zoukankan      html  css  js  c++  java
  • cf452A Eevee

    A. Eevee
    time limit per test
    1 second
    memory limit per test
    256 megabytes
    input
    standard input
    output
    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.

    Input

    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).

    Output

    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).

    Sample test(s)
    input
    7
    j......
    
    output
    jolteon
    
    input
    7
    ...feon
    
    output
    leafeon
    
    input
    7
    .l.r.o.
    
    output
    flareon
    
    Note

    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;
    	  	}
    	  }
    }
    

      

    ——by zhber,转载请注明来源
  • 相关阅读:
    lintcode 最长上升连续子序列 II(二维最长上升连续序列)
    lintcode 滑动窗口的最大值(双端队列)
    windows下实现Git在局域网使用
    eclipse导入web项目之后项目中出现小红叉解决办法
    HTTP协议基础
    c++面试常用知识(sizeof计算类的大小,虚拟继承,重载,隐藏,覆盖)
    Codeforces Round #327 (Div. 2) B. Rebranding C. Median Smoothing
    c语言表白程序代码
    矩阵的逆矩阵,伴随矩阵
    document.getElementById()与 $()区别
  • 原文地址:https://www.cnblogs.com/zhber/p/4035915.html
Copyright © 2011-2022 走看看