代码高尔夫系列似乎相当受欢迎。我遇到了一些将数字转换为单词表示的代码。一些例子是（编程乐趣2的权力）：
我的同事出现的算法差不多有两百行。似乎会有更简洁的方法来做到这一点。
当前指南：
答案 0 :(得分：70)
好的，我认为是时候在Windows BATCH脚本中自己实现了（应该适用于Windows 2000或更高版本）。
以下是代码：
@echo off
set zero_to_nineteen=Zero One Two Three Four Five Six Seven Eight Nine Ten Eleven Twelve Thirteen Fourteen Fifteen Sixteen Seventeen Eighteen Nineteen
set twenty_to_ninety=ignore ignore Twenty Thirty Forty Fifty Sixty Seventy Eighty Ninety
set big_numbers=ignore Thousand Million Billion Trillion Quadrillion Quintillion Sextillion Septillion Octillion Nonillion Decillion Undecillion Duodecillion Tredecillion Quattuordecillion Quindecillion Sexdecillion Septendecillion Octodecillion Novemdecillion Vigintillion
rem 10^0 10^3 10^6 10^9 10^12 10^15 10^18 10^21 10^24 10^27 10^30 10^33 10^36 10^39 10^42 10^45 10^48 10^51 10^54 10^57 10^60 10^63
call :parse_numbers %*
exit /B 0
:parse_numbers
:parse_numbers_loop
if "$%~1" == "$" goto parse_numbers_end
call :parse_number %~1
echo %~1 -^> %parse_number_result%
shift
goto parse_numbers_loop
:parse_numbers_end
exit /B 0
:parse_number
call :get_sign %~1
set number_sign=%get_sign_result%
call :remove_groups %get_sign_result_number%
call :trim_leading_zeros %remove_groups_result%
set number=%trim_leading_zeros_result%
if "$%number%" == "$0" (
set parse_number_result=Zero
exit /B 0
)
set counter=0
set parse_number_result=
:parse_number_loop
set last_three=%number:~-3%
set number=%number:~0,-3%
call :parse_three %last_three%
call :get_from %counter% %big_numbers%
if "$%get_from_result%" == "$" (
set parse_number_result=* ERR: the number is too big! Even wikipedia doesn't know how it's called!
exit /B 0
)
if not "$%parse_three_result%" == "$Zero" (
if %counter% == 0 (
set parse_number_result=%parse_three_result%
) else (
if not "$%parse_number_result%" == "$" (
set parse_number_result=%parse_three_result% %get_from_result% %parse_number_result%
) else (
set parse_number_result=%parse_three_result% %get_from_result%
)
)
)
set /A counter+=1
if not "$%number%" == "$" goto parse_number_loop
if "$%parse_number_result%" == "$" (
set parse_number_result=Zero
exit /B 0
) else if not "$%number_sign%" == "$" (
set parse_number_result=%number_sign% %parse_number_result%
)
exit /B 0
:parse_three
call :trim_leading_zeros %~1
set three=%trim_leading_zeros_result%
set /A three=%three% %% 1000
set /A two=%three% %% 100
call :parse_two %two%
set parse_three_result=
set /A digit=%three% / 100
if not "$%digit%" == "$0" (
call :get_from %digit% %zero_to_nineteen%
)
if not "$%digit%" == "$0" (
if not "$%get_from_result%" == "$Zero" (
set parse_three_result=%get_from_result% Hundred
)
)
if "$%parse_two_result%" == "$Zero" (
if "$%parse_three_result%" == "$" (
set parse_three_result=Zero
)
) else (
if "$%parse_three_result%" == "$" (
set parse_three_result=%parse_two_result%
) else (
set parse_three_result=%parse_three_result% %parse_two_result%
)
)
exit /B 0
:parse_two
call :trim_leading_zeros %~1
set two=%trim_leading_zeros_result%
set /A two=%two% %% 100
call :get_from %two% %zero_to_nineteen%
if not "$%get_from_result%" == "$" (
set parse_two_result=%get_from_result%
goto parse_two_20_end
)
set /A digit=%two% %% 10
call :get_from %digit% %zero_to_nineteen%
set parse_two_result=%get_from_result%
set /A digit=%two% / 10
call :get_from %digit% %twenty_to_ninety%
if not "$%parse_two_result%" == "$Zero" (
set parse_two_result=%get_from_result% %parse_two_result%
) else (
set parse_two_result=%get_from_result%
)
goto parse_two_20_end
:parse_two_20_end
exit /B 0
:get_from
call :trim_leading_zeros %~1
set idx=%trim_leading_zeros_result%
set /A idx=0+%~1
shift
:get_from_loop
if "$%idx%" == "$0" goto get_from_loop_end
set /A idx-=1
shift
goto get_from_loop
:get_from_loop_end
set get_from_result=%~1
exit /B 0
:trim_leading_zeros
set str=%~1
set trim_leading_zeros_result=
:trim_leading_zeros_loop
if not "$%str:~0,1%" == "$0" (
set trim_leading_zeros_result=%trim_leading_zeros_result%%str%
exit /B 0
)
set str=%str:~1%
if not "$%str%" == "$" goto trim_leading_zeros_loop
if "$%trim_leading_zeros_result%" == "$" set trim_leading_zeros_result=0
exit /B 0
:get_sign
set str=%~1
set sign=%str:~0,1%
set get_sign_result=
if "$%sign%" == "$-" (
set get_sign_result=Minus
set get_sign_result_number=%str:~1%
) else if "$%sign%" == "$+" (
set get_sign_result_number=%str:~1%
) else (
set get_sign_result_number=%str%
)
exit /B 0
:remove_groups
set str=%~1
set remove_groups_result=%str:'=%
exit /B 0
这是我使用的测试脚本：
@echo off
rem 10^x:x= 66 63 60 57 54 51 48 45 42 39 36 33 30 27 24 21 18 15 12 9 6 3 0
call number 0
call number 2
call number -17
call number 30
call number 48
call number -256
call number 500
call number 874
call number 1'024
call number -17'001
call number 999'999
call number 1'048'576
call number -1'000'001'000'000
call number 912'345'014'587'957'003
call number -999'912'345'014'587'124'337'999'999
call number 111'222'333'444'555'666'777'888'999'000'000'000'001
call number -912'345'014'587'912'345'014'587'124'912'345'014'587'124'337
call number 999'999'999'999'999'999'999'999'999'999'999'999'999'999'999'999'999'999'999'999'999'999
call number 1'000'000'000'000'000'000'000'000'000'000'000'000'000'000'000'000'000'000'000'000'000'000
