zoukankan      html  css  js  c++  java
  • hdu 1039 Easier Done Than Said?

    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.

     细节问题,第一个代码就是找不出错误在哪里?

    #include<stdio.h>
    #include<string.h>
      int main ()
        {
        	char s[25];
        	int i,sum,k;
           while(gets(s))
             {
             	if(strcmp(s,"end")==0)  break;
             	i=0;
             	sum=0;
             	k=1;
              while(s[i]!='')
             	{
             	  if(s[i]=='a'||s[i]=='e'||s[i]=='o'||s[i]=='u'||s[i]=='i')
             	     {
             	     	sum++; 
             	     }
    			     
    			  if(s[i]=='a'||s[i]=='e'||s[i]=='o'||s[i]=='u'||s[i]=='i')
    			    {
    			    	if(s[i+1]=='a'||s[i+1]=='e'||s[i+1]=='o'||s[i+1]=='u'||s[i+1]=='i'||s[i+1]!='')
    			    	  {
    			    	  	if(s[i+2]=='a'||s[i+2]=='e'||s[i+2]=='o'||s[i+2]=='u'||s[i+2]=='i'||s[i+2]!='')
    			    	  	   {
    			    	  	   	k=0;
    			    	  	    i++;
    			    	  	    continue;
    			    	  	   }
    			    	  	   
    			    	  }
    			    }
    			      
    			  if(!(s[i]=='a'||s[i]=='e'||s[i]=='o'||s[i]=='u'||s[i]=='i'))
    			    {
    			    	if(!(s[i+1]=='a'||s[i+1]=='e'||s[i+1]=='o'||s[i+1]=='u'||s[i+1]=='i')||s[i+1]!='')
    			    	  {
    			    	  	if(!(s[i+2]=='a'||s[i+2]=='e'||s[i+2]=='o'||s[i+2]=='u'||s[i+2]=='i')||s[i+2]!='')
    			    	  	 {
    			    	  	 	k=0;
    			    	  	    i++;
    			    	  	    continue;
    			    	  	   
    			    	  	 }  
    			    	  }
    			    }
    			   if(s[i]==s[i+1]&&s[i]!='e'&&s[i]!='o'&&s[i+1]!='')
    			       {
    			       	k=0;
    			       	i++;
    			       	continue;
    			       }
    			    i++;
             	}
                    if(sum>=1&&k==1)   
                       {
                       	printf("<%s> is acceptable.
    ",s);
                       }
                     else
                        printf("<%s> is not acceptable
    ",s);
             }
             
            return 0;
        }
    /*#include<stdio.h>
    #include<string.h>
      int number_v(char *s)
        {   int i;
            int sum;
             i=0;
             sum=0;
        	while(s[i]!='')
        	  {
        	  	if(s[i]=='a'||s[i]=='e'||s[i]=='o'||s[i]=='u'||s[i]=='i')
        	  	  sum++;
        	  	  i++;
        	  }
           return sum;
        }
       int is_v(char c)
          {
          	if(c=='a'||c=='e'||c=='i'||c=='o'||c=='u')
                return 1;
            return 0;
          }
       int is_three(char *s)
         {  int i;
    	    int len;
    	    int t1,t2;
    	      t1=t2=0;
               i=0;
               
            
             len=strlen(s);
         	for(i=0;i<len-2;i++)
         	  {
         	  	if(is_v(s[i])&&is_v(s[i+1])&&is_v(s[i+2]))
         	  	    t1=1;
         	  	if(!is_v(s[i])&&!is_v(s[i+1])&&!is_v(s[i+2]))
         	  	    t1=1;
         	  	if(s[i]==s[i+1]&&s[i+1]!='e'&&s[i+1]!='o')
         	  	    t2=1;
         	  }
         	  if(!t2&&(s[len-1]==s[len-2]&&s[len-2]!='e'&&s[len-2]!='o'))
         	        t2=1;
         	if(!t1&&!t2)     return 1;
         	    return 0;
         }
       
        int main()
           {
           	  char s[25];
           	  while(gets(s))
           	    {
           	    	if(strcmp(s,"end")==0)   break;
           	    	
           	    	if(number_v(s)&&is_three(s))
           	    	    printf("<%s> is acceptable.
    ",s);
           	    	else
           	    	    printf("<%s> is not acceptable.
    ",s);
           	    }
           	return 0;
           }*/

  • 相关阅读:
    [转]Amazon DynamoDB – a Fast and Scalable NoSQL Database Service Designed for Internet Scale Applications
    [转]CIDR简介
    [转]Amazon AWS亚马逊云服务免费一年VPS主机成功申请和使用方法
    [转]SQL Server 连接串关键字别名
    [转]各种符号的英文单词
    【解决】SharePoint 2013 当鼠标悬停在用户名称上时页面会崩溃
    【HOW】如何限制Reporting Services报表导出功能中格式选项
    【解决】“不能手工移动或复制项到放置库”
    【解决】SharePoint 2013 with SP1安装问题及解决
    [转]安装SharePoint 2013时安装AppFabric失败(错误码:1603)
  • 原文地址:https://www.cnblogs.com/jhcelue/p/7210822.html
Copyright © 2011-2022 走看看