思路：
中文对金额的描述以四位为一组，
只考虑一万亿以内的数字则每组内以千、百、十和[亿万元]区分各位
连续的零按一个处理，组内最低位的零可略去
无角无分说整，有角无分只说角，无角有分说零X分，有角有分...
代码：
1 using System; 2 using System.Collections.Generic; 3 using System.Linq; 4 using System.Text; 5 6 namespace MorrisSpace 7 { 8 /// <summary> 9 /// 中文金额字符串辅助类。Helper for Amount string in Chinese 10 /// </summary> 11 public class AmountStringHelper 12 { 13 static private readonly char[] units = { '分', '角', '拾', '佰', '仟', '圆', '万', '亿', '整' }; 14 // 0 1 2 3 4 5 6 7 8 15 static private readonly char[] numbers = { '零', '壹', '贰', '叁', '肆', '伍', '陆', '柒', '捌', '玖' }; 16 17 /// <summary> 18 /// 数字金额转大写金额 19 /// </summary> 20 /// <param name="num">金额数字</param> 21 /// <returns>大写金额字符串</returns> 22 public static string GetAmountInWords(double num) 23 { 24 double amount = Math.Round(num, 2); 25 long integ = (int)amount; 26 double fract = Math.Round(amount - integ, 2); 27 if (integ.ToString().Length > 12) 28 { 29 return null; 30 } 31 string result = ""; 32 if (fract - 0.0 != 0) 33 { 34 string tempstr = fract.ToString(); 35 if (tempstr.Length == 3) 36 { 37 result += numbers[(int)(fract * 10)]; 38 result += units[1]; 39 } 40 else 41 { 42 int frist = (int)(fract * 10); 43 int second = (int)(fract * 100 - frist * 10); 44 if (frist != 0) 45 { 46 result += numbers[frist]; 47 result += units[1]; 48 result += numbers[second]; 49 result += units[0]; 50 } 51 else 52 { 53 result += numbers[0]; 54 result += numbers[second]; 55 result += units[0]; 56 } 57 } 58 } 59 else 60 { 61 result += units[8]; 62 } 63 64 for (int temp = (int)(integ % 10000), secnum = 1; temp != 0; temp = (int)(integ % 10000), secnum++) 65 { 66 result = FourBitTrans(temp) + units[secnum + 4] + result; 67 integ /= 10000; 68 if (integ != 0 && temp < 1000) 69 { 70 result = numbers[0] + result; 71 } 72 } 73 return result; 74 } 75 76 /// <summary> 77 /// 进行四位数字转换的辅助函数 78 /// </summary> 79 /// <param name="num">四位以下数字</param> 80 /// <returns>大写金额四位节</returns> 81 public static string FourBitTrans(int num) 82 { 83 string tempstr = num.ToString(); 84 if (tempstr.Length > 4) 85 { 86 return null; 87 } 88 string result = string.Empty; 89 int i = tempstr.Length; 90 int j = 0; 91 bool zeromark = true; 92 while (--i >= 0) 93 { 94 j++; 95 96 if (tempstr[i] == '0') 97 { 98 if (zeromark == true) 99 { 100 continue; 101 } 102 zeromark = true; 103 result = numbers[0] + result; 104 continue; 105 } 106 zeromark = false; 107 if (j > 1) 108 { 109 result = units[j] + result; 110 } 111 int temp = tempstr[i] - '0'; 112 result = numbers[temp] + result; 113 } 114 return result; 115 } 116 117 118 } 119 }
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这个代码只适合一亿以内的金额，但相信以满足绝大多数情况