1082. Read Number in Chinese (25)

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

用读中文的方法读出给出的数字，要考虑的情况比较多。

首先将数字分成四位为一组（开头的可以不满4位），因为每四位的读法是一样的。有如下的情况：

1.0算是一种特殊的情况；

2.1234这种是最容易的情况，直接每一位都读出来，即yi Qian er Bai san Shi si；

3.1023这种带有0的，零的那一位只读出零就行了，即yi Qian ling er Shi san；

4.1002这种两个0连着的，只读出一个零，即yi Qian ling er；

5.0012这种是0开头的，零的地方只读出零且只读出一个，即ling yi Shi er；

6.1200这种是0结尾的，不用读出零，即yi Qian er Bai；

7.其他情况是还不满四位的，也是按照上面的规律。

这样得到每四位的读法，然后如果有9位数，则用“Yi“和“Wan“连接；如果是5到8位数，则用“Wan”来连接；如果小于等于4位，则按照四位数的读法直接读出。注意最后不能留有空格。

代码：

#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <vector>
using namespace std;

string read1[10]={"ling ","yi ","er ","san ","si ","wu ","liu ","qi ","ba ","jiu "};
string read2[5]={"","","Shi ","Bai ","Qian "};
string read3[4]={"","","Wan ","Yi "};

string read4(string s)
{
int i=0,j=0,n=s.size();
string ret="";
while(true)
{
j=i;
while(s[i]=='0'&&i<n) i++;
if(i==n) break;
if(i!=j) ret+="ling ";
else
{
ret+=read1[s[i]-'0']+read2[n-i];
i++;
}
if(i==n) break;
}
return ret;
}

int main()
{
string num;
cin>>num;
if(num=="0")
{
cout<<"ling"<<endl;
return 0;
}
string res="";
if(num[0]=='-')
{
res+="Fu ";
num=num.substr(1);
}
int n=num.size(),m,k=(n%4==0)?4:n%4;
if(n==9) m=3;
else if(n>=5) m=2;
else m=1;
for(int i=m;i>0;i--)
{
if(i<m) k=4;
res+=read4(num.substr(0,k));
num=num.substr(k);
res+=read3[i];
}
res=res.substr(0,res.size()-1);
cout<<res<<endl;
}