[編程題]學英語

Talk is cheap, show me the code.

一、問題描述

Jessi初學英語，爲了快速讀出一串數字，編寫程序將數字轉換成英文：
如22：twenty two，123：one hundred and twenty three。

說明：

數字爲正整數，長度不超過九位，不考慮小數，轉化結果爲英文小寫；

輸出格式爲twenty two；

非法數據請返回“error”；
關鍵字提示：and，billion，million，thousand，hundred。

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

輸入描述：

輸入一個long型整數

輸出描述：

輸出相應的英文寫法

輸入例子：

2356

輸出例子：

two thousand three hundred and fifty six

二、問題分析

這道題就是字符串處理，但是很多細節需要注意，比如空格的問題，and的位置問題，單詞的拼寫問題。主要思路是，三位數三位數地處理。

#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;
}