D1164:Easier Done Than Said?

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描述

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.

输入

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.

输出

For each password, output whether or not it is acceptable, using the precise format shown in the example.

样例输入

a tv ptoui bontres zoggax wiinq eep houctuh end

样例输出

is acceptable.
is not acceptable.
is not acceptable.
is not acceptable.
is not acceptable.
is not acceptable.
is acceptable.
is acceptable.

思路：输入数据，每输入一个数据判断是否接受，如果有元音j++;a判断连续的元音个数，b判断连续的辅音个数，c判断除了“ee”或“oo”外是否有连续相同的两个字母，最后判断该单词是否可被接受。

代码：

#include
#include
using namespace std;
int main()
{
        char a[5]="end";
        char data[20];
        while(cin>>data)
        {
                if(strcmp(data,a)==0)
                        return 0;
                int i,j=0,a=0,b=0,c=0;
                int n=strlen(data);
                for(i=0;i
                {
                       if(data[i]==‘a‘||data[i]==‘e‘||data[i]==‘i‘||data[i]==‘o‘||data[i]==‘u‘)
                        {
                                j++;
                                a++;
                                b=0;
                                if(data[i]==data[i+1]&&data[i]!=‘e‘&&data[i]!=‘o‘)
                                        c++;
                        }
                        else
                        {
                                b++;
                                if(data[i]==data[i+1]&&data[i]!=‘e‘&&data[i]!=‘o‘)
                                        c++;
                                a=0;
                        }
                        if(a==3||b==3)
                                break;
                }
                if(j!=0&&c==0&&a<3&&b<3)
                        cout<<"<"<<data<<">"<<" is acceptable."<<endl;
                else
                        cout<<"<"<<data<<">"<<" is not acceptable."<<endl;        
        }
        
        return 0;
}

D1164:Easier Done Than Said?

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原文地址：http://www.cnblogs.com/Joazer/p/5239827.html