uva--401 Palindromes

题目：Palindromes

解题思路：判断是否为回文：判断字符串前n/2的第i个字符与第n-1-i个字符是否相同，都相同就是回文；判断是否为mirrored，判断字符串前n/2字符的第n-1-i个字符是否与第i个的镜像字符相同，都相同则是mirrored；要注意每个输出后面都要再打印一行空白行；

#include<stdio.h>#include<string.h>const int N = 50;char *s1 = "ABCDEFGHIJKLMNOPQRSTUVWXYZ123456789";char *s2 = "A000300HIL0JM0O0002TUVWXY51SE0Z0080";bool ispalindrome(char * s) {int i, n = strlen(s);for (i = 0;i <= n/2; i++)if (s[i] != s[n-1-i])return false;return true;}bool ismirrored (char * s) {int i,j,n = strlen(s);for (i = 0; i <= n/2; i++) {for (j = 0;s1[j] && s[i] != s1[j]; j++);if (s1[j])if (s[n-1-i] != s2[j])return false;}return true;}int main() {char s[N];bool m, p;int i,j;while (scanf("%s", &s) == 1) {p=ispalindrome(s);m=ismirrored(s);if(m && p )printf("%s -- is a mirrored palindrome.\n",s);else if (p)printf("%s -- is a regular palindrome.\n",s);else if (m)printf("%s -- is a mirrored string.\n",s);else printf("%s -- is not a palindrome.\n",s);printf("\n");}return 0;}