String Task CodeForces - 118A

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本文解析了CodeForces-118A的字符串任务，详细介绍了如何删除元音字母，将大写字母转换为小写，并在辅音字母前插入点号的过程。提供了一个C语言的解决方案。

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String Task CodeForces - 118A

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.

Examples
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/CodeForces-118A)

问题简述：
输入字符串a,然后把字符串中的元音字母删除，大写字母转为小写字母，每个辅音字母前输出一个.

问题分析：
利用ASCII把大写加32转换成小写，判断元音字母，后一个覆盖到前一个实现在数组中删除元音字母，在输出辅音字母前输出一个.

程序说明：使用gets_s(,)函数读入字符串

AC通过的C语言程序如下：

#include<iostream>
using namespace std;
#include<cstdio>
int main()
{
	char L[105];
	gets_s(L, 105);                                            //内存要偏大
	for (int i = 0; L[i] != '\0'; i++)
	{
            while (L[i] == 'a' || L[i] == 'i' || L[i] == 'o' || L[i] == 'u' || L[i] == 'e' || L[i] == 'y' || L[i] == 'A' || L[i] == 'E' || L[i] == 'I' || L[i] == 'O' || L[i] == 'U' || L[i] == 'Y')                                       //要求不输出元音字母  那就挑选掉
		{
			for (int j = i; L[j] != '\0'; j++)
				L[j] = L[j + 1];
		}
		if (L[i] > 64 && L[i] < 91)
			L[i] += 32;
		if (L[i]>96&&L[i]<123)
			cout << "." << L[i];
	}
	return 0;
}