num % 10;
        return tens[leftDigit - 2] + " " + ones[rightDigit];
    }
    else if (num >= 0)
    {
        return ones[num];
    }
    return "number too large or smaller than 1";
}
int main()
{
    int num;
    cout << "Enter a Number ";
    cin >> num;

    cout << endl << intToWord(num) << endl;

    system("PAUSE");
    return 0;
}

此代码经过测试并正常运行。如果您愿意，可以进一步扩展它。它使用递归，这使得这非常容易。

此代码一直有效至999999999。

将3位数字转换为单词C ++

3 个答案: