UVa-401 Palindromes回文词——水

Palindromes

Description

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.

Hint

use the C++'s class of string will be convenient, but not a must

Source

Root :: Competitive Programming: Increasing the Lower Bound of Programming Contests (Steven & Felix Halim) :: Chapter 6. String Processing ::   Ad Hoc String Processing
Root :: AOAPC I: Beginning Algorithm Contests (Rujia Liu) :: Volume 1. Elementary Problem Solving ::   String
Root :: AOAPC II: Beginning Algorithm Contests (Second Edition) (Rujia Liu) :: Chapter 3. Arrays and Strings ::   Examples
Root :: Competitive Programming 3: The New Lower Bound of Programming Contests (Steven & Felix Halim) :: Introduction :: Ad Hoc Problems ::   Palindrome
Root :: Competitive Programming 2: This increases the lower bound of Programming Contests. Again (Steven & Felix Halim) :: Introduction :: Ad Hoc Problems - Part 1 ::   Palindrome / Anagram

Submit Status

水题，要点是常量数组的应用。
贴下自己的代码。

#include<iostream>
#include<cctype>
#include<cstring>
using namespace std;

char zmrev[99] = "A***3**HIL*JM*O***2TUVWXY5";          //常数数组有时可以极大的简化代码。
char surev[99] = "1SE*Z**8*";
char s[1000];

char diu(char c)
{
	if (c >= 'A'&&c <= 'Z') return zmrev[c - 'A'];      //diu函数返回处理过后的字符串，便于后面判断。
	if (c >= '1'&&c <= '9') return surev[c - '1'];
}

int main()
{
	while (cin >> s)
	{
		int len = strlen(s);
		int ispalindrome = 1;
		int ismirrored = 1;
		for (int i = 0; i<=(len/ 2); i++)          //要注意是等于号，因为开始没注意wrong了几次。
		{
			if (s[i] != s[len - i - 1]) ispalindrome = 0;
			if (s[i] != diu(s[len - i - 1])) ismirrored = 0;
		}
		
		if (ispalindrome == 0 && ismirrored == 0) cout << s << " -- is not a palindrome." << endl;
		else if (ispalindrome == 1 && ismirrored == 0) cout << s << " -- is a regular palindrome." << endl;
		else if (ispalindrome == 0 && ismirrored == 1) cout << s << " -- is a mirrored string." << endl;
		else if (ispalindrome == 1 && ismirrored == 1) cout << s << " -- is a mirrored palindrome." << endl;
		cout << endl;                           //格式。
	}
}