HDU 1039 Easier Done Than Said?
Easier Done Than Said?
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12751 Accepted Submission(s): 6165
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.
判读条件
1:有元音字母
2:不能三个连续元音或辅音
3.不能连续两个相同的字母,除非ee或oo
下面给出AC代码:
1 #include <bits/stdc++.h> 2 using namespace std; 3 int main() 4 { 5 char str[25]; 6 int a[25]; 7 while(scanf("%s",str)&&strcmp("end",str)) 8 { 9 int len=strlen(str); 10 int flag=1; 11 int sum=0;//sum 标记元音的个数 12 for(int i=0;i<len;i++) 13 { 14 switch(str[i]) 15 { 16 case 'a': 17 case 'e': 18 case 'i': 19 case 'o': 20 case 'u':a[i]=1;sum++;break; 21 default :a[i]=0; 22 } 23 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 24 { 25 flag=0; 26 break; 27 } 28 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, 29 { 30 flag=0; 31 break; 32 } 33 } 34 if(sum&&flag) 35 printf("<%s> is acceptable.\n",str); 36 else printf("<%s> is not acceptable.\n",str); 37 } 38 return 0; 39 }
作 者:Angel_Kitty
出 处:https://www.cnblogs.com/ECJTUACM-873284962/
关于作者:阿里云ACE,目前主要研究方向是Web安全漏洞以及反序列化。如有问题或建议,请多多赐教!
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