Problem:
Score : 200 points

Problem Statement

Given are a positive integer N and a string S of length N consisting of lowercase English letters.

Determine whether the string is a concatenation of two copies of some string.
That is, determine whether there is a string T such that S = T + T.

Constraints

1 \leq N \leq 100

S consists of lowercase English letters.

|S| = N

Input

Input is given from Standard Input in the following format:

N
S

Output

If S is a concatenation of two copies of some string, print Yes; otherwise, print No.

Sample Input 1

6
abcabc

Sample Output 1

Yes

Let T = abc, and S = T + T.

Sample Input 2

6
abcadc

Sample Output 2

No

Sample Input 3

1
z

Sample Output 3

No

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

//
//  atcoder.cpp
//
//
///
#include <bits/stdc++.h>
#include <boost/optional.hpp>
#include <boost/range/irange.hpp>
#include <boost/range/algorithm.hpp>
#include <boost/range/numeric.hpp>
#include <boost/bimap/bimap.hpp>
#include <boost/bimap/multiset_of.hpp>
#include <boost/multi_array.hpp>
#include <boost/math/common_factor.hpp>
#include <boost/icl/interval_map.hpp>
#include <boost/algorithm/string.hpp>
#include <boost/pending/disjoint_sets.hpp>
#include <boost/graph/graph_traits.hpp>
#include <boost/graph/adjacency_list.hpp>
#include <boost/graph/dijkstra_shortest_paths.hpp>
#include <boost/graph/kruskal_min_spanning_tree.hpp>
#include <boost/graph/graph_utility.hpp>
#include <boost/numeric/ublas/vector.hpp>
#include <boost/multiprecision/cpp_int.hpp>
// ↓これいる？
#include <boost/graph/strong_components.hpp>

#define REP(i, N) for(int i=0; i<(N); i++)
#define REC(f, ...) (void)0; const auto f = [&](const auto f, __VA_ARGS__)
using namespace std;
using namespace boost::math;
using namespace boost::numeric;
using namespace boost::multiprecision;
using ll = long long;
namespace bst = boost;
template<class T> void fill(T& v) { for (auto&& i : v) cin >> i; }

//#include "atcoder.hpp"

int main() {
    int N;string S;
    
    cin >> N >> S;
  
    if (N % 2 == 1) {
      cout << "No" << endl;
      return 0;
    }
  
    int i = 0, half = N / 2;
  
    while (half < N)
      if (S[i++] != S[half++]) {
        cout << "No" << endl;
      }
    
    cout << "Yes" << endl;
    
    return 0;
}