回文词

来源：互联网发布：儿童学古诗软件编辑：程序博客网时间：2024/04/29 23:52

#include <stdio.h>#include <string.h>#include <ctype.h>const char* rev = "A   3  HIL JM O   2TUVWXY51SE Z  8 ";const char* msg[] = {"not a palindrome","a regular palindrome","a mirrored string","a mirrored palindrome"};//常量指针，指向的常量不能变//指针常量，int *const b = &a;指针是一个常量，指针不变char r(char ch){    if(isalpha(ch)) //isalpha(),是否为英文，大写返回2，小写返回1，不是返回0        return rev[ch - 'A']; //单个字符单引号    else        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-i-1]) p = 0;            if(r(s[i]) != s[len-1-i]) m = 0;        }        printf("%s -- is %s.\n\n", s, msg[m*2+p]); //注意！巧妙    }    return 0;}结果：NOTAPALINDROMENOTAPALINDROME -- is not a palindrome.ISAPALINILAPASIISAPALINILAPASI -- is a regular palindrome.2A3MEAS2A3MEAS -- is a mirrored string.ATOYOTAATOYOTA -- is a mirrored palindrome.

输入一个字符串，判断它是否为回文串（abba）以及镜像串（3AIAE）。

CharacterReverseCharacterReverseCharacterReverseAAMMYYB N Z5C OO11D P 2SE3Q 3EF R 4 G S25ZHHTT6 IIUU7 JLVV88K WW9 LJXX

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