A. Tavas and Nafas

Time Limit: 1 Sec  Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/535/problem/A

Description

Today Tavas got his test result as an integer score and he wants to share it with his girlfriend, Nafas.

His phone operating system is Tavdroid, and its keyboard doesn't have any digits! He wants to share his score with Nafas via text, so he has no choice but to send this number using words.

He ate coffee mix without water again, so right now he's really messed up and can't think.

Your task is to help him by telling him what to type.

Input

The first and only line of input contains an integer s (0 ≤ s ≤ 99), Tavas's score.

1000000000.

Output

In the first and only line of output, print a single string consisting only from English lowercase letters and hyphens ('-'). Do not use spaces.

Sample Input

6

Sample Output

six

HINT

题意

给你100以下数字,输出英文

题解:

沙比提

代码:

//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 200001
#define mod 10007
#define eps 1e-9
//const int inf=0x7fffffff; //无限大
const int inf=0x3f3f3f3f;
/* int buf[10];
inline void write(int i) {
int p = 0;if(i == 0) p++;
else while(i) {buf[p++] = i % 10;i /= 10;}
for(int j = p-1; j >=0; j--) putchar('0' + buf[j]);
printf("\n");
}
*/
//**************************************************************************************
inline ll read()
{
int x=,f=;char ch=getchar();
while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
return x*f;
}
map<int,string> s; int main()
{
int n=read();
s[]="one";
s[]="two";
s[]="three";
s[]="four";
s[]="five";
s[]="six";
s[]="seven";
s[]="eight";
s[]="nine";
s[]="ten";
s[]="eleven";
s[]="twelve";
s[]="thirteen";
s[]="fourteen";
s[]="fifteen";
s[]="sixteen";
s[]="seventeen";
s[]="eighteen";
s[]="nineteen";
s[]="twenty";
s[]="thirty";
s[]="forty";
s[]="fifty";
s[]="sixty";
s[]="seventy";
s[]="eighty";
s[]="ninety";
if(n==)
cout<<"zero"<<endl;
else if(n<=)
cout<<s[n]<<endl;
else if(n%==)
cout<<s[n]<<endl;
else
cout<<s[n-n%]<<"-"<<s[n%]<<endl;
}

Codeforces Round #299 (Div. 2) A. Tavas and Nafas 水题的更多相关文章

  1. Codeforces Round #299 (Div. 1) A. Tavas and Karafs 水题

    Tavas and Karafs Time Limit: 1 Sec  Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/536/prob ...

  2. Codeforces Round #299 (Div. 2) B. Tavas and SaDDas 水题

    B. Tavas and SaDDas Time Limit: 1 Sec  Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/535/p ...

  3. 水题 Codeforces Round #299 (Div. 2) A. Tavas and Nafas

    题目传送门 /* 很简单的水题,晚上累了,刷刷水题开心一下:) */ #include <bits/stdc++.h> using namespace std; ][] = {" ...

  4. 二分搜索 Codeforces Round #299 (Div. 2) C. Tavas and Karafs

    题目传送门 /* 题意:给定一个数列,求最大的r使得[l,r]的数字能在t次全变为0,每一次可以在m的长度内减1 二分搜索:搜索r,求出sum <= t * m的最大的r 详细解释:http:/ ...

  5. DFS Codeforces Round #299 (Div. 2) B. Tavas and SaDDas

    题目传送门 /* DFS:按照长度来DFS,最后排序 */ #include <cstdio> #include <algorithm> #include <cstrin ...

  6. Codeforces Round #297 (Div. 2)A. Vitaliy and Pie 水题

    Codeforces Round #297 (Div. 2)A. Vitaliy and Pie Time Limit: 2 Sec  Memory Limit: 256 MBSubmit: xxx  ...

  7. Codeforces Round #290 (Div. 2) A. Fox And Snake 水题

    A. Fox And Snake 题目连接: http://codeforces.com/contest/510/problem/A Description Fox Ciel starts to le ...

  8. Codeforces Round #322 (Div. 2) A. Vasya the Hipster 水题

    A. Vasya the Hipster Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/581/p ...

  9. Codeforces Round #373 (Div. 2) B. Anatoly and Cockroaches 水题

    B. Anatoly and Cockroaches 题目连接: http://codeforces.com/contest/719/problem/B Description Anatoly liv ...

随机推荐

  1. 面试中关于Redis的问题看这篇就够了

    昨天写了一篇自己搭建redis集群并在自己项目中使用的文章,今天早上看别人写的面经发现redis在面试中还是比较常问的(笔主主Java方向).所以查阅官方文档以及他人造好的轮子,总结了一些redis面 ...

  2. java正则: 忽略大小写匹配

    import java.util.regex.Matcher; import java.util.regex.Pattern; import com.sun.org.apache.xerces.int ...

  3. 【Learn】CSS定义

    CSS基础语法 本文用于介绍CSS相关的知识,用于记录自己的学习笔记.由于我已经熟悉了部分的HTML,所以相关的概念也不在这里进行描述了,直接写自己的一些心得感悟. 1.CSS规则 CSS是由两个主要 ...

  4. windows下phpstrom中xdebug的使用

    https://laravel-china.org/articles/16425/windows-phpstorm-xdebug-breakpoint-debugging

  5. UTF-8和GB2312互转的最简单快捷的方法

    一.如果你想把utf-8转为GB2312 1.用记事本打开源码,把<meta http-equiv="Content-Type" content="text/htm ...

  6. html5弹性布局两则,有交互。

    要开发一个后台管理框架,要求如下效果. 然后开始找各种弹性布局啊什么的,用了flex写了一个,但是觉得不好,首先是兼容,其次它会破坏掉里面子元素的一些css特性,为了不给自己找麻烦我还是用传统写法吧. ...

  7. 【LOJ】#2350. 「JOI 2017/2018 决赛」月票购买

    题解 首先求一个最短路图出来,最短路图就是这条边在最短路上就保留,否则就不保留,注意最短路图是一个有向图,一条边被保留的条件是 dis(S,u) + val(u,v) = dis(v,T)我们需要求两 ...

  8. KVM调整cpu和内存

    一.修改kvm虚拟机的配置 1.virsh edit centos7 找到“memory”和“vcpu”标签,将 <name>centos7</name> <uuid&g ...

  9. Django学习笔记-2018.11.16

    知识储备: 1 Python基础 2 数据库SQL 3 HTTP协议 4 HTML&&CSS 5 正则表达式 Django启动 django-admin startproject pr ...

  10. linux中使用rm命令将文件移到回收站的方法

    今天在终端下,看到我的用户目录下有个-的文件夹(maven生成),相要删除收回点空间,习惯性的用命令 rm -rf ~ ,一回车,猛然想起的时候已经来不及了,世界一下子清静了,想死的心都有了! 没错, ...