<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[101][101][101];
    double rec(vector<int>& nums, int k, int idx, int size)
    {
        if(k==0)
        {
            if(idx==nums.size())
            {
                return 0.0;
            }
            else
            {
                return -1e9/2.0;
            }
        }
        if(idx==nums.size())
        return -1e9/2.0;
        if(dp[k][idx][size]!=-1)
        return dp[k][idx][size];
        // cut or non-cut
        double ans=0;
        if(idx-(size)>=0)
        {
            ans=max(ans, (nums[idx]-nums[idx-size])/(double)size+(rec(nums, k-1, idx+1, 1)));
        }
        else
        {
            ans= max(ans, (nums[idx])/(double)size+rec(nums, k-1, idx+1, 1));
        }
        ans=max(ans, rec(nums, k, idx+1, size+1));
        return dp[k][idx][size]=ans;
    }
    double largestSumOfAverages(vector<int>& nums, int k) {
        for(int i=0; i<=nums.size(); ++i)
        {
            for(int j=0; j<=nums.size(); ++j)
            {
                for(int k=0; k<=nums.size(); ++k)
                dp[i][j][k]=-1;
            }
        }
        for(int i=1; i<nums.size(); ++i)
        {
            nums[i]+=nums[i-1];
        }
        return rec(nums, k, 0, 1);
    }
};
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