s-palindrome

Let’s call a string “s-palindrome” if it is symmetric about the middle of the string. For example, the string “oHo” is “s-palindrome”, but the string “aa” is not. The string “aa” is not “s-palindrome”, because the second half of it is not a mirror reflection of the first half.

English alphabet
You are given a string s. Check if the string is “s-palindrome”.

Input
The only line contains the string s (1 ≤ |s| ≤ 1000) which consists of only English letters.

Output
Print “TAK” if the string s is “s-palindrome” and “NIE” otherwise.

Examples
Input
oXoxoXo
Output
TAK
Input
bod
Output
TAK
Input
ER
Output
NIE

题意是说给你一个字符串，看是不是完全对称的，这里需要注意的是这个对称是完全对称的，并不是对应位置是同一个字母，用一个二维数组就可以解决，只要对应位置上是两个中的一个，对称位置上就只能是对应的字母。

细节之处在于把所有对称字母找出来。

AC代码

#include<iostream>
#include<stdio.h>
using namespace std;
char c[19][2]={'A','A','b','d','d','b','H','H','I','I','M','M','O','O','o','o','p','q','q','p','U','U','V','V','v','v','W','W','w','w','X','X','x','x','Y','Y','T','T'};
int main()
{string str;cin>>str;int len=str.length();int f=0;for(int i=0;i<=len/2;i++){int flag=0;for(int j=0;j<19;j++){if(str[i]==c[j][0]&&str[len-i-1]==c[j][1]){flag=1;break;}   }if(flag==0){cout<<"NIE"<<endl;f=1;break;}}if(f==0) cout<<"TAK"<<endl;return 0;
}

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