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 <iostream>
#include <string>
#include <algorithm>
#include <vector>
#include <numeric>
//#include <stdio.h>
//#include <stack>
//#include <queue>
//#include <cstdio>
#include <cmath>
#include <iterator>
#include <map>
#include <fstream>
//#include <list>
#include <iomanip>
//#include <stdlib.h>
//#include <cctype>

using namespace std;

const int MAX_N = 1000000;
#define IOS ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
#define prec std::cout << std::fixed << std::setprecision(15);
#define endl "\n"
#define MOD 1000000007
#define Int int64_t
#define PI 3.14159265358979
#define rep(i, n) for (int i = 0; i < int(n); i++)
#define ssort(z) sort(z.begin(), z.end())
#define rsort(z) sort(z.rbegin(), z.rend())
#define eerase(z) z.erase(unique(z.begin(), z.end()), z.end())
#define ccnt(z, w) count(z.begin(), z.end(), w)
template <typename T>
void printV(const std::vector<T>& v, const char * const separator = " ")
{
    if(!v.empty())
    {
        std::copy(v.begin(),
                  --v.end(),
                  std::ostream_iterator<T>(std::cout, separator));
        std::cout << v.back() << "\n";
    }
}


Int gcd(Int a, Int b) {
    return b != 0 ? gcd(b, a % b) : a;
}

Int lcm(Int a, Int b) {
    return a / gcd(a, b) * b;
    //a*bは64bit integer overflow
}

int Max(int a, int b, int c) {
    int temp = max(a, b);
    return max(temp, c);
}

int Min(int a, int b, int c) {
    int temp = min(a, b);
    return min(temp, c);
}

bool integer(double num) {
    return floor(num) == num;
}

Int fact(int num) {
    if (num == 0)
        return 1;
    else
        return num * fact(num - 1);
}

void y() {
    cout << "Yes" << endl;
    exit(0);
}

void n() {
    cout << "No" << endl;
    exit(0);
}

void Y() {
    cout << "YES" << endl;
    exit(0);
}

void N() {
    cout << "NO" << endl;
    exit(0);
}

Int yakusu(int n) {
    int cnt = 0;
    for (int i = 1; i * i <= n; i++) {
        if (n % i == 0) {
            cnt++;
            if (i * i != n) cnt++;
        }
    }
    return cnt;
}

/*   kokomade  */

int cnt = 0;
bool is_prime[ 50000 + 1 ];
vector<int> P;
void Eratosthenes( const int N )
{
    for( int i = 0; i <= N; i++ )
    {
        is_prime[ i ] = true;//初期化
    }
    for( int i = 2; i <= N; i++ )
    {
        if( is_prime[ i ] )
        {
            for( int j = 2 * i; j <= N; j += i )
            {
                is_prime[ j ] = false;
            }
            P.emplace_back( i );
        }
    }
}
bool compare_by_b(pair<int, int> a, pair<int, int> b) {
    if(a.second != b.second){
        return a.second < b.second;
    }else{
        return a.first < b.first;
    }
}
vector< int64_t > divisor(int64_t n) {
    vector< int64_t > ret;
    for(int64_t i = 1; i * i <= n; i++) {
        if(n % i == 0) {
            ret.push_back(i);
            if(i * i != n) ret.push_back(n / i);
        }
    }
    sort(begin(ret), end(ret));
    return (ret);
}


void euclid(Int N, Int a, Int b)
{
    for (Int d = a+b; d <= N; d += b) {
        //cout << b << ", " << d << endl;
        euclid(N, b, d);
    }
}
int main() {
    IOS;
    prec;
    Int a,b,n,k,x,ans=0;
    string str,s="",t="";
    cin>>n;
    cin>>str;
    if (n%2!=0){
        cout<<"No";
        return 0;
    }
    for (int i = 0; i < n/2; ++i) {
        s+=str[i];
        t+=str[n/2+i];
    }
    if(s==t)cout<<"Yes";
    else cout<<"No";













    cout<<endl;
    return 0;
}

/*
テレビも無ェ ラジオも無ェ
自動車もそれほど走って無ェ
ピアノも無ェ バーも無ェ
巡査 毎日ぐーるぐる
朝起ぎで 牛連れで
二時間ちょっとの散歩道
電話も無ェ 瓦斯も無ェ
バスは一日一度来る
俺らこんな村いやだ 俺らこんな村いやだ
東京へ出るだ 東京へ出だなら
銭コァ貯めで 東京でベコ(牛)飼うだ
*/