题目链接：https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=342
题目意思：给出一段字符串（大写字母+数字组成）。判断是否为回文串 or 镜像串 or 回文镜像串 or 什么都不是。每个字母的镜像表格如下
| 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 |
注意是没有数字0的哦。（该题，数字 0 与字母 O 看成是一样的）
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
using namespace std; const int maxn = + ;
char mirror[] = "A 3 HIL JM O 2TUVWXY51SE Z 8 ";
const char *msg[] = {" -- is not a palindrome.", " -- is a regular palindrome.", " -- is a mirrored string.", " -- is a mirrored palindrome."};
char s[maxn]; char change(char ch)
{
if (ch >= 'A' && ch <= 'Z')
return mirror[ch-'A'];
return mirror[ch-''+];
} int main()
{
#ifndef ONLINE_JUDGE
freopen("in.txt", "r", stdin);
#endif // ONLINE_JUDGE while (scanf("%s", s) != EOF) {
int len = strlen(s);
int p = , m = ; for (int i = ; i < len; i++) {
if (s[i] != s[len-i-]) p = ;
if (change(s[i]) != s[len-i-]) m = ;
}
printf("%s%s\n\n", s, msg[p+m*]);
}
return ;
}