rem 10^x:x= 66 63 60 57 54 51 48 45 42 39 36 33 30 27 24 21 18 15 12 9 6 3 0
这是我从测试脚本中得到的输出：
0 -> Zero
2 -> Two
-17 -> Minus Seventeen
30 -> Thirty
48 -> Forty Eight
-256 -> Minus Two Hundred Fifty Six
500 -> Five Hundred
874 -> Eight Hundred Seventy Four
1'024 -> One Thousand Twenty Four
-17'001 -> Minus Seventeen Thousand One
999'999 -> Nine Hundred Ninety Nine Thousand Nine Hundred Ninety Nine
1'048'576 -> One Million Forty Eight Thousand Five Hundred Seventy Six
-1'000'001'000'000 -> Minus One Trillion One Million
912'345'014'587'957'003 -> Nine Hundred Twelve Quadrillion Three Hundred Forty Five Trillion Fourteen Billion Five Hundred Eighty Seven Million Nine Hundred Fifty Seven Thousand Three
-999'912'345'014'587'124'337'999'999 -> Minus Nine Hundred Ninety Nine Septillion Nine Hundred Twelve Sextillion Three Hundred Forty Five Quintillion Fourteen Quadrillion Five Hundred Eighty Seven Trillion One Hundred Twenty Four Billion Three Hundred Thirty Seven Million Nine Hundred Ninety Nine Thousand Nine Hundred Ninety Nine
111'222'333'444'555'666'777'888'999'000'000'000'001 -> One Hundred Eleven Undecillion Two Hundred Twenty Two Decillion Three Hundred Thirty Three Nonillion Four Hundred Forty Four Octillion Five Hundred Fifty Five Septillion Six Hundred Sixty Six Sextillion Seven Hundred Seventy Seven Quintillion Eight Hundred Eighty Eight Quadrillion Nine Hundred Ninety Nine Trillion One
-912'345'014'587'912'345'014'587'124'912'345'014'587'124'337 -> Minus Nine Hundred Twelve Tredecillion Three Hundred Forty Five Duodecillion Fourteen Undecillion Five Hundred Eighty Seven Decillion Nine Hundred Twelve Nonillion Three Hundred Forty Five Octillion Fourteen Septillion Five Hundred Eighty Seven Sextillion One Hundred Twenty Four Quintillion Nine Hundred Twelve Quadrillion Three Hundred Forty Five Trillion Fourteen Billion Five Hundred Eighty Seven Million One Hundred Twenty Four Thousand Three Hundred Thirty Seven
999'999'999'999'999'999'999'999'999'999'999'999'999'999'999'999'999'999'999'999'999'999 -> Nine Hundred Ninety Nine Vigintillion Nine Hundred Ninety Nine Novemdecillion Nine Hundred Ninety Nine Octodecillion Nine Hundred Ninety Nine Septendecillion Nine Hundred Ninety Nine Sexdecillion Nine Hundred Ninety Nine Quindecillion Nine Hundred Ninety Nine Quattuordecillion Nine Hundred Ninety Nine Tredecillion Nine Hundred Ninety Nine Duodecillion Nine Hundred Ninety Nine Undecillion Nine Hundred Ninety Nine Decillion Nine Hundred Ninety Nine Nonillion Nine Hundred Ninety Nine Octillion Nine Hundred Ninety Nine Septillion Nine Hundred Ninety Nine Sextillion Nine Hundred Ninety Nine Quintillion Nine Hundred Ninety Nine Quadrillion Nine Hundred Ninety Nine Trillion Nine Hundred Ninety Nine Billion Nine Hundred Ninety Nine Million Nine Hundred Ninety Nine Thousand Nine Hundred Ninety Nine
1'000'000'000'000'000'000'000'000'000'000'000'000'000'000'000'000'000'000'000'000'000'000 -> * ERR: the number is too big! Even wikipedia doesn't know how it's called!
如果我能找到更多names of large numbers，脚本将支持更大的数字。但是，目前，该脚本将使用从 - （10 ^ 66-1）到（10 ^ 66-1）的所有数字。
我必须提一下，我在BATCH中解决这个问题很有意思。 ：）
答案 1 :(得分：45)
C＃ - 30行包含方法声明和{} s ：
考虑前面提到的所有逗号，连字符和连字符。我只包括octillion因为decimal.MaxValue只在octillions中。对于更大的整数，您需要将相应的项添加到thou []数组中，并且可能将数字作为字符串传递，修改该行以使用最后3个字符提取块，而不是像我在这里使用模数。
static string wordify(decimal v)
{
if (v == 0) return "zero";
var units = " one two three four five six seven eight nine".Split();
var teens = " eleven twelve thir# four# fif# six# seven# eigh# nine#".Replace("#", "teen").Split();
var tens = " ten twenty thirty forty fifty sixty seventy eighty ninety".Split();
var thou = " thousand m# b# tr# quadr# quint# sext# sept# oct#".Replace("#", "illion").Split();
var g = (v < 0) ? "minus " : "";
var w = "";
var p = 0;
v = Math.Abs(v);
while (v > 0)
{
int b = (int)(v % 1000);
if (b > 0)
{
var h = (b / 100);
var t = (b - h * 100) / 10;
var u = (b - h * 100 - t * 10);
var s = ((h > 0) ? units[h] + " hundred" + ((t > 0 | u > 0) ? " and " : "") : "")
+ ((t > 0) ? (t == 1 && u > 0) ? teens[u] : tens[t] + ((u > 0) ? "-" : "") : "")
+ ((t != 1) ? units[u] : "");
s = (((v > 1000) && (h == 0) && (p == 0)) ? " and " : (v > 1000) ? ", " : "") + s;
w = s + " " + thou[p] + w;
}
v = v / 1000;
p++;
}
return g + w;
}
使用：
调用static void Main(string[] args)
{
Console.WriteLine(wordify(decimal.MaxValue));
}
输出：
七十九个八十，二百 和二十八个septillion，一个 一百六十二性别，五 一百一十四个quintillion，两个 一百六十四千万亿， 三百三十七 万亿，五百和 九十三亿，五百和 四千三百万，九百和 五万，三百和 35
答案 2 :(得分：40)
在A86 assember中 - 组装成.COM可执行文件：
dd 0ba02c6bfh, 0b8bd10c1h, 0e808b512h, 0ea870100h, 08700e9e8h, 010273eah
dd 0e0e8c2h, 06b51872h, 0c000ebe8h, 0b3c02e8h, 03368067dh, 0b2e901h
dd 0baaa5004h, 0fd8110c1h, 0cd7c1630h, 0bf3031bbh, 0a0571000h, 0ec880080h
dd 0c581c589h, 023c0081h, 0e7f087ch, 0823e38h, 027b00875h, 0e901d068h
dd 0b6400080h, 04f6f603h, 080d08a1ch, 0b60f80c4h, 07f06c7f4h, 088303000h
dd 0ac00813eh, 087ef828h, 0b00056e8h, 051e81dh, 0d83850adh, 0e7f157ch
dd 0a74fc38h, 0262ce088h, 0e901a368h, 01d2c003bh, 0580036e8h, 0b7efc38h
dd 0774d838h, 0f828e088h, 0800026e8h, 0127e1dfah, 0afd448ah, 0440afe44h
dd 074f838ffh, 0e8c28a05h, 0cafe000fh, 0ab7cee39h, 05a2405c6h, 021cd09b4h
dd 05e856c3h, 020b05e00h, 0c5bec3aah, 074c00a02h, 03c80460ah, 0fefa755bh
