本文共 1387 字，大约阅读时间需要 4 分钟。
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 #include #include using namespace std;int main(){ map 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< <<" -- is a mirrored palindrome."< <
转载地址：http://mxxli.baihongyu.com/