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>

int main()
{
	char s[100],s1[50],s2[50];
	int n,n1,n2,i,a=0,b=0,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*2<n1-1;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*2<n2-1;i++)
		if(s[i]==s[n2-1-i])
			c++;
	if((a==n/2)&&(b==(n1-1)/2)&&(c==(n2-1)/2))
		printf("Yes\n");
	else
		printf("No\n");
	return 0;
}