dd 046f675c8h, 0745b3cach, 0f8ebaae8h, 0eec1d689h, 08a3c8a03h, 07e180cah
dd 0cfd2c1feh, 0ebe8c342h, 0fed8d0ffh, 0c3f775cdh, 01e581e8fh, 0303c5ea8h
dd 0df6f652ah, 078bde03ch, 05e027500h, 01ec1603ch, 07d40793dh, 0603c8080h
dd 09f6f2838h, 040f17a3dh, 080f17a22h, 0403d7264h, 0793cdee1h, 0140740f1h
dd 01e2f7d32h, 02f488948h, 0a7c43b05h, 0a257af9bh, 0be297b6ch, 04609e30ah
dd 0b8f902abh, 07c21e13eh, 09a077d9eh, 054f82ab5h, 0fabe2af3h, 08a6534cdh
dd 0d32b4c97h, 035c7c8ceh, 082bcc833h, 0f87f154fh, 0650ff7eah, 02f143fdfh
dd 0a1fd687fh, 0c3e687fdh, 0c6d50fe0h, 075f13574h, 0898c335bh, 0e748ce85h
dd 08769676fh, 0ad2cedd3h, 0928c77c7h, 077e2d18eh, 01a77e8f6h
db 0bah, 01bh
这是一个454字节的可执行文件。
这是（略小）代码。由于A86只是8086的汇编程序，我不得不手动编写32位扩展代码：
mov di,strings
mov dx,tree_data * 8 + 1
mov bp,code_data * 8
l1:
mov ch,8
call extract_bits
xchg dx,bp
call extract_bit
xchg dx,bp
jnc l2
add dx,ax
l2:
call extract_bit
jc l3
mov ch,6
call extract_bits
shr al,2
cmp al,11
push l27
jl get_string
l25:
add al,48+32
stosb
l27:
mov dx,tree_data * 8 + 1
l3:
cmp bp,end_data * 8
jl l1
convert:
mov bx,'01'
mov di,01000h
push di
mov al,[80h]
mov ah,ch
mov bp,ax
add bp,81h
cmp al,2
jl zero
jg l90
cmp byte ptr [82h],bh
jne l90
zero:
mov al,39
push done
get_string:
mov si,strings-1
or al,al
je l36
l35:
inc si
cmp byte ptr [si],';'+32
jne l35
dec al
jnz l35
l36:
inc si
l37:
lodsb
cmp al,';'+32
je ret
stosb
jmp l37
l90:
inc ax
mov dh,3
div dh
add al,28
mov dl,al
add ah,80h
db 0fh, 0b6h, 0f4h ; movzx si,ah
mov word ptr [80h],'00'
l95:
lodsb
sub al,bh
jle l100
call get_string2
mov al,29
call get_string2
l100:
lodsw
push ax
cmp al,bl
jl l150
jg l140
cmp ah,bh
je l140
mov al,ah
sub al,'0'-10
push l150
get_string2:
push si
call get_string
pop si
mov al,' '
stosb
ret
l140:
sub al,'0'-19
call get_string2
l150:
pop ax
cmp ah,bh
jle l200
cmp al,bl
je l200
mov al,ah
sub al,bh
call get_string2
l200:
cmp dl,29
jle l300
mov al,[si-3]
or al,[si-2]
or al,[si-1]
cmp al,bh
je l300
mov al,dl
call get_string2
l300:
dec dl
cmp si,bp
jl l95
done:
mov byte ptr [di],'$'
pop dx
mov ah,9
int 21h
int 20h
l41:
rcr al,1
dec ch
jz ret
extract_bits:
push l41
extract_bit:
mov si,dx
shr si,3
mov bh,[si]
mov cl,dl
and cl,7
inc cl
ror bh,cl
inc dx
ret
tree_data:
dw 01e8fh, 01e58h, 05ea8h, 0303ch, 0652ah, 0df6fh, 0e03ch, 078bdh
dw 07500h, 05e02h, 0603ch, 01ec1h, 0793dh, 07d40h, 08080h, 0603ch
dw 02838h, 09f6fh, 07a3dh, 040f1h, 07a22h, 080f1h, 07264h, 0403dh
dw 0dee1h, 0793ch, 040f1h, 01407h, 07d32h, 01e2fh, 08948h
db 048h
code_data:
dw 052fh, 0c43bh, 09ba7h, 057afh, 06ca2h, 0297bh, 0abeh, 09e3h
dw 0ab46h, 0f902h, 03eb8h, 021e1h, 09e7ch, 077dh, 0b59ah, 0f82ah
dw 0f354h, 0be2ah, 0cdfah, 06534h, 0978ah, 02b4ch, 0ced3h, 0c7c8h
dw 03335h, 0bcc8h, 04f82h, 07f15h, 0eaf8h, 0ff7h, 0df65h, 0143fh
dw 07f2fh, 0fd68h, 0fda1h, 0e687h, 0e0c3h, 0d50fh, 074c6h, 0f135h
dw 05b75h, 08c33h, 08589h, 048ceh, 06fe7h, 06967h, 0d387h, 02cedh
dw 0c7adh, 08c77h, 08e92h, 0e2d1h, 0f677h, 077e8h, 0ba1ah
db 01bh
end_data:
strings:
使用霍夫曼编码存储文本。命令行作为字符串传递，因此转换它很简单 - 将字符串拆分为三个组，并使用当前乘数（数百万，数千等）解析每个组（数百，数十和单位）。
答案 3 :(得分：36)
Lisp，仅使用标准函数：
(format nil "~r" 1234) ==> "one thousand two hundred thirty-four"
加成：
(format nil "~@r" 1234) ==> "MCCXXXIV"
答案 4 :(得分：28)
C ++，15行：
#include <string>
using namespace std;
string Thousands[] = { "zero", "thousand", "million", "billion", "trillion", "quadrillion", "quintillion", "sexillion", "septillion", "octillion", "nonillion", "decillion" };
string Ones[] = { "zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine", "ten", "eleven", "twelve", "thirteen", "fourteen", "fifteen", "sixteen", "seventeen", "eighteen", "nineteen" };
string Tens[] = { "zero", "ten", "twenty", "thirty", "forty", "fifty", "sixty", "seventy", "eighty", "ninety" };
string concat(bool cond1, string first, bool cond2, string second) { return (cond1 ? first : "") + (cond1 && cond2 ? " " : "") + (cond2 ? second : ""); }
string toStringBelowThousand(unsigned long long n) {
return concat(n >= 100, Ones[n / 100] + " hundred", n % 100 != 0, (n % 100 < 20 ? Ones[n % 100] : Tens[(n % 100) / 10] + (n % 10 > 0 ? " " + Ones[n % 10] : "")));
}
string toString(unsigned long long n, int push = 0) {
return n == 0 ? "zero" : concat(n >= 1000, toString(n / 1000, push + 1), n % 1000 != 0, concat(true, toStringBelowThousand(n % 1000), push > 0, Thousands[push]));
}
用法：
cout << toString(51351); // => fifty one thousand three hundred fifty one
答案 5 :(得分：23)
perl -MNumber::Spell -e 'print spell_number(2);'
答案 6 :(得分：18)
Paul Fischer和Darius：你们有一些很棒的想法，但我不愿意看到它们以过于冗长的方式实施。 :)开个玩笑，你的解决方案很棒，但是我将＃strike> 14 挤出了30多个字节，同时保留在79列内并维持python 3的兼容性。
所以这是我在79列中的416字节python :(谢谢你们，我站在你的肩膀上）
w=lambda n:_(n,["","thousand "]+p("m b tr quadr quint","illion"))[:-1]or"zero"
_=lambda n,S:n*"x"and _(n//M,S[1:])+(Z[n%M//C]+"hundred ")*(n%M//C>0)+(n%C>19
and p("twen thir fo"+R,"ty")[n%C//10-2]+Z[n%10]or Z[n%C])+S[0]*(n%M>0)
p=lambda a,b="":[i+b+" "for i in a.split()]
R="r fif six seven eigh nine"
M=1000
C=100
Z=[""]+p("one two three four five%st nine ten eleven twelve"%R[5:20])+p(
"thir fou"+R,"teen")
测试：
if __name__ == "__main__":
import sys
assert(w(0)=="zero")
assert(w(100)=="one hundred")
assert(w(1000000)=="one million")
