我是靠谱客的博主 复杂唇膏,这篇文章主要介绍CodeForces-118A String Task,现在分享给大家,希望可以做个参考。

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

题目解析:
先将所有的大写字母转化为小写字母,再利用循环在输出的时候把元音跳过,相当于删除。n是a的长度,本来是用的sizeof(a),一直都不行,后来才发现无论string里放多长的字符串,它的sizeof()都是固定的,于是就换成了a.size()。

#include <iostream>
#include <string>
using namespace std;
int main()
{
	string a;
	cin >> a;
	int n;
	n = a.size();
	for (int i = 0; i<n; i++)
	{
		if (a[i] >= 65 && a[i] <= 90)
			a[i] += 32;
		if (a[i] == 97 || a[i] == 101 || a[i] == 105 || a[i] == 111 || a[i] == 117 || a[i] == 121)
			continue;
		else cout << "."<< a[i];
	}
	cout << endl;
}

最后

以上就是复杂唇膏最近收集整理的关于CodeForces-118A String Task的全部内容,更多相关CodeForces-118A内容请搜索靠谱客的其他文章。

本图文内容来源于网友提供,作为学习参考使用,或来自网络收集整理,版权属于原作者所有。
点赞(139)

评论列表共有 0 条评论

立即
投稿
返回
顶部