PAT 1082 Read Number in Chinese[难]

1082 Read Number in Chinese （25 分）

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

题目大意：给出一个数，用中文给读出来，并且0要正确处理。

//我一看见这个题就蒙了，完全没有思路。

代码来自：https://www.liuchuo.net/archives/2204

#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 / ;//亿以上的
    part[]= (n % ) / ;//千万级
    part[] = n % ;//万级
    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 ;
}

//这个逻辑太难了。我是写不出来的，勉强看懂。