assert(w(1024)=="one thousand twenty four")
assert(w(1048576)=="one million forty eight thousand five hundred seventy six")
答案 7 :(得分：17)
请参阅recursive的更好答案。这是更好的方式。
疯狂道具Darius获取灵感。你的大 - W（现在我的p）特别聪明。
w=lambda n:["zero"," ".join(_(n,0))][n>0]
_=lambda n,l:_(n//M,l+1)+[E,Z[n%M//C]+["hundred"]][n%M//C>0]+\
(p("twen thir fo"+R,"ty")[n%C//10-2]+Z[n%10]if n%C>19 else Z[n%C])+\
[E,([E,["thousand"]]+p("m b tr quadr quint","illion"))[l]][n%M>0]if n else E
p=lambda a,b:[[i+b]for i in a.split()]
E=[];R="r fif six seven eigh nine";M=1000;C=100
Z=[E]+p("one two three four five six seven eight nine ten eleven twelve","")+\
p("thir fou"+R,"teen")
我用它测试它：
if __name__ == "__main__":
import sys
print w(int(sys.argv[1]))
assert(w(100)=="one hundred")
assert(w(1000000)=="one million")
assert(w(1024)=="one thousand twenty four")
assert(w(1048576)=="one million forty eight thousand five hundred seventy six")
此时，这是Darius当前解决方案的调整，这反过来是我的旧解决方案的调整，受到他的启发，他在评论中给出了一些错误提示。这也是对Python的犯罪。
下面的剧透，rot13'为了你的保护，因为高尔夫的一半乐趣 弄清楚如何。我强烈推荐mnenhy Firefox 扩展来内联解码这个（和其他简单的编码方案）。
Pbafgnagf（V eranzrq gurz guvf erivfvba gb ubcrshyyl znxr gurz pyrnere。）
R：Gur rzcgl frg。E：Gung juvpu vf va pbzzba orgjrra pbhagvat va gur“grraf”（ e grra，
svs grra， fvk grra ...）naq va gur graf（ e gl， svs gl， FVK GL ....）Z，P：Jung gurl ner va Ebzna ahzrenyf。M：Nyy gur ahzoref sebz bar gb gjragl。Shapgvbaf（fbzr nyfb eranzrq guvf ebhaq）
j：Gur choyvp-snpvat shapgvba，juvpu gheaf n ahzore vagb jbeqf。_：Erphefviryl gheaf gur ahzore vagb jbeqf，gubhfnaq-ol-gubhfnaq。 a vf
gur ahzore，y vf ubj sne guebhtu gur cbjref bs 1000 jr ner。 Ergheaf n
yvfg bs fvatyrgba yvfgf bs rnpu jbeq va gur ahzore，r.t。
[['bar'],['gubhfnaq'],['gjragl'],['sbhe']]。c：sbe rnpu jbeq va gur fcnpr-frcnengrq jbeq yvfg n，nccraqf o nf n
fhssvk naq chgf gurz rnpu vagb n fvatyrgba yvfg。 Sbe rknzcyr，
c("z o ge","vyyvba") == [['zvyyvba'],['ovyyvba'],['gevyyvba']]。答案 8 :(得分：10)
Python，446个字节。 80列以下的所有行都是该死的。这是Paul Fisher's solution几乎每行都有编码调整，低于他的488字节版本;他已经挤掉了几个字节，我承认了。投票支持他的答案！
g=lambda n:["zero"," ".join(w(n,0))][n>0]
w=lambda n,l:w(n//m,l+1)+[e,z[n%m//100]+["hundred"]][n%m//100>0]+\
(p("twen thir fo"+r,"ty")[n%100//10-2]+z[n%10]if n%100>19 else z[n%100])+\
[e,k[l]][n%m>0]if n else e
p=lambda a,b:[[i+b]for i in a.split()]
e=[];r="r fif six seven eigh nine";m=1000
k=[e,["thousand"]]+p("m b tr quadr quint","illion")
z=[e]+p("one two three four five six seven eight nine ten eleven twelve","")+\
p("thir fou"+r,"teen")
历史变得复杂。我从下面的未经混淆的代码开始，它支持负数和范围检查，加上一些数字的破折号以获得更好的英语：
>>> n2w(2**20)
'one million forty-eight thousand five hundred seventy-six'
def n2w(n):
if n < 0: return 'minus ' + n2w(-n)
if n < 10: return W('zero one two three four five six seven eight nine')[n]
if n < 20: return W('ten eleven twelve',
'thir four fif six seven eigh nine',
'teen')[n-10]
if n < 100:
tens = W('', 'twen thir for fif six seven eigh nine', 'ty')[n//10-2]
return abut(tens, '-', n2w(n % 10))
if n < 1000:
return combine(n, 100, 'hundred')
for i, word in enumerate(W('thousand', 'm b tr quadr quint', 'illion')):
if n < 10**(3*(i+2)):
return combine(n, 10**(3*(i+1)), word)
assert False
def W(b, s='', suff=''): return b.split() + [s1 + suff for s1 in s.split()]
def combine(n, m, term): return abut(n2w(n // m) + ' ' + term, ' ', n2w(n % m))
def abut(w10, sep, w1): return w10 if w1 == 'zero' else w10 + sep + w1
然后我通过模糊处理（对我来说是新的）将其压缩到大约540个字节，而Paul Fisher发现了一个更短的算法（删除破折号）以及一些奇妙可怕的Python编码技巧。我偷了编码技巧，以达到508（仍然没有获胜）。我尝试用一种新算法重新开始，这种算法无法击败费舍尔。最后，这是他的代码的调整。对于！
混淆代码已针对干净代码进行了测试，该代码已通过眼球检查了一堆案例。
答案 9 :(得分：6)
好的，这是F＃，试图保持可读性，大约830字节：
#light
let thou=[|"";"thousand";"million";"billion";"trillion";"quadrillion";"quintillion"|]
let ones=[|"";"one";"two";"three";"four";"five";"six";"seven";"eight";"nine";"ten";"eleven";
"twelve";"thirteen";"fourteen";"fifteen";"sixteen";"seventeen";"eighteen";"nineteen"|]
let tens=[|"";"";"twenty";"thirty";"forty";"fifty";"sixty";"seventy";"eighty";"ninety"|]
let (^-) x y = if y="" then x else x^"-"^y
let (^+) x y = if y="" then x else x^" "^y
let (^?) x y = if x="" then x else x^+y
let (+^+) x y = if x="" then y else x^+y
let Tiny n = if n < 20 then ones.[n] else tens.[n/10] ^- ones.[n%10]
let Small n = (ones.[n/100] ^? "hundred") +^+ Tiny(n%100)
let rec Big n t = if n = 0UL then "" else
(Big (n/1000UL) (t+1)) +^+ (Small(n%1000UL|>int) ^? thou.[t])
let Convert n = if n = 0UL then "zero" else Big n 0
以下是单元测试
let Show n =
printfn "%20u -> \"%s\"" n (Convert n)
let tinyTests = [0; 1; 10; 11; 19; 20; 21; 30; 99] |> List.map uint64
let smallTests = tinyTests @ (tinyTests |> List.map (fun n -> n + 200UL))
let MakeTests t1 t2 =
List.map (fun n -> n * (pown 1000UL t1)) smallTests
|> List.map_concat (fun n -> List.map (fun x -> x * (pown 1000UL t2) + n) smallTests)
for n in smallTests do
Show n
for n in MakeTests 1 0 do
Show n
for n in MakeTests 5 2 do
Show n
Show 1000001000678000001UL
Show 17999999999999999999UL
答案 10 :(得分：5)
这是一个Scala解决方案。我不高兴尝试让它看起来很短 - 我牺牲了一点可读性:(
object NumSpeller {
val digits = Array("","one","two","three","four","five","six","seven","eight","nine")
val teens = Array("ten", "eleven", "twelve", "thirteen", "fourteen", "fifteen", "sixteen", "seventeen", "eighteen", "nineteen")
val tens = Array("", "ten", "twenty", "thirty", "fourty", "fifty", "sixty", "seventy", "eighty", "ninety")
