A1082 Read Number in Chinese (25 分)
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
#include<bits/stdc++.h>
using namespace std; string Week[]={
"MON","TUE","WED","THU","FRI","SAT","SUN"
}; string num[]={
"ling","yi","er","san","si","wu","liu","qi","ba","jiu"
}; string wei[]={
"Shi","Bai","Qian","Wan","Yi"
}; int main(){ string str; cin>>str; int len=str.size(); int left=,right=len-; if(str[]=='-')
{
left++;
cout<<"Fu";
} while(left+<=right) //注意等于号
right-=; while(left<len){ bool flag=false;
bool isPrint=false; while(left<=right){ if(left>&&str[left]==''){
flag=true;
}else{
if(flag){
cout<<" ling";
flag=false;
}
isPrint=true; if(left>)cout<<" "; //注意格式空格
cout<<num[str[left]-'']; if(left!=right)
cout<<" "<<wei[right-left-]; } left++;
} if(isPrint==true&&right!=len-){
cout<<" "<<wei[(len--right)/+];
} right+=; } return ;
}
A1082 Read Number in Chinese (25 分)的更多相关文章
- A1082 Read Number in Chinese (25)(25 分)
A1082 Read Number in Chinese (25)(25 分) Given an integer with no more than 9 digits, you are suppose ...
- 【PAT甲级】1082 Read Number in Chinese (25 分)
题意: 输入一个九位整数,输出它的汉字读法(用拼音表示). trick: 字符串数组""其实会输出一个空格,而不是什么都不输出,导致测试点0和4格式错误. AAAAAccepted ...
- 1082 Read Number in Chinese (25分)
// 1082.cpp : 定义控制台应用程序的入口点. // #include <iostream> #include <string> #include <vecto ...
- pat1082. Read Number in Chinese (25)
1082. 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, Yu ...
- A1082. Read Number in Chinese
Given an integer with no more than 9 digits, you are supposed to read it in the traditional Chinese ...
- 1082. Read Number in Chinese (25)
题目如下: Given an integer with no more than 9 digits, you are supposed to read it in the traditional Ch ...
- PAT甲级——A1082 Read Number in Chinese
Given an integer with no more than 9 digits, you are supposed to read it in the traditional Chinese ...
- 1082. Read Number in Chinese (25)-字符串处理
题意就是给出9位以内的数字,按照汉子的读法读出来. 读法请看下方的几个例子: 5 0505 0505 伍亿零伍佰零伍万零伍佰零伍 5 5050 5050 伍亿伍仟零伍拾万伍仟零伍拾 (原本我以为这个 ...
- PAT (Advanced Level) 1082. Read Number in Chinese (25)
模拟题. #include<cstdio> #include<cstring> #include<cmath> #include<vector> #in ...
随机推荐
- Python多进程编程-进程间协作(Queue、Lock、Semaphore、Event、Pipe)
进程与进程之间是相互独立的,互不干扰.如果多进程之间需要对同一资源操作,就需要进程间共享变量,上一篇文章介绍了进程间共享数据的三大类Value.Array.Manager,这三种类的主要区别在于管理的 ...
- Ascii Chart
Char Dec Oct Hex | Char Dec Oct Hex | Char Dec Oct Hex | Char Dec Oct Hex -------------------------- ...
- Pikachu漏洞练习平台实验——暴力破解(一)
概述 一个有效的字典可以大大提高暴力破解的效率 比如常用的用户名/密码TOP500 脱裤后的账号密码(社工库) 根据特定的对象(比如手机.生日和银行卡号等)按照指定的规则来生成密码 暴力破解流程 确认 ...
- [Java] 缓存池
new Integer(123) 与 Integer.valueOf(123) 的区别在于: new Integer(123) 每次都会新建一个对象: Integer.valueOf(123) 会使用 ...
- redis 入门之集合
sadd 将一个或多个 member 元素加入到集合 key 当中,已经存在于集合的 member 元素将被忽略.假如 key 不存在,则创建一个只包含 member 元素作成员的集合.当 key 不 ...
- HDU - 2181 C - 哈密顿绕行世界问题(DFS
题目传送门 C - 哈密顿绕行世界问题 一个规则的实心十二面体,它的 20个顶点标出世界著名的20个城市,你从一个城市出发经过每个城市刚好一次后回到出发的城市. Input 前20行的第i行有3个数, ...
- go语言从例子开始之Example20.错误处理
Go 语言使用一个独立的·明确的返回值来传递错误信息的.这与使用异常的 Java 和 Ruby 以及在 C 语言中经常见到的超重的单返回值/错误值相比,Go 语言的处理方式能清楚的知道哪个函数返回了错 ...
- Codeforces 356D 倍增优化背包
题目链接:http://codeforces.com/contest/356/problem/D 思路(官方题解):http://codeforces.com/blog/entry/9210 此题需要 ...
- python安装pika模块rabbitmq
1.pip install pika 2.如找不到 拷贝 D:\python\testmq\venv\Lib\site-packages \pika目录
- CF963E Circles of Waiting
Circles of Waiting 求一个整点四连通随机游⾛,离原点距离超过R期望步数.R≤50. 带状矩阵法 本质上就是网格图的随机游走. \[ E_x=\sum_y P_{x,y}E_y+1 \ ...