C#纯数学方法递归实现货币数字转换中文

概述最近由于项目的原因，需要写一个货币数字转换中文的算法，先在网了找了一下，结果发现无一列外都是用(Replace)替换的方式来实现的，所以想写个另外的算法；因为本人是学数学出身的，所以用纯数学的方法实现。注意：本文中的算法支持小于1023 (也就是9999亿兆)货币数字转化。 货币中文说明： 在说明代码之前，首先让我们回顾一下货币的读法。10020002.23  读为 壹仟零贰万零贰元贰角叁分1020         读为 壹仟零贰拾元整。100000       读为 拾万元整0.13         读为 壹角叁分代码：测试工程  static void Main(string[] args){Console.WriteLine("请输入金额");string inputNum = Console.ReadLine();while (inputNum != "exit"){//货币数字转化类NumCast nc = new NumCast();if (nc.IsValidated<string>(inputNum)){try{string chineseCharacter = nc.ConvertToChinese(inputNum);Console.WriteLine(chineseCharacter);}catch (Exception er){Console.WriteLine(er.Message);}}else{Console.WriteLine("不合法的数字或格式");}Console.WriteLine("n请输入金额");inputNum = Console.ReadLine();}Console.ReadLine();}测试结果如下：货币转化类（NumCast类）功能介绍1 常量的规定 /// <summary>/// 数位/// </summary>public enum NumLevel { Cent, Chiao, Yuan, Ten, Hundred, Thousand, TenThousand, hundredMillon, Trillion };/// <summary>/// 数位的指数/// </summary>private int[] NumLevelExponent = new int[] { -2, -1, 0, 1, 2, 3, 4, 8, 12 };/// <summary>/// 数位的中文字符/// </summary>private string[] NumLeverChineseSign = new string[] { "分", "角", "元", "拾", "佰", "仟", "万", "亿", "兆" };/// <summary>/// 大写字符/// </summary>private string[] NumChineseCharacter = new string[] {"零","壹","贰","叁","肆","伍","陆","柒","捌","玖"};/// <summary>/// 整(当没有 角分 时)/// </summary>private const string EndOfInt = "整";2：数字合法性验证，采用正则表达式验证 /// <summary>/// 正则表达验证数字是否合法/// </summary>/// <param name="Num"></param>/// <returns></returns>public bool IsValidated<T>(T Num){Regex reg = new Regex(@"^(([0])|([1-9]d{0,23}))(.d{1,2})?$");if (reg.IsMatch(Num.ToString())){return true;}return false;}3： 获取数位 例如 1000的数位为 NumLevel.Thousand/// <summary>/// 获取数字的数位使用log/// </summary>/// <param name="Num"></param>/// <returns></returns>private NumLevel GetNumLevel(double Num){double numLevelLength;NumLevel NLvl = new NumLevel();if (Num > 0){numLevelLength = Math.Floor(Math.Log10(Num));for (int i = NumLevelExponent.Length - 1; i >= 0; i--){if (numLevelLength >= NumLevelExponent[i]){NLvl = (NumLevel)i;break;}}}else{NLvl = NumLevel.Yuan;}return NLvl;}4：判断数字之间是否有跳位，也就是中文中间是否要加零，例如1020 就应该加零。 /// <summary>/// 是否跳位/// </summary>/// <returns></returns>private bool IsDumpLevel(double Num){ if (Num > 0){NumLevel? currentLevel = GetNumLevel(Num);NumLevel? nextLevel = null;int numExponent = this.NumLevelExponent[(int)currentLevel];double postfixNun = Math.Round(Num % (Math.Pow(10, numExponent)),2);if(postfixNun> 0)nextLevel = GetNumLevel(postfixNun);if (currentLevel != null && nextLevel != null){if (currentLevel > nextLevel + 1){return true;}}}return false;}5 把长数字分割为两个较小的数字数组，例如把9999亿兆，分割为9999亿和0兆，因为计算机不支持过长的数字。 /// <summary>/// 是否大于兆，如果大于就把字符串分为两部分，/// 一部分是兆以前的数字/// 另一部分是兆以后的数字/// </summary>/// <param name="Num"></param>/// <returns></returns>private bool IsBigThanTillion(string Num){bool isBig = false;if (Num.IndexOf('.') != -1){//如果大于兆if (Num.IndexOf('.') > NumLevelExponent[(int)NumLevel.Trillion]){isBig = true;}}else{//如果大于兆if (Num.Length > NumLevelExponent[(int)NumLevel.Trillion]){isBig = true;}}return isBig;}/// <summary>/// 把数字字符串由‘兆’分开两个/// </summary>/// <returns></returns>private double[] SplitNum(string Num){//兆的开始位double[] TillionLevelNums = new double[2];int trillionLevelLength;if (Num.IndexOf('.') == -1)trillionLevelLength = Num.Length - NumLevelExponent[(int)NumLevel.Trillion];elsetrillionLevelLength = Num.IndexOf('.') - NumLevelExponent[(int)NumLevel.Trillion];//兆以上的数字TillionLevelNums[0] = Convert.ToDouble(Num.Substring(0, trillionLevelLength));//兆以下的数字TillionLevelNums[1] = Convert.ToDouble(Num.Substring(trillionLevelLength ));return TillionLevelNums;}6 是否以“壹拾”开头，如果是就可以把它变为“拾”  bool isStartOfTen = false;while (Num >=10){if (Num == 10){isStartOfTen = true;break;}//Num的数位NumLevel currentLevel = GetNumLevel(Num);int numExponent = this.NumLevelExponent[(int)currentLevel];Num = Convert.ToInt32(Math.Floor(Num / Math.Pow(10, numExponent)));if (currentLevel == NumLevel.Ten && Num == 1){isStartOfTen = true;break;}}return isStartOfTen;7 合并大于兆连个数组转化成的货币字符串/// <summary>/// 合并分开的数组中文货币字符/// </summary>/// <param name="tillionNums"></param>/// <returns></returns>private string ContactNumChinese(double[] tillionNums){string uptillionStr = CalculateChineseSign(tillionNums[0], NumLevel.Trillion, true, IsStartOfTen(tillionNums[0]));string downtrillionStr = CalculateChineseSign(tillionNums[1], null, true,false);string chineseCharactor = string.Empty;//分开后的字符是否有跳位if (GetNumLevel(tillionNums[1] * 10) == NumLevel.Trillion){chineseCharactor = uptillionStr + NumLeverChineseSign[(int)Nu

