2016-07-12

UVA - 401 Palindromes

题目大意：判断是否回文、镜面。回文即从左往右与从右往左相同。镜面根据所提供的表格，可能存在非法字符。一开始理解错题意，以为不存在非法字符就是镜面，实际上是从左往右根据表格将全部字符化为镜像后，要与从右往左相同。

解题思路：遍历每个元素，进行回文的判断，是回文的话tag[0] = 1 ,不是的话tag[0] = 0 ，是回文的话，对回文的前半部分进行判断，判断是否是镜像回文，不是回文的话，对字符串前半部分进行判断，判断是否是镜像字符串。

注意：对字符串的中间部分要进行判断。

#include <iostream>#include <cstring>using namespace std;int main () {    char str[1000];    while ( cin >> str ) {        int tag[2] = {1,0};        int len = strlen(str);        for ( int i = 0; i < len/2; i++) {            if ( str[i] != str[len-i-1] ) {                tag[0] = 0;                 break;            }            else    tag[0] = 1;            if ( tag[0] == 1 && ( str[i] == 'A' || str[i] == 'T' || str[i] == 'Y' || str[i] == 'H' || str[i] == 'I' || str[i] == 'M' || str[i] == 'O' || str[i] == 'U' || str[i] == 'V' || str[i] == 'W' || str[i] == 'X' || str[i] == '1' || str[i] == '8' ) )                tag[0] = 2;            else {                tag[0] = 1;                break;            }        }        if ( tag[0] == 2 ) {            if ( str[len/2] == 'A' || str[len/2] == 'T' || str[len/2] == 'Y' || str[len/2] == 'H' || str[len/2] == 'I' || str[len/2] == 'M' || str[len/2] == 'O' || str[len/2] == 'U' || str[len/2] == 'V' || str[len/2] == 'W' || str[len/2] == 'X' || str[len/2] == '1' || str[len/2] == '8' )                 tag[0] = 2;            else    tag[0] = 1;        }        for ( int i = 0; i < len/2; i++) {            if ( str[i] == str[len-i-1] || ( str[i] == 'E' && str[len-i-1] == '3' ) || ( str[i] == 'J' && str[len-i-1] == 'L' ) || ( str[i] == 'L' && str[len-i-1] == 'J' ) || ( str[i] == 'S' && str[len-i-1] == '2' ) || ( str[i] == 'Z' && str[len-i-1] == '5' ) || ( str[i] == '2' && str[len-i-1] == 'S' ) || ( str[i] == '3' && str[len-i-1] == 'E' ) || ( str[i] == '5' && str[len-i-1] == 'Z' ) && (tag[0] == 0) )                tag[1] = 1;            else {                tag[1] = 0;                break;            }        }        if ( len == 1 && ( str[0] == 'A' || str[0] == 'T' || str[0] == 'Y' || str[0] == 'H' || str[0] == 'I' || str[0] == 'M' || str[0] == 'O' || str[0] == 'U' || str[0] == 'V' || str[0] == 'W' || str[0] == 'X' || str[0] == '1' || str[0] == '8' ) )                tag[0] = 2;        switch ( tag[0] ) {            case 1 :    cout << str << " -- is a regular palindrome." << endl;    break;            case 2 :    cout << str << " -- is a mirrored palindrome." << endl;    break;                case 0 :    if ( tag[1] == 0 ) {                        cout << str << " -- is not a palindrome." << endl;                        break;                    }                    else {                        cout << str << " -- is a mirrored string." << endl;                        break;                    }        }                    cout << endl;        memset ( str, '\0', sizeof(str));    }    return 0;}