Maximize Score After N Operations
<p>You are given <code>nums</code>, an array of positive integers of size <code>2 * n</code>. You must perform <code>n</code> operations on this array.</p>

<p>In the <code>i<sup>th</sup></code> operation <strong>(1-indexed)</strong>, you will:</p>

<ul>
	<li>Choose two elements, <code>x</code> and <code>y</code>.</li>
	<li>Receive a score of <code>i * gcd(x, y)</code>.</li>
	<li>Remove <code>x</code> and <code>y</code> from <code>nums</code>.</li>
</ul>

<p>Return <em>the maximum score you can receive after performing </em><code>n</code><em> operations.</em></p>

<p>The function <code>gcd(x, y)</code> is the greatest common divisor of <code>x</code> and <code>y</code>.</p>

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

<pre>
<strong>Input:</strong> nums = [1,2]
<strong>Output:</strong> 1
<strong>Explanation:</strong>&nbsp;The optimal choice of operations is:
(1 * gcd(1, 2)) = 1
</pre>

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

<pre>
<strong>Input:</strong> nums = [3,4,6,8]
<strong>Output:</strong> 11
<strong>Explanation:</strong>&nbsp;The optimal choice of operations is:
(1 * gcd(3, 6)) + (2 * gcd(4, 8)) = 3 + 8 = 11
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,2,3,4,5,6]
<strong>Output:</strong> 14
<strong>Explanation:</strong>&nbsp;The optimal choice of operations is:
(1 * gcd(1, 5)) + (2 * gcd(2, 4)) + (3 * gcd(3, 6)) = 1 + 4 + 9 = 14
</pre>

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

<ul>
	<li><code>1 &lt;= n &lt;= 7</code></li>
	<li><code>nums.length == 2 * n</code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>6</sup></code></li>
</ul>
Bit Manipulation; Array; Math; Dynamic Programming; Backtracking; Bitmask; Number Theory
class Solution {
public:
    int maxScore(vector<int>& nums) {
        int m = nums.size();
        int g[m][m];
        for (int i = 0; i < m; ++i) {
            for (int j = i + 1; j < m; ++j) {
                g[i][j] = gcd(nums[i], nums[j]);
            }
        }
        int f[1 << m];
        memset(f, 0, sizeof f);
        for (int k = 0; k < 1 << m; ++k) {
            int cnt = __builtin_popcount(k);
            if (cnt % 2 == 0) {
                for (int i = 0; i < m; ++i) {
                    if (k >> i & 1) {
                        for (int j = i + 1; j < m; ++j) {
                            if (k >> j & 1) {
                                f[k] = max(f[k], f[k ^ (1 << i) ^ (1 << j)] + cnt / 2 * g[i][j]);
                            }
                        }
                    }
                }
            }
        }
        return f[(1 << m) - 1];
    }
};