cf452A Eevee
1 second
256 megabytes
standard input
standard output
You are solving the crossword problem K from IPSC 2014. You solved all the clues except for one: who does Eevee evolve into? You are not very into pokemons, but quick googling helped you find out, that Eevee can evolve into eight different pokemons: Vaporeon, Jolteon, Flareon, Espeon, Umbreon, Leafeon, Glaceon, and Sylveon.
You know the length of the word in the crossword, and you already know some letters. Designers of the crossword made sure that the answer is unambiguous, so you can assume that exactly one pokemon out of the 8 that Eevee evolves into fits the length and the letters given. Your task is to find it.
First line contains an integer n (6 ≤ n ≤ 8) – the length of the string.
Next line contains a string consisting of n characters, each of which is either a lower case english letter (indicating a known letter) or a dot character (indicating an empty cell in the crossword).
Print a name of the pokemon that Eevee can evolve into that matches the pattern in the input. Use lower case letters only to print the name (in particular, do not capitalize the first letter).
7
j......
jolteon
7
...feon
leafeon
7
.l.r.o.
flareon
Here's a set of names in a form you can paste into your solution:
["vaporeon", "jolteon", "flareon", "espeon", "umbreon", "leafeon", "glaceon", "sylveon"]
{"vaporeon", "jolteon", "flareon", "espeon", "umbreon", "leafeon", "glaceon", "sylveon"}
什么奇怪的东西……意思是给一个不完整的串,问是那8个串中的哪个
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
using namespace std;
inline int read()
{
int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
int n;
const string ch[8]={"vaporeon", "jolteon", "flareon", "espeon", "umbreon", "leafeon", "glaceon", "sylveon"};
char c[10];
int main()
{
scanf("%d",&n);
scanf("%s",c);
for (int i=0;i<8;i++)
{
bool mrk=1;
for (int j=0;j<8;j++)
if(c[j]!='.'&&c[j]!=ch[i][j])mrk=0;
if (mrk)
{
cout<<ch[i];
return 0;
}
}
}
cf452A Eevee的更多相关文章
- CF---(452)A. Eevee
A. Eevee time limit per test 1 second memory limit per test 256 megabytes input standard input outpu ...
- Codeforces 452A Eevee
#include<bits/stdc++.h> using namespace std; string m[]={"vaporeon","jolteon&qu ...
- 【Henu ACM Round#20 C】 Eevee
[链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] 处理处所有的字符串可能的样子. 存在map里面就好. [代码] #include <bits/stdc++.h> usi ...
- MemSQL Start[c]UP 2.0 - Round 1
A. Eevee http://codeforces.com/contest/452/problem/A 字符串水题 #include<cstdio> #include<cstrin ...
- 为什么开发人员对于PHP语言褒贬不一
PHP 语言,作为服务器端开发的脚本语言,在网站开发方面非常有名.从1995年 Rasmus Lerdorf 创建之后,W3Techs 的调查显示在已知的服务端编程语言中,PHP 占了82%.其中不乏 ...
- Pencil OJ 01 开发的准备
操作系统 ubuntu-12.04.5-desktop-amd64.iso 基本应用 Node 0.12.7 MongoDB 3.0.4 Robomongo 0.8.4 Atom 参考资料 OJ hu ...
- CF 452A(Eevee-直接试)
A. Eevee time limit per test 1 second memory limit per test 256 megabytes input standard input outpu ...
- Ubuntu 16.04下为Android编译OpenCV 3.2.0 Manager
http://johnhany.net/2016/07/build-opencv-manager-for-android-on-ubuntu/ 最近想在Android上尝试一下SIFT和SURF匹配算 ...
- [TypeScript] Creating a Class in TypeScript
Typescript classes make traditional object oriented programming easier to read and write. In this le ...
随机推荐
- SqlConnection类
一.常用属性 ConnectionString 获取或设置用于打开 SQL Server 数据库的字符串. (重写 DbConnection.ConnectionString.) Connectio ...
- JAVA的类加载器,详细解释
JVM规范定义了两种类型的类装载器:启动内装载器(bootstrap)和用户自定义装载器(user-defined class loader). 一. ClassLoader基本概念 1.ClassL ...
- Luci流程分析(openwrt下)
1. 页面请求: 1.1. 代码结构 在openwrt文件系统中,lua语言的代码不要编译,类似一种脚本语言被执行,还有一些uhttpd服务器的主目录,它们是: /www/index.html cgi ...
- Linux系统编程(17)——正则表达式进阶
C的变量和Shell脚本变量的定义和使用方法很不相同,表达能力也不相同,C的变量有各种类型,而Shell脚本变量都是字符串.同样道理,各种工具和编程语言所使用的正则表达式规范的语法并不相同,表达能力也 ...
- 【转】 linux内核移植和网卡驱动(二)
原文网址:http://blog.chinaunix.net/uid-29589379-id-4708911.html 一,内核移植步骤: 1, 修改顶层目录下的Makefile ARCH ...
- 解决mongodb连接失败问题
错误提示: MongoDB shell version: 2.4.9 connecting to: test Mon Mar 3 23:45:09.491 Error: couldn't conne ...
- cc150 Chapter 2 | Linked Lists 2.6 Given a circular linked list, implement an algorithm which returns node at the beginning of the loop.
2.6Given a circular linked list, implement an algorithm which returns the node at the beginning of ...
- Java Base64编码与图片互转
import java.io.FileInputStream; import java.io.FileOutputStream; import java.io.IOException; import ...
- kettle工具二次开发-代码启动JOB
kettle工具是一款优秀的数据同步.数据处理的BI工具,收到了很多人的青睐.kettle软件通过可视化的图标可以让我们很轻易的能完成数据同步.处理的开发工作.但是使用kettle可视化界面在跑JOB ...
- android之ListPreference的用法_PreferenceActivity用法
首先,我们明确,preference是和数据存储相关的. 其次,它能帮助我们方便的进行数据存储!为什么这个地方一定要强调下方便的这个词呢?原因是,我们可以根本就不使用,我们有另外的N种办 ...