UVa 401

题意

输入一个字符串，判断它是否为回文串以及镜像串。输入字符串保证不含数字0。所谓回文串，就是反转以后和原串相同，如abba和madam。所有镜像串，就是左右镜像之后和原串相同，如2S和3AIAE。注意，并不是每个字符在镜像之后都能得到一个合法字符。

思路

水题
用字符串记录一下镜像表即可

但是要注意几个细节:
1. 当已经判断某串为镜像/镜像回文时 , 如果串长为奇数, 要特判最中间元素是否镜像, 如果最中间元素不镜像那么这就是一个普通回文串
2. 特判串长为1的 , 如果该元素镜像则是镜像回文, 否则只是回文

AC代码

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

const int maxn = 30;
char s[maxn];

char ms[] = " SE Z               3    L J      2      5";
char mp[] = "1      8        A      HI   M O    TUVWXY ";

int main()
{
    int len;
    while( ~scanf("%s",s) )
    {
        len = strlen(s);
        bool MS = false, MP = false, OR = false;
        if( len == 1 ){
            if( mp[s[0]-'1'] != ' ' )   MP = true;
            else OR = true;
        }
        for( int i = 0; i < len/2; i++ ){
            if( s[i] == s[len-i-1]){
                if( mp[s[i]-'1'] != ' ' ){
                    MP = true;
                    if(MS) MP = false;
                    if(OR) MP = false;
                }
                else{
                    OR = true;
                    MP = false;
                }
            }
            else{
                if( ms[s[i]-'1'] == s[len-i-1] ){
                    MS = true;
                    if(MP) MP = false;
                    if(OR) MS = false;
                }
                else{
                    OR = false;
                    break;
                }
            }
            //printf("%d & %d
",i,len-i-1);
            //printf("MP:%d   MS:%d   OR:%d
",MP,MS,OR);
        }
        if( len % 2 != 0 ){
            if( MP && mp[s[len/2]-'1'] == ' ' ){
                MP = false;
                OR = true;
            }
            else if( MS && mp[s[len/2]-'1'] == ' ' ){
                MS = false;
                OR = false;
            }
        }
        if( MS )     cout << s << " -- is a mirrored string." << endl << endl;
        else if( MP )     cout << s << " -- is a mirrored palindrome." << endl << endl;
        else if( OR )    cout << s << " -- is a regular palindrome." << endl << endl;
        else     cout << s << " -- is not a palindrome." << endl << endl;
    }
    return 0;
}