UVA - 401 Palindromes

题目大意：给出一个字符串，判断他是否镜像和回文

解题思路：设置数组判断是否镜像，根据相等判断是否相等

#include<cstdio>#include<cstring>char m1[36]="ABCDEFGHIJKLMNOPQRSTUVWXYZ123456789";char m2[36]="A   3  HIL JM O   2TUVWXY51SE Z  8 ";int judge_p(char *str, int len) {if( len == 1)//如果长度为1，表示回文return 1;for( int i = 0; i < len / 2; i++)//按照对称进行比较，相等继续，不等返回if(str[i] == str[len - 1 - i])continue;else return 0;return 1;//如果全部符合则返回1}int judge_m(char *str, int len) {int j;int mark = 0;if(len == 1) {for(int i = 0; i < 36; i++)if(str[0] == m2[i] )return 1;return 0;}if(len % 2 == 0)mark = 1;elsemark = 0;char t = str[len / 2];if(mark == 0){if(t == 'A' || t == 'H' || t == 'I' || t == 'M' || t == 'O' || t == 'T' || t == 'U' || t == 'V' || t == 'W'|| t == 'X' || t == 'Y' || t == '1' || t == '8') ;elsereturn 0;}for(int i = 0; i < len / 2; i++) {for( j = 0; j < 36; j++)if(str[i] == m1[j] && str[len - i - 1] == m2[j]) break;if(j == 36)return 0;}return 1;}/*int judge_m(char *str, int len) {int mark ;for(int i = 0; i < len / 2; i++) {switch(str[i]) {case 'A': if(str[len - 1 - i] == 'A')  mark = 0;  else  mark = 1;  break;case 'E': if(str[len - 1 - i] == '3')  mark = 0;  else  mark = 1;  break;case 'H': if(str[len - 1 - i] == 'H')  mark = 0;  else  mark = 1;  break;case 'I': if(str[len - 1 - i] == 'I')  mark = 0;  else  mark = 1;  break;case 'J': if(str[len - 1 - i] == 'L')  mark = 0;  else  mark = 1;  break;case 'L': if(str[len - 1 - i] == 'J')  mark = 0;  else  mark = 1;  break;case 'M': if(str[len - 1 - i] == 'M')  mark = 0;  else  mark = 1;  break;case 'O': if(str[len - 1 - i] == 'O')  mark = 0;  else  mark = 1;  break;case 'S': if(str[len - 1 - i] == '2')  mark = 0;  else  mark = 1;  break;case 'T': if(str[len - 1 - i] == 'T')  mark = 0;  else  mark = 1;  break;case 'U': if(str[len - 1 - i] == 'U')  mark = 0;  else  mark = 1;  break;case 'V': if(str[len - 1 - i] == 'V')  mark = 0;  else  mark = 1;  break;case 'W': if(str[len - 1 - i] == 'W')  mark = 0;  else  mark = 1;  break;case 'X': if(str[len - 1 - i] == 'X')  mark = 0;  else  mark = 1;  break;case 'Y': if(str[len - 1 - i] == 'Y')  mark = 0;  else  mark = 1;  break;case 'Z': if(str[len - 1 - i] == '5')  mark = 0;  else  mark = 1;  break;case '1': if(str[len - 1 - i] == '1')  mark = 0;  else  mark = 1;  break;case '2': if(str[len - 1 - i] == 'S')  mark = 0;  else  mark = 1;  break;case '3': if(str[len - 1 - i] == 'E')  mark = 0;  else  mark = 1;  break;case '5': if(str[len - 1 - i] == 'Z')  mark = 0;  else  mark = 1;  break;case '8': if(str[len - 1 - i] == '8')  mark = 0;  else  mark = 1;  break;deafult: mark = 1;}if(mark == 1) //如果发现一个不回文，则返回领return 0;}return 1;//如果全部检测完了表示过了.返回1} */int main() {char str[50];int m;int p;int len;char str2[50];while(gets(str)) {len = strlen(str);strcpy(str2,str);for(int i = 0; i < len; i++) if(str[i] == '0')str[i] = 'O';m = judge_m(str,len);p = judge_p(str,len);if(m == 0 && p == 0)printf("%s -- is not a palindrome.\n\n", str2);if(m == 0 && p == 1)printf("%s -- is a regular palindrome.\n\n", str2);if(m == 1 && p == 0)printf("%s -- is a mirrored string.\n\n", str2);if(m == 1 && p == 1)printf("%s -- is a mirrored palindrome.\n\n",str2);memset(str,'\0',sizeof(str));}return 0;}