例如一个自然数n，0< =n< =2^31-1。
输出描述
输出这个数的英文，最后不要有多余的空格。
用例输入 1
1111111111
用例输出 1
one billion one hundred and eleven million one hundred and eleven thousand one hundred and eleven
该回答引用ChatGPT
可以使用递归的方法来将给定的自然数转换成英文。具体的转换过程可以根据数值的范围和数位来分类讨论，下面是一种实现方式：
#include <iostream>
#include <string>
using namespace std;
// 将0~19的数字转换成英文
string ones[] = {"", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine",
"ten", "eleven", "twelve", "thirteen", "fourteen", "fifteen", "sixteen", "seventeen", "eighteen", "nineteen"};
// 将20, 30, 40, ..., 90的数字转换成英文
string tens[] = {"", "", "twenty", "thirty", "forty", "fifty", "sixty", "seventy", "eighty", "ninety"};
// 将数字的数位转换成英文
string scale[] = {"", "thousand", "million", "billion", "trillion", "quadrillion", "quintillion", "sextillion", "septillion", "octillion", "nonillion"};
// 将3位数以内的数字转换成英文
string convert(int num) {
if (num == 0) {
return "";
} else if (num < 20) {
return ones[num];
} else if (num < 100) {
return tens[num / 10] + " " + ones[num % 10];
} else {
return ones[num / 100] + " hundred " + convert(num % 100);
}
}
// 将整数转换成英文
string toEnglish(int num) {
if (num == 0) {
return "zero";
}
string result = "";
int i = 0;
while (num > 0) {
int threeDigits = num % 1000;
if (threeDigits > 0) {
result = convert(threeDigits) + " " + scale[i] + " " + result;
}
num /= 1000;
i++;
}
return result;
}
int main() {
int n;
cin >> n;
cout << toEnglish(n) << endl;
return 0;
}
https://blog.csdn.net/iloveyouwh/article/details/105870765
def num_to_words(num):
# 定义单词的数值映射
ones = ["", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine", "ten",
"eleven", "twelve", "thirteen", "fourteen", "fifteen", "sixteen", "seventeen", "eighteen", "nineteen"]
tens = ["", "", "twenty", "thirty", "forty", "fifty", "sixty", "seventy", "eighty", "ninety"]
scales = ["", "thousand", "million", "billion"]
if num == 0:
return "zero"
words = ""
i = 0
while num > 0:
# 将每三个数字的分组处理，一组最多有3个数字，例如123456789会分成三组，分别是(123,456,789)
group = num % 1000
num //= 1000
# 处理分组的英文表示
group_words = ""
if group > 99:
group_words += ones[group // 100] + " hundred"
group %= 100
if group > 0:
group_words += " and "
if group > 19:
group_words += tens[group // 10] + " "
group %= 10
if group > 0:
group_words += ones[group] + " "
# 将分组的英文表示和对应的量级（thousand/million/billion）组合起来
if group_words != "":
words = group_words + scales[i] + " " + words
i += 1
return words.strip()
# 测试代码
num = int(input())
print(num_to_words(num))