String Task

Time limit1000 ms
Memory limit262144 kBSourceCodeforces
Round #267 (Div. 2)
Tagsimplementation *800
EditorialAnnouncement

Problem Description

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.

Example

input

tour

output

.t.r

Input

Codeforces

Output

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

Input

aBAcAba

Output

.b.c.b

问题链接：https://vjudge.net/problem/24454/origin
问题简述：输入一串字符，将所有元音删掉，大写换成小写，辅音前加“.”
问题分析：大小写转换，将元音部分附‘0’实现“删除元音”，删除多余项，将数组放大两倍实现前赋‘.’
程序说明：通过ASCII的大小更改字母大小写的转换

AC通过的c++语言程序如下:

//
#include<iostream>
#include<string>
using namespace std;
int main()
{
    char letter[200];
	char print[200];
	char vowel[] = { 'a','o','y','e','u','i','A','O','Y','E','U','I' };
	cin>>(letter);
	int len;
	len = strlen(letter);
	for (int i = 0; i < len; i++)
	{
		if (letter[i] >= 65 && letter[i] <= 90)
			letter[i] += 32;
	}
	for (int i = 0; i < len; i++)
		for (int j = 0; j < 11; j++)
		{
			if (letter[i] == vowel[j])
			{
				letter[i] = '0';
			}
		}
	int t = 0;
	for (int i = 0; i < len; i++)
	{
		if (letter[i] >= 97 && letter[i] <= 122||letter[i]>65&&letter[i]<=90)
		{
			print[t] = letter[i];
			t++;
		}
	}
	int sum = 0;
	for (int i = 0; i < 2 * t; i++)
	{
		if (i % 2 == 0)
			letter[i] = '.';
		else
		{
			letter[i] = print[sum];
			sum++;
		}
	}
	for (int i = 0; i < 2 * sum; i++)
		cout << letter[i];
}

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