杭电1039

Easier Done Than Said?

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4246    Accepted Submission(s): 2142

Problem Description

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.

 

Input

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.

 

Output

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

 

Sample Input

a
tv
ptoui
bontres
zoggax
wiinq
eep
houctuh
end

 

Sample Output

<a> is acceptable.
<tv> is not acceptable.
<ptoui> is not acceptable.
<bontres> is not acceptable.
<zoggax> is not acceptable.
<wiinq> is not acceptable.
<eep> is acceptable.
<houctuh> is acceptable.

 

//杭电1039
#include <iostream>
#include <cstring>
using namespace std;

inline bool isVowel(char ch)
{
    return (ch == 'a' || ch == 'e' || ch == 'i' || ch == 'o' || ch == 'u');
}

int main(int argc, char **argv)
{
    string str;
    while (cin >> str && str != "end") {
        int count1 = 0;//元音连续的个数
        int count2 = 0;//辅音连续的个数
        int count3 = 0;//元音总个数
        char ch = '\0';//记录上一个字符
        
        bool isAcceptable = true;//表示一个passward是否被接受
        
        for (int i = 0; i < str.length() && isAcceptable; ++ i)
        {
            if (ch == str[i] && ch != 'e' && ch != 'o') //是否有两个重复的字符，如果有是不是ee或者oo如果不是则不被接受
            {
                isAcceptable = false;
                break;
            }
            else
            {
                ch = str[i];//记录前一个字符
            }
            if (isVowel(str[i]))//如果是元音，则元音计数加1，同时辅音计数清零
            {
                ++ count1;
                ++ count3;//元音个数加1
                count2 = 0;
            }
            else//辅音计数加1，同时元音计数清零
            {
                ++ count2;
                count1 = 0;
            }
        
            if (count1 > 2 || count2 > 2)//是否有连个以上连续的元音或者连续的辅音，如果有则不被接受
            {
                isAcceptable = false;
                break;
            }
            
        }
        if (count3 == 0)//是否有元音出现，如果没有元音出现那么不被接受
            isAcceptable = false;
        
        //打印结果
        if (isAcceptable)
            cout <<"<"<<str << ">" <<" is acceptable."<<endl;
        else
            cout <<"<"<<str << ">" <<" is not acceptable."<<endl;

    }
    return 0;
}