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: Accepted)

#include <bits/stdc++.h>

#define mp make_pair
#define fst first
#define snd second
#define forn(i,n) for (size_t i = 0; i < size_t(n); i++)
#define forn1(i,n) for (size_t i = 1; i <= size_t(n); i++)
#define popcnt __builtin_popcount
#define ffs __builtin_ffs
#define ctz __builtin_ctz
#define clz __builtin_clz
#define all(a) (a).begin(), (a).end()

using namespace std;

using uint = unsigned int;
using ll = long long;
using ull = unsigned long long;
using pii = pair<int,int>;
using pli = pair<ll,int>;
using pil = pair<int,ll>;
using pll = pair<ll,ll>;
template <typename T> using vec = vector<T>;
using vi = vec<int>;
using vl = vec<ll>;
template <typename T> using que = queue<T>;
template <typename T> using deq = deque<T>;

template <typename T> T id(T b) {return b;};
template <typename T> void chmax(T &x, T y) {if (x < y) x = y;}
template <typename T> void chmin(T &x, T y) {if (x > y) x = y;}
template <typename S, typename K> bool contains(S s, K k) { return s.find(k) != s.end(); }
void fastio() { ios_base::sync_with_stdio(false); cin.tie(nullptr); }
constexpr ll TEN(int n) { if (n == 0) return 1LL; else return 10LL*TEN(n-1); }

int n;
string s;

int main()
{
  fastio();

  cin >> n >> s;
  if (n % 2) {
    cout << "No\n";
    return 0;
  }

  bool ok = true;
  forn(i,n/2) {
    ok &= s[i] == s[i+n/2];
  }

  if (ok)
    cout << "Yes\n";
  else
    cout << "No\n";

  return 0;
}