<p>Suppose you have <code>n</code> integers labeled <code>1</code> through <code>n</code>. A permutation of those <code>n</code> integers <code>perm</code> (<strong>1-indexed</strong>) is considered a <strong>beautiful arrangement</strong> if for every <code>i</code> (<code>1 &lt;= i &lt;= n</code>), <strong>either</strong> of the following is true:</p>

<ul>
	<li><code>perm[i]</code> is divisible by <code>i</code>.</li>
	<li><code>i</code> is divisible by <code>perm[i]</code>.</li>
</ul>

<p>Given an integer <code>n</code>, return <em>the <strong>number</strong> of the <strong>beautiful arrangements</strong> that you can construct</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> n = 2
<strong>Output:</strong> 2
<b>Explanation:</b> 
The first beautiful arrangement is [1,2]:
    - perm[1] = 1 is divisible by i = 1
    - perm[2] = 2 is divisible by i = 2
The second beautiful arrangement is [2,1]:
    - perm[1] = 2 is divisible by i = 1
    - i = 2 is divisible by perm[2] = 1
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 1
<strong>Output:</strong> 1
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 15</code></li>
</ul>

class Solution {
public:
vector<int> mapp;
map<pair<int,vector<int>>,int> dp;

    int cal(int index, int n, vector<int>* possible) {

        if(index>n)
        return 1;
        if(dp.find({index,mapp})!=dp.end()){
            return dp[{index,mapp}];
        }
        int ret=0;
        for(int j:possible[index]){
            if(mapp[j]!=0){
                mapp[j]=0;
                ret+= cal(index+1,n,possible);
                mapp[j]=1;
            }
        }
        return dp[{index,mapp}] =ret;
    }

    int countArrangement(int n) {
        
        vector<int> possible[n+1];
        mapp.resize(n+1);
        for(int i=1;i<n+1;i++)
        mapp[i]++;

        for(int i=1;i<n+1;i++){
            for(int j=1;j<n+1;j++){
                if(i%j==0 or j%i==0){
                    possible[i].push_back(j);
                }
            }
        }

        return cal(1,n,possible);
    }
};
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