回文词

Posted 2020-09-29 Intro1997

题目来自于刘汝佳的《算法竞赛入门经典（第二版）》

题目描述：
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，就是反转以后跟原串相同，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。在本题中，每个字符的镜像如图所示

　　输入的每行包含一个字符串（保证只有上述字符，不含空白字符），判断它是否为回文串和镜像串（共四种组合）。每组数据之后输出一个空行。

　　样例输入：

　　NOTAPALINDROME

　　ISAPALINILAPASI

　　2A3MEAS

　　ATOYOTA

　　样例输出：
　　NOTAPALINDROME -- is not a palindrome.

　　ISAPALINILAPASI -- is a regular palindrome.

　　2A3MEAS -- is a mirrored string.

　　ATOYOTA -- is a mirrored palindrome.

我的代码：

#include<iostream>
#include<cstring>
#include<ctype.h>
using namespace std;

int main() {
    char a[100], chat[]="A   3  HIL J O   2TUVWX51SE Z  8 ";
    int i, j, note = 0, no = 0;
    while (cin >> a)
    {
        note = 0;
        for (i = 0; i < strlen(a) / 2; i++) 
        {
            if (a[i] == a[strlen(a) - 1 - i]) 
            {
                if (isalpha(a[i]) && chat[a[i] - 65] != a[strlen(a) - 1 - i])
                {
                    no = 1;
                    break;
                }
            }
            if (a[i] != a[strlen(a) - 1 - i])
            {
                if (isalpha(a[i]) && chat[a[i] - 65] != a[strlen(a) - 1 - i])
                {
                    no = 1;
                    break;
                }
                note = 1;
                break;
            }
        }
        if (note == 0 && no == 0)
        {
            cout << a << "-- is a mirrored palindrome.\\n";
            continue;
        }
        if (note == 1)
        {
            cout << a << " -- is not a palindrome.\\n";
            continue;
        }
        if (note == 0)
        {
            cout << a << " -- is a regular palindrome.\\n";
            continue;
        }
        if (no == 0)
        {
            cout << a << "-- is a mirrored string.\\n";
            continue;
        }
        if (no == 1)
        {
            cout << a << "-- is not a mirrored string.\\n";
            continue;
        }
    }
    return 0;
}

答案的代码：

#include<stdio.h>
#include<string.h>
#include<ctype.h>
const char* rev = "A   3  HIL J O   2TUVWX51SE 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\'];
    return rev[ch - \'0\' + 25];
}
int main() {
    char s[30];
    while (scanf("%s", s) == 1) {
        int len = strlen(s);
        int p = 1, m = 1;
        for (int i = 0; i < (len + 1) / 2; i++) {
            if (s[i] != s[len - 1 - i])p = 0;
            if (r(s[i]) != s[len - 1 - i]) m = 0;
        }
        printf("%s -- is %s.\\n\\n", s, msg[m * 2 + p]);
    }
    return 0;
}