Problem   Three

palindromes

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 elementsof the string is changed to its reverse (if it has a reverse) andthe string is read backwards the result is the same as the originalstring. For example, the string "3AIAE" is a mirroredstring because "A" and "I" are their ownreverses, and "3" and "E"are each others‘reverses.

A mirrored palindrome is a string that meets the criteria of aregular palindrome and the criteria of a mirrored string. Thestring "ATOYOTA" is a mirrored palindrome because if thestring is read backwards, the string is the same as the originaland because if each of the characters is replaced by its reverseand the result is read backwards, the result is the same as theoriginal string. Of course, "A", "T","O", and "Y"are all their own reverses.

A list of all valid characters and their reverses is asfollows.

Note that O (zero) and 0 (the letter) areconsidered the same character and thereforeONLYthe letter "0" is a valid character.

Input 

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

Output 

For each input string, you should print the string starting incolumn 1 immediately followed by exactly one of the followingstrings.

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

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

SampleInput 

NOTAPALINDROME ISAPALINILAPASI 2A3MEAS ATOYOTA

Sample Output 

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


CODE
      
#include"stdio.h"
#include

char const jxhw[]="A   3  HIL JM O   2TUVWXY51SE Z  8 ";

char jh(char ch)
{
    char a;
    a=ch-‘A‘;
    if(a>=0)
        return jxhw[ch-‘A‘];
    return jxhw[ch-‘0‘+25];
}

int main()
{
    char inp[99];
    while(scanf("%s",inp)!=EOF)
    {
        int len=strlen(inp);
        int i,jingxiang=1,huiwen=1;
        for(i=0; i<(len+1)/2; i++)
        {
            if(inp[i]!=inp[len-i-1])
                huiwen=0;
            if(jh(inp[i])!=inp[len-i-1])
                jingxiang=0;
        }
        if(jingxiang==0&&huiwen==0)
            printf("%s -- is not a palindrome.\n\n",inp);
        if(jingxiang==0&&huiwen==1)
            printf("%s -- is a regular palindrome.\n\n",inp);
        if(jingxiang==1&&huiwen==0)
            printf("%s -- is a mirrored string.\n\n",inp);
        if(jingxiang==1&&huiwen==1)
            printf("%s -- is a mirrored palindrome.\n\n",inp);
    }
    return 0;
}



My Way
    这道题想了挺长时间的，在最初的时候，我没有头绪。后来经过一番思考，理清头绪后，开始完成这道题目。
    首先，用const强制定义了一个数组，将所用的镜像全部都列了出来（其实不用const也可以，保险起见）。程序开始，定义了一个足够长的数组，用来储存输入。用while与EOF搭配来循环输入。然后测定输入字符串的长度，通过长度来计算for循环中所需计算的次数。在循环之前，定义了jingxiang和huiwen两个变量都为一（是）。在循环中，两个if分别是用来判断是否是回文或者是镜像的。最后，通过判断jingxiang和huiwen的是非便可以分别出是否是镜像还是回文，亦或是两者都是。
    题目不难，但是由于不能熟练的应用知识，导致自己做了很长时间。看看那些秒AC的大神，自己要走的路还有很长，很长。

Problem 003——palindromes