Extra Krunch

我会带着你远行 2022-06-15 22:53 48阅读 0赞

Description

A krunched word has no vowels ("A", "E", "I", "O", and "U") and no repeated letters. Removing vowels and letters that appear twice or more from MISSISSIPPI yields MSP. In a krunched word, a letter appears only once, the first time it would appear in the unkrunched word. Vowels never appear.   
  
Krunched phrases similarly have no vowels and no repeated letters. Consider this phrase:

RAILROAD CROSSING

and its krunched version:

RLD CSNG

Blanks are krunched differently. Blanks are removed so that a krunched phrase has no blanks on its beginning or end, never has two blanks in a row, and has no blanks before punctuation. Otherwise, blanks not removed. If we represent blanks by "\_",

MADAM_I_SAY_I_AM_ADAM__

krunches to:

MD_SY

where the single remaining blank is shown by "\_".   
  
Write a program that reads a line of input (whose length ranges from 2 to 70 characters), and krunches it. Put the krunched word or phrase in the output file. The input line has only capital letters, blanks, and the standard punctuation marks: period, comma, and question mark.

Input

A single line to be krunched.

Output

A single krunched line that follows the rules above.

题意：给出一个只含大写字母，空格 ，逗号，句号，问号的字符串，最后输出的字符串首尾不能有空格，不能有连续的空格，不能有元音字母，字母只能出现一次，标点符号前面不可以空格

Sample Input

NOW IS THE TIME FOR ALL GOOD MEN TO COME TO THE AID OF THEIR COUNTRY.

Sample Output

NW S TH M FR L GD C Y.

#include<iostream>
    #include<algorithm>
    #include<string>
    #include<cstdio>
    #include<cstring>
    using namespace std;
    int main()
    {
        bool vis[26];
        char ch[75],res[75];
        int i,cnt;
        while(gets(ch))
        {
            cnt=0;
            memset(vis,0,sizeof(vis));
            vis['A'-'A']=true;
            vis['E'-'A']=true;
            vis['I'-'A']=true;
            vis['O'-'A']=true;
            vis['U'-'A']=true;
            for(int i=0;ch[i]!='\0';i++)
            {
                if(isalpha(ch[i]))
                {
                    if(!vis[ch[i]-'A'])
                    {
                        res[cnt++]=ch[i];
                        vis[ch[i]-'A']=true;
                    }
                }
                else if(ch[i]==','||ch[i]=='.'||ch[i]=='?')
                {
                    if(res[cnt-1]==' ')
                        res[cnt-1]=ch[i];
                    else
                        res[cnt++]=ch[i];
                }
                else if(res[cnt-1]!=' '&&cnt!=0)
                {
                    if(ch[i]==' ')
                    {
                        res[cnt++]=ch[i];
                    }
                }
            }
            res[cnt]='\0';
            printf("%s\n",res);
        }
        return 0;
    }