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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<string.h>
//強い回文だと確かめるには全体と前半分もしくは後ろ半分だけを判定すればよい。後ろ半分は前半分を反転させたものだから
int main()
{
	char s[100],s1[50],s2[50];
	int n,n1,i,a=0,b=0;
	//int n2,c=0;

	scanf("%s",s);
	n=strlen(s);
	for(i=0;i<(n-1)/2;i++)
		if(s[i]==s[n-1-i])//strcmp(s[i],s[n-1-i])==0一文字分ならstrcmpを使わなくとも==だけで比較すればよい
			a++;
	strncpy(s1,s,n/2);
	n1=strlen(s1);
	//s1[n1]='\0';
	for(i=0;i<n1/2;i++)
		if(s[i]==s[n1-1-i])
			b++;
	/*strncpy(s2,s+(n/2+1),n/2);
	n2=strlen(s2);
	//s2[n2]='\0';
	for(i=0;i<(n2-1)/2;i++)
		if(s[i]==s[n2-1-i])
			c++;*/
	if((a==n/2)&&(b==n1/2))
		printf("Yes\n");
	else
		printf("No\n");
	return 0;
}