/* 
.Net/C#/T-SQL/VB/Java 实现: 将天文数字转换成中文大写 
最近这里流行这个大写金额转换 
我看了几个,感觉思路都不如这个 2000 年左右用 VB6 写的: 
《VB6 之数据格式化对象使用技巧》 
http://search.csdn.net/Expert/topic/38/38970.xml?temp=.5078089 
<<精华: 将金额小写转大写的代码可转到亿位>> 
http://www.dev-club.com/club/bbs/showEssence.asp?id=20684&page=1 
思路非常简单,且没有任何位数限制! 
例如: 401,0103,1013 
读作: 肆佰零壹[亿]零壹佰零叁[万]壹仟零壹拾叁 
咱们先按每四位一组 从左到右,高位到低位分别"大声朗读"一下: 
"肆佰零壹" 单位是: "[亿]" 
"壹佰零叁" 单位是: "[万]" 
"壹仟零壹拾叁" 单位是 "" (相当于没有单位) 
很容易发现,每四位: 只有 千位,百位,十位,个位 这四种情况! 
我们把 [万],[亿] 当作单位就可以了! 
这就是规律了!简单吧! 
依据该思路,只用区区不到 50 行代码就可以搞定: 
只要你能够提供足够多的"单位" 
任何天文数字都可以正确转换! 
*/ 
 
namespace Microshaoft 
{ 
  using System; 
 
  public class Util 
  { 
    public static string ConvertNumberToChinese(string x, string[] Nums, string[] Digits, string[] Units) 
    { 
      string S = ""; //返回值 
      int p = 0; //字符位置指针 
      int m = x.Length % 4; //取模 
 
      // 四位一组得到组数 
      int k = (m > 0 ? x.Length / 4 + 1 : x.Length / 4); 
 
      // 外层循环在所有组中循环 
      // 从左到右 高位到低位 四位一组 逐组处理 
      // 每组最后加上一个单位: "[万亿]","[亿]","[万]" 
      for (int i = k; i > 0; i--) 
      { 
        int L = 4; 
        if (i == k && m != 0) 
        { 
          L = m; 
        } 
        // 得到一组四位数 最高位组有可能不足四位 
        string s = x.Substring(p, L); 
        int l = s.Length; 
 
        // 内层循环在该组中的每一位数上循环 从左到右 高位到低位 
        for (int j = 0; j < l; j++) 
        { 
          //处理改组中的每一位数加上所在位: "仟","佰","拾",""(个) 
          int n = Convert.ToInt32(s.Substring(j, 1)); 
          if (n == 0) 
          { 
            if (j < l - 1 
              && Convert.ToInt32(s.Substring(j + 1, 1)) > 0 //后一位(右低) 
              && !S.EndsWith(Nums[n])) 
            { 
              S += Nums[n]; 
            } 
          } 
          else 
          { 
            //处理 1013 一千零"十三", 1113 一千一百"一十三" 
            if (!(n == 1 && (S.EndsWith(Nums[0]) | S.Length == 0) && j == l - 2)) 
            { 
              S += Nums[n]; 
            } 
            S += Digits[l - j - 1]; 
          } 
        } 
        p += L; 
        // 每组最后加上一个单位: [万],[亿] 等 
        if (i < k) //不是最高位的一组 
        { 
          if (Convert.ToInt32(s) != 0) 
          { 
            //如果所有 4 位不全是 0 则加上单位 [万],[亿] 等 
            S += Units[i - 1]; 
          } 
        } 
        else 
        { 
          //处理最高位的一组,最后必须加上单位 
          S += Units[i - 1]; 
        } 
      } 
      return S; 
    } 
 
    // 测试程序 
    private static void Main() 
    { 
      //数字 数组 
      string[] Nums = new string[] {"零", "壹", "贰", "叁", "肆", "伍", "陆", "柒", "捌", "玖"}; 
      //位 数组 
      string[] Digits = new string[] {"", "拾", "佰", "仟"}; 
      //单位 数组 
      string[] Units = new string[] {"", "[万]", "[亿]", "[万亿]"}; 
 
