Largest Sum of Averages
<p>You are given an integer array <code>nums</code> and an integer <code>k</code>. You can partition the array into <strong>at most</strong> <code>k</code> non-empty adjacent subarrays. The <strong>score</strong> of a partition is the sum of the averages of each subarray.</p>

<p>Note that the partition must use every integer in <code>nums</code>, and that the score is not necessarily an integer.</p>

<p>Return <em>the maximum <strong>score</strong> you can achieve of all the possible partitions</em>. Answers within <code>10<sup>-6</sup></code> of the actual answer will be accepted.</p>

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

<pre>
<strong>Input:</strong> nums = [9,1,2,3,9], k = 3
<strong>Output:</strong> 20.00000
<strong>Explanation:</strong> 
The best choice is to partition nums into [9], [1, 2, 3], [9]. The answer is 9 + (1 + 2 + 3) / 3 + 9 = 20.
We could have also partitioned nums into [9, 1], [2], [3, 9], for example.
That partition would lead to a score of 5 + 2 + 6 = 13, which is worse.
</pre>

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

<pre>
<strong>Input:</strong> nums = [1,2,3,4,5,6,7], k = 4
<strong>Output:</strong> 20.50000
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 100</code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
	<li><code>1 &lt;= k &lt;= nums.length</code></li>
</ul>
Array; Dynamic Programming; Prefix Sum
class Solution {
public:
    double largestSumOfAverages(vector<int>& nums, int k) {
        int n = nums.size();
        int s[n + 1];
        double f[n][k + 1];
        memset(f, 0, sizeof(f));
        s[0] = 0;
        for (int i = 0; i < n; ++i) {
            s[i + 1] = s[i] + nums[i];
        }
        auto dfs = [&](this auto&& dfs, int i, int k) -> double {
            if (i == n) {
                return 0;
            }
            if (k == 1) {
                return (s[n] - s[i]) * 1.0 / (n - i);
            }
            if (f[i][k] > 0) {
                return f[i][k];
            }
            double ans = 0;
            for (int j = i + 1; j < n; ++j) {
                ans = max(ans, (s[j] - s[i]) * 1.0 / (j - i) + dfs(j, k - 1));
            }
            return f[i][k] = ans;
        };
        return dfs(0, k);
    }
};
