PAT甲级——【牛客练习题1002】

题目描述

Given an integer with no more than 9 digits, you are supposed to read it in the traditional Chinese way.  Output "Fu" first if it is negative.  For example, -123456789 is read as "Fu yi Yi er Qian san Bai si Shi wu Wan liu Qian qi Bai ba Shi jiu".  Note: zero ("ling") must be handled correctly according to the Chinese tradition.  For example, 100800 is "yi Shi Wan ling ba Bai".

输入描述:

Each input file contains one test case, which gives an integer with no more than 9 digits.

输出描述:

For each test case, print in a line the Chinese way of reading the number.  The characters are separated by a space and there must be no extra space at the end of the line.

输入例子:

-123456789

输出例子:

Fu yi Yi er Qian san Bai si Shi wu Wan liu Qian qi Bai ba Shi jiu

 #include <iostream>
 #include <vector>
 #include <string>

 using namespace std;

 int main()
 {
     vector<string> level = { "Fu","Shi","Bai","Qian" };
     vector<string> Wei = { "","Wan","Yi" };
     vector<string> numbers = { "ling","yi","er","san","si","wu","liu","qi","ba","jiu" };
     vector<string> res;
     string Num;
     cin >> Num;
     if(Num[] == '-')//如果是负数
     {
         res.push_back(level[]);
         Num.erase(, );
     }
     int n = Num.length();
     if (n == )//如果只有一位，则直接输出即可并结束
     {
         cout << numbers[Num[] - ''] << endl;
         return ;
     }
     int f = ;
     for (int i = ; i < n; ++i)
     {
         int a = Num[i] - '';//取出数字
         int p = (n - i - ) % ;//判断是否是4位间隔
         if (a > )
         {
             if (f)//中间有零存在
             {
                 res.push_back(numbers[]);
                 f = ;
             }
             res.push_back(numbers[a]);//输入数字
             if (p > )//不是各位
                 res.push_back(level[p]);//输入位
         }
         else if (p != )//当中间有0且不是0不是在个位上
             f = ;
         if (p ==  && res[res.size() - ] != "Yi")//是4位间隔且中间不是全为0，例如100000004，就不用输出wan
             res.push_back(Wei[(n - i) / ]);
     }
     for (int i = ; i < res.size() - ; ++i)
         cout << res[i] << " ";
     cout << res[res.size() - ] << endl;//最后一位不用输出空格
     return ;
 }