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: Wrong Answer)

#include <bits/stdc++.h>
#include <string.h>
using namespace std;

int main() {
  string S;
  int N1, N2, N3, C;
  cin >> S;
  N1 = S.size(); // 文字列Sの文字数
  N2 = (N1 - 1) / 2; // 2番目の条件について判定する文字数
  N3 = N1 - ((N1 + 3) / 2); // 3番目の条件について判定する文字数
  C = 0;
  
  for (int i = 0; i < N1 / 2; i++) {
    if (S.at(i) != S.at(N1 - i - 1)) { // 条件1の回文判定 No
      break;
    }
    else if (S.at(i) == S.at(N1 - i - 1)) { // 条件1の回文判定 Yes
      for (int j = 0; j < N2 / 2; j++) {
        if (S.at(j) != S.at(N2 - j - 1)) { // 条件2の回文判定 No
          break;
        }
        else if (S.at(j) == S.at(N2 - j - 1)) { // 条件2の回文判定 Yes
          for (int k = 0; k < N3 / 2; k++) {
            if (S.at(((N1 + 3) / 2 - 1) + k) != S.at(N1 - k - 1)) { // 条件3の回文判定 No
              break;
            }
            else if (S.at(((N1 + 3) / 2 - 1) + k) == S.at(N1 - k - 1)) { // 条件3の回文判定 Yes
              C = C + 1;
              cout << C << endl;
            }
          }
        }
      }   
    }
  }

  if (C == (N1 - 1) / 2) {
    cout << "Yes";
  }
  else {
    cout << "No";
  }

}