PAT甲级——A1082 Read Number in Chinese

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.

Input Specification:

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

Output Specification:

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.

Sample Input 1:

-123456789

Sample Output 1:

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

Sample Input 2:

100800

Sample Output 2:

yi Shi Wan ling ba Bai

吐槽一波，这就不能算是算法题，纯属是字符串硬处理

 #include <iostream>
 #include <string>
 #include <vector>
 using namespace std;
 string num[] = { "ling","yi", "er", "san", "si", "wu", "liu", "qi", "ba", "jiu" };
 string c[] = { "Ge","Shi", "Bai", "Qian", "Yi", "Wan" };
 int J[] = { , , , , , , , ,  };
 vector<string> res;
 int main() {
     int n;
     cin >> n;
     if (n == ) {
         cout << "ling";
         return ;
     }
     if (n < ) {
         cout << "Fu ";
         n = -n;
     }
     int part[];
     part[] = n / ;
     part[] = (n % ) / ;
     part[] = n % ;
     bool zero = false; //是否在非零数字前输出合适的ling
     int printCnt = ; //用于维护单词前没有空格，之后输入的单词都在前面加一个空格。
     for (int i = ; i < ; i++) {
         int temp = part[i]; //三个部分，每部分内部的命名规则都一样，都是X千X百X十X
         for (int j = ; j >= ; j--) {
             int curPos =  - i *  + j; //当前数字的位置
             if (curPos >= ) continue; //最多九位数
             int cur = (temp / J[j]) % ;//取出当前数字
             if (cur != ) {
                 if (zero) {
                     printCnt++ ==  ? cout << "ling" : cout << " ling";
                     zero = false;
                 }
                 if (j == )
                     printCnt++ ==  ? cout << num[cur] : cout << ' ' << num[cur]; //在个位，直接输出
                 else
                     printCnt++ ==  ? cout << num[cur] << ' ' << c[j] : cout << ' ' << num[cur] << ' ' << c[j]; //在其他位，还要输出十百千
             }
             else {
                 if (!zero&&j !=  && n / J[curPos] >= ) zero = true;   //注意100020这样的情况
             }
         }
         if (i !=  && part[i] > ) cout << ' ' << c[i + ]; //处理完每部分之后，最后输出单位，Yi/Wan
     }
     return ;
 }