1082. Read Number in Chinese (25)

1. 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

这题拿到手，一看文字描述好像很少，觉得很水。
结果想半天都觉得很棘手，情况太多，完全找不到一种合适的方法。
无奈之下，去搜AC代码
参考了下面这位大神的代码
但是其中有一个不足
http://blog.csdn.net/apie_czx/article/details/48270285
下面贴一下我的代码：

#include <bits/stdc++.h>using namespace std;int main(void){    string num[] = { "ling", "yi", "er", "san", "si", "wu", "liu", "qi", "ba", "jiu" };    string wei[] = { "", "Shi", "Bai", "Qian", "Wan", "Shi", "Bai", "Qian", "Yi" };    vector<string> res;    vector<int> temp;    int number, i, j;    cin >> number;    if (number == 0)    {        cout << "ling";;        return 0;    }    else if (number < 0)    {        cout << "Fu ";        number = -number;    }    while (number)    {        temp.push_back(number % 10);        number /= 10;    }    for (j = 0; j < temp.size(); j++)    {        if (temp[j] != 0) break;    }    for (i = j; i < temp.size(); i++)    {        if (i != 0 &&(temp[i] != 0 || i == 8 || i == 4))//之所以一定要 i != 0，是因为格式问题         {                                               //一行末尾不能有空格，为了下面的输出做铺垫             res.push_back(wei[i]);        }        res.push_back(num[temp[i]]);    }    for (i = res.size() - 1; i >= 0; i--)    {        if (i != res.size() - 1) cout << " ";        bool ispush = false;        while (i >= 0 && res[i] == "ling")        {            i--;            ispush = true;        }        if (res[i] != "Wan" && ispush) cout << "ling ";        cout << res[i];    }    if (j > 4 && j < 8) cout << " Wan";//为了应对10000000这种情况 }

针对第一个非0数高于万位的情况
pat里面的测试点没有针对这种情况的测试
翁恺老师教我们说
一个oj只能证明你的程序是错的，但是不能证明你的程序是对的
继续加油！