对于那些喜欢这种事情的人来说，这是一个编码问题。让我们看一下函数的实现（当然是用你选择的语言），它返回一个指定整数的人类可读字符串表示。例如：
特别聪明/优雅的解决方案的奖励积分！
这似乎是一种毫无意义的练习，但是这种算法有许多现实世界的应用（虽然支持高达10亿的数字可能有点过分： - ）
答案 0 :(得分：8)
已经有一个问题： Convert integers to written numbers
答案是C＃，但我认为你可以搞清楚。
答案 1 :(得分：3)
import math
def encodeOnesDigit(num):
return ['', 'one', 'two', 'three', 'four', 'five', 'six', 'seven', 'eight', 'nine'][num]
def encodeTensDigit(num):
return ['twenty', 'thirty', 'forty', 'fifty', 'sixty', 'seventy', 'eighty', 'ninety'][num-2]
def encodeTeens(num):
if num < 10:
return encodeOnesDigit(num)
else:
return ['ten', 'eleven', 'twelve', 'thirteen', 'fourteen', 'fifteen', 'sixteen', 'seventeen', 'eighteen', 'nineteen'][num-10]
def encodeTriplet(num):
if num == 0: return ''
str = ''
if num >= 100:
str = encodeOnesDigit(num / 100) + ' hundred'
tens = num % 100
if tens >= 20:
if str != '': str += ' '
str += encodeTensDigit(tens / 10)
if tens % 10 > 0:
str += '-' + encodeOnesDigit(tens % 10)
elif tens != 0:
if str != '': str += ' '
str += encodeTeens(tens)
return str
def zipNumbers(numList):
if len(numList) == 1:
return numList[0]
strList = ['', ' thousand', ' million', ' billion'] # Add more as needed
strList = strList[:len(numList)]
strList.reverse()
joinedList = zip(numList, strList)
joinedList = [item for item in joinedList if item[0] != '']
return ', '.join(''.join(item) for item in joinedList)
def humanReadable(num):
if num == 0: return 'zero'
negative = False
if num < 0:
num *= -1
negative = True
numString = str(num)
tripletCount = int(math.ceil(len(numString) / 3.0))
numString = numString.zfill(tripletCount * 3)
tripletList = [int(numString[i*3:i*3+3]) for i in range(tripletCount)]
readableList = [encodeTriplet(num) for num in tripletList]
readableStr = zipNumbers(readableList)
return 'negative ' + readableStr if negative else readableStr
答案 2 :(得分：1)
支持高达9.99亿，但没有负数：
String humanReadable(int inputNumber) {
if (inputNumber == -1) {
return "";
}
int remainder;
int quotient;
quotient = inputNumber / 1000000;
remainder = inputNumber % 1000000;
if (quotient > 0) {
return humanReadable(quotient) + " million, " + humanReadable(remainder);
}
quotient = inputNumber / 1000;
remainder = inputNumber % 1000;
if (quotient > 0) {
return humanReadable(quotient) + " thousand, " + humanReadable(remainder);
}
quotient = inputNumber / 100;
remainder = inputNumber % 100;
if (quotient > 0) {
return humanReadable(quotient) + " hundred, " + humanReadable(remainder);
}
quotient = inputNumber / 10;
remainder = inputNumber % 10;
if (remainder == 0) {
//hackish way to flag the algorithm to not output something like "twenty zero"
remainder = -1;
}
if (quotient == 1) {
switch(inputNumber) {
case 10:
return "ten";
case 11:
return "eleven";
case 12:
return "twelve";
case 13:
return "thirteen";
case 14:
return "fourteen";
case 15:
return "fifteen";
case 16:
return "sixteen";
case 17:
return "seventeen";
case 18:
return "eighteen";
case 19:
return "nineteen";
}
}
switch(quotient) {
case 2:
return "twenty " + humanReadable(remainder);
case 3:
return "thirty " + humanReadable(remainder);
case 4:
return "forty " + humanReadable(remainder);
case 5:
return "fifty " + humanReadable(remainder);
case 6:
return "sixty " + humanReadable(remainder);
case 7:
return "seventy " + humanReadable(remainder);
case 8:
return "eighty " + humanReadable(remainder);
case 9:
return "ninety " + humanReadable(remainder);
}
switch(inputNumber) {
case 0:
return "zero";
case 1:
return "one";
case 2:
return "two";
case 3:
return "three";
case 4:
return "four";
case 5:
return "five";
case 6:
return "six";
case 7:
return "seven";
case 8:
return "eight";
case 9:
return "nine";
}
}
答案 3 :(得分：1)
using System;
namespace HumanReadable
{
public static class HumanReadableExt
{
private static readonly string[] _digits = {
"", "one", "two", "three", "four", "five",
"six", "seven", "eight", "nine", "eleven", "twelve",
"thirteen", "fourteen", "fifteen", "sixteen", "seventeen",
"eighteen", "nineteen"
};
private static readonly string[] _teens = {
"", "", "twenty", "thirty", "forty", "fifty",
"sixty", "seventy", "eighty", "ninety"
};
private static readonly string[] _illions = {
"", "thousand", "million", "billion", "trillion"
};
private static string Seg(int number)
{
var work = string.Empty;
if (number >= 100)
work += _digits[number / 100] + " hundred ";
if ((number % 100) < 20)
work += _digits[number % 100];
else
work += _teens[(number % 100) / 10] + "-" + _digits[number % 10];
return work;
}
public static string HumanReadable(this int number)
{
if (number == 0)
return "zero";
var work = string.Empty;
var parts = new string[_illions.Length];
for (var ind = 0; ind < parts.Length; ind++)
parts[ind] = Seg((int) (number % Math.Pow(1000, ind + 1) / Math.Pow(1000, ind)));
for (var ind = 0; ind < parts.Length; ind++)
if (!string.IsNullOrEmpty(parts[ind]))
work = parts[ind] + " " + _illions[ind] + ", " + work;
work = work.TrimEnd(',', ' ');
var lastSpace = work.LastIndexOf(' ');
if (lastSpace >= 0)
work = work.Substring(0, lastSpace) + " and" + work.Substring(lastSpace);
return work;
}
}
class Program
{
static void Main(string[] args)
{
Console.WriteLine(1.HumanReadable());
Console.WriteLine(53.HumanReadable());
Console.WriteLine(723603.HumanReadable());
Console.WriteLine(1456376562.HumanReadable());
Console.ReadLine();
}
}
}
答案 4 :(得分：0)
这个函数实现有一个很大的问题。它是未来的本地化。这个由英语母语人士编写的函数很可能不适用于除英语以外的任何其他语言。在世界上为任何人类语言方言编写一般易于本地化的函数几乎是不可能的，除非你真的需要保持它的一般性。实际上在现实世界中，您不需要使用大整数运算，因此您可以将所有数字保存在一个大的（甚至不是那么大的）字符串数组中。
答案 5 :(得分：0)
同意有许多现实世界的应用程序。 因此，已有许多现实世界的实施。
它已成为bsdgames的一部分，因为它几乎永远......
> man number