高橋君とカード / Tak and Cards

Time limit : 2sec / Stack limit : 256MB / Memory limit : 256MB

Score : 300 points

Problem Statement

Tak has N cards. On the i-th (1≤i≤N) card is written an integer xi. 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≤N≤50
1≤A≤50
1≤xi≤50
N, A, xi are integers.

Partial Score

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

Input

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

N A
x1 x2 … xN

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

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.

分析：二维背包；

代码：

#include <iostream>

#include <cstdio>

#include <cstdlib>

#include <cmath>

#include <algorithm>

#include <climits>

#include <cstring>

#include <string>

#include <set>

#include <map>

#include <queue>

#include <stack>

#include <vector>

#include <list>

#define rep(i,m,n) for(i=m;i<=n;i++)

#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)

#define mod 1000000000

#define inf 0x3f3f3f3f

#define vi vector<int>

#define pb push_back

#define mp make_pair

#define fi first

#define se second

#define ll long long

#define pi acos(-1.0)

#define pii pair<int,int>

#define Lson L, mid, rt<<1

#define Rson mid+1, R, rt<<1|1

const int maxn=1e5+;

using namespace std;

ll gcd(ll p,ll q){return q==?p:gcd(q,p%q);}

ll qpow(ll p,ll q){ll f=;while(q){if(q&)f=f*p;p=p*p;q>>=;}return f;}

int n,m,k,t;

ll dp[][];

int a,now;

ll ans;

int main()

{

    int i,j;

    dp[][]=;

    scanf("%d%d",&n,&a);

    rep(i,,n)

    {

        scanf("%d",&k);

        now+=k;

        for(j=i;j>=;j--)

            for(t=now;t>=k;t--)

                dp[j][t]+=dp[j-][t-k];

    }

    rep(i,,n)ans+=dp[i][a*i];

    printf("%lld\n",ans);

    //system("Pause");

    return ;

}