题目来源：http://acm.hdu.edu.cn/showproblem.php?pid=1039
Easier Done Than Said?
Time Limit: 2000/1000 MS
(Java/Others) Memory Limit: 65536/32768 K
(Java/Others)
Total Submission(s):
6855 Accepted Submission(s):
3396
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.
代码如下：
1 #include<iostream>
2 #include<stdio.h>
3 #include<string>
4 #include<string.h>
5 #include<map>
6 #include<math.h>
7 #include<algorithm>
8 #define N 25
9 using namespace std;
10
11 int main()
12 {
13 char str[N];
14 int a[N];
15 while(scanf("%s",str) && strcmp("end" , str))
16 {
17 int len=strlen(str);
18 int flag=1,sum=0; // sum 标记元音的个数，flag标记
19 int i;
20 for(i=0;i<len;i++)
21 {
22 switch(str[i])
23 {
24 case ‘a‘:
25 case ‘e‘:
26 case ‘i‘:
27 case ‘o‘:
28 case ‘u‘: a[i]=1; sum++;break;
29 default : a[i]=0;
30 }
31 if(i>0 && str[i]== str[i-1] && str[i]!= ‘e‘&& str[i]!= ‘o‘) //2个连续的相同音，排除e/o
32 {
33 flag=0;
34 break;
35 }
36 if(i>1 && a[i]== a[i-1] && a[i-1] == a[i-2] ) // 连续三个相连的元音或辅音 注意 i满足的条件，
37 {
38 flag=0;
39 break;
40 }
41 }
42 if(sum && flag)
43 printf("<%s> is acceptable.\n",str);
44 else
45 printf("<%s> is not acceptable.\n",str);
46
47 }
48 return 0 ;
49 }
hdu 1039 连续字符串处理,布布扣,bubuko.com
hdu 1039 连续字符串处理
原文：http://www.cnblogs.com/zn505119020/p/3582184.html
