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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <stdbool.h>

#define lli long long int
#define min(a,b) ((a)<(b)?(a):(b))
#define max(a,b) ((a)>(b)?(a):(b))

int na2;
int na;

typedef lli type;
typedef struct dynamic {
	type val;
	bool bl;
} dynamic;

lli solve(int j,int t,int *y,dynamic **dp){
	//printf("j = %2d t = %2d y = %2d\n", j,t,y[j]);
	if(dp[j][t].bl){
		return dp[j][t].val;
	}
	dp[j][t].bl = true;
	if(j == 0 && t == 0){
		return dp[j][t].val = 1;
	}else if(j > 0 && t - y[j] < -na && t - y[j] > na){
		return dp[j][t].val = solve(j-1,t,y,dp);
	}else if(j > 0 && t - y[j] >= -na && t - y[j] <= na){
		return dp[j][t].val = solve(j-1,t,y,dp) + solve(j-1,t-y[j],y,dp);
	}
	return dp[j][t].val = 0;
}

int main(void){
	int i,j;
	int n,a;
	dynamic **dp;
	int x;
	int *y;
	scanf("%d%d",&n,&a);
	dp = calloc(sizeof(dynamic*),n+1);
	y = calloc(sizeof(int),n+1);
	na2 = n*a*2;
	na = n*a;
	for(i=1;i<=n;i++){
		scanf("%d",&x);
		y[i] = x - a;
	}
	for(i=0;i<=n;i++){
		dp[i] = calloc(sizeof(dynamic),na2+3);
		dp[i] += na + 1;
	}
	printf("%lld\n", solve(n,0,y,dp)-1);

	return 0;
}
/*
想定解法2
*/