UVaOJ 401 - Palindromes

镜像回文判断程序

最新推荐文章于 2020-02-19 00:14:29 发布

原创最新推荐文章于 2020-02-19 00:14:29 发布 · 497 阅读

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#UVaOJ401 #Palindromes

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本文介绍了一个使用C语言编写的程序，该程序能够判断输入的字符串是否为镜像回文或者普通的回文。通过定义两个字符数组来分别存放可形成镜像的字符及其对应的镜像字符，程序实现了对字符串的双向比较。

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

char mirror[22] = {'A','E','H','I','J','L','M','O','S','T','U','V','W','X','Y','Z','1','2','3','5','8','\0'};
char raw[22] = {'A','3','H','I','L','J','M','O','2','T','U','V','W','X','Y','5','1','S','E','Z','8','\0'};

int look_up(char c);
int main(){
    //freopen("in.txt","r",stdin);
    char s[30];
    while(gets(s)!=NULL){
        int len = strlen(s);
        bool isPalindrome = true;
        for(int i =0;i<=len/2;i++){
            if(s[i]!=s[len-1-i]) {isPalindrome = false;break;}
        }
        bool isMirrored = true;
        for(int i = 0;i<len;i++){
            int pos = look_up(s[i]);
            if(pos != -1){
                if(raw[pos] != s[len-1-i]){
                    isMirrored = false;
                    break;
                }
            }else{
                isMirrored = false;
                break;
            }
        }
        printf("%s -- is ",s);
        if(isPalindrome){
            if(isMirrored){
                printf("a mirrored palindrome.\n");
            }else{
                printf("a regular palindrome.\n");
            }
        }else{
            if(isMirrored){
                printf("a mirrored string.\n");
            }else{
                printf("not a palindrome.\n");
            }
        }
        printf("\n");
    }
    return 0;
}


int look_up(char c){
    for(int i = 0;i<22;i++){
        if(mirror[i] == c)
            return i;
    }
    return -1;
}