1.程序说明：将一个浮点数转换成人命币的读法字符串

2.程序流程介绍以及注意事项

流程：

*接收用户输入的一个double数值

*将这个double数值进行处理，得到整数部分和小数部分

*分别对整数部分和小数部分进行处理

*合并整数和小数字符串结果，输出结果，结束。

注意事项：

利用math.round()四舍五入

利用string.tochar()转换字符串数组

主要解决问题

字符串中零的处理，将它分为两种情况

（1）首位是否为0

不处理，一般输入的数值的首位不为0

（2）末尾的0

直接不翻译

（3）中间的单个0和连续出现的0的处理

我们每次处理4位数字，分首部的最前4位和另外的中间以及末尾4位

a.首部4位：中间单个0译为零；连续的0译为一个零，末尾的0不需要翻译

？100，连续的0到末尾的话应该也是不翻译的。

b.其他的4位：目前应该跟a.情况一样

c.全部为0的处理

翻译为零元

（5）整数和小数部分都为0的情况

（4）对于超大位数的数值进行怎样处理

目前默认处理12位

（5）根据单元的不同将数值分成每4位进行处理

1-4 单位 元

5-8 万

9-12 亿

*代码

import java.util.Scanner; import com.sun.corba.se.spi.extension.ZeroPortPolicy; import com.sun.xml.internal.ws.api.streaming.XMLStreamReaderFactory.Zephyr; public class Num2Rmb { private double digitalValue; private boolean zero; String[] strHan = {"零", "壹","贰","叁","肆","伍","陆","柒","捌","玖"}; String[] moneyCount = {"","拾","佰","仟"}; String[] moneyUnit = {"元","万","亿"}; //the default constructer Num2Rmb() { digitalValue = 0.0; zero = false; } //the definite constructed with the @param Num2Rmb(double digitalRmbValue,boolean flag) { digitalValue = digitalRmbValue; zero = flag; } //divide the digital value into the integer part and decimal part //return the array of String[] //string[0] is the integer part //string[1] is the decimal part private String[] divide(double digitalRmbValue) { //get the integer part of the digital money long intPart = (long)(digitalRmbValue); //get the decimal part of the digital money long decimalPart = Math.round((digitalRmbValue - intPart)*100); //another method use Integer.toString(int a) //if the decimal part is equal to 0, then we will give 0 to the string String strIntPart = ""; String strDecimalPart = ""; if(intPart == 0) strIntPart = "0"; else { strIntPart = ""+ intPart; } if(decimalPart<10 && decimalPart >= 0) strDecimalPart = "0" +decimalPart; else { strDecimalPart = "" + decimalPart; } if (intPart==0 && decimalPart == 0) { zero = true; } return new String[] {strIntPart,strDecimalPart }; } //process the basic unit string with at most 12 characters private String proDecimalRmb(String str) { String strResult; char[] strRmb = new char[20]; strRmb = str.toCharArray(); if(strRmb[0]== \'0\' && strRmb[1] == \'0\') { strResult = ""; } else { int m = strRmb[0] - \'0\'; int n = strRmb[1] - \'0\'; strResult = strHan[m]+"角"+ strHan[n]+"分"; } return strResult; } //process the 4 first bit number as the form:{"456"} //the first kind situation private String proFirstFourBit(String str) { char[] firstFourBitRmb = new char[20]; firstFourBitRmb = str.toCharArray(); int len = firstFourBitRmb.length; String strResult = ""; for(int i = 0; i < len; i++) { //the first zero //if the last bit is 0 , we process it if(firstFourBitRmb[i] == \'0\'&& i < len) { boolean zeroFlag = false; //这个while循环表达式可能会发生错误，指针越界访问，注意表达式语句的顺序 while(i<len && firstFourBitRmb[i] ==\'0\' ) { zeroFlag = true; i++; } if(i != len && zeroFlag == true ) strResult += "零"; i--; } //if non-zero bit else { int m = firstFourBitRmb[i] - \'0\'; strResult += strHan[m] + moneyCount[len-i-1]; //the following } } //end for return strResult; }//proFirstFourBit() //process the middle or the last four bit situation /*private String proSecondFourBit(String str) { // int i =0; String strResult= ""; char[] secondFourBitRmb = new char[20]; secondFourBitRmb = str.toCharArray(); int len = secondFourBitRmb.length; for(i=0; i< secondFourBitRmb.length; i++) { if(secondFourBitRmb[i] == \'0\') { boolean zeroFlag = false; while(secondFourBitRmb[i] ==\'0\'&& i< len) { if(zeroFlag == false) { zeroFlag = true; strResult += "零"; } i++; } i--; } //if non-zero bit else { int m = secondFourBitRmb[i] - \'0\'; strResult += strHan[m] + moneyCount[len-i-1]; //the following } } //end for return strResult; }//proSecondFourBit() */ public static void main(String[] args) { String strIndicator = "begin to test the digital RMB"; System.out.println(strIndicator); System.out.println("please input the digital money:"); String strTestRmb = "2356.782"; String[] strTempRmb = new String[2]; double strValue = 0D; Scanner sc = new Scanner(System.in); strValue = sc.nextDouble(); Num2Rmb nr = new Num2Rmb(); //divide the digital Rmb strTempRmb = nr.divide(strValue); //process the decimal ditial money String result1 =nr.proDecimalRmb(strTempRmb[1]); //process the interger part money String result2=""; int len = strTempRmb[0].length(); if(len >12) System.out.println("the data you input is too large to translate"); else { if(len>=9) { String strBit1 = strTempRmb[0].substring(0,len-8); String strBit2 = strTempRmb[0].substring(len-8,len-4); String strBit3 = strTempRmb[0].substring(len-4,len); String strResult1 = nr.proFirstFourBit(strBit1)+"亿"; String strResult2 = nr.proFirstFourBit(strBit2)+"万"; String strResult3 = nr.proFirstFourBit(strBit3)+"元"; result2 = strResult1 + strResult2 +strResult3; } else if(len>=5) { String strBit1 = strTempRmb[0].substring(0,len-4); String strBit2 = strTempRmb[0].substring(len-4,len); String strResult1 = nr.proFirstFourBit(strBit1)+"万"; String strResult2 = nr.proFirstFourBit(strBit2)+"元"; result2 = strResult1 + strResult2; } else if(len>=1) { if (nr.zero == true) { result2 ="零元"; } else result2 = nr.proFirstFourBit(strTempRmb[0])+"元"; } System.out.println("translate the form of the money is:"); System.out.println(result2+result1); } }//main() }

//这个while循环表达式可能会发生错误，指针越界访问，注意表达式语句的顺序
while(firstFourBitRmb[i] ==\'0\'&& i<len)