HDU 1039Problem DescriptionPassword security is a tricky thi

HDU 1039
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
is acceptable.
is not acceptable.
is not acceptable.
is not acceptable.
is not acceptable.
is not acceptable.
is acceptable.
is acceptable.
我的代码：
#include
#include
using namespace std;
char str[32];
int r[256]={0},len,i,sv,f;
int main(){
r['a']=r['e']=r['i']=r['o']=r['u']=1;
while(scanf("%s",str)!=EOF){
len=strlen(str);sv=0;
if(3==len&&'e'==str[0]&&'n'==str[1]&&'d'==str[2]) break;
for(i=0,f=0;i0){
if(i>1&&r[str[i-2]]==r[str[i-1]]&&r[str[i-1]]==r[str[i]]){
f++;break;
}
if(str[i-1]==str[i]&&'e'!=str[i]&&'u'!=str[i]){
f++;break;
}
}
}
i==len&&!sv?f++:0;
printf(" is ",str);
puts(f?"not acceptable.":"acceptable.");
}
return 0;
}

这个程序对三个条件的检查应该分开独立进行,这样就不容易出错了,提供一个AC的程序,可以对照一下：
#include
#include
int isvowel(const char a) /*检查 a 字符是不是元音字母*/
{
return (a=='a'||a=='e'||a=='i'||a=='o'||a=='u');
}
int chk1(const char *a) /*检查是否满足第一个条件,即至少有一个元音字母*/
{
int i;
for (i = 0; a[i]; i++)
if (isvowel(a[i]))
return 1;
return 0;
}
int chk2(const char *a) /*检查是否满足第二个条件,即不能有三个连续的元音字母或者三个连续的铺音字母*/
{
unsigned int i;
if (strlen(a)