将数字金额转换为英文单词的最有效方法是什么
e.g。 12到12 127到一百二十七个
答案 0 :(得分：16)
没过多久。这是一个用Java编写的实现。
http://snippets.dzone.com/posts/show/3685
public class IntToEnglish {
static String[] to_19 = { "zero", "one", "two", "three", "four", "five", "six",
"seven", "eight", "nine", "ten", "eleven", "twelve", "thirteen",
"fourteen", "fifteen", "sixteen", "seventeen", "eighteen", "nineteen" };
static String[] tens = { "twenty", "thirty", "forty", "fifty", "sixty", "seventy", "eighty", "ninety"};
static String[] denom = { "",
"thousand", "million", "billion", "trillion", "quadrillion",
"quintillion", "sextillion", "septillion", "octillion", "nonillion",
"decillion", "undecillion", "duodecillion", "tredecillion", "quattuordecillion",
"sexdecillion", "septendecillion", "octodecillion", "novemdecillion", "vigintillion" };
public static void main(String[] argv) throws Exception {
int tstValue = Integer.parseInt(argv[0]);
IntToEnglish itoe = new IntToEnglish();
System.out.println(itoe.english_number(tstValue));
/* for (int i = 0; i < 2147483647; i++) {
System.out.println(itoe.english_number(i));
} */
}
// convert a value < 100 to English.
private String convert_nn(int val) throws Exception {
if (val < 20)
return to_19[val];
for (int v = 0; v < tens.length; v++) {
String dcap = tens[v];
int dval = 20 + 10 * v;
if (dval + 10 > val) {
if ((val % 10) != 0)
return dcap + "-" + to_19[val % 10];
return dcap;
}
}
throw new Exception("Should never get here, less than 100 failure");
}
// convert a value < 1000 to english, special cased because it is the level that kicks
// off the < 100 special case. The rest are more general. This also allows you to
// get strings in the form of "forty-five hundred" if called directly.
private String convert_nnn(int val) throws Exception {
String word = "";
int rem = val / 100;
int mod = val % 100;
if (rem > 0) {
word = to_19[rem] + " hundred";
if (mod > 0) {
word = word + " ";
}
}
if (mod > 0) {
word = word + convert_nn(mod);
}
return word;
}
public String english_number(int val) throws Exception {
if (val < 100) {
return convert_nn(val);
}
if (val < 1000) {
return convert_nnn(val);
}
for (int v = 0; v < denom.length; v++) {
int didx = v - 1;
int dval = new Double(Math.pow(1000, v)).intValue();
if (dval > val) {
int mod = new Double(Math.pow(1000, didx)).intValue();
int l = val / mod;
int r = val - (l * mod);
String ret = convert_nnn(l) + " " + denom[didx];
if (r > 0) {
ret = ret + ", " + english_number(r);
}
return ret;
}
}
throw new Exception("Should never get here, bottomed out in english_number");
}
}
答案 1 :(得分：4)
此链接提供了一个看起来很好的方法的详细说明： http://www.blackwasp.co.uk/NumberToWords.aspx
答案 2 :(得分：2)
实现此目的的一种方法是使用查找表。这将是一种蛮力，但易于设置。例如，你可以在一个表中使用0-99的单词，然后是数十，数百，数千等的表。一些简单的数学将为你提供每个表所需单词的索引。
答案 3 :(得分：2)
这是我硬盘上的一些旧python代码。可能存在错误，但应该显示基本想法：
class Translator:
def transformInt(self, x):
translateNumbers(0,[str(x)])
def threeDigitsNumber(self,number):
snumber=self.twoDigitsNumber(number/100)
if number/100!=0:
snumber+=" hundred "
snumber+self.twoDigitsNumber(number)
return snumber+self.twoDigitsNumber(number)
def twoDigitsNumber(self,number):
snumber=""
if number%100==10:
snumber+="ten"
elif number%100==11:
snumber+="eleven"
elif number%100==12:
snumber+="twelve"
elif number%100==13:
snumber+="thirteen"
elif number%100==14:
snumber+="fourteen"
elif number%100==15:
snumber+="fifteen"
elif number%100==16:
snumber+="sixteen"
elif number%100==17:
snumber+="seventeen"
elif number%100==18:
snumber+="eighteen"
elif number%100==19:
snumber+="nineteen"
else:
if (number%100)/10==2:
snumber+="twenty-"
elif (number%100)/10==3:
snumber+="thirty-"
elif (number%100)/10==4:
snumber+="forty-"
elif (number%100)/10==5:
snumber+="fifty-"
elif (number%100)/10==6:
snumber+="sixty-"
elif (number%100)/10==7:
snumber+="seventy-"
elif (number%100)/10==8:
snumber+="eighty-"
elif (number%100)/10==9:
snumber+="ninety-"
if (number%10)==1:
snumber+="one"
elif (number%10)==2:
snumber+="two"
elif (number%10)==3:
snumber+="three"
elif (number%10)==4:
snumber+="four"
elif (number%10)==5:
snumber+="five"
elif (number%10)==6:
snumber+="six"
elif (number%10)==7:
snumber+="seven"
elif (number%10)==8:
snumber+="eight"
elif (number%10)==9:
snumber+="nine"
elif (number%10)==0:
if snumber!="":
if snumber[len(snumber)-1]=="-":
snumber=snumber[0:len(snumber)-1]
return snumber
def translateNumbers(self,counter,words):
if counter+1<len(words):
self.translateNumbers(counter+1,words)
else:
if counter==len(words):
return True
k=0
while k<len(words[counter]):
if (not (ord(words[counter][k])>47 and ord(words[counter][k])<58)):
break
k+=1
if (k!=len(words[counter]) or k==0):
return 1
number=int(words[counter])
from copy import copy
if number==0:
self.translateNumbers(counter+1,copy(words[0:counter]+["zero"]+words[counter+1:len(words)]))
self.next.append(copy(words[0:counter]+["zero"]+words[counter+1:len(words)]))
return 1
if number<10000:
self.translateNumbers(counter+1,copy(words[0:counter]
+self.seperatewords(self.threeDigitsNumber(number))
+words[counter+1:len(words)]))
self.next.append(copy(words[0:counter]
+self.seperatewords(self.threeDigitsNumber(number))
+words[counter+1:len(words)]))
if number>999:
snumber=""
if number>1000000000:
snumber+=self.threeDigitsNumber(number/1000000000)+" billion "
number=number%1000000000
if number>1000000:
snumber+=self.threeDigitsNumber(number/1000000)+" million "
number=number%1000000
if number>1000:
snumber+=self.threeDigitsNumber(number/1000)+" thousand "
number=number%1000
snumber+=self.threeDigitsNumber(number)
self.translateNumbers(counter+1,copy(words[0:counter]+self.seperatewords(snumber)
+words[counter+1:len(words)]))
self.next.append(copy(words[0:counter]+self.seperatewords(snumber)
+words[counter+1:len(words)]))
答案 4 :(得分：2)
此解决方案不会尝试考虑尾随空格，但速度非常快。
typedef const char* cstring;
using std::string;
using std::endl;
std::ostream& GetOnes(std::ostream &output, int onesValue)
{
cstring ones[] = { "zero", "one", "two", "three", "four", "five", "six",
"seven", "eight", "nine" };
output << ones[onesValue];
return output;
}
std::ostream& GetSubMagnitude(std::ostream &output, int subMagnitude)
{
cstring tens[] = { "zeroty", "ten", "twenty", "thirty", "fourty", "fifty",
"sixty", "seventy", "eighty", "ninety"};
if (subMagnitude / 100 != 0)
{
GetOnes(output, subMagnitude / 100) << " hundred ";
GetSubMagnitude(output, subMagnitude - subMagnitude / 100 * 100);
}
else
{
if (subMagnitude >= 20)
{
output << tens[subMagnitude / 10] << " ";
GetOnes(output, subMagnitude - subMagnitude / 10 * 10);
}
else if (subMagnitude >= 10)
{
cstring teens[] = { "ten", "eleven", "twelve", "thirteen",
"fourteen", "fifteen", "sixteen", "seventeen",
"eighteen", "nineteen" };
output << teens[subMagnitude - 10] << " ";
}
else
{
GetOnes(output, subMagnitude) << " ";
}
}
return output;
}
std::ostream& GetLongNumber(std::ostream &output, double input)
{
cstring magnitudes[] = {"", "hundred", "thousand", "million", "billion",
"trillion"};
double magnitudeTests[] = {1, 100.0, 1000.0, 1000000.0, 1000000000.0,
1000000000000.0 };
int magTestIndex = 0;
while (floor(input / magnitudeTests[magTestIndex++]) != 0);
magTestIndex -= 2;
if (magTestIndex >= 0)
{
double subMagnitude = input / magnitudeTests[magTestIndex];
GetSubMagnitude(output, (int)subMagnitude);
if (magTestIndex) {
output << magnitudes[magTestIndex] << " ";
double remainder = input - (floor(input /
magnitudeTests[magTestIndex]) *
magnitudeTests[magTestIndex]);
if (floor(remainder) > 0)
{
GetLongNumber(output, remainder);
}
}
}
else
{
output << "zero";
}
return output;
}
答案 5 :(得分：2)
首先解决1-99，使用1-20的数字列表，然后使用30,40，...，90。然后添加数百以获得1-999。然后使用该例程给出每个功率1000的数量，以达到你想要的最高值（我认为最高的标准命名是十亿分之十，即10 ^ 33）。
一个小小的警告是，如果你试图在没有多余空白的情况下尝试开始和结束，那么在所有情况下让空白正确是有点棘手的。简单的解决方案是在每个单词后面留空，然后在完成所有操作后剥去尾随空白。如果您在构建字符串时尝试更精确，则可能最终会丢失空白或空白。
答案 6 :(得分：1)
。）创建一个包含所有数字和库的库。位置（例如1表示除10之外的其他表示法，而不是100表示等） 。）列出例外情况（例如12）并注意，在你的algorythm中，相同的例外是112,1012等。
如果您想要更快的速度，请制作一组您需要的缓存数字。
答案 7 :(得分：1)
请注意一些规则：
digit place喜欢“三千”的某种组合。您可以使用整数的分区来获得数字的位置：532/100 - ＆gt; 5
答案 8 :(得分：0)
Here's how GNU CLISP does it和here's how CMUCL does it（更容易阅读，恕我直言）。
为“格式为百万亿”做一个code search将会发现很多。