最近由于项目的原因，需要写一个货币数字转换中文的算法，先在网了找了一下，结果发现无一列外都是用(Replace)替换的方式来实现的，所以想写个另外的算法；因为本人是学数学出身的，所以用纯数学的方法实现。

　　注意：本文中的算法支持小于1023 (也就是9999亿兆)货币数字转化。

　　货币中文说明： 在说明代码之前，首先让我们回顾一下货币的读法。

　　10020002.23  读为 壹仟零贰万零贰元贰角叁分　　1020         读为 壹仟零贰拾元整。　　100000       读为 拾万元整　　0.13         读为 壹角叁分

　

　　代码：

测试工程  static voID Main(string[] args){

Console.Writeline("请输入金额");string inputNum = Console.Readline();while (inputNum != "exit"){

//货币数字转化类NumCast nc = new NumCast();

if (nc.IsValIDated(inputNum)){try{string chineseCharacter = nc.Converttochinese(inputNum);Console.Writeline(chineseCharacter);}catch (Exception er){Console.Writeline(er.Message);}}else{Console.Writeline("不合法的数字或格式");}

Console.Writeline("\n请输入金额");inputNum = Console.Readline();

}Console.Readline();}

测试结果如下：

货币转化类（NumCast类）功能介绍

1 常量的规定 ///

/// 数位///

public enum NumLevel { Cent,Chiao,Yuan,Ten,Hundred,Thousand,TenThousand,hundredMillon,Trillion };

///

/// 数位的指数///

private int[] NumLevelExponent = new int[] { -2,-1,1,2,3,4,8,12 };

///

/// 数位的中文字符///

private string[] NumLeverChineseSign = new string[] { "分","角","元","拾","佰","仟","万","亿","兆" };

///

/// 大写字符///

private string[] NumChineseCharacter = new string[] {"零","壹","贰","叁","肆","伍","陆","柒","捌","玖"};

