題目背景

給一個長為 $n$的字串，當有兩個母音字母相鄰時，刪除後一個母音，輸出修改後的字串。（注： $a$, $e$

題目描述

Victor tries to write his own text editor, with word correction included. However, the rules of word correction are really strange.

Victor thinks that if a word contains two consecutive vowels, then it’s kinda weird and it needs to be replaced. So the word corrector works in such a way: as long as there are two consecutive vowels in the word, it deletes the first vowel in a word such that there is another vowel right before it. If there are no two consecutive vowels in the word, it is considered to be correct.

You are given a word s s . Can you predict what will it become after correction?

In this problem letters a, e, i, o, u and y are considered to be vowels.

輸入輸出格式

輸入格式：

The first line contains one integer n n ( 1<=n<=100 1<=n<=100 ) — the number of letters in word s s before the correction.

The second line contains a string s s consisting of exactly n n lowercase Latin letters — the word before the correction.

輸出格式：

Output the word s s after the correction.

輸入輸出樣例

輸入樣例#1：

5
weird

輸出樣例#1：

werd

輸入樣例#2：

4
word

輸出樣例#2：

word

輸入樣例#3：

5
aaeaa

輸出樣例#3：

a

思想：如果當前字母與前一個都是母音字母，就什麼也不做，否則輸出當前字母

話不多說，程式碼有解釋

如下：

#include<iostream>
using namespace std;
char s;
int main(){
    int n,m=0;
    cin>>n;
    for(int i=1;i<=n;i++){
        cin>>s;
        if(s!='a' && s!='e'&& s!='i' && s!='o' && s!='u' && s!='y')
            cout<<s,m=0;
        else if(m) continue;
        else cout<<s,m=1;
    }
    return 0;
}