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。

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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;
}




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