///

/// 整(当没有 角分 时)///

private const string EndOfInt = "整";

2：数字合法性验证，采用正则表达式验证 ///

/// 正则表达验证数字是否合法///

/// /// public bool IsValIDated(T Num){Regex reg = new Regex(@"^(([0])|([1-9]\d{0,23}))(\.\d{1,2})?$");if (reg.IsMatch(Num.ToString())){return true;

}return false;}

3： 获取数位 例如 1000的数位为 NumLevel.Thousand

///

/// 获取数字的数位　使用log///

/// /// private NumLevel GetNumLevel(double Num){double numLevelLength;NumLevel NLvl = new NumLevel();if (Num > 0){numLevelLength = Math.Floor(Math.Log10(Num));for (int i = NumLevelExponent.Length - 1; i >= 0; i--){if (numLevelLength >= NumLevelExponent[i]){NLvl = (NumLevel)i;break;}}}else{NLvl = NumLevel.Yuan;}return NLvl;

}

4：判断数字之间是否有跳位，也就是中文中间是否要加零，例如1020 就应该加零。 ///

/// 是否跳位///

/// private bool IsDumpLevel(double Num){ if (Num > 0){NumLevel? currentLevel = GetNumLevel(Num);NumLevel? nextLevel = null;int numExponent = this.NumLevelExponent[(int)currentLevel];

double postfixnun = Math.Round(Num % (Math.Pow(10,numExponent)),2);if(postfixnun> 0)nextLevel = GetNumLevel(postfixnun);if (currentLevel != null && nextLevel != null){if (currentLevel > nextLevel + 1){return true;}}}return false;

}

5 把长数字分割为两个较小的数字数组，例如把9999亿兆，分割为9999亿和0兆，

因为计算机不支持过长的数字。 ///

/// 是否大于兆，如果大于就把字符串分为两部分，/// 一部分是兆以前的数字/// 另一部分是兆以后的数字///

/// /// private bool IsBigThanTillion(string Num){bool isBig = false;if (Num.IndexOf('.') != -1){//如果大于兆if (Num.IndexOf('.') > NumLevelExponent[(int)NumLevel.Trillion]){isBig = true;}}else{//如果大于兆if (Num.Length > NumLevelExponent[(int)NumLevel.Trillion]){isBig = true;}}return isBig;}

///

/// 把数字字符串由‘兆’分开两个///

/// private double[] SplitNum(string Num){//兆的开始位double[] TillionLevelNums = new double[2];int trillionLevelLength;if (Num.IndexOf('.') == -1)trillionLevelLength = Num.Length - NumLevelExponent[(int)NumLevel.Trillion];elsetrillionLevelLength = Num.IndexOf('.') - NumLevelExponent[(int)NumLevel.Trillion];//兆以上的数字TillionLevelNums[0] = Convert.Todouble(Num.Substring(0,trillionLevelLength));//兆以下的数字TillionLevelNums[1] = Convert.Todouble(Num.Substring(trillionLevelLength ));return TillionLevelNums;}

6 是否以“壹拾”开头，如果是就可以把它变为“拾”  bool isstartOfTen = false;while (Num >=10){if (Num == 10){isstartOfTen = true;break;}//Num的数位NumLevel currentLevel = GetNumLevel(Num);

int numExponent = this.NumLevelExponent[(int)currentLevel];Num = Convert.ToInt32(Math.Floor(Num / Math.Pow(10,numExponent)));if (currentLevel == NumLevel.Ten && Num == 1){isstartOfTen = true;break;}}return isstartOfTen;

7 合并大于兆连个数组转化成的货币字符串

///

/// 合并分开的数组中文货币字符///

/// /// private string ContactNumChinese(double[] tillionNums){string uptillionStr = CalculateChineseSign(tillionNums[0],NumLevel.Trillion,true,IsstartOfTen(tillionNums[0]));string downtrillionStr = CalculateChineseSign(tillionNums[1],null,false);string chineseCharactor = string.Empty;//分开后的字符是否有跳位if (GetNumLevel(tillionNums[1] * 10) == NumLevel.Trillion){chineseCharactor = uptillionStr + NumLeverChineseSign[(int)NumLevel.Trillion] + downtrillionStr;}else{chineseCharactor = uptillionStr + NumLeverChineseSign[(int)NumLevel.Trillion];if (downtrillionStr != "零元整"){chineseCharactor += NumChineseCharacter[0] + downtrillionStr;}else{chineseCharactor += "元整";}}return chineseCharactor;

