UVaOJ 401 - Palindromes

//假如题目说字符串的字符有20个，就至少要预多两个空位，一个留给'\n'，一个留给'\0'#include <stdio.h>#include <string.h>#include <ctype.h>#include <math.h>int main(){char ct[26] = { 'A', '\0', '\0', '\0', '3', '\0', '\0', 'H', 'I', 'L', '\0', 'J', 'M','\0', 'O', '\0', '\0', '\0', '2', 'T', 'U', 'V', 'W', 'X', 'Y', '2' };char nt[10] = { 'O', '1', 'S', 'E', '\0', 'Z', '\0', '\0', '8', '\0' };char buf[22];memset(buf, '\0', sizeof(buf));while (fgets(buf, sizeof(buf), stdin) != NULL){int len = strlen(buf) - 1;buf[len] = '\0';bool isMirrored = true;bool isPalindrome = true;if (len == 1)//一开始忘记处理只有一个字符的字符串了{if (isdigit(buf[0]) && buf[0] != nt[buf[0] - '0'])isMirrored = false;else if (!isdigit(buf[0]) && buf[0] != ct[buf[0] - 'A'])isMirrored = false;}for (int i = 0; i < len / 2; ++i){if (!isPalindrome && !isMirrored)break;if (isPalindrome && buf[i] != buf[len - i - 1])//一开始没有加isPalindrome && isPalindrome = false;//下面一行一开始写为isdigit(buf[i])，后来发现我要转换的是buf[len - i - 1]，才发现我自己逻辑混乱了……if (isMirrored && isdigit(buf[len - i - 1]) && buf[i] != nt[buf[len - i - 1] - '0'])isMirrored = false;//下面一行代码一开始也忘记加!isdigit(buf[len - i - 1]) &&，导致随意进入这个分支，导致错误else if (isMirrored && !isdigit(buf[len - i - 1]) && buf[i] != ct[buf[len - i - 1] - 'A'])isMirrored = false;if (isMirrored && i == len / 2 - 1 && len % 2 != 0)if (isdigit(buf[i + 1]) && buf[i + 1] != nt[buf[i + 1] - '0'])isMirrored = false;else if (buf[i + 1] != ct[buf[i + 1] - 'A'])isMirrored = false;}if (isPalindrome && isMirrored)printf("%s -- is a mirrored palindrome.\n\n", buf);else if (isPalindrome)printf("%s -- is a regular palindrome.\n\n", buf);else if (isMirrored)printf("%s -- is a mirrored string.\n\n", buf);elseprintf("%s -- is not a palindrome.\n\n", buf);memset(buf, '\0', sizeof(buf));}return 0;}//总体来说就是我自己的逻辑思维不够严密，编程习惯不够好，要慢慢改进