AtCoder Beginner Contest 044 C

C - 高橋君とカード / Tak and Cards

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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<bits/stdc++.h>
using namespace std;
const int N=55;
long long dp[N][N*N];      //dp值会爆int
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--)    //每一次个数向下减少
            for(int k=0;k<=N*j;k++)    //优化时间  由于k最大为N*j次
                dp[j+1][k+x]+=dp[j][k];       //个数加1 则找到前面个数的值相加
    }
    long long ans=0;       
    for(int i=1;i<=n;i++)
            ans+=dp[i][i*a];    //累计所有个数会出现平均值满足a的情况
    cout<<ans<<endl;
    return 0;
}