Problem:
Score : 200 points

Problem Statement

A string S of an odd length is said to be a strong palindrome if and only if all of the following conditions are satisfied:

S is a palindrome.

Let N be the length of S. The string formed by the 1-st through ((N-1)/2)-th characters of S is a palindrome.

The string consisting of the (N+3)/2-st through N-th characters of S is a palindrome.

Determine whether S is a strong palindrome.

Constraints

S consists of lowercase English letters.

The length of S is an odd number between 3 and 99 (inclusive).

Input

Input is given from Standard Input in the following format:

S

Output

If S is a strong palindrome, print Yes;
otherwise, print No.

Sample Input 1

akasaka

Sample Output 1

Yes

S is akasaka.

The string formed by the 1-st through the 3-rd characters is aka.

The string formed by the 5-th through the 7-th characters is aka.
All of these are palindromes, so S is a strong palindrome.

Sample Input 2

level

Sample Output 2

No

Sample Input 3

atcoder

Sample Output 3

No

Solution in C++14 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<stack>
#include<cmath>
#include<algorithm>
#include<map>
typedef long long ll;
using namespace std;
#define INF 0x3f3f3f3f
/**
*超时有可能是输入的问题
*注意边界
*仔细读题
*没思路试试逆向思维
**加油，奥利给！！！ 
**/
int main()
{
	int m,n,i,j,t,k,s1,s2;
	char a[10000];
	scanf("%s",a+1);
	k=strlen(a+1);
	if(k%2==0) cout<<"No"; 
	else {
		for(i=1;i<=k/2;i++)
	{
		if(a[i]!=a[k-i+1]) {
			cout<<"No"; return 0;
		}
	}
	for(i=1;i<=(k-1)/2;i++)
	{
		if(a[i]!=a[(k-1)/2-i+1]) {
			cout<<"No"; return 0;
		}
		
	}
	for(i=k/2+1;i<=n;i++)
	{
		if(a[i]!=a[n-i+k/2+1]) {
			cout<<"No"; return 0;
		}
	
	}	
	cout<<"Yes";
	}
 }