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

 

 

1 #include 
     
       2 #include 
      
        3 #include 
       
         4  5  6 using namespace std; 7  8 int main() 9 {10     vector
        
          level = { "Fu","Shi","Bai","Qian" };11     vector
         
           Wei = { "","Wan","Yi" };12     vector
          
            numbers = { "ling","yi","er","san","si","wu","liu","qi","ba","jiu" };13 vector
           
             res;14 string Num;15 cin >> Num;16 if(Num[0] == '-')//如果是负数17 {18 res.push_back(level[0]);19 Num.erase(0, 1);20 }21 int n = Num.length(); 22 if (n == 1)//如果只有一位，则直接输出即可并结束23 {24 cout << numbers[Num[0] - '0'] << endl;25 return 0;26 }27 int f = 0;28 for (int i = 0; i < n; ++i)29 {30 int a = Num[i] - '0';//取出数字31 int p = (n - i - 1) % 4;//判断是否是4位间隔32 if (a > 0)33 {34 if (f)//中间有零存在35 {36 res.push_back(numbers[0]);37 f = 0;38 }39 res.push_back(numbers[a]);//输入数字40 if (p > 0)//不是各位41 res.push_back(level[p]);//输入位42 }43 else if (p != 0)//当中间有0且不是0不是在个位上44 f = 1;45 if (p == 0 && res[res.size() - 1] != "Yi")//是4位间隔且中间不是全为0，例如100000004，就不用输出wan46 res.push_back(Wei[(n - i) / 4]);47 }48 for (int i = 0; i < res.size() - 1; ++i)49 cout << res[i] << " ";50 cout << res[res.size() - 1] << endl;//最后一位不用输出空格51 return 0;52 }