每日编程-20170315

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。

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

解答：

主要在于分析，英文数字是每3位一组的，

1-3位是几百几十几，

4-6位是几百几十几k，

6-9位是几百几十几million,

如果长度不超过9，用不上billion

所以使用一个函数输出几百几十几，4-6加上thousand，6-9加上billion就好

11-19有对应的单词，存起来就好。

#include <iostream>
#include <string>
#include <vector>
using namespace std;
vector<vector<string>> dict{
    { "one", "two", "three" , "four" , "five" , "six", "seven", "eight", "nine",
    "ten", "eleven", "twelve", "thirteen" , "fourteen" , "fifteen" , "sixteen", "seventeen",
    "eighteen", "nineteen"},　　//dict[0]有19个元素
    { "twenty", "tirty", "fourty" , "fifty" , "sixty" , "senevnty" , "eighty" , "ninety"}};
vector<string> answer;
void three0(int n) {
    string s;
    if ((n % 100 > 20))　　//后两位20以上的要使用dict[0]和dict[1]组合
    {
        answer.push_back(dict[0][(n % 10) - 1 ]);
        answer.push_back(dict[1][((n / 10) % 10) - 2]);
    }
    else
    {
        answer.push_back(dict[0][n - 1]);　　//后两位20以下的，直接查dict[0]
    }
    if (n > 99)
    {
        answer.push_back("hundred and");
        answer.push_back(dict[0][(n / 100) % 10 -1]);
    }

}
int main() {
    long number;
    cin >> number;
    if (number > 999999999)　　//检查
    {
        cout << "error" << endl;
        return 0;
    }
    three0(number);
    if (number/1000)
    {
        number /= 1000;
        answer.push_back("thousand");
        three0(number);
    }
    if (number / 1000)
    {
        number /= 1000;
        answer.push_back("million");
        three0(number);
    }
    for (auto end = answer.end(); end != answer.begin();)
    {
        --end;
        cout << *end << " ";
    }
    cout << endl;
}

 上一个版本错误百出，还在那洋洋得意，让大神教训了

这个版本目前还没有试出来Bug,我回头再想想

#include <iostream>
#include <string>
#include <vector>
using namespace std;
vector<vector<string>> dict{
    { "one", "two", "three" , "four" , "five" , "six", "seven", "eight", "nine",
    "ten", "eleven", "twelve", "thirteen" , "fourteen" , "fifteen" , "sixteen", "seventeen",
    "eighteen", "nineteen"},
    { "twenty", "tirty", "fourty" , "fifty" , "sixty" , "senevnty" , "eighty" , "ninety"}};
vector<string> answer;
long number;
void three0(int n) {
    n %= 1000;
    if (n != 0)
    {
        if ((n % 100 > 20) && ((n % 100) != 0))
        {
            answer.push_back(dict[0][(n % 10) - 1]);
            answer.push_back(dict[1][((n / 10) % 10) - 2]);
            if (number > 999){ answer.push_back("and"); }
        }
        else
        {
            if ((n % 100) != 0)
            {
                answer.push_back(dict[0][(n % 100) - 1]);
                if (number > 999) { answer.push_back("and"); }
            }
        }
        if (n > 99)
        {
            answer.push_back("hundred");
            answer.push_back(dict[0][(n / 100) % 10 - 1]);
        }
    }
}
int main() {
    cin >> number;
    if (number > 999999999)
    {
        cout << "error" << endl;
        return 0;
    }
    if ((number% 1000) != 0)
    {
        three0(number);
    }
    number /= 1000;
    if ((number% 1000) != 0)
    {
        answer.push_back("thousand");
        three0(number);
    }
    if (number / 1000)
    {
        number /= 1000;
        answer.push_back("million");
        three0(number);
    }
    for (auto end = answer.end(); end != answer.begin();)
    {
        --end;
        cout << *end << " ";
    }
    cout << endl;
}