POJ 1590解题报告

这道题按照题意处理就可以了。似乎一开始就想得和题目要求比较一致（比如字符串末端的空格不算，奇数长度的字符串中间的那个字符如果不能reverse的话字符串不算mirrored）。bug之处在于构建那个reverse表的时候少输入了个Y。还好样例中有Y。。。

代码如下：

#include <iostream>#include <map>#include <string>using namespace std;int main(){map<char, char> reverses;reverses['A'] = 'A';reverses['E'] = '3';reverses['H'] = 'H';reverses['I'] = 'I';reverses['J'] = 'L';reverses['L'] = 'J';reverses['M'] = 'M';reverses['O'] = 'O';reverses['S'] = '2';reverses['T'] = 'T';reverses['U'] = 'U';reverses['V'] = 'V';reverses['W'] = 'W';reverses['X'] = 'X';reverses['Y'] = 'Y';reverses['Z'] = '5';reverses['1'] = '1';reverses['2'] = 'S';reverses['3'] = 'E';reverses['5'] = 'Z';reverses['8'] = '8';string str;while(true){cin>>str;if(cin.eof())break;bool ispalindrome = true, ismirrored = true;for(int i = 0; i <= str.size() / 2; ++i){if(!ispalindrome && !ismirrored){break;}if(i == str.size() / 2){if(ismirrored && (str.size() % 2) && reverses.find(str[i]) == reverses.end())ismirrored = false;}else{if(ispalindrome && str[i] != str[str.size() - 1 - i]){ispalindrome = false;}if(ismirrored && (reverses.find(str[i]) == reverses.end() || reverses[str[i]] != str[str.size() - 1 - i])){ismirrored = false;}}}if(ispalindrome && ismirrored)cout<<str<<" -- is a mirrored palindrome."<<endl<<endl;else if(ispalindrome)cout<<str<<" -- is a regular palindrome."<<endl<<endl;else if(ismirrored)cout<<str<<" -- is a mirrored string."<<endl<<endl;elsecout<<str<<" -- is not a palindrome."<<endl<<endl;}return 0;}