uva401 - Palindromes

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 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.

 

Character	Reverse	Character	Reverse	Character	Reverse
A	A	M	M	Y	Y
B		N		Z	5
C		O	O	1	1
D		P		2	S
E	3	Q		3	E
F		R		4
G		S	2	5	Z
H	H	T	T	6
I	I	U	U	7
J	L	V	V	8	8
K		W	W	9
L	J	X	X

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

Input 

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.

Output 

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

 

STRING	CRITERIA
" -- is not a palindrome."	if the string is not a palindrome and is not a mirrored string
" -- is a regular palindrome."	if the string is a palindrome and is not a mirrored string
" -- is a mirrored string."	if the string is not a palindrome and is a mirrored string
" -- is a mirrored palindrome."	if the string is a palindrome and is a mirrored string

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.

Sample Input 

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.

#include<stdio.h>                     //做了一天
#include<string.h>
/*比if……else简洁多了，值得学习:
为了实现对字符串的判断，首先定义了两个字串m和n,
m存储了镜像字符，n存储了镜像后的字符。
这种方法在做字符串类的题目时比较[常见也比较实用]，[方便]了之后对字符串的操作。*/
char m[]="AEHIJLMOSTUVWXYZ12358";         
char n[]="A3HILJMO2TUVWXY51SEZ8";  
int huiwen(char a[],int len)    
{
    int i,j;
    for(i = 0,j = len-1;i <= j;)
    {
        if(a[i] == a[j])
        {
            i++;
            j--;
        }
        else                          //千万别忘了这个，不然就死循环，输不出来了
            break;
    }
    if(i > j)
        return 1;
    else
        return 0;
}
int jing(char a[],int len)           //一个函数用来判断是否回文，一个函数用来判断是否镜像。这样，显得整个程序都很简明
{
    int i,j,flag;
    for(i = 0;i < len;i++)                  //一个字母一个字母的看               方法十分巧妙，你想到了吗？
    {
        flag = 0;                         //每次循环前先让flag=0，再做判断
        for(j = 0;m[j] != '\0';j++)
        {
            if(a[i] == m[j])              //如果输入的字母中有镜像字母
            {
                if(a[len-1-i] == n[j])     //就看它对称的字母是否等于char n[]中对应的字母
                {
                    flag = 1;
                    break;                  //跳出一层循环，再看下一个字母
                }
            }
        }
        if(flag == 0)                            //如果有一个不满足，就return 0  !!      好好理解这一块，十分欠缺
            return 0;
    }
    return 1;
}
int main()
{
    char a[25];
    int len,flag1,flag2;
    while(gets(a) != 0)
    {
        len = strlen(a);
        flag1 = huiwen(a,len);
        flag2 = jing(a,len);
        if(!flag1 && !flag2)
            printf("%s -- is not a palindrome.",a);               //这样岂不是清楚多了？毫不费力，记得一定要用函数,不要都写在主函数里
        else if(flag1 && !flag2)
            printf("%s -- is a regular palindrome.",a);
        else if(!flag1 && flag2)
            printf("%s -- is a mirrored string.",a);
        else if(flag1 && flag2)
            printf("%s -- is a mirrored palindrome.",a);
        printf("\n\n");
    }
    return 0;
}