UVA401 word palindrome Palindromes

Title Description

Topic Link

Input Format

Output Format

Translation of the meaning of problems

The input is a string, the string is determined whether it is a palindromic sequence, and mirror. 00 digital input string is guaranteed to be free. The so-called palindromic sequence, that is, after reversing the original string and the same, and as abbaabba madammadam. The so-called image string, and the original image is left after the same string, such as 2S2S and 3AIAE3AIAE. Note that not every character can get a legal character after the mirror. In this problem, each mirror legal characters shown in the following table:

Character   Reverse
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
5           Z
8           8

Sample input and output

Input # 1

NOTAPALINDROME
ISAPALINILAPASI
2A3MEAS
ATOYOTA

Output # 1

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

code show as below:

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<iostream>
#include<cstdio>
using namespace std;
#pragma warning(disable:4996)
int n, l, i, a[10000];

const char* rev = "A   3  HIL JM O   2TUVWXY51SE Z  8 ";
const char* msg[] = {"not a palindrome","a regular palindrome","a mirrored string","a mirrored palindrome"};

char r(char ch) {
	if (isalpha(ch))return rev[ch-'A'];//是字母，返回反转的字母/数字
	return rev[ch-'0'+25];//如果是数字，返回字母（字母26+数字9）
}

int main()
{
	char s[30];
	while (scanf("%s",s)==1)
	{
		int len = strlen(s);//计算长度，二分，左边右边相比较
		int p = 1,m = 1;//标志
		for (int i=0;i<(len+1)/2;i++)//左右相比
		{
			if (s[i] != s[len - 1 - i])p = 0;//不是回文串
			if (r(s[i]) != s[len - 1 - i])m = 0;//不是镜像串
		}
		printf("%s -- is %s.\n\n",s,msg[m*2+p]);//按格式打印

	}
	return 0;
}