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2013
11-09

POJ 1575 Easier Done Than Said? [解题报告] Java

Easier Done Than Said?

问题描述 :

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.

输入:

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.

输出:

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

样例输入:

a
tv
ptoui
bontres
zoggax
wiinq
eep
houctuh
end

样例输出:

<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.

解题代码:

//* @author: [email protected]
import java.io.*;
public class Main
{
 static int[][] sorce;
 public static void main(String[] args) throws IOException
 {
  InputStreamReader is=new InputStreamReader(System.in);
  BufferedReader in=new BufferedReader(is);
  while(true)
  {
	String s=in.readLine();
	if(s.equals("end"))break;
	boolean bb=true,b1=false,b2;
	for(int i=0;i< s.length();i++)
	{
		if(isv(s.charAt(i)))
		{
	          b1=true;
		  break;
		}
	}
	if(!b1){
	  System.out.println("<"+s+"> is not acceptable.");
	  continue;
	}
	int cv=0,cf=0,mav=0,maf=0;
	for(int i=0;i< s.length();i++)
	{
	   if(cv!=0||cf!=0){
	    if(s.charAt(i)==s.charAt(i-1)&&s.charAt(i)!='e'&&s.charAt(i)!='o')
		{
                    mav=100;
		    break;
	        }
	    }
	    if(isv(s.charAt(i)))
	    {
		cv++;
		if(maf< cf) maf=cf;
		cf=0;
	     }
	    else
	    {
		cf++;
		if(mav< cv) mav=cv;
		cv=0;
	    }
	    if(cv>mav) mav=cv;
	    if(cf>maf) maf=cf;
	    if(mav>2||maf>2) break;
	}
	if(mav>2||maf>2)
	{
		System.out.println("<"+s+"> is not acceptable.");
		continue;
	}
	System.out.println("<"+s+"> is acceptable.");
   }
 }

  public static boolean isv(char c)
   {
    switch(c)
    {
	case 'a':
	case 'e':
	case 'i':
	case 'o':
	case 'u':
	  return true;
    }
    return false;
   }
}

  1. 代码是给出了,但是解析的也太不清晰了吧!如 13 abejkcfghid jkebfghicda
    第一步拆分为 三部分 (bejk, cfghi, d) * C(13,3),为什么要这样拆分,原则是什么?