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>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef vector<bool> vb;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef map<int,int> mii;
typedef map<ll,ll> mll;
typedef vector<pii> vii;
typedef vector<pll> vll;
#define fi first
#define se second
#define pi 3.141592653589793
#define mod 1e9 + 7
#define pb push_back
#define all(v) v.begin(),v.end()
#define rall(v) v.begin(), v.end(), greater<ll>()
#define tc int t;cin>>t;while(t--)
#define pqmax priority_queue<int>
#define pqmin priority_queue<int,vi,greater<int>>
#define fast_io ios_base::sync_with_stdio(0), cin.tie(NULL)
#define tc_g int tt;cin>>tt;for(int ti=1;ti<=tt;ti++)
#define case_g "Case #"<<ti<<": "
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
typedef tree<int, null_type, less_equal<int>, rb_tree_tag, tree_order_statistics_node_update> ordered_multiset;
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;
typedef tree<int, int, less<int>, rb_tree_tag, tree_order_statistics_node_update> ordered_map;
using namespace std;
int main() {
    int n;
    string ss;
    cin >> n;
    cin >> ss;
    if(n & 1) {
        cout << "No";
        return 0;
    }
    string mid = ss.substr(0, n / 2);
    string nd = ss.substr(n / 2, n / 2);
    // << mid << nd;
    if(mid == nd)
        cout << "Yes";
    else
        cout << "No";
}