Password security is a tricky thing. Users prefer simple passwords that are easy to remember (like buddy), but such passwords are often insecure. Some sites use random computer-generated passwords (like xvtpzyo), but users have a hard time remembering them and sometimes leave them written on notes stuck to their computer. One potential solution is to generate "pronounceable" passwords that are relatively secure but still easy to remember.
FnordCom is developing such a password generator. You work in the quality control department, and it's your job to test the generator and make sure that the passwords are acceptable. To be acceptable, a password must satisfy these three rules:
It must contain at least one vowel.
It cannot contain three consecutive vowels or three consecutive consonants.
It cannot contain two consecutive occurrences of the same letter, except for 'ee' or 'oo'.
(For the purposes of this problem, the vowels are 'a', 'e', 'i', 'o', and 'u'; all other letters are consonants.) Note that these rules are not perfect; there are many common/pronounceable words that are not acceptable.

The input consists of one or more potential passwords, one per line, followed by a line containing only the word 'end' that signals the end of the file. Each password is at least one and at most twenty letters long and consists only of lowercase letters.

For each password, output whether or not it is acceptable, using the precise format shown in the example.

 #include<iostream>

 #include<cstdio>

 #include<cmath>

 #include<algorithm>

 #include<cstring>

 #include<string>

 #define LL long long

 #define PI acos(-1.0)

 using namespace std;

 char t[]={'a','e','i','o','u'};

 bool isT(char a)//先扫一遍看看有没有元音字母,有的话直接返回真

 {

     for(int i=;i<;i++)

         if(a==t[i])return true;

     return false;

 }

 int main()

 {

     int len;//表示字符串的长度

     char a[]={};

     bool f[],g,h;//f数组是对字母进行标记

     while(cin>>a,strcmp(a,"end")){

         len=strlen(a);

         memset(f,false,sizeof(f));//全部初始化为false

         for(int i=;i<len;i++)//标记元音

             if(isT(a[i]))f[i]=true;//是元音则为true

         g=true,h=false;//g来判断条件，而h是用来标记至少有一个元音字母这个条件

         for(int i=;i<len;i++){

             if(f[i])h=true;//标记一下是否有元音字母

             if(i>&&a[i]==a[i-]&&a[i]!='e'&&a[i]!='o'){g=false;break;}//标记连续相等的两个除了是e和o的

             if(i>&&f[i]&&f[i-]&&f[i-]){g=false;break;}//标记连续三个是元音的

             if(i>&&!f[i]&&!f[i-]&&!f[i-]){g=false;break;}//标记连续三个是辅音的

         }

         if(!h)g=false;//没有元音字母就置g为假,便于下一出口统一操作（单一出口）

         if(g)cout<<'<'<<a<<'>'<<" is acceptable."<<endl;

         else cout<<'<'<<a<<'>'<<" is not acceptable."<<endl;

     }

     return ;

 }