UVA Palindromes(模拟)

本文介绍了一种判断字符串是否为回文串或镜像串的方法,并提供了一个C语言实现的示例代码。通过对输入字符串进行逐字符对比,可以准确判断其是否同时满足回文和镜像条件。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

Palindromes
Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu

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.

CharacterReverseCharacterReverseCharacterReverse
AAMMYY
B N Z5
C OO11
D P 2S
E3Q 3E
F R 4 
G S25Z
HHTT6 
IIUU7 
JLVV88
K WW9 
LJXX  


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.

STRINGCRITERIA
" -- 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



      一共只有四种情况,1.既是回文串也是镜像串  2.是回文串但不是镜像串   3.不是回文串但是镜像串   4.既不是回文串也不是镜像串

#include<stdio.h>
#include<string.h>
#include<stdlib.h>

int main()
{
    char a[110];
    while(scanf("%s",a)!=EOF)
    {
        int n;
        n = strlen(a);
        int flag = 0;
        ///flag = 1  是回文不是镜子
        ///flag = 2  既是回文也是镜子
        ///flag = 3  不是回文是镜子
        ///flag = 4  既不是回文也不是镜子
        for(int i=0;i<(n/2)+1;i++)
        {
            if(a[i] == a[n-i-1])
            {
                if(a[i]!='A' && a[i]!='H' && a[i]!='I' && a[i]!='M' && a[i]!='O' && a[i]!='T' && a[i]!='U' && a[i]!='V' && a[i]!='W' && a[i]!='X' && a[i]!='Y' && a[i]!='1' && a[i]!='8')
                {
                    if(flag == 0 || flag == 1 || flag == 2)
                    {
                        flag = 1;///是回文串不是镜子
                    }
                    else
                    {
                        flag = 4;///既不是回文串也不是一面镜子
                        break;
                    }

                }
                else
                {
                    if(flag == 0 || flag ==2)
                    {
                        flag=2;///既是回文串也是一面镜子
                    }
                    else if(flag == 1)
                    {
                        flag == 1;
                    }
                    else if(flag == 3)
                    {
                        flag = 3;
                    }
                    else
                    {
                        flag = 4;
                        break;
                    }

                }
            }
            else
            {
                if((a[i] == 'E' && a[n-1-i] == '3') || (a[n-i-1] == 'E' && a[i] == '3') || (a[i] == 'J' && a[n-1-i] == 'L') || (a[i] == 'L' && a[n-1-i] == 'J') || (a[i] == 'S' && a[n-1-i] == '2') || (a[i] == '2' && a[n-1-i] == 'S') || (a[i]== 'Z' && a[n-1-i] == '5') || (a[i]== '5' && a[n-1-i] == 'Z'))
                {
                    if(flag == 0 || flag == 3 || flag == 2)
                    {
                        flag = 3;///不是回文串是镜子
                    }
                    else
                    {
                        flag = 4;
                        break;
                    }

                }
                else
                {
                    flag = 4;
                    break;
                }
            }
            if(flag == 4)
            {
                break;
            }
        }
        if(flag == 1)
        {
            printf("%s -- is a regular palindrome.\n\n",a);
        }
        else if(flag == 2)
        {
            printf("%s -- is a mirrored palindrome.\n\n",a);
        }
        else if(flag == 3)
        {
            printf("%s -- is a mirrored string.\n\n",a);
        }
        else if(flag == 4)
        {
            printf("%s -- is not a palindrome.\n\n",a);
        }
    }
    return 0;
}




<think>嗯,用户想了解计算机科学中的回文定义和例子。首先,我需要回忆回文的基本概念,确保定义准确。回文是指正读反读都相同的字符串,比如“madam”或“racecar”。在计算机科学中,回文常用于算法和数据结构的问题,比如判断字符串是否为回文,或者寻找最长回文子串。 接下来,用户可能需要一些具体的例子,比如常见的回文字符串,或者数字回文。例如“121”在数字中是回文。此外,还要考虑回文在计算机中的应用场景,比如数据校验、字符串处理等。 然后,用户的问题可能涉及到算法实现。比如如何用代码检测回文,可能需要比较字符串和其反转后的版本是否相同,或者使用双指针法来检查对称性。例如,Python中可以用s == s[::-1]来判断。但需要注意效率问题,特别是处理长字符串时的优化方法。 另外,用户可能对回文相关的问题感兴趣,比如最长回文子串问题,这在LeetCode等平台上常见。动态规划或者Manacher算法是常用的解决方案。这里可能需要引用到LeetCode的题目,如第5题,引用[1]提到LeetCode的简单题,虽然最长回文子串可能属于中等难度,但可以作为扩展。 还要考虑回文的其他应用,比如在数据结构中的回文链表问题,判断链表是否对称。这时通常需要找到中间节点,反转后半部分进行比较,这也是常见的面试题。 用户可能还想了解回文在计算机科学中的重要性,比如用于数据压缩、生物信息学中的DNA序列分析等。这些实际应用能帮助用户理解回文的价值。 最后,根据用户提供的引用,引用[1]提到LeetCode简单题,可能相关的问题如判断回文的算法题,而引用[2]中的代码示例虽然不直接相关,但可以忽略,因为用户明确询问回文。需要确保回答中的引用正确对应,比如算法题可能来自LeetCode,所以引用[1]合适。 总结,回答应包括回文的定义、例子、检测方法、相关算法问题以及应用场景,同时生成相关问题引导进一步学习。需要检查LaTeX格式是否符合要求,比如变量用$...$,独立公式用$$...$$,并确保中文回答结构清晰。</think>### 回文的定义与示例 在计算机科学中,**回文(Palindrome)**指正向和反向读取完全相同的字符串、数字序列或其他数据结构。例如: - 字符串:$s = \text{"madam"}$,$s_{\text{reverse}} = \text{"madam"}$; - 数字:$n = 121$,$n_{\text{reverse}} = 121$; - 链表:节点值按顺序排列对称(如 $1 \rightarrow 2 \rightarrow 3 \rightarrow 2 \rightarrow 1$)。 回文在算法设计中常用于字符串处理、数据校验和对称性分析[^1]。 --- ### 回文检测的代码示例 Python实现字符串回文检测: ```python def is_palindrome(s): s = ''.join(filter(str.isalnum, s)).lower() return s == s[::-1] ``` 此代码通过过滤非字母数字字符并统一大小写后,比较字符串与其反转结果是否一致。 --- ### 计算机科学中的回文应用 1. **数据校验**:如校验传输数据是否因错误导致非对称。 2. **字符串算法**:最长回文子串问题(LeetCode第5题)需动态规划或Manacher算法。 3. **链表对称性检测**:判断链表是否为回文结构(如LeetCode第234题)。 ---
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包

打赏作者

叶孤心丶

你的鼓励将是我创作的最大动力

¥1 ¥2 ¥4 ¥6 ¥10 ¥20
扫码支付:¥1
获取中
扫码支付

您的余额不足,请更换扫码支付或充值

打赏作者

实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值