HDU1039 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): 11141    Accepted Submission(s): 5381

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
  
  

 

题目大意：

与其说不如去做些简单的事情；意思是说这是关于密码安全的问题，人们喜欢用些简单的密码（比如 rubby），因为它比较好记，但是有些简单的好记的密码安全性并不高。有些使用者喜欢用电脑随机的密码（比如 xvtpzyo），但是使用者可能很难记住这些密码，他们更可能的是把这些密码记录在笔记本上。有些使用者就喜欢用发音去给出密码，这样的密码又好记又不因为简单而不安全，你的工作就是去编写一个程序去给出那些密码的发音的密码是可以被电脑接受的，那些事不可以的。这里给出几项规则：

1：能被接受的密码至少有一个元音字母；

2：包含连续的三个元音字母或者三个连续的辅音字母都是不被接受的；

3：包含两个连续的相同的字母，除了 “ee”或者“oo之外都是不被接受的。

由于程序本身的思想并不完美，所以有些普通的发音也是不能被接受的；

分析：

我们只要按照这三个规则，去遍历我们的字符串看看其中是否有不满足的地方做个标记，然后给出答案就好了；使用标记法；

给出AC代码：

#include<iostream>
#include<cstring>
using namespace std;
int main()
{
	char str[1000];
	int f1, f2, f3, f4;
	while (cin >> str,strcmp(str, "end"))
	{
		f1 = f2 = f3 = f4 = 0;
		int len = strlen(str);
		for (int i = 0; i < len; i++) //判断是否有元音；
		{
			if (str[i] == 'a' || str[i] == 'e' || str[i] == 'i' || str[i] == 'o' || str[i] == 'u')
			{
				f1 = 1; 
				break;
			}
		}
		if (len >= 2)
		{
			for (int i = 0; i < len-1; i++) //判断是否有两个连续的元音；
			{
				if (str[i] == str[i + 1] )
				{
					if (str[i] != 'o'&& str[i] != 'e')
					{
						f2 = 1;
						break;
					}
				}
			}
		}
		if (len >= 3)//判断是否有三个连续的元音或者辅音；
		{
			for (int i = 0; i < len-2; i++)
			{
				if (str[i] == 'a' || str[i] == 'e' || str[i] == 'i' || str[i] == 'o' || str[i] == 'u')
				{
					if (str[i + 1] == 'a' || str[i + 1] == 'e' || str[i + 1] == 'i' || str[i + 1] == 'o' || str[i + 1] == 'u')
					if (str[i + 2] == 'a' || str[i + 2] == 'e' || str[i + 2] == 'i' || str[i + 2] == 'o' || str[i + 2] == 'u')
						f3 = 1;
				}
				else if (str[i] != 'a' && str[i] != 'e' && str[i] != 'i' && str[i] != 'o' && str[i] != 'u')
				{
					if (str[i + 1] != 'a' && str[i + 1] != 'e' && str[i + 1] != 'i' && str[i + 1] != 'o' && str[i + 1] != 'u')
					if (str[i + 2] != 'a' && str[i + 2] != 'e' && str[i + 2] != 'i' && str[i + 2] != 'o' && str[i + 2] != 'u')
						f4 = 1;
				}
			}
		}
		if (f1 == 1 && f2 == 0 && f3 == 0 && f4 == 0)
			cout << "<" << str << ">" << " is acceptable." << endl;
		else cout << "<" << str << ">" << " is not acceptable." << endl;
	}
	return 0;
}