Problem:
Score : 300 points

Problem Statement

Tak has N cards. On the i-th (1 \leq i \leq N) card is written an integer x_i.
He is selecting one or more cards from these N cards, so that the average of the integers written on the selected cards is exactly A.
In how many ways can he make his selection?

Constraints

1 \leq N \leq 50

1 \leq A \leq 50

1 \leq x_i \leq 50

N,\,A,\,x_i are integers.

Partial Score

200 points will be awarded for passing the test set satisfying 1 \leq N \leq 16.

Input

The input is given from Standard Input in the following format:

N A
x_1 x_2 ... x_N

Output

Print the number of ways to select cards such that the average of the written integers is exactly A.

Sample Input 1

4 8
7 9 8 9

Sample Output 1

5

The following are the 5 ways to select cards such that the average is 8:

Select the 3-rd card.

Select the 1-st and 2-nd cards.

Select the 1-st and 4-th cards.

Select the 1-st, 2-nd and 3-rd cards.

Select the 1-st, 3-rd and 4-th cards.

Sample Input 2

3 8
6 6 9

Sample Output 2

0

Sample Input 3

8 5
3 6 2 8 7 6 5 9

Sample Output 3

19

Sample Input 4

33 3
3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3

Sample Output 4

8589934591

The answer may not fit into a 32-bit integer.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=2e6+20;
const ll mod=1e9+7;
char s[N],t[N];
int fail[N],suf[N];
void getfail(int *fail,char *s)
{
	fail[0]=-1;
	int i=0,j=-1;
	while(s[i])
	{
		if(j==-1||s[i]==s[j])
		{
			fail[i+1]=j+1;
			i++,j++;
		}
		else
			j=fail[j];
	}
}
int main()
{
	scanf("%s",s);
	ll ans=0;
	int len=strlen(s);
	for(int i=0;s[i];i++)
	t[i]=s[len-i-1];
	t[len]='\0';
	getfail(fail,s);
	getfail(suf,t);
	int m;
	if(fail[len]==0||len%(len-fail[len]))
	m=1;
	else
	m=2;
	bool flag=true;
	for(int i=1;s[i];i++)
	{
		if(s[i]!=s[i-1])
			flag=false;
	}	
	if(flag)
	m=len;
	if(m==1||flag)
	ans=1;
	else
	{
		for(int i=0;i<len-1;i++)
		{
			int k=len-i-1;
			if((fail[i+1]==0||(i+1)%((i+1)-fail[i+1])) &&(suf[k]==0||(k)%(k-suf[k])))
			ans++;
		}
	}	
	printf("%d\n%d\n",m,ans);
	return 0;
}