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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int len,i,j,len1,len2,count=0,count1=0,count2=0;
    char s[99];
    cin>>s;
    len=strlen(s);
    char s1[(len-1)/2],s2[len-((len+3)/2)+1];
    for(i=0;i<(len-1)/2;i++)
    {
        if(s[i]==s[len-1-i])
        count++;
    }
    for(i=0;i<(len-1)/2;i++)
    s1[i]=s[i];
    len1=((len-1)/2);
    for(i=0;i<(len1-1)/2;i++)
    {
        if(s1[i]==s1[len1-1-i])
        count1++;
    }
    j=0;
    for(i=(len+3)/2-1;i<len;i++)
    {
        s2[j]=s[i];
        j++;
    }
    len2=j-1;
    for(i=0;i<(len2-1)/2;i++)
    {
        if(s2[i]==s2[len2-1-i])
        count2++;
    }
     if((count==(len-1)/2)&&(count1==(len1-1)/2)&&(count2==(len2-1)/2))
    cout<<"Yes";
    else
    cout<<"No";
}