val thousands = Array("", "thousand", "million", "billion", "trillion", "quadrillion", "quintillion")
def spellGroup(num:Int) = {
val (v3, v2, v1) = ((num / 100) % 10, (num / 10) % 10, num % 10)
val hs = v3 match { case 0 => ""; case d => digits(d) + " hundred " }
val ts = v2 match {
case 0 => digits(v1)
case 1 => teens(v1)
case _ => v3 match { case 0 => tens(v2); case _ => tens(v2) + "-" + digits(v1) }
}
hs + ts
}
def numberGroups(num:Long) = {
def _numberGroups(num:Long, factor:Int):List[(Double,Int)] = factor match {
case 0 => List((num % 1000,0))
case _ => ((num / Math.pow(1000, factor)) % 1000, factor) :: _numberGroups(num, factor - 1)
}
val ints = _numberGroups(num, 6) map (x => (x._1.asInstanceOf[Int],x._2))
ints dropWhile (x => x._1 == 0.0)
}
def spell(num:Long) = num match { case 0 => "zero"; case _ => (numberGroups(num) map { x => spellGroup(x._1) + " " + thousands(x._2) + " " }).mkString.trim }
}
用法是：
NumSpeller.spell(458582)
答案 11 :(得分：5)
T-SQL（SQL Server 2005）函数，包括测试用例：
if exists (select 1 from sys.objects where object_id = object_id(N'dbo.fnGetNumberString'))
drop function fnGetNumberString
go
/*
Tests:
declare @tests table ( testValue bigint )
insert into @tests select -43213 union select -5 union select 0 union select 2 union select 15 union select 33 union select 100 union select 456 union select 1024 union select 10343 union select 12345678901234 union select -3434343434343
select testValue, dbo.fnGetNumberString(testValue) as textValue
from @tests
*/
create function dbo.fnGetNumberString
(
@value bigint
)
returns nvarchar(1024)
as
begin
if @value = 0 return 'zero' -- lets me avoid special-casing this later
declare @isNegative bit
set @isNegative = 0
if @value < 0
select @isNegative = 1, @value = @value * -1
declare @groupNames table ( groupOrder int, groupName nvarchar(15) )
insert into @groupNames select 1, '' union select 2, 'thousand' union select 3, 'million' union select 4, 'billion' union select 5, 'trillion' union select 6, 'quadrillion' union select 7, 'quintillion' union select 8, 'sextillion'
declare @digitNames table ( digit tinyint, digitName nvarchar(10) )
insert into @digitNames select 0, '' union select 1, 'one' union select 2, 'two' union select 3, 'three' union select 4, 'four' union select 5, 'five' union select 6, 'six' union select 7, 'seven' union select 8, 'eight' union select 9, 'nine' union select 10, 'ten' union select 11, 'eleven' union select 12, 'twelve' union select 13, 'thirteen' union select 14, 'fourteen' union select 15, 'fifteen' union select 16, 'sixteen' union select 17, 'seventeen' union select 18, 'eighteen' union select 19, 'nineteen'
declare @tensGroups table ( digit tinyint, groupName nvarchar(10) )
insert into @tensGroups select 2, 'twenty' union select 3, 'thirty' union select 4, 'forty' union select 5, 'fifty' union select 6, 'sixty' union select 7, 'seventy' union select 8, 'eighty' union select 9, 'ninety'
declare @groups table ( groupOrder int identity, groupValue int )
declare @convertedValue varchar(50)
while @value > 0
begin
insert into @groups (groupValue) select @value % 1000
set @value = @value / 1000
end
declare @returnValue nvarchar(1024)
set @returnValue = ''
if @isNegative = 1 set @returnValue = 'negative'
select @returnValue = @returnValue +
case when len(h.digitName) > 0 then ' ' + h.digitName + ' hundred' else '' end +
case when len(isnull(t.groupName, '')) > 0 then ' ' + t.groupName + case when len(isnull(o.digitName, '')) > 0 then '-' else '' end + isnull(o.digitName, '') else case when len(isnull(o.digitName, '')) > 0 then ' ' + o.digitName else '' end end +
case when len(n.groupName) > 0 then ' ' + n.groupName else '' end
from @groups g
join @groupNames n on n.groupOrder = g.groupOrder
join @digitNames h on h.digit = (g.groupValue / 100)
left join @tensGroups t on t.digit = ((g.groupValue % 100) / 10)
left join @digitNames o on o.digit = case when (g.groupValue % 100) < 20 then g.groupValue % 100 else g.groupValue % 10 end
order by g.groupOrder desc
return @returnValue
end
go
答案 12 :(得分：5)
这是C中相对简单的实现（52行）。
注意：这不执行任何边界检查;调用者必须确保调用缓冲区足够大。
#include <stdio.h>
#include <string.h>
const char *zero_to_nineteen[20] = {"", "One ", "Two ", "Three ", "Four ", "Five ", "Six ", "Seven ", "Eight ", "Nine ", "Ten ", "Eleven ", "Twelve ", "Thirteen ", "Fourteen ", "Fifteen ", "Sixteen ", "Seventeen ", "Eighteen ", "Nineteen "};
const char *twenty_to_ninety[8] = {"Twenty ", "Thirty ", "Forty ", "Fifty ", "Sixty ", "Seventy ", "Eighty ", "Ninety "};
const char *big_numbers[7] = {"", "Thousand ", "Million ", "Billion ", "Trillion ", "Quadrillion ", "Quintillion "};
void num_to_word(char *buf, unsigned long long num)
{
unsigned long long power_of_1000 = 1000000000000000000ull;
int power_index = 6;
if(num == 0)
{
strcpy(buf, "Zero");
return;
}
buf[0] = 0;
while(power_of_1000 > 0)
{
int group = num / power_of_1000;
if(group >= 100)
{
strcat(buf, zero_to_nineteen[group / 100]);
strcat(buf, "Hundred ");
group %= 100;
}
if(group >= 20)
{
strcat(buf, twenty_to_ninety[group / 10 - 2]);
group %= 10;
}
if(group > 0)
strcat(buf, zero_to_nineteen[group]);
if(num >= power_of_1000)
strcat(buf, big_numbers[power_index]);