      Console.WriteLine(ConvertNumberToChinese("1100000013", Nums, Digits, Units)); 
      Console.WriteLine(ConvertNumberToChinese("2100000000", Nums, Digits, Units)); 
      Console.WriteLine(ConvertNumberToChinese("1000000000", Nums, Digits, Units)); 
      Console.WriteLine(ConvertNumberToChinese("40000000013", Nums, Digits, Units)); 
      Console.WriteLine(ConvertNumberToChinese("40000000001", Nums, Digits, Units)); 
      Console.WriteLine(ConvertNumberToChinese("400000010000", Nums, Digits, Units)); 
      Console.WriteLine(ConvertNumberToChinese("40101031013", Nums, Digits, Units)); 
      Console.WriteLine(ConvertNumberToChinese("40101031113", Nums, Digits, Units)); 
      Console.WriteLine(ConvertNumberToChinese("101140101031013", Nums, Digits, Units)); 
      Console.WriteLine(ConvertNumberToChinese("100000001000013", Nums, Digits, Units)); 
      Console.WriteLine(ConvertNumberToChinese("100000001000113", Nums, Digits, Units)); 
      Console.WriteLine(ConvertNumberToChinese("100011003", Nums, Digits, Units)); 
      Console.WriteLine(ConvertNumberToChinese("10010103", Nums, Digits, Units)); 
      Console.WriteLine(ConvertNumberToChinese("10110013", Nums, Digits, Units)); 
      Console.WriteLine(ConvertNumberToChinese("130000", Nums, Digits, Units)); 
      Console.ReadLine(); 
    } 
  } 
}

create function ConvertNumberToChinese(@ varchar(100)) 
returns varchar(100) 
as 
begin 
--declare @ varchar(100) 
--set @ = '101140101031013' 
declare @s varchar(100) 
set @s = '' 
 
declare @p integer 
set @p = 0 
 
declare @m integer 
set @m = len(@) % 4 
 
declare @k integer 
set @k = len(@)/4 
 
select @k = @k + 1 
 where @m > 0 
 
declare @i integer 
set @i = @k 
 
while (@i > 0) 
begin --outer 
  declare @L integer 
  set @L = 4 
 
  select @L = @m 
  where @i = @k and @m != 0 
 
  declare @ss varchar(4) 
  set @ss = substring(@,@p+1,@L) 
 
  declare @ll integer 
  set @ll = len(@ss) 
 
  --inner 
  declare @j integer 
  set @j = 0 
 
  while (@j < @ll) --inner 
  begin --inner 
   declare @n integer 
   set @n = cast(substring(@ss,@j+1,1) as integer) 
 
   declare @num varchar(2)  
   select @num = Num 
    from ( 
       select 0 as id,'零' as Num 
       union all select 1,'壹' 
       union all select 2,'贰' 
       union all select 3,'叁' 
       union all select 4,'肆' 
       union all select 5,'伍' 
       union all select 6,'陆' 
       union all select 7,'柒' 
       union all select 8,'捌' 
       union all select 9,'玖' 
       ) Nums 
   where id = @n 
 
   if @n = 0 
   begin 
     select @s = @s + @num 
     where @j < @ll - 1 
        and cast(substring(@ss,(@j+1)+1,1) as integer) > 0 
        and right(@ss,1) != @num 
   end 
   else 
   begin 
     declare @jj integer 
     set @jj = 1 
 
     select @jj = @j - 1 
     where @j > 1 
 
     select @s = @s + @num 
     where not (@n = 1 
           and @j = @ll - 2 
           and (len(@s) = 0 
             or right(@s,1) = '零' 
/* 
                      (  select Num 
                        from 
                        ( 
                        select 0 as id,'零' as Num 
                        union all select 1,'壹' 
                        union all select 2,'贰' 
                        union all select 3,'叁' 
                        union all select 4,'肆' 
                        union all select 5,'伍' 
                        union all select 6,'陆' 
                        union all select 7,'柒' 
                        union all select 8,'捌' 
                        union all select 9,'玖' 
                        ) Nums 
                        where id = 0 
                      ) 
*/ 
                      ) 
             ) 
 
     select @s = @s + digit 
      from ( 
         select 0 as id,'' as digit 
         union all select 1,'拾' 
         union all select 2,'佰' 
         union all select 3,'仟' 
        ) digits 
     where id = @ll - @j - 1 
 
   end 
   set @j = @j + 1 --inner 
  end --inner 
  set @p = @p + @L 
 
  declare @unit varchar(10) 
  select @unit = Unit 
   from ( 
      select 0 as id,'' as Unit 
      union all select 1,'[万]' 
      union all select 2,'[亿]' 
      union all select 3,'[万亿]' 
     ) Units 
  where id = @i - 1 
 
  if @i < @k 
  begin 
   select @s = @s + @unit 
    where cast(@ss as integer) != 0 
  end 
  else 
  begin 
   set @s = @s + @unit 
  end 
  set @i = @i - 1 -- outer 
end --out 
return @s 
/* 
--Test: 
select dbo.ConvertNumberToChinese('1011111112101013') 
,dbo.ConvertNumberToChinese('40000000001') 
,dbo.ConvertNumberToChinese('400000010000') 
,dbo.ConvertNumberToChinese('40101031013') 
,dbo.ConvertNumberToChinese('101140101031013') 
,dbo.ConvertNumberToChinese('100000001000003') 
,dbo.ConvertNumberToChinese('10011003') 
,dbo.ConvertNumberToChinese('10010103') 
,dbo.ConvertNumberToChinese('10010013') 
*/ 
end

public class Class1 
{ 
  public static String ConvertNumberToChinese(String x, String[] Nums, String[] Digits, String[] Units) 
  { 
    String S = ""; //返回值 
    int p = 0; //字符位置指针 
    int m = x.length() % 4; //取模 
 
