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: Wrong Answer)

#include<bits/stdc++.h>
#include<iomanip>
#define C char
#define D double
#define F float
#define LL long long
#define ULLI unsigned long long int
#define LLI long long int
#define LLD long long double
#define LD long double
#define S string
#define PI (2.0*acos(0.0))
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define READ 		freopen("input.txt","r",stdin)
#define WRITE  		freopen("output.txt","w",stdout)
using namespace std;

int main()

{
    IOS;
    LL t,count=0;
    cin>>t;
    LL z = t/2;
    string s;
    cin>>s;
    char xrr[z];
    char yrr[z];
    if(t%2==1) cout<<"NO\n";
        else
        {
    for(int i=0; i<z; i++)
    {
        xrr[i] = s[i];
    }
    LL j=0;
    for(int i=z; i<t; i++)
    {
        yrr[j] = s[i];
        j++;
    }
    for(int i=0; i<z; i++)
    {
        if(xrr[i]==yrr[i]) count++;
    }
    if(count==z) cout<<"Yes\n";
    else cout<<"No\n";
        }
    return 0;
}