题目：与以往的回文不同，只判断字母是不是构成回文，不是字母的一律当作不存在。
Sample Input
radar
rAd. Ar
radiar
able was I ere I saw elba!
b
bb
ab
aabaa
aabbaa
aab2aa
aabaaa
This is a palindrome
radar is a palindrome
a man, a plan, a canal, panama!
a bell was eye, ear I saw elbow
go hang a salami, bub, i’m a lasagna hog!!
 
Sample Output
YES
YES
NO
YES
YES
YES
NO
YES
YES
YES
NO
NO
NO
YES
NO
YES
 

代码，用指针实现
#include "stdio.h"
#include "string.h"
int main()
{
    int i;
    char a[101];
    char *p,*q;
  while(gets(a)!=EOF)   
   { i=strlen(a);
    p=&a[0];
    q=&a[i-1];     
    while(p<q)
    {
      
      if(*p==*q)
     {
        p++;
        q--;        
      }
   
     else if(!((*p<='A'&&*p>='Z')||(*p>='a'&&*p<='z') ) )   *p++;
       else if(!((*q>='A'&&*q<='Z')||(*q>='a'&&*q<='z') ) ) *q--;
        else  {
                  printf("N0\n");
                  break;
                }   
    }
    if(p>=q) printf("YES\n");
  } return 0;
}

不是这个问题，我改了交过，也超时。