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 <algorithm>
#include <cmath>
#include <cstdio>
#include <iostream>
#include <list>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>

using namespace std;

typedef long long ll;

#define pb(a) push_back(a)
#define all(x) (x).begin(), (x).end()

#define FOR(i, a, n) for (int i = (int)(a); i < (int)(n); ++i)
#define REP(i, n) FOR(i, 0, n)
#define SORT(a) sort(all(a))
#define REVERSE(a) reverse(all(a))

#define PRINT(str) printf(#str "\n")
#define yOUT PRINT(Yes)
#define nOUT PRINT(No)
#define YOUT PRINT(YES)
#define NOUT PRINT(NO)

//関数定義
//最大公約数
ll gcd(ll a, ll b);

//最小公倍数
ll lcm(ll a, ll b);

int main() {
  // ========================================================
  int n;
  string s;
  cin >> n >> s;
  string a, b;
  a = s.substr(0, n / 2);
  b = s.substr(n / 2, n / 2);

  if (a == b && n % 2 == 0)
    yOUT;
  else
    nOUT;
}

//最大公約数
ll gcd(ll a, ll b) {
  if (!b) return a;
  return gcd(b, a % b);
}

//最小公倍数
ll lcm(ll a, ll b) { return a * b / gcd(a, b); }