学步园

UVa OJ

A regular palindrome is a string of numbers or letters that is the same forward as backward. For example, the string "ABCDEDCBA" is a palindrome because it is the same when the string is read from left
to right as when the string is read from right to left.

A mirrored string is a string for which when each of the elements of the string is changed to its reverse (if it has a reverse) and the string is read backwards the result is the same as the original string. For example, the string "3AIAE" is a mirrored
string because "A" and "I" are their own reverses, and "3" and "E"are each others' reverses.

A mirrored palindrome is a string that meets the criteria of a regular palindrome and the criteria of a mirrored string. The string "ATOYOTA" is a mirrored palindrome because if the string is read backwards, the string is the same as the original and
because if each of the characters is replaced by its reverse and the result is read backwards, the result is the same as the original string. Of course, "A", "T", "O", and "Y"are all their own reverses.

A list of all valid characters and their reverses is as follows.

Note that O (zero) and 0 (the letter) are considered the same character and therefore ONLY the letter "0" is a valid character.

Input consists of strings (one per line) each of which will consist of one to twenty valid characters. There will be no invalid characters in any of the strings. Your program should read to the end of file.

For each input string, you should print the string starting in column 1 immediately followed by exactly one of the following strings.

Note that the output line is to include the -'s and spacing exactly as shown in the table above and demonstrated in the Sample Output below.

In addition, after each output line, you must print an empty line.

INPUT

NOTAPALINDROME 
ISAPALINILAPASI 
2A3MEAS 
ATOYOTA

OUTPUT

NOTAPALINDROME -- is not a palindrome.
 
ISAPALINILAPASI -- is a regular palindrome.
 
2A3MEAS -- is a mirrored string.
 
ATOYOTA -- is a mirrored palindrome.

这道题主要分为两个部分。

一个部分是字符串是否为回文字符串。比较简单

还有一个部分是判断是否为mirrored string 这其实是在回文的基础上有了新的对称规则。

那么，首先就要判断该字符是否为有对称的字符，然后再看相对应位置的字符是否为mirrored alpha。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
char str[30],
mirror[40][2]={'A','A','E','3','H','H','I','I','J','L','L','J','M','M',
               'O','O','S','2','T','T','U','U','V','V','W','W','X','X',
               'Y','Y','Z','5','1','1','2','S','3','E','5','Z','8','8'};
using namespace std;
int rev()
{
    int i,j,len;
    len=strlen(str);
    for (i=0,j=len-1; i<j; i++,j--)
        if (str[i]!=str[j])
            return 0;
    return 1;
}
int mirr()
{
    int i,j,len,k,p=0;
    len=strlen(str);
    for (i=0,j=len-1; i<=j; i++,j--) //i必须小于等于j 要不然在判断mirror的时候，如果串为基数，则最中间不判了
    {
        p=0;
        for (k=0; k<30; k++)
            if (str[i]==mirror[k][0])
            {
                p=1;
                if (str[j]!=mirror[k][1])
                    return 0;
            }
        if (p==0) return 0;
    }
    return 1;
}
int main ()
{
    int k,p;
    while(cin>>str)
    {
        k=rev();
        p=mirr();
        if (k==1 && p==1) cout<<str<<" -- is a mirrored palindrome."<<endl;
        if (k==1 && p!=1) cout<<str<<" -- is a regular palindrome."<<endl;
        if (k!=1 && p==1) cout<<str<<" -- is a mirrored string."<<endl;
        if (k!=1 && p!=1) cout<<str<<" -- is not a palindrome."<<endl;
        cout<<endl;
    }
    return 0;
}