Easier Done Than Said?
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
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.
tv
ptoui
bontres
zoggax
wiinq
eep
houctuh
end
<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.
#include<stdio.h>
#include<string.h>
int main()
{
int i,flag,len,vowel;
char str[25];
while(gets(str))
{
if(!strcmp(str,"end")) break;
len=strlen(str);
vowel=0; flag=1;
for(i=0;i<=1;i++) /*特判str[0]和str[1]*/
if(str[i]=='a'||str[i]=='e'||str[i]=='i'||str[i]=='o'||str[i]=='u')
vowel++;
if(str[0]==str[1]&&str[0]!='o'&&str[0]!='e')
{
flag=0;
printf("<%s> is not acceptable.\n",str);
continue;
}
for(i=2;i<len;i++)
{
if((str[i]==str[i-1])&&(str[i]!='e'&&str[i]!='o')) /*两个连续*/
{
flag=0;
break;
}
if(str[i]=='a'||str[i]=='e'||str[i]=='i'||str[i]=='o'||str[i]=='u') /*元音字母*/
vowel++;
if((str[i]=='a'||str[i]=='e'||str[i]=='i'||str[i]=='o'||str[i]=='u')&&
(str[i-1]=='a'||str[i-1]=='e'||str[i-1]=='i'||str[i-1]=='o'||str[i-1]=='u')&&
(str[i-2]=='a'||str[i-2]=='e'||str[i-2]=='i'||str[i-2]=='o'||str[i-2]=='u')) /*三个连续元音*/
{
flag=0;
break;
}
if((str[i]!='a'&&str[i]!='e'&&str[i]!='i'&&str[i]!='o'&&str[i]!='u')&&
(str[i-1]!='a'&&str[i-1]!='e'&&str[i-1]!='i'&&str[i-1]!='o'&&str[i-1]!='u')&&
(str[i-2]!='a'&&str[i-2]!='e'&&str[i-2]!='i'&&str[i-2]!='o'&&str[i-2]!='u')) /*三个连续辅音*/
{
flag=0;
break;
}
}
if(flag&&vowel>0)
printf("<%s> is acceptable.\n",str);
else
printf("<%s> is not acceptable.\n",str);
}
return 0;
}