num %= power_of_1000;
power_of_1000 /= 1000;
power_index--;
}
buf[strlen(buf) - 1] = 0;
}
这是一个更加混淆的版本（682个字符）。如果我真的尝试的话，它可能会减少一点。
#include <string.h>
#define C strcat(b,
#define U unsigned long long
char*z[]={"","One","Two","Three","Four","Five","Six","Seven","Eight","Nine","Ten","Eleven","Twelve","Thirteen","Fourteen","Fifteen","Sixteen","Seventeen","Eighteen","Nineteen"},*t[]={"Twenty ","Thirty ","Forty ","Fifty ","Sixty ","Seventy ","Eighty ","Ninety "},*q[]={"","Thousand ","Million ","Billion ","Trillion ","Quadrillion ","Quintillion "};
void W(char*b,U n){U p=1000000000000000000ull;int i=6;*b=0;if(!n)strcpy(b,"Zero ");else while(p){int g=n/p;if(g>99){C z[g/100]);C " ");C "Hundred ");g%=100;}if(g>19){C t[g/10-2]);g%=10;}if(g)C z[g]),C " ");if(n>=p)C q[i]);n%=p;p/=1000;i--;}b[strlen(b)-1]=0;}
答案 13 :(得分：4)
Perl 5.10 my %expo=(0,'',
qw'1 thousand 2 million 3 billion 4 trillion 5 quadrillion 6 quintillion
7 sextillion 8 septillion 9 octillion 10 nonillion 11 decillion 12 undecillion
13 duodecillion 14 tredecillion 15 quattuordecillion 16 quindecillion
17 sexdecillion 18 septendecillion 19 octodecillion 20 novemdecillion
21 vigintillion'
);
my %digit=(0,'',
qw'1 one 2 two 3 three 4 four 5 five 6 six 7 seven 8 eight 9 nine 10 ten
11 eleven 12 twelve 13 thirteen 14 fourteen 15 fifteen 16 sixteen 17 seventeen
18 eighteen 19 nineteen 2* twenty 3* thirty 4* forty 5* fifty 6* sixty
7* seventy 8* eighty 9* ninety'
);
sub spell_number(_){
local($_)=@_;
($_,@_)=split/(?=(?:.{3})*+$)/;
$_=0 x(3-length).$_;
unshift@_,$_;
my @o;
my $c=@_;
for(@_){
my $o='';
/(.)(.)(.)/;
$o.=$1?$digit{$1}.' hundred':'';
$o.=$2==1?
' '.$digit{$2.$3}
:
($2?' '.$digit{"$2*"}:'').
($2&&$3?' ':'').
$digit{$3}
;
$o.=--$c?($o?' '.$expo{$c}.', ':''):'';
push@o,$o;
}
my $o;
$o.=$_ for@o;
$o=~/^\s*+(.*?)(, )?$/;
$o?$1:'zero';
}
注意：
split()似乎是主要问题。现在，它们占据了大部分角色。my和local来缩短它，并将其全部放在一行上。strict和warnings下工作。答案 14 :(得分：3)
无论如何，为了记录，我在PHP中为法语提供了一个众所周知的解决方案（不是我的！）：Écriture des nombres en français。 ： - ）
请注意您的措辞含糊不清（自愿或不自愿）：“以任何语言提交的提交内容”
我先把它当作“自然语言”，然后才明白你的意思是“编程语言......”
该算法在英语中可能更简单（并且具有较少的区域变体......）。
答案 15 :(得分：3)
string Number(ulong i)
{
static string[] names = [
""[],
" thousand",
" million",
" billion",
" trillion",
" quadrillion",
];
string ret = null;
foreach(mult; names)
{
if(i%1000 != 0)
{
if(ret != null) ret = ret ~ ", "
ret = Cent(i%1000) ~ mult ~ ret;
}
i /= 1000;
}
return ret;
}
string Cent(int i)
{
static string[] v =
[""[], "one", "two", "three", "four",
"five", "six", "seven", "eight", "nine"];
static string[] tens =
["!"[], "!", "twenty", "thirty", "forty",
"fifty", "sixty", "seventy", "eighty", "ninety"];
string p1, p2, p3 = "";
if(i >= 100)
{
p1 = v[i/100] ~ " hundred";
p3 = (i % 100 != 0) ? " and " : ""; //optional
}
else
p1 = "";
i %= 100;
switch(i)
{
case 0, 1, 2, 3, 4, 5, 6, 7, 8, 9:
p2 = v[i];
break;
case 10: p2 = "ten"; break;
case 11: p2 = "eleven"; break;
case 12: p2 = "twelve"; break;
case 13: p2 = "thirteen"; break;
case 14: p2 = "fourteen"; break;
case 15: p2 = "fifteen"; break;
case 16: p2 = "sixteen"; break;
case 17: p2 = "seventeen"; break;
case 18: p2 = "eighteen"; break;
case 19: p2 = "nineteen"; break;
default:
p2 = tens[i/10] ~ "-" ~ v[i%10];
break;
}
return p1 ~ p3 ~ p2;
}
import std.stdio;
void main()
{
writef("%s\n", Number(8_000_400_213));
}
答案 16 :(得分：2)
是否有人计划在短时间内添加相应的逗号和'和'？或者连字符二十一到九十九？没有多大意义，恕我直言：）
“九十九九十九九十九”VS
'九九九九九九'
（不，我的不行。但是。）
答案 17 :(得分：2)
#!/usr/bin/env perl
my %symbols = (
1 => "One", 2 => "Two", 3 => "Three", 4 => "Four", 5 => "Five",
6 => "Six", 7 => "Seven", 8 => "Eight", 9 => "Nine", 10 => "Ten",
11 => "Eleven", 12 => "Twelve", 13 => "Thirteen", 14 => "Fourteen",
15 => "Fifteen", 16 => "Sixteen", 17 => "Seventeen", 18 => "Eighteen",
19 => "Nineteen", 20 => "Twenty", 30 => "Thirty", 40 => "Forty",
50 => "Fifty", 60 => "Sixty", 70 => "Seventy", 80 => "Eighty",
90 => "Ninety", 100 => "Hundred");
my %three_symbols = (1 => "Thousand", 2 => "Million", 3 => "Billion" );
sub babo {
my ($input) = @_;
my @threes = split(undef, $input);
my $counter = ($#threes + 1);
my $remainder = $counter % 3;
my @result;
while ($counter > 0){
my $digits = "";
my $three;
my $full_match = 0;
if ($remainder > 0){
while ($remainder > 0) {
$digits .= shift(@threes);
$remainder--;
$counter--;
}
}
else {
$digits = join('',@threes[0,1,2]);
splice(@threes, 0, 3);
$counter -= 3;
}
if (exists($symbols{$digits})){
$three = $symbols{$digits};
$full_match = 1;
}
elsif (length($digits) == 3) {
$three = $symbols{substr($digits,0,1)};
$three .= " Hundred";
$digits = substr($digits,1,2);
if (exists($symbols{$digits})){
$three .= " " . $symbols{$digits};
$full_match = 1;
}
}
if ($full_match == 0){
$three .= " " . $symbols{substr($digits,0,1)."0"};
$three .= " " . $symbols{substr($digits,1,1)};
}
push(@result, $three);
if ($counter > 0){
push(@result, "Thousand");
}
}
my $three_counter = 0;
my @r = map {$_ eq "Thousand" ? $three_symbols{++$three_counter}:$_ }
reverse @result;
return join(" ", reverse @r);
}
print babo(1) . "\n";
print babo(12) . "\n";
print babo(120) . "\n";
print babo(1234) . "\n";
print babo(12345) . "\n";
print babo(123456) . "\n";
print babo(1234567) . "\n";
print babo(1234567890) . "\n";
答案 18 :(得分：1)
