/*
Name: Palindromes
Copyright: UVaOJ-401
Author: LHD
Date: 17/11/13 09:59
Description: 字符串处理
可以编写两个函数：Mirror(判断是否为镜像) Pal(判断是否为回文)
*/
#include <stdio.h>
#include <string.h>
#define MAXN 2000
bool Pal(char a[MAXN])
{
int i,lengtha;
lengtha=strlen(a)-1;
for(i=0;i<=lengtha;i++)
if(a[i]!=a[lengtha-i]) return false;
return true;
}
bool Mirror(char a[MAXN])
{
int i,lengtha;
lengtha=strlen(a)-1;
for(i=0;i<=lengtha;i++)
if(a[i]!='A' && a[i]!='3' && a[i]!='H' && a[i]!='I' && a[i]!='L' && a[i]!='J' && a[i]!='M' && a[i]!='O' && a[i]!='2' && a[i]!='T' && a[i]!='U' && a[i]!='V' && a[i]!='W' && a[i]!='X' && a[i]!='Y' && a[i]!='5' && a[i]!='1' && a[i]!='S' && a[i]!='E' && a[i]!='Z' && a[i]!='8')
return false;
for(i=0;i<=lengtha;i++)
if(a[i]=='E'||a[i]=='J'||a[i]=='L'||a[i]=='S'||a[i]=='Z'||a[i]=='2'||a[i]=='3'||a[i]=='5'||a[i]!=a[lengtha-i])
{
switch(a[i])
{
case 'E':if(a[lengtha-i]!='3') return false;break;
case 'J':if(a[lengtha-i]!='L') return false;break;
case 'L':if(a[lengtha-i]!='J') return false;break;
case 'S':if(a[lengtha-i]!='2') return false;break;
case 'Z':if(a[lengtha-i]!='5') return false;break;
case '2':if(a[lengtha-i]!='S') return false;break;
case '3':if(a[lengtha-i]!='E') return false;break;
case '5':if(a[lengtha-i]!='Z') return false;break;
default:return false;
}
}
return true;
}
int main()
{
char a[MAXN]="0";
int i;
while(scanf("%s",a)!=NULL)
{
for(i=0;i<strlen(a);i++) //把'0'转换成'O'
if(a[i]=='0')
a[i]='O';
if(Pal(a))
{
if(Mirror(a))
printf("%s -- is a mirrored palindrome.\n",a);
else
printf("%s -- is a regular palindrome.\n",a);
}
else if(Mirror(a))
printf("%s -- is a mirrored string.\n",a);
else printf("%s -- is not a palindrome.\n",a);
printf("\n");
}
return 0;
}
