题目描述：
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。
本题含有多组输入数据。
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
示例：
输入：
2356
输出：
two thousand three hundred and fifty six
解题思路：
建立两个字典，分别记录0-19的单词和十倍数的单词。输入数字后，按照其位数从低到高写判断，0-19就直接返回字典里的单词；20-99就先判断是不是正好的十倍数，分情况返回；100-999也分两个情况，如果不是正好的，则返回X百+English（几十）；再往上就是千、百万、十亿，特点就是间隔了1000，比如千，就将数据分为多少个1000和几百几十几，百万就分为多少个百万、多少个千和几百几十几，十亿就以此类推。English函数主要通过递归来实现高位数的分析。
测试代码：
#include <iostream> #include <string> #include <vector> using namespace std; vector<string> T19={"zero","one","two","three","four", \ "five","six","seven","eight","nine",\ "ten","eleven","twelve","thirteen","fourteen",\ "fifteen","sixteen","seventeen","eighteen","ninteen"}; // 前二十个数 vector<string> Tenmul={"none","ten","twenty","thirty","forty","fifty","sixty","seventy","eighty","ninety"}; // 十的倍数 string English(int number) { if(number>=0&&number<=19) { return T19[number]; } if(number>=20&&number<=99) { if(number%10==0) { return Tenmul[number/10]; } return Tenmul[number/10]+" "+English(number%10); } if(number>=100&&number<=999) { if(number%100==0) { return T19[number/100]+" hundred"; } return T19[number/100]+" hundred and "+English(number%100); } if(number>=1000&&number<=999999) { if(number%1000==0) { return English(number/1000)+" thousand"; } return English(number/1000)+" thousand "+English(number%1000); } if(number>=1000000&&number<=999999999) { if(number%1000000==0) { return English(number/1000000)+" million"; } return English(number/1000000)+" million "+English(number%1000000/1000)+" thousand "+English(number%1000); } if(number>=1000000000) { if(number%1000000000==0) { return English(number/1000000000)+" billion"; } return English(number/1000000000)+" billion "+English(number%1000000000/1000000)+" million "+English(number%1000000/1000)+" thousand "+English(number%1000); } return "error"; } int main() { int number; while(cin>>number) { cout<<English(number)<<endl; } return 0; }
