uvaoj401.2

第二種方法，AC，時間勉強過得去 runtime 16ms 有交流的同仁請留言！

#include<stdio.h>
#include<string.h>
char a[25]; 
char b[100];
char characterArr[40]="ABCDEFGHIJKLMNOPQRSTUVWXYZ123456789";
char   reverseArr[40]="A   3  HIL JM O   2TUVWXY51SE Z  8 ";

int Is_paLindrome(char *s)
{
	int i;
	int n = strlen(s);
	for(i = 0; i <= n/2-1; i++)
	{  
		if(s[i] != s[n-1-i]) 
		{
			return 0; 
		}
	}
	return 1;
}

int Is_mirrored(char *s)
{
	int n = strlen(s);
	int l,j,h,k;
	int m = strlen(characterArr);
	for(l = 0;l <= n-1; l++)
	{ 
		a[l] = s[l];
	}
	for(j = 0; j <= n-1; j++)
	{  
		for(k = 0; k <= m-1; k++) 
	if(a[j] == characterArr[k])  
	{    
		  a[j] = reverseArr[k];   
		  break;   
	 }
	}
	for(h = 0; h <= n/2; h++)     
		if(s[h] != a[n-1-h])   
			return 0;
	return 1;
}

int main()
{
	while(gets(b))
{
  if(Is_paLindrome(b))
  {
   if(!Is_mirrored(b))
    printf("%s -- is a regular palindrome.\n",b);
   else
    printf("%s -- is a mirrored palindrome.\n",b);
  }
  else 
  {
  if(!Is_mirrored(b))
   printf("%s -- is not a palindrome.\n",b);
  else
   printf("%s -- is a mirrored string.\n",b);
  }
  printf("\n");
  memset(b,0,sizeof(b));
 }

 return 0;
}