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=;
 const int mod=1e9+;
 ll dp[N][N*N];
 int main()
 {
     int n,a;
     cin>>n>>a;
     dp[][]=;
     for(int i=;i<=n;i++){
         int x;
         cin>>x;
         for(int j=i-;j>=;j--)
             for(int k=;k<=N*j;k++)
                 dp[j+][k+x]+=dp[j][k];
     }
     ll ans=;
     for(int i=;i<=n;i++)
         ans+=dp[i][i*a];
     cout<<ans<<endl;
     return ;
 }