Educational Codeforces Round 38 (Rated for Div. 2) A. Word Correction

A. Word Correction
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. 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.InputThe first line contains one integer n (1 ≤ n ≤ 100) — the number of letters in word s before the correction.The second line contains a string s consisting of exactly n lowercase Latin letters — the word before the correction.OutputOutput the word s after the correction.ExamplesinputCopy

5
weird

output

werd

inputCopy

4
word

output

word

inputCopy

5
aaeaa

output

a

题意：给你一个单词，从左到右进行统计，当出现连续的元音字母时删掉后一个，问你最后的结果。
思路：模拟，每碰到两个
4000
连续时，删掉后一个，同时新接合出来的位置可能还会连续，所以要继续删。
这里我用了string的erase()字符删除函数。#include<bits/stdc++.h>
#define ll long long
using namespace std;
ll vis[30],n,flag;
string s;
int main()
{
memset(vis,0,sizeof(vis));
vis[1]=vis[5]=vis[9]=vis[15]=vis[21]=vis[25]=1; //标记元音
while(cin>>n>>s)
{
while(1) //反复遍历这个字符串，直到没有连续元音时break
{
flag=0;
for(ll i=1;i<n;i++)
{
if(vis[s[i-1]-'a'+1]&&vis[s[i]-'a'+1])
{
flag=1;
s.erase(i,1); //字符串删除函数，i是起始位置，1是要删除几个元素
}
}
if(flag==0)break; //没有连续元音break
}
cout<<s<<endl;
}
return 0;
}