【简单过滤】#89 A. String Task

最新推荐文章于 2022-05-04 15:08:44 发布

原创最新推荐文章于 2022-05-04 15:08:44 发布 · 933 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#Codeforces #Beta Round #89 #Div. 2 #String Task

CF-A题终结者专栏收录该内容

96 篇文章

订阅专栏

本文介绍了一个简单的编程任务，要求处理给定的字符串，删除其中的所有元音，插入一个点字符在每个辅音之前，并将所有大写字母转换为小写。提供了一个具体的例子和代码实现。

Petya started to attend programming lessons. On the first lesson his task was to write a simple program. The program was supposed to do the following: in the given string, consisting if uppercase and lowercase Latin letters, it:

deletes all the vowels,
inserts a character "." before each consonant,
replaces all uppercase consonants with corresponding lowercase ones.

Vowels are letters "A", "O", "Y", "E", "U", "I", and the rest are consonants. The program's input is exactly one string, it should return the output as a single string, resulting after the program's processing the initial string.

Help Petya cope with this easy task.

Input

The first line represents input string of Petya's program. This string only consists of uppercase and lowercase Latin letters and its length is from 1 to 100, inclusive.

Output

Print the resulting string. It is guaranteed that this string is not empty.

Sample test(s)

input
tour

output
.t.r

input
Codeforces

output
.c.d.f.r.c.s

input
aBAcAba

output
.b.c.b

题意是说给一个字符串，要求输出一个序列，只包含原字符串的非元音字母的小写形式，每个字母输出前要加一个点

首先处理大小写，我们直接把源字符串变成全小写的就好，tolower(char a)函数返回的就是当前字母的小写形式，然后我们判断一下这个字母是否元音字母，不是的话我们就输出点和这个字母即可

Code：

#include <cstdio>
#include <string>
#include <cstring> 
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;
// http://codeforces.com/contest/118
// String Task

bool judge(char c)
{
	if(c=='a'||c=='e'||c=='i'||c=='o'||c=='u'||c=='y')return false;
	return true;
}

int main()
{
	string s;	cin>>s;
	for(int i=0;i<s.length();i++)
	{
		char now=tolower(s[i]);
		if(judge(now))cout<<'.'<<now;
	} 
	return 0;
}