<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:
    int countArrangement(int n) {
        vector<int> vis(n + 1);
        map<pair<int,vector<int>>,int> dp;
        return solve(n, 1, vis, dp);
    }

    int solve(int n, int idx, vector<int> vis, map<pair<int,vector<int>>,int>& dp){
        if(idx > n) {
            return 1;
        }
        if(dp.count({idx, vis})){
            return dp[{idx, vis}];
        }
        //what number to put here
        int ret = 0;
        for(int i = 1;i <= n;i ++) {
            if(vis[i] == 0) {
                vis[i] = 1;
                if(i%idx == 0 || idx % i == 0) {
                    ret += solve(n, idx + 1, vis, dp);
                }
                vis[i] = 0;
            }
        }

        return dp[{idx, vis}] = ret;
    }
};
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