<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>

class Solution {
public:
    double dp[105][105][105];
    double partition(vector<int> &prefix, int n, int k, int prev, int pos) {
        if(pos == n) return 0;
        if(k == 1) {
            return (prefix[n-1] - prefix[prev-1])/(double)(n-prev);
        }
        if(dp[pos][prev][k] != -1.0) return dp[pos][prev][k];
        double ma = 0;
        for(int i=pos;i<n;i++) {
            if(n - i <= k) {
                ma = max(ma, (prefix[i]-prefix[prev-1])/(double)(i-prev+1) + partition(prefix, n, k-1, i+1, i+1));
            }
            else {
                double pick = (prefix[i]-prefix[prev-1])/(double(i-prev+1)) + partition(prefix, n, k-1, i+1, i+1);
                double not_pick = partition(prefix, n, k, prev, i+1);
                ma = max(ma, max(pick,not_pick));
            }
        }
        return dp[pos][prev][k] = ma;
    }
    double largestSumOfAverages(vector<int>& nums, int k) {
        int n = nums.size();
        vector<int> prefix(n+1,0);
        for(int i=1;i<=n;i++) {
            prefix[i] = prefix[i-1] + nums[i-1];
        }
        n++;
        for(int i=0;i<105;i++) {
            for(int j=0;j<105;j++) {
                for(int k=0;k<105;k++) {
                    dp[i][j][k] = -1.0;
                }
            }
        }
        return partition(prefix, n, k, 1, 1);
    }
};
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