}

8：递归计算货币数字的中文 ///

/// 计算中文字符串///

/// /// /// /// 中文大写public string CalculateChineseSign(double Num,NumLevel? NL,bool IsDump,bool IsExceptTen){Num = Math.Round(Num,2);bool isDump = false;//Num的数位NumLevel? currentLevel = GetNumLevel(Num);int numExponent = this.NumLevelExponent[(int)currentLevel];

string Result = string.Empty;

//整除后的结果int prefixnum;//余数 当为小数的时候 分子分母各乘100double postfixnun ;if (Num >= 1){prefixnum = Convert.ToInt32(Math.Floor(Num / Math.Pow(10,numExponent)));postfixnun = Math.Round(Num % (Math.Pow(10,2);}else{prefixnum = Convert.ToInt32(Math.Floor(Num*100 / Math.Pow(10,numExponent+2)));postfixnun = Math.Round(Num * 100 % (Math.Pow(10,numExponent + 2)),2);postfixnun *= 0.01;}

if (prefixnum < 10 ){//避免以‘壹拾’开头if (!(NumChineseCharacter[(int)prefixnum] == NumChineseCharacter[1]&& currentLevel == NumLevel.Ten && IsExceptTen)){Result += NumChineseCharacter[(int)prefixnum];}else{IsExceptTen = false;}//加上单位if (currentLevel == NumLevel.Yuan ){////当为 “元” 位不为零时 加“元”。if (NL == null){Result += NumLeverChineseSign[(int)currentLevel];//当小数点后为零时 加 "整"if (postfixnun == 0){Result += EndOfInt;}}}else{Result += NumLeverChineseSign[(int)currentLevel];} //当真正的个位为零时　加上“元”if (NL == null && postfixnun < 1 && currentLevel > NumLevel.Yuan && postfixnun > 0){Result += NumLeverChineseSign[(int)NumLevel.Yuan];

}

}else{//当 前缀数字未被除尽时， 递归下去NumLevel? NextNL = null;if ((int)currentLevel >= (int)(NumLevel.TenThousand))NextNL = currentLevel;

Result += CalculateChineseSign((double)prefixnum,NextNL,isDump,IsExceptTen);if ((int)currentLevel >= (int)(NumLevel.TenThousand)){Result += NumLeverChineseSign[(int)currentLevel];}}

//是否跳位// 判断是否加零， 比如302 就要给三百 后面加零，变为 三百零二。if (IsDumpLevel(Num)){Result += NumChineseCharacter[0];isDump = true;

}

//余数是否需要递归if (postfixnun > 0){Result += CalculateChineseSign(postfixnun,NL,false);}else if (postfixnun == 0 && currentLevel > NumLevel.Yuan ){//当数字是以零元结尾的加上 元整 比如1000000一百万元整if (NL == null){Result += NumLeverChineseSign[(int)NumLevel.Yuan];Result += EndOfInt;}}

return Result;}

9：外部调用的转换方法。

 ///

/// 外部调用的转换方法///

/// /// public string Converttochinese(string Num){

if (!IsValIDated(Num)){throw new OverflowException("数值格式不正确，请输入小于9999亿兆的数字且最多精确的分的金额！");}string chineseCharactor = string.Empty;if (IsBigThanTillion(Num)){double[] tillionNums = SplitNum(Num);chineseCharactor = ContactNumChinese(tillionNums);}else{double dNum = Convert.Todouble(Num);chineseCharactor = CalculateChineseSign(dNum,IsstartOfTen(dNum));}return chineseCharactor;}

　　小结：

　　个人认为程序的灵魂是算法，大到一个系统中的业务逻辑，小到一个货币数字转中文的算法，处处都体现一种逻辑思想。

　　是否能把需求抽象成一个好的数学模型，直接关系到程序的实现的复杂度和稳定性。在一些常用功能中想些不一样的算法，对我们开拓思路很有帮助。                                                         本文转载自‘中国IT实验室’

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