HDU-1039-Easier Done Than Said?

Easier Done Than Said?

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11280 Accepted Submission(s): 5450

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.

题意：给出字符串s1，满足下面三个条件的，输出<s1> is acceptable.否则输出<s1> is not acceptable.
输入end则表示文件结束，end不用处理
一，至少一个元音
二，不能有三个连续的元音或者三个连续的辅音
三，不能有连续相同的两个字母，ee,oo,除外

思路：三个条件分别写出对应的判断函数，只要思路清晰，一般不会错。
字符串的水题，为了练习string才做的。。。。。
string用法http://blog.csdn.net/qq_32680617/article/details/51122395

代码

#include<stdio.h>
#include<string.h>
#include<string>
#include<algorithm>
#include<iostream>
#include<math.h>
#include<stack>
using namespace std;
bool is_char(char c)//元音返回真，辅音返回假
{
    if(c=='a'||c=='e'||c=='i'||c=='o'||c=='u')
        return 1;
    return false;
}
bool judge1(string s1)//至少有一个元音则返回真，否则返回假
{
    if(s1.find('a')!=string::npos||s1.find('e')!=string::npos||s1.find('i')!=string::npos||s1.find('o')!=string::npos||s1.find('u')!=string::npos)
        return 1;
    return false;
}
bool judge2(string s1)//有三个连续的元音或辅音返回假，否则返回真
{
    for(string::size_type i=2; i<s1.size(); i++)
    {
        if(is_char(s1[i])&&is_char(s1[i-1])&&is_char(s1[i-2]))//三个连续元音
            return false;
        if(!is_char(s1[i])&&!is_char(s1[i-1])&&!is_char(s1[i-2]))//三个连续辅音
            return false;
    }
    return 1;
}
bool judge3(string s1)//有两个除ee,oo,外连续相同的字母返回假，否则返回真
{
    for(string::size_type i=1; i<s1.size(); i++)
    {
        if(s1[i]==s1[i-1]&&s1[i-1]!='e'&&s1[i-1]!='o')
            return false;
    }
    return 1;
}

int main()
{
    string s1;
    while(cin>>s1)
    {
        if(s1=="end")
            break;
        if(judge1(s1)&&judge2(s1)&&judge3(s1))
        {
            printf("<");
            cout<<s1;
            printf("> is acceptable.\n");
        }
        else
        {
            printf("<");
            cout<<s1;
            printf("> is not acceptable.\n");
        }
    }
    return 0;
}