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++ (Clang 10.0.0) (judge verdict: Accepted)

#include <algorithm>
#include <iostream>
#include <math.h>
#include <stdio.h>
#include <string>
#include <stack>
#include <sstream>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <cmath>
#define rep(i, n) for(int i = 0; i < (int)(n); i++)

using namespace std;
typedef long long int ll;
ll dp[50][51][2501];	// i番目以降で構成される配列で、j個の数字を使って、kという数字を作る場合の数
int numList[50];

void init(){
	for(int i = 0; i < 50; i++){
		for(int j = 0; j < 51; j++){
			for(int k = 0; k < 2501; k++)dp[i][j][k] = -1;
		}
	}
}

ll dfs(int target,int n,int startIndex,int N){	// n 個の数を使って targetを作る場合の数を返す
	if(startIndex == N) return 0;
	if(n == 0) return 0;
	if(target <= 0) return 0;
	if(dp[startIndex][n][target] != -1) return dp[startIndex][n][target];
	if(n == 1){
		int count = 0;
		for(int i = startIndex; i < N; i++){
			if( numList[i] == target) count++;
		}
		dp[startIndex][n][target] = count;
		return count;
	}
	
	if( numList[startIndex] > target){
		dp[startIndex][n][target] = 0;
		return 0;
	}
	ll ans = 0;
	//numList[startIndex] を使う場合と使わない場合の二通りの和を返す
	int a = numList[startIndex];
	ans += dfs(target,n,startIndex + 1,N); //使わない場合
	ans += dfs(target-a,n-1,startIndex+1,N);
	dp[startIndex][n][target] = ans;
	return ans;
}

int main()
{
	init();
	ll N,A;
	cin>>N>>A;
	rep(i,N) cin>>numList[i];
	sort(numList,numList+N);
	ll ans = 0;
	for(int i = 1 ; i <= N; i++){
		ans += dfs(A*i,i,0,N);
	}
	cout<<ans<<endl;
	return 0;
}