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 <iostream>
2 #include <vector>
3 #include <string>
4
5
6 using namespace std;
7
8 int main()
9 {
10 vector<string> level = { "Fu","Shi","Bai","Qian" };
11 vector<string> Wei = { "","Wan","Yi" };
12 vector<string> numbers = { "ling","yi","er","san","si","wu","liu","qi","ba","jiu" };
13 vector<string> 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wan
46 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 }
