Easier Done Than Said?杭电ACM1039

Easier Done Than Said?

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12815    Accepted Submission(s): 6188

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

atvptouibontreszoggaxwiinqeephouctuhend

 

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.
3个条件，1至少一个元音。2不能连续3元或3辅音。3不能连续2个一样字母，除了oo ee。
解体思路：
先第一个条件不符合，直接输出not acceptable，用continue跳出循环，输入下一个字符串
  第二个条件不符合，直接输出not acceptable，用continue跳出循环，输入下一个字符串
  第三个条件不符合，直接输出not acceptable，用continue跳出循环，输入下一个字符串
  都符合，        直接输出acceptable，自动下一个循环         
#include<stdio.h> #include<string.h>int main(){char a[10000]={0},b[10000]={0};while(~scanf("%s",a)){    if(strcmp(a,"end")==0) break;int i,k,l,point=0,flag=1,count=0;l=strlen(a);for(i=0;i<l;i++){if(a[i]=='a'||a[i]=='e'||a[i]=='i'||a[i]=='o'||a[i]=='u'){point=1;break;}}if(point==0){printf("<%s> is not acceptable.\n",a);continue;}strcpy(b,a);for(i=0;i<l;i++){if(a[i]=='a'||a[i]=='i'||a[i]=='u')a[i]=1;else if(a[i]=='e'||a[i]=='o')a[i]=2;else a[i]=0;}for(i=0;i<l-2;i++){if((a[i]&&a[i+1]&&a[i+2])||(a[i]==0&&a[i+1]==0&&a[i+2]==0))flag=0;}if(flag==0){printf("<%s> is not acceptable.\n",b);continue;}for(i=0;i<l-1;i++){    if((b[i]==b[i+1])&&(a[i]!=2))    count=1;}if(count==1){printf("<%s> is not acceptable.\n",b);continue;}printf("<%s> is acceptable.\n",b);}}
wa了一次，是把a[]当成了没有改变的a[]，用从a[]复制过来的b[]就ac了。