A case study of C # translation of the Chinese function of currency digital conversion based on the pure mathematics recursive method in detail

This article mainly introduces C # based on the pure mathematical method recursive implementation of currency digital conversion Chinese function, involving C # for string traversal, conversion and mathematical operations related operation skills, the need for friends can refer to the following

In this paper, we describe the function of C # Recursive implementation of Chinese currency digital conversion based on pure mathematical method. Share to everyone for your reference, as follows:

Recently because of the project, the need to write a currency number conversion Chinese algorithm, first in the net to find a bit, the results found that no column is replaced with (replace) the way to achieve, so want to write a different algorithm, because I was learning mathematics, so the method of pure mathematics to achieve.

Note: the algorithm in this article supports conversion of currency numbers less than 1023 (i.e. 999.9 billion trillion).

Chinese note: Before explaining the code, let's review the method of currency reading first.

10020002.23 reading for the first 0 of the Zero II three
1020 reading for the first 0 and the whole.
100000 read for pickup million whole
0.13 read as one corner three points

Code:

Test Engineering

static void Main (string[] args) {Console.WriteLine ("Enter Amount"); string inputnum = Console.ReadLine (); while (inputnum! = " Exit ") {//currency digital conversion class 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 ("Illegal numbers or formats");} Console.WriteLine ("\ n Please enter Amount"); inputnum = Console.ReadLine ();} Console.ReadLine ();}

Introduction to the function of currency conversion class (Numcast Class)

1. Provisions of constants

<summary>///digits//</summary>public enum Numlevel {Cent, Chiao, Yuan, Ten, Hundred, Thousand, Tenthousand , Hundredmillon, trillion};///<summary>///digit index///</summary>private int[] numlevelexponent = new int[] {-  2,-1, 0, 1, 2, 3, 4, 8,};///<summary>///Chinese characters//</summary>private string[] numleverchinesesign = new String[] {"Min", "Jiao", "Yuan", "Pick", "Bai", "Qian", "million", "billion", "mega"};///<summary>///uppercase///</summary>private string[] Numchinesecharacter = new string[] {"0", "one", "II", "three", "Restaurant", "WU", "Lu", "Qi", "ba", "JIU"};///<summary>///Whole (when no corner ticks)///< /summary>private Const string endofint = "integer";

2. Verification of digital legitimacy using regular expression validation

<summary>///Regular expression verifies whether the number is legal///</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. Get digital digits such as 1000 for Numlevel.thousand

<summary>///get digital digits using 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 (num)), for (int i = math.log10. Length-1; I >= 0; i--) {if (numlevellength >= numlevelexponent[i]) {nlvl = (numlevel) I;break;}}} ELSE{NLVL = Numlevel.yuan;} return NLVL;}

4. Determine if there is a jump between the numbers, that is, whether to add 0 in the middle of the Chinese, for example 1020 should be added 0.

<summary>///whether to jump///</summary>///<returns></returns>private bool IsDumpLevel (double Num {if (num > 0) {numlevel? currentlevel = Getnumlevel (num); Numlevel? Nextlevel = Null;int Numexponent = this. numlevelexponent[(int) currentlevel];d ouble Postfixnun = Math.Round (Num% (Math.pow (numexponent)), 2); if ( postfixnun> 0) Nextlevel = getnumlevel (Postfixnun); if (currentlevel! = NULL && Nextlevel! = null) {if (currentle Vel > nextlevel + 1) {return true;}}} return false;}

5. Divide the long digits into two smaller arrays of numbers , for example 999.9 billion megabytes, divided into 999.9 billion and 0 megabytes, because the computer does not support too long numbers.

Whether the <summary>///is greater than the trillion, if greater than the string is divided into two parts,////part is the previous number of megabytes///or the other part is the number of megabytes///</summary>///<param name= "Num" ></param>///<returns></returns>private bool Isbigthantillion (string Num) {bool IsBig = false;if ( Num.indexof ('. ')! =-1) {//if greater than Num.indexof ('. ') > numlevelexponent[(int) numlevel.trillion]) {IsBig = true;}} else{//if greater than (Num.length > numlevelexponent[(int) numlevel.trillion]) {IsBig = true;}} return IsBig;} <summary>///the number string from ' Mega ' two///</summary>///<returns></returns>private double[] Splitnum (String Num) {//Gigabit start bit 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];//more than the number tillionlevelnums[0] = convert.todouble (num.substring (0, trillionlevellength));//megabits below tillionlevelnums[1] = convert.todouble (num.substring (trillionlEvellength)); return tillionlevelnums;} 

6. If you want to start with "one pick", you can change it to "pick up"

BOOL Isstartoften = false;while (num >=10) {if (num = =) {Isstartoften = True;break;} Num's digit Numlevel currentlevel = getnumlevel (num); int numexponent = this. numlevelexponent[(int) currentlevel]; Num = Convert.ToInt32 (Math.floor (Num/math.pow, numexponent)); if (CurrentLevel = = Numlevel.ten && Num = = 1) {i Sstartoften = True;break;}} return isstartoften;

