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 rep(i,n) for(int i=0;i<n;i++)
#define FOR(i,start,end) for(int i=start;i<=end;i++)
const int INF = 1001001001;
typedef long long ll;
const ll MOD=1000000007;
using namespace std;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
template<class T>auto MAX(const T& a) { return *max_element(a.begin(),a.end()); }
template<class T>auto MIN(const T& a) { return *min_element(a.begin(),a.end()); }
template<class T, class U>U SUM(const T& a, const U& v) { return accumulate(a.begin(),a.end(), v); }
template<class T, class U>U COUNT(const T& a, const U& v) { return count(a.begin(),a.end(), v); }
template<class T, class U>int LOWER(const T& a, const U& v) { return lower_bound(a.begin(),a.end(), v) - a.begin(); }
template<class T, class U>int UPPER(const T& a, const U& v) { return upper_bound(a.begin(),a.end(), v) - a.begin(); }
int GCD(int a, int b) { return b ? GCD(b, a%b) : a; }
int LCM(int a, int b) { int g = GCD(a, b); return a / g * b; }

int main(void){
    // Your code here!
    int n; cin >> n;
    string s; cin >> s;
    string first ;
    string last;
    rep(i,n/2) {
        first += s[i];
    }
    FOR(i,n/2,n-1){
        last += s[i];
    }
    if (first == last) cout << "Yes" << endl;
    else cout << "No" << endl;
}