Convert the numeric string RMB into a uppercase string RMB

/*
* Creation date: 2005-3-8
*
* To change the template of the generated file in todo, go
* Window-preferences-Java-code style-Code Template
*/
Package com. test;

/**
* @ Author Administrator
*
* To change the template of the type comment generated by todo, go
* Window-preferences-Java-code style-Code Template
*/
Public class testrmb {
/**
* Test the numberconvert method.
* @ Param ARGs string []
*/
Public static void main (string [] ARGs ){
// This is the largest number that can be converted
String RMB = "90909.99 ";
System. Out. println (numberconvert (RMB ));
}

/**
* Converts string numbers to Chinese characters.
* @ Param in number
* @ Return (uppercase)
*/
Public static string numberconvert (string in ){
Stringbuffer out = new stringbuffer ();
String CHN [] = {
"Zero", "one", "two", "San", "Si", "Wu", "Lu", "Wu ", "success "};
String end [] = {
"", "Pick", "success", "fail ",};
String mid [] = {
"", "MB ",};

Double ind = double. parsedouble (in );
Long ind_left = (long) IND;
Long [] inte = new long [13]; // stores the entire number of digits.
Int [] floa = new int [4]; // stores decimal places

/* Place the integer part in inte [] */
For (INT I = 12; I> = 0; I --){
Inte [I] = (long) (ind_left/(long) (math. Pow (10, I )));
Ind_left % = (long) (math. Pow (10, I ));
}
/* Put the fractional part of the image into floa [] */
Int Index = in. indexof ('.');
If (index! =-1 ){
String STR = in. substring (index + 1 );
For (INT I = 0; I <Str. Length (); I ++ ){
Floa [I] = integer. parseint (Str. substring (I, I + 1 ));
}
}
/** If the integer is not 0, the example of processing the entire number is 111.1234 **/
If (INT) ind! = 0 ){
Int I = 12;
While (inte [I] = 0) {file: // search for the first place
I --;
}
/* Output to out */
For (Int J = I/4; j> = 0; j --){
If (j = 3) {// if the number is 13 digits, process 1 MB separately
Out. append (CHN [(INT) (inte [12]). append (mid [3]);
J --;
}
For (int K = 3; k> = 0; k --){
Int num = 4 * j + K;
If (inte [num]! = 0) {// no more than 0 bits last time add
Out. append (CHN [(INT) (inte [num]). append (end [k]);
}
Else {
If (k = 0 | K = 0 & K = 1 | K = 0 & K = 1 & K = 2 |
K = 0 & K = 1 & K = 2 & K = 3) {} // if the last few digits of a trillion are all 0, no output is made.
Else {
If (inte [num-1] = 0) {} // judge whether to add or not 0 based on the following assumption: 1003
Else {
Out. append (CHN [0]);
}
}
}
}
If (! (Inte [J * 4 + 0] = 0 & inte [J * 4 + 1] = 0 &&
Inte [J * 4 + 2] = 0 & inte [J * 4 + 3] = 0) {// if there is a segment that is all 0, it will not be outputted. Example: 1000, 0000, 0003, 100 billion, and
Out. append (mid [J]);
}
}
If (Out. charat (0) = '0') {// because the last digit does not have the next one, if it is 0, delete it.
Out. deletecharat (0 );
} // Example 30
}
/** If the integer is 0, add 0 directly to the integer. Example: 0.1234 **/
Else {
Out. append ('0 ');
}

/* Process the fractional part */
/* If there are decimal places */
If (index! =-1 ){
Out. append ('point'); // process decimal places
For (INT m = 0; m <= 3; m ++ ){
Out. append (CHN [floa [m]);
}
Int zero = out. Length ()-1;
While (Out. charat (zero) = '0') {// Delete the last few digits if they are 0
Out. deletecharat (zero );
Zero --;
}
}
Return out. tostring ();
}
}