1082. Read Number in Chinese (25)

 

时间限制  400 ms

内存限制   65536 kB

代码长度限制  16000 B

判题程序  Standard

作者

CHEN, Yue

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

解读：就是输出一个绝对值在9位以内的整数；

①读入数后，按每一个位读入数据，存入digit数组中，这里用vector，因为不知道初始的数据大小。

note  预先设置string a[ ]为数字大小（0-9），b[ ]为shi bai qian wan shi bai qian yi；

②读入的数，若为0，直接输出0；若为负数，先输出fu，然后将数字置反，以正数形式，即取绝对值进行之后的比较

③按位存入，num%10--digit.push_back,将最后一位数据存入数组的最后一位，

num/10，即循环下一位；

//④开始判定位数，digit.size 。；8位输出亿

⑤对于非首位的0，进行判断，如果有8个0，输出亿

对于几个0，我们赋值i，i非0，且数字非0.则返回数制，例如  i为2，返回百，

返回其他则返回0，这里我们返回的digit[ i ]是从后向前的，所以1500，digit[2]=5，

#include <string>  
#include <vector>  
#include <iostream>  
  
using namespace std;  
  
int main()  
{  
    string a[] = {"ling", "yi", "er", "san", "si", "wu", "liu", "qi", "ba", "jiu"};  
    string b[] = {"", "Shi", "Bai", "Qian", "Wan", "Shi", "Bai", "Qian", "Yi"};  
    vector<string> res;  
    vector<int> digit;  
    int num, e;  
    cin >> num;  
  
    if (num == 0)  
    {  
        cout << "ling";  
        return 0;  
    } else if (num < 0)  
    {  
        cout << "Fu ";  
        num = -num;  
    }  
    while (num != 0)  
    {  
        digit.push_back(num%10);  
        num /= 10;  
    }  
    for (e=0; e<digit.size()&& digit[e]==0; ++ e)
   // for (e=0; e<digit.size() && digit[e]==0; ++ e)
    {
        cout<<e<<endl;
        cout<<digit[e]<<endl;
    }  
    if (e == 8)  
    {  
        cout << a[digit[e]] << " Yi";  
        return 0;  
    }  
    for (int i = e; i < digit.size(); ++ i)  
    {  
        if (i!=0 && (digit[i]!=0 || i==4 || i==8))  
        {  
            res.push_back( b[i] ); 
            cout<<"1st"<<i<<endl;
            cout<<b[i]<<endl; 
        }  
        res.push_back(a[ digit[i] ]);  
        cout<<"2nd"<<i<<endl;
        cout<<digit[i]<<endl;
        cout<<a[digit[i]]<<endl;
    }  
    for (int i = res.size()-1; i >= 0; -- i)  
    {  
        if (i != res.size()-1)  
        {  
            cout << " ";  
        }  
        int cnt = 0;  
        while (i>=0 && res[i]=="ling")  
        {  
            -- i;  
            ++ cnt;  
        }  
        if (cnt!=0 && res[i]!="Wan") // 即当前的res[i]为数字或位数，自己本身(res[i+cnt])是ling  
        {  
        cout<<"cnt"<<cnt<<res[i]<<endl;
        
            cout << "ling ";  
        }  
        cout << res[i];  
    }  
  
    return 0;  
}