7. Merge a currency string that is larger than the number of megabytes converted into

<summary>///Merge separate array Chinese currency character////</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;//If the character is separated by a skip if (Getnumlevel (tillionnums[1] *) = = numlevel.trillion) {chinesecharactor = Uptillionstr + numleverchinesesign[(int) numlevel.trillion] + downtrillionstr;} Else{chinesecharactor = uptillionstr + numleverchinesesign[(int) numlevel.trillion];if (downtrillionstr! = "0-dollar whole") { Chinesecharactor + = numchinesecharacter[0] + downtrillionstr;} Else{chinesecharactor + = "meta-integer";}} return chinesecharactor;}

8. Recursive calculation of currency numbers in Chinese

<summary>///Calculating Chinese strings///</summary>///<param name= "Num" > Digital </param>///<param name= "NL" > Digital level such as 10 million of the digital level is million </param>///<param name= "Isexceptten" > whether to start with ' one pick ' </param>///<returns > Chinese capital </returns>public string calculatechinesesign (double Num, numlevel?) NL, bool Isdump,bool isexceptten) {num = Math.Round (num, 2); bool isdump = false;//num digit numlevel? currentlevel = Getnumleve L (Num); int numexponent = this. numlevelexponent[(int) currentlevel];string Result = string. empty;//the result of an int prefixnum;//remainder when it is fractional, the numerator denominator is multiplied by 100double Postfixnun; if (Num >= 1) {prefixnum = Convert.ToInt32 ( Math.floor (Num/math.pow (numexponent)));p Ostfixnun = Math.Round (Num% (Math.pow (numexponent)), 2);} Else{prefixnum = Convert.ToInt32 (Math.floor (Num*100/math.pow (Ten, numexponent+2)));p Ostfixnun = Math.Round (Num * 100 ( Math.pow (Ten, numexponent + 2)), 2);p Ostfixnun *= 0.01;} if (Prefixnum < 10) {//Avoid starting with ' one pick ' if (! ( numchinesecharacter[(int) prefixnum] = = Numchinesecharacter[1]&& CurrentLevel = = Numlevel.ten && isexceptten) {Result + = numchinesecharacter[( int) prefixnum];} Else{isexceptten = false;} Add the unit if (currentlevel = = Numlevel.yuan) {////when the "meta" bit is not zero plus "meta". if (NL = = null) {result + = numleverchinesesign[(int) currentlevel];//When the decimal point is zero plus "whole" if (Postfixnun = = 0) {result + = Endofint;} }}else{result + = numleverchinesesign[(int) currentlevel];}//When the true digit is zero plus "meta" if (NL = = null && Postfixnun < 1 &am p;& currentlevel > Numlevel.yuan && postfixnun > 0) {Result + = numleverchinesesign[(int) Numlevel.yuan] ;}} else{//the current prefix number is not removed, recursion goes down 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];}} Whether to jump bit//to determine whether to add 0, such as 302 will give 300 after adding 0, into 302. if (Isdumplevel (Num)) {Result + = Numchinesecharacter[0];isDump = true;} Whether the remainder requires recursion if (Postfixnun > 0) {Result + = Calculatechinesesign (Postfixnun, NL, Isdump, false);} else if (Postfixnun = = 0 && currentlevel > Numlevel.yuan) {//When the number is a 0-dollar end plus a meta-integer such as 10,000,001,000,000-yuan integer if (NL = = null) {ResU Lt + = numleverchinesesign[(int) Numlevel.yuan]; Result + = Endofint;}} return Result;}

9. The conversion method of the external call.

<summary>///conversion method for external calls///</summary>///<param name= "Num" ></param>///<returns> </returns>public string Converttochinese (String Num) {if (! Isvalidated<string> (Num)) {throw new OverflowException ("The numeric format is incorrect, please enter a number less than 999.9 billion trillion and the maximum amount of accurate cent!") ");} String chinesecharactor = String. EMPTY;IF (isbigthantillion (num)) {double[] tillionnums = splitnum (num); chinesecharactor = Contactnumchinese ( Tillionnums);} Else{double dnum = convert.todouble (Num); chinesecharactor = Calculatechinesesign (Dnum, NULL, True, Isstartoften (DNum)) ;} return chinesecharactor;}

Summary:

Personally, the soul of the program is the algorithm, large to a system of business logic, small to a currency number to Chinese algorithm, everywhere embodies a logic idea.

Whether the requirements can be abstracted into a good mathematical model is directly related to the complexity and stability of the program implementation. In some commonly used functions to think of a different algorithm, it is very helpful for us to develop ideas.