AtCoder Beginner Contest 044 C - 高橋君とカード / Tak and Cards

题目链接：http://abc044.contest.atcoder.jp/tasks/arc060_a

Time limit : 2sec / 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.

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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.

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Sample Input 1

Copy

4 8
7 9 8 9

Sample Output 1

Copy

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.

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Sample Input 2

Copy

3 8
6 6 9

Sample Output 2

Copy

0

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Sample Input 3

Copy

8 5
3 6 2 8 7 6 5 9

Sample Output 3

Copy

19

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Sample Input 4

Copy

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

Copy

8589934591

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

题意：给定一串数字，问能够组成多少种不连续子串使得子串的平均数为某个数

题解：用动态规划，i表示相加的个数，j表示加起来后的值

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <cstdlib>
#include <iomanip>
#include <cmath>
#include <ctime>
#include <map>
#include <set>
#include <queue>
using namespace std;
#define lowbit(x) (x&(-x))
#define max(x,y) (x>y?x:y)
#define min(x,y) (x<y?x:y)
#define MAX 100000000000000000
#define MOD 1000000007
#define pi acos(-1.0)
#define ei exp(1)
#define PI 3.141592653589793238462
#define INF 0x3f3f3f3f3f
#define mem(a) (memset(a,0,sizeof(a)))
typedef long long ll;
ll gcd(ll a,ll b){
    return b?gcd(b,a%b):a;
}
bool cmp(int x,int y)
{
    return x>y;
}
const int N=55;
const int mod=1e9+7;
ll dp[N][N*N];
int main()
{
    int n,a;
    cin>>n>>a;
    dp[0][0]=1;
    for(int i=1;i<=n;i++){
        int x;
        cin>>x;
        for(int j=i-1;j>=0;j--)
            for(int k=0;k<=N*j;k++)
                dp[j+1][k+x]+=dp[j][k];
    }
    ll ans=0;
    for(int i=1;i<=n;i++)
        ans+=dp[i][i*a];
    cout<<ans<<endl;
    return 0;
}