1082. Read Number in Chinese (25)
时间限制
400 ms
内存限制
32000 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
简单的模拟，只是过程描述有点啰嗦。
首先这句话什么意思 Note: zero ("ling") must be handled correctly according to the Chinese tradition
什么叫Chinese tradition，其实就是对0的特殊处理。
【1】尾部的0不发音
【2】多个连续的0，只发一个音
【3】多个不连续的0，都要发音
【4】大于4位的数肯定要发 Wan 音
【5】大于8位的数一定要发 Yi 音
而且还有个规律： Shi Bai Qian 这三个音循环出现。
Ac代码：
// pat-1082.cpp : 定义控制台应用程序的入口点。
//
#include "stdafx.h"
#include "iostream"
#include "string"
#include "algorithm"
#include "vector"
#include "stack"
using namespace std;
stack<string> ans;
string num[]={"ling", "yi" ,"er" , "san" , "si" , "wu", "liu" , "qi", "ba", "jiu"};
string pos[]={ "Shi","Bai","Qian", "Wan","Yi" };
int main()
{
long int n=0;
bool firstout=true;
cin>>n;
if (n==0)
{
cout<<"ling";
goto end;
}
if (n<0)
{
cout<<"Fu";
firstout=false;
n=-(n);
}
int cnt=0;
bool wan_flag=false;
bool zero=false;
bool first=true;
while(n)
{
int temp=n%10;
//0 特殊处理
if (temp==0)
{
if (cnt==3)
{
wan_flag=true;
}
n/=10;
if (!first)
{
cnt++;
}
first=false;
if (zero)//第一次遇到0
{
ans.push(num[0]);
zero=false;
continue;
}
else
{
continue;
}
}
zero=true;
if(first)//忽略个位
{
ans.push(num[temp]);//yi er san si ~~~~~
n/=10;
first=false;
continue;
}
if(cnt<7)
{
if (wan_flag)
{
wan_flag=false;
ans.push(pos[3]);
}
ans.push(pos[cnt%4]);//shi bai qian wan
}
else
{
ans.push(pos[4]);//yi
}
ans.push(num[temp]);//yi er san si ~~~~~
n/=10;
cnt++;
}
while (!ans.empty())
{
string temp=ans.top();
ans.pop();
if (firstout)
{
firstout=false;
cout<<temp;
continue;
}
cout<<" "<<temp;
}
end: return 0;
}