.net 将一串价格数字转化为中文大写形式

#region 数字转大写

        public static string ToAmountWords(double money)
        {
            string temp = "";
            string resu = "";
            string jf = "";
            int j = 0;
            int j_1 = 0;
            int jiao = 0;
            int fen = 0;
            int len = 0;
            List<string> Num = new List<string>() { "零", "壹", "贰", "叁", "肆", "伍", "陆", "柒", "捌", "玖" };
            List<string> A = new List<string>() { "分", "角", "元", "拾", "佰", "仟", "万", "拾", "佰", "仟", "亿", "拾", "佰", "仟", "兆", "拾", "佰", "仟" };

            temp = ((Math.Truncate(Math.Round(money * 100))).ToString()).Trim();
            len = temp.Length;
            resu = "";
            if (len > 13 || len == 0)
                return "";
            jiao = Convert.ToInt32(temp.Substring(len - 2, 1));
            fen = Convert.ToInt32(temp.Substring(len - 1, 1));
            if (fen == 0)
            {
                if (jiao == 0)
                    jf = "整";
                else
                    jf = Num[jiao] + "角整";
            }
            else
            {
                if (jiao == 0)
                    jf = "零" + Num[fen] + "分";
                else
                    jf = Num[jiao] + "角" + Num[fen] + "分";
            }
            for (int i = 0; i < len - 2; i++)
            {
                j = Convert.ToInt32(temp.Substring(i, 1));//取第一位数字
                if (j == 0)
                {
                    j_1 = Convert.ToInt32(temp.Substring(i + 1, 1));//取第二位数字
                    if (j_1 == 0)
                        continue;
                    if (A[len - i - 1] == "萬" || A[len - i - 1] == "億")
                        resu = resu + A[len - i - 1] + Num[j];
                    else
                    {
                        if (A[len - i - 1] == "元")
                            resu = resu + "元";
                        else
                            resu = resu + Num[j];
                    }
                }
                else
                    resu = resu + Num[j] + A[len - i - 1];
            }
            return resu + jf;
        }
        #endregion