<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<vector<int>> permutation;
    void get_permute(vector<int>& nums, vector<int>& curr, vector<bool>& freq) {
        if (curr.size() == nums.size()) {
            permutation.push_back(curr);
            return;
        }
        for (int i = 0; i < nums.size(); i++) {
            if (!freq[i]&&(nums[i]%(curr.size()+1)==0||(curr.size()+1)%(nums[i])==0)) {
                curr.push_back(nums[i]);
                freq[i] = true;
                get_permute(nums, curr, freq);
                curr.pop_back();
                freq[i] = false;
            }
        }
    }
    int countArrangement(int n) {
        vector<bool> freq(n + 1, false);
        vector<int> nums(n , 0);
        for (int i = 0; i < nums.size(); i++) {
            nums[i] = i+1;
        }
        vector<int> curr;
        get_permute(nums, curr, freq);
        int res=0;
        for(int i=0;i<permutation.size();i++)
        {
            bool flag=0;
            for(int j=0;j<permutation[i].size();j++)
            {
                if(permutation[i][j]%(j+1)!=0&&(j+1)%permutation[i][j]!=0)
                {
                    flag=true;
                }
            }
            if(!flag)
            {
                res++;
            }
        }

        return res;
    }
};
17727