我现在找不到该文件，但这是我去学校的编程简介（学期后期）。我们必须能够将浮动转换为有效的书面号码，以便在支票上使用。
任务完成后，教授展示了一些C ++代码，只使用我们已经介绍过的概念来解决问题。它只运行了43行，并且记录良好。
答案 19 :(得分：1)
这是PHP中的一个，来自 Convert Numbers to Words ：
convert_number(2850)
返回
二千八百五十
如果你想要一个更加精彩的处理逗号和数字的警告，请查看zac hesters在 Language Display Functions 工作：
function convert_number($number)
{
if (($number < 0) || ($number > 999999999))
{
throw new Exception("Number is out of range");
}
$Gn = floor($number / 1000000); /* Millions (giga) */
$number -= $Gn * 1000000;
$kn = floor($number / 1000); /* Thousands (kilo) */
$number -= $kn * 1000;
$Hn = floor($number / 100); /* Hundreds (hecto) */
$number -= $Hn * 100;
$Dn = floor($number / 10); /* Tens (deca) */
$n = $number % 10; /* Ones */
$res = "";
if ($Gn)
{
$res .= convert_number($Gn) . " Million";
}
if ($kn)
{
$res .= (empty($res) ? "" : " ") .
convert_number($kn) . " Thousand";
}
if ($Hn)
{
$res .= (empty($res) ? "" : " ") .
convert_number($Hn) . " Hundred";
}
$ones = array("", "One", "Two", "Three", "Four", "Five", "Six",
"Seven", "Eight", "Nine", "Ten", "Eleven", "Twelve", "Thirteen",
"Fourteen", "Fifteen", "Sixteen", "Seventeen", "Eightteen",
"Nineteen");
$tens = array("", "", "Twenty", "Thirty", "Fourty", "Fifty", "Sixty",
"Seventy", "Eigthy", "Ninety");
if ($Dn || $n)
{
if (!empty($res))
{
$res .= " and ";
}
if ($Dn < 2)
{
$res .= $ones[$Dn * 10 + $n];
}
else
{
$res .= $tens[$Dn];
if ($n)
{
$res .= "-" . $ones[$n];
}
}
}
if (empty($res))
{
$res = "zero";
}
return $res;
}
答案 20 :(得分：1)
#!/usr/bin/perl
use strict;
use warnings;
use Lingua::EN::Numbers qw(num2en);
print num2en($_), "\n" for 2, 1024, 1024*1024;
C:\Temp> n.pl two one thousand and twenty-four one million, forty-eight thousand, five hundred and seventy-six
答案 21 :(得分：0)
几年前，我在C＃中为多语言应用程序创建了这个。
这是基类：
public abstract class ValueSource
{
public abstract object Value { get; }
}
这个是用于wordify ..
public abstract class NumberTextValueSource:ValueSource
{
public abstract decimal Number { get; }
public abstract string Format { get; }
public abstract string Negative { get; }
public abstract bool UseValueIfZero { get; }
public abstract string N0 { get; }
public abstract string N1 { get; }
public abstract string N2 { get; }
public abstract string N3 { get; }
public abstract string N4 { get; }
public abstract string N5 { get; }
public abstract string N6 { get; }
public abstract string N7 { get; }
public abstract string N8 { get; }
public abstract string N9 { get; }
public abstract string N10 { get; }
public abstract string N11 { get; }
public abstract string N12 { get; }
public abstract string N13 { get; }
public abstract string N14 { get; }
public abstract string N15 { get; }
public abstract string N16 { get; }
public abstract string N17 { get; }
public abstract string N18 { get; }
public abstract string N19 { get; }
public abstract string N20 { get; }
public abstract string N30 { get; }
public abstract string N40 { get; }
public abstract string N50 { get; }
public abstract string N60 { get; }
public abstract string N70 { get; }
public abstract string N80 { get; }
public abstract string N90 { get; }
public abstract string N100 { get; }
public abstract string NHundred { get; }
public abstract string N1000 { get; }
public abstract string NThousand { get; }
public abstract string NMillion { get; }
public abstract string NBillion { get; }
public abstract string NTrillion { get; }
public abstract string NQuadrillion { get; }
string getOne(Type t, string v)
{
if (v[0] == '0' && !UseValueIfZero)
return "";
return (string)t.GetProperty("N" + v[0].ToString()).GetValue(this, null);
}
string getTwo(Type t, string v)
{
if (v[0] == '0')
if (v[1] != '0')
return getOne(t, v.Substring(1));
else
return "";
if (v[1] == '0' || v[0] == '1')
return (string)t.GetProperty("N" + v).GetValue(this, null);
return (string)t.GetProperty("N" + v[0].ToString() + "0").GetValue(this, null) +
getOne(t, v.Substring(1));
}
string getThree(Type t, string v)
{
if(v[0] == '0')
return getTwo(t,v.Substring(1));
if (v[0] == '1')
return
N100 +
getTwo(t, v.Substring(1));
return
getOne(t, v[0].ToString()) +
NHundred +
getTwo(t, v.Substring(1));
}
string getFour(Type t, string v)
{
if (v[0] == '0')
return getThree(t, v.Substring(1));
if (v[0] == '1')
return
N1000 +
getThree(t, v.Substring(1));
return
getOne(t, v[0].ToString()) +
NThousand +
getThree(t, v.Substring(1));
}
string getFive(Type t, string v)
{
if (v[0] == '0')
return getFour(t, v.Substring(1));
return
getTwo(t, v.Substring(0, 2)) +
NThousand +
getThree(t, v.Substring(2));
}
string getSix(Type t, string v)
{
if (v[0] == '0')
return getFive(t, v.Substring(1));
return
getThree(t, v.Substring(0, 3)) +
NThousand +
getThree(t, v.Substring(3));
}
string getSeven(Type t, string v)
{
if (v[0] == '0')
return getSix(t, v.Substring(1));
return
getOne(t, v[0].ToString()) +
NMillion +
getSix(t, v.Substring(3));
}
string getEight(Type t, string v)
{
if (v[0] == '0')
return getSeven(t, v.Substring(1));
return
getTwo(t, v.Substring(0, 2)) +
NMillion +
getSix(t, v.Substring(2));
}
string getNine(Type t, string v)
{
if (v[0] == '0')
return getEight(t, v.Substring(1));
return
getThree(t, v.Substring(0, 3)) +
NMillion +
getSix(t, v.Substring(3));
}
string getTen(Type t, string v)
{
if (v[0] == '0')
