Read Number in Chinese (25)

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

这道题看似简单，实则非常困难，需要填的坑特别多。

我的代码：

（注：在牛客网上十九个测试点都通过了。但是在PAT甲级真题网上还有一个测试点尚未通过）

#include<iostream>
#include<vector>
#include<string>
#include<algorithm>
using namespace std;
vector<char>v;
vector<char>::iterator it,ot;
char a[20];
int i,j=0,k=0,flag=0,n;
string b[20],c[9]={"yi","er","san","si","wu","liu","qi","ba","jiu"};
int main()
{
    gets(a);
    n=atoi(a);
    for(i=0;a[i];i++) v.push_back(a[i]);
    reverse(v.begin(),v.end());
    for(it=v.begin();it!=v.end();it++)
    {
        j++;
        if(*it=='0')
        {
            if(n==0) b[k++]="ling";
            else if(flag==0) continue;
            else if(j==5)
            {
                for(ot=it;ot!=v.end();ot++)
                {
                    j++;
                    if(*ot!='0') break;
                }
                if(j<=9)
                {
                    flag=0;
                    b[k++]="Wan";
                }
                j=5;
            }
            else
            {
                flag=0;
                b[k++]="ling";
            }
        }
        else if(*it=='-') b[k++]="Fu";
        else
        {
            flag=1;
            if(j==2 || j==6) b[k++]="Shi";
            else if(j==3 || j==7) b[k++]="Bai";
            else if(j==4 || j==8) b[k++]="Qian";
            else if(j==5) b[k++]="Wan";
            else if(j==9) b[k++]="Yi";
            b[k++]=c[*it-'0'-1];
        }
    }
    while(k-- && k>=1) cout<<b[k]<<" ";
    cout<<b[0]<<endl;
    return 0;
}