hdu 1039 连续字符串处理

题目来源：http://acm.hdu.edu.cn/showproblem.php?pid=1039

Easier Done Than Said?

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6855    Accepted Submission(s): 3396

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.

 

 

代码如下：

#include<iostream>
#include<stdio.h>
#include<string>
#include<string.h>
#include<map>
#include<math.h>
#include<algorithm>
#define N 25
using namespace std;

int main()
{
    char str[N];
    int a[N];
    while(scanf("%s",str) && strcmp("end" , str))
    {
        int len=strlen(str);
        int flag=1,sum=0; // sum 标记元音的个数，flag标记
        int i;
        for(i=0;i<len;i++)
        {
            switch(str[i])
            {
                case 'a':
                case 'e':
                case 'i':
                case 'o':
                case 'u': a[i]=1; sum++;break;
                default : a[i]=0;
            }
            if(i>0 && str[i]== str[i-1] && str[i]!= 'e'&& str[i]!= 'o')  //2个连续的相同音，排除e/o
            {
                flag=0;
                break;
            }
            if(i>1 && a[i]== a[i-1] && a[i-1] == a[i-2] )    // 连续三个相连的元音或辅音   注意 i满足的条件， 
            {
                flag=0;
                break;
            }
        }
        if(sum && flag)
            printf("<%s> is acceptable.
",str);
        else
            printf("<%s> is not acceptable.
",str);

    }
    return 0 ;
}