<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[100][100][100];
    vector<double>pre;
    double solve(vector<int>&nums,int ind,int prev,int k){
        if(ind==nums.size()-1){
            double sub=(prev-1<0 ? 0:pre[prev-1]);
            int len=ind-prev+1;
            double sum=(pre[ind]-sub)/len;
            return sum;
        }
        if(dp[ind][prev][k]!=0) return dp[ind][prev][k];
        double notake=0,take=0;
        if(k>0){
            double sub=(prev-1<0 ? 0:pre[prev-1]);
            double val=(pre[ind]-sub)/(ind-prev+1);
            take=val+solve(nums,ind+1,ind+1,k-1);
        } 
        notake=solve(nums,ind+1,prev,k);
        return dp[ind][prev][k]=max(take,notake);
    }
    double largestSumOfAverages(vector<int>& nums, int k) {
        
        memset(dp,0,sizeof(dp));
        double sum=0;
        for(int i=0;i<nums.size();i++){
            sum+=(double)nums[i];
            pre.push_back(sum);
        }
        
        return solve(nums,0,0,k-1);
        return 5;
    }
};
18000