Problem:
Score : 200 points

Problem Statement

A string S of an odd length is said to be a strong palindrome if and only if all of the following conditions are satisfied:

S is a palindrome.

Let N be the length of S. The string formed by the 1-st through ((N-1)/2)-th characters of S is a palindrome.

The string consisting of the (N+3)/2-st through N-th characters of S is a palindrome.

Determine whether S is a strong palindrome.

Constraints

S consists of lowercase English letters.

The length of S is an odd number between 3 and 99 (inclusive).

Input

Input is given from Standard Input in the following format:

S

Output

If S is a strong palindrome, print Yes;
otherwise, print No.

Sample Input 1

akasaka

Sample Output 1

Yes

S is akasaka.

The string formed by the 1-st through the 3-rd characters is aka.

The string formed by the 5-th through the 7-th characters is aka.
All of these are palindromes, so S is a strong palindrome.

Sample Input 2

level

Sample Output 2

No

Sample Input 3

atcoder

Sample Output 3

No

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <string>

using namespace std;

bool kaibun(string s, int l, int n){
  string s1, s2;
  if(n==0){
    for(int i=0; i<(l-1)/2; i++) s1+=s[i];
    //cout << s1 << " ";
    for(int i=l-1; i>(l+3)/2-2; i--) s2+=s[i];
    //cout << s2 << endl;
    if(s1==s2){
      //cout << "true0" << endl;
      return (kaibun(s1, s1.length(), 1) && kaibun(s2, s2.length(), 1));
    }
  }
  else if(n==1){
    if(l%2!=0){
      for(int i=0; i<(l-1)/2; i++) s1+=s[i];
      //cout << s1 << " ";
      for(int i=l-1; i>l/2; i--) s2+=s[i];
      //cout << s2 << endl;
    } else{
      for(int i=0; i<(l-1)/2+1; i++) s1+=s[i];
      //cout << s1 << " ";
      for(int i=l-1; i>l/2-1; i--) s2+=s[i];
      //cout << s2 << endl;
    }
    if(l==1 || s1==s2){
      //cout << "true1" << endl;
      return true;
    }
  }
  return false;
}

int main(void){

  string S;
  int N;

  cin >> S;
  N = S.length();

  //cout << N << endl;

  if(kaibun(S, N, 0)) cout << "Yes" << endl;
  else cout << "No" << endl;

  return 0;
}