    // 四位一组得到组数 
    int k = (m > 0 ? x.length() / 4 + 1 : x.length() / 4); 
 
    // 外层循环在所有组中循环 
    // 从左到右 高位到低位 四位一组 逐组处理 
    // 每组最后加上一个单位: "[万亿]","[亿]","[万]" 
    for (int i = k; i > 0; i--) 
    { 
      int L = 4; 
      if (i == k && m != 0) 
      { 
        L = m; 
      } 
      // 得到一组四位数 最高位组有可能不足四位 
      String s = x.substring(p, p + L); 
      int l = s.length(); 
 
      // 内层循环在该组中的每一位数上循环 从左到右 高位到低位 
      for (int j = 0; j < l; j++) 
      { 
        //处理改组中的每一位数加上所在位: "仟","佰","拾",""(个) 
        int n = java.lang.Integer.parseInt(s.substring(j, j+1)); 
        if (n == 0) 
        { 
          if ((j < l - 1) 
            && (java.lang.Integer.parseInt(s.substring(j + 1, j + 1+ 1)) > 0) //后一位(右低) 
            && !S.endsWith(Nums[n])) 
          { 
            S += Nums[n]; 
          } 
        } 
        else 
        { 
          //处理 1013 一千零"十三", 1113 一千一百"一十三" 
          if (!(n == 1 && (S.endsWith(Nums[0]) | S.length() == 0) && j == l - 2)) 
          { 
            S += Nums[n]; 
          } 
          S += Digits[l - j - 1]; 
        } 
      } 
      p += L; 
      // 每组最后加上一个单位: [万],[亿] 等 
      if (i < k) //不是最高位的一组 
      { 
        if (java.lang.Integer.parseInt(s) != 0) 
        { 
          //如果所有 4 位不全是 0 则加上单位 [万],[亿] 等 
          S += Units[i - 1]; 
        } 
      } 
      else 
      { 
        //处理最高位的一组,最后必须加上单位 
        S += Units[i - 1]; 
      } 
    } 
    return S; 
  } 
 
  // 测试程序 
  public static void main(String[] args) throws Exception 
  { 
    //数字 数组 
    String[] Nums = new String[] {"零", "壹", "贰", "叁", "肆", "伍", "陆", "柒", "捌", "玖"}; 
    //位 数组 
    String[] Digits = new String[] {"", "拾", "佰", "仟"}; 
    //单位 数组 
    String[] Units = new String[] {"", "[万]", "[亿]", "[万亿]"}; 
 
    System.out.println(ConvertNumberToChinese("111112100113", Nums, Digits, Units)); 
    System.out.println(ConvertNumberToChinese("1100000000", Nums, Digits, Units)); 
    System.out.println(ConvertNumberToChinese("1000000000", Nums, Digits, Units)); 
    System.out.println(ConvertNumberToChinese("40000000013", Nums, Digits, Units)); 
    System.out.println(ConvertNumberToChinese("40000000001", Nums, Digits, Units)); 
    System.out.println(ConvertNumberToChinese("400000010000", Nums, Digits, Units)); 
    System.out.println(ConvertNumberToChinese("40101031013", Nums, Digits, Units)); 
    System.out.println(ConvertNumberToChinese("40101031113", Nums, Digits, Units)); 
    System.out.println(ConvertNumberToChinese("101140101031013", Nums, Digits, Units)); 
    System.out.println(ConvertNumberToChinese("100000001000013", Nums, Digits, Units)); 
    System.out.println(ConvertNumberToChinese("100000001000113", Nums, Digits, Units)); 
    System.out.println(ConvertNumberToChinese("100011003", Nums, Digits, Units)); 
    System.out.println(ConvertNumberToChinese("10010103", Nums, Digits, Units)); 
    System.out.println(ConvertNumberToChinese("10110013", Nums, Digits, Units)); 
    System.out.println(ConvertNumberToChinese("130000", Nums, Digits, Units)); 
 
    //System.in.read(); 
  } 
}