return getNine(t, v.Substring(1));
return
getOne(t, v.Substring(0, 1)) +
NBillion +
getNine(t, v.Substring(1));
}
string getEleven(Type t, string v)
{
if (v[0] == '0')
return getTen(t, v.Substring(1));
return
getTwo(t, v.Substring(0, 2)) +
NBillion +
getNine(t, v.Substring(2));
}
string getTwelve(Type t, string v)
{
if (v[0] == '0')
return getEleven(t, v.Substring(1));
return
getThree(t, v.Substring(0, 3)) +
NBillion +
getNine(t, v.Substring(3));
}
string getThirteen(Type t, string v)
{
if (v[0] == '0')
return getTwelve(t, v.Substring(1));
return
getOne(t, v.Substring(0, 1)) +
NTrillion +
getTwelve(t, v.Substring(1));
}
string getForteen(Type t, string v)
{
if (v[0] == '0')
return getThirteen(t, v.Substring(1));
return
getTwo(t, v.Substring(0, 2)) +
NTrillion +
getTwelve(t, v.Substring(2));
}
string getFifteen(Type t, string v)
{
if (v[0] == '0')
return getForteen(t, v.Substring(1));
return
getThree(t, v.Substring(0, 3)) +
NTrillion +
getTwelve(t, v.Substring(3));
}
string getSixteen(Type t, string v)
{
if (v[0] == '0')
return getFifteen(t, v.Substring(1));
return
getOne(t, v.Substring(0, 1)) +
NQuadrillion +
getFifteen(t, v.Substring(1));
}
string getSeventeen(Type t, string v)
{
if (v[0] == '0')
return getSixteen(t, v.Substring(1));
return
getTwo(t, v.Substring(0, 2)) +
NQuadrillion +
getFifteen(t, v.Substring(2));
}
string getEighteen(Type t, string v)
{
if (v[0] == '0')
return getSeventeen(t, v.Substring(1));
return
getThree(t, v.Substring(0, 3)) +
NQuadrillion +
getFifteen(t, v.Substring(3));
}
string convert(Type t, string hp)
{
switch (hp.Length)
{
case 1:
return getOne(t, hp);
case 2:
return getTwo(t, hp);
case 3:
return getThree(t, hp);
case 4:
return getFour(t, hp);
case 5:
return getFive(t, hp);
case 6:
return getSix(t, hp);
case 7:
return getSeven(t, hp);
case 8:
return getEight(t, hp);
case 9:
return getNine(t, hp);
case 10:
return getTen(t, hp);
case 11:
return getEleven(t, hp);
case 12:
return getTwelve(t, hp);
case 13:
return getThirteen(t, hp);
case 14:
return getForteen(t, hp);
case 15:
return getFifteen(t, hp);
case 16:
return getSixteen(t, hp);
case 17:
return getSeventeen(t, hp);
case 18:
return getEighteen(t, hp);
}
return "";
}
public override object Value
{
get
{
decimal d = Number;
decimal highPoint, lowPoint;
bool isNeg = d < 0;
d = Math.Abs(d);
highPoint = Math.Floor(d);
lowPoint = d - highPoint;
Type t = this.GetType();
string strHigh = convert(t, highPoint.ToString()),
strLow =
lowPoint > 0 ?
convert(t, lowPoint.ToString().Substring(2)) :
UseValueIfZero ? N0 : "";
if (isNeg) strHigh = Negative + " " + strHigh;
return string.Format(Format, strHigh, strLow);
}
}
}
这个是土耳其语（TRY）：
public class TRYNumberTextValueSource:NumberTextValueSource
{
decimal num;
public TRYNumberTextValueSource(decimal value)
{
num = Math.Round(value, 2);
}
public override decimal Number
{
get { return num; }
}
public override string Format
{
get
{
if (num == 0)
return N0 + " YTL";
if (num > -1 && num < 1)
return "{0}{1} Kurus";
return "{0} YTL {1} Kurus";
}
}
public override string Negative
{
get { return "-"; }
}
public override bool UseValueIfZero
{
get { return false; }
}
public override string N0
{
get { return "sifir"; }
}
public override string N1
{
get { return "bir"; }
}
public override string N2
{
get { return "iki"; }
}
public override string N3
{
get { return "üç"; }
}
public override string N4
{
get { return "dört"; }
}
public override string N5
{
get { return "bes"; }
}
public override string N6
{
get { return "alti"; }
}
public override string N7
{
get { return "yedi"; }
}
public override string N8
{
get { return "sekiz"; }
}
public override string N9
{
get { return "dokuz"; }
}
public override string N10
{
get { return "on"; }
}
public override string N11
{
get { return "onbir"; }
}
public override string N12
{
get { return "oniki"; }
}
public override string N13
{
get { return "onüç"; }
}
public override string N14
{
get { return "ondört"; }
}
public override string N15
{
get { return "onbes"; }
}
public override string N16
{
get { return "onalti"; }
}
public override string N17
{
get { return "onyedi"; }
}
public override string N18
{
get { return "onsekiz"; }
}
public override string N19
{
get { return "ondokuz"; }
}
public override string N20
{
get { return "yirmi"; }
}
public override string N30
{
get { return "otuz"; }
}
public override string N40
{
get { return "kirk"; }
}
public override string N50
{
get { return "elli"; }
}
public override string N60
{
get { return "altmis"; }
}
public override string N70
{
get { return "yetmis"; }
}
public override string N80
{
get { return "seksen"; }
}
public override string N90
{
get { return "doksan"; }
}
public override string N100
{
get { return "yüz"; }
}
public override string NHundred
{
get { return "yüz"; }
}
public override string N1000
{
get { return "bin"; }
}
public override string NThousand
{
get { return "bin"; }
}
public override string NMillion
{
get { return "milyon"; }
}
public override string NBillion
{
get { return "milyar"; }
}
public override string NTrillion
{
get { return "trilyon"; }
}
public override string NQuadrillion
{
get { return "trilyar"; }
}
}
它以这种方式使用：
MessageBox.show((string)(new TRYNumberTextValueSource(12345)).Value);