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。仅仅是过程描写叙述有点啰嗦。
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，事实上就是对0的特殊处理。
【1】尾部的0不发音
【2】多个连续的0，仅仅发一个音
【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
并且还有个规律： 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;stackans;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<