[编程题]学英语

墨蓝 2022-07-13 03:24 243阅读 0赞

> Talk is cheap, show me the code.

### 一、问题描述 ###

Jessi初学英语，为了快速读出一串数字，编写程序将数字转换成英文：  
如22：twenty two，123：one hundred and twenty three。

说明：  
  
数字为正整数，长度不超过九位，不考虑小数，转化结果为英文小写；  
  
输出格式为twenty two；  
  
非法数据请返回“error”；  
关键字提示：and，billion，million，thousand，hundred。

方法原型：public static String parse(long num)

**输入描述：**

> 输入一个long型整数

**输出描述：**

> 输出相应的英文写法

**输入例子：**

> 2356

**输出例子：**

> two thousand three hundred and fifty six

### 二、问题分析 ###

这道题就是字符串处理，但是很多细节需要注意，比如空格的问题，and的位置问题，单词的拼写问题。主要思路是，三位数三位数地处理。

#include <iostream>
    #include <string>
    using namespace std;
    
    string convert(int n)
    {
        string change[9] = {"one", "two", "three", "four", "five", "six", "seven", "eight", "nine"};
        string ten[10] = {"ten", "eleven", "twelve", "thirteen", "fourteen", "fifteen", "sixteen", "seventeen", "eightteen", "nineteen"};
        string a[8] = {"twenty", "thirty", "forty", "fifty", "sixty", "seventy", "eighty", "ninety"};
        int m = 0, s = 0, t = 0;
        m = n % 10;
        n = n / 10;
        if (n != 0)
            s = n % 10;
        n = n / 10;
        if (n != 0)
            t = n % 10;
        string str;
        if (t != 0)
            str = str + change[t - 1] + " hundred";
        if (s != 0 || m != 0)
        {
            if (t != 0)
                str = str + " and ";
            if (s == 1)
            {
                str = str + ten[m];
            } else if (s != 0) {
                str = str + a[s - 2];
            }
            if (s != 1 && m != 0)
            {
                if (s != 0)
                    str = str + " " + change[m - 1];
                else
                    str = str + change[m - 1];
            }
        }
        return str;
    }
    
    int main()
    {
        long num;
        while (cin >> num)
        {
            int n1 = 0, n2 = 0, n3 = 0, n4 = 0;
            n1 = num % 1000;
            num = num / 1000;
            if (num != 0)
                n2 = num % 1000;
            num /= 1000;
            if (num != 0)
                n3 = num % 1000;
            num /= 1000;
            if (num != 0)
                n4 = num % 1000;
            string s;
            if (n4 != 0)
                s = s + convert(n4) + " billion ";
            if (n3 != 0)
                s = s + convert(n3) + " million ";
            if (n2 != 0)
                s = s + convert(n2) + " thousand ";
            s = s + convert(n1);
            cout << s << endl;
        }
    
        return 0;
    }