<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 f(int start,int len,int k,vector<int> &nums,vector<int> &pre){
    int n=nums.size();
     if(start+len-1 == nums.size()-1)
       return (pre[n-1]-(start>0?pre[start-1]:0))*1.0/(len*1.0);
       if(dp[start][len][k]>-0.9)
          return dp[start][len][k];
     double take=0;
     take=f(start,len+1,k,nums,pre);
     double no=0;
     if(k>1){
         no=(pre[start+len-1]- (start>0?pre[start-1]:0))*1.0/(len*1.0);
         no+= f(start+len,1,k-1,nums,pre);
     }
         return dp[start][len][k]=max(take,no);
   }
    double largestSumOfAverages(vector<int>& nums, int k) {
        int n=nums.size();
        memset(dp,-1.0,sizeof(dp));
        vector<int> pre(n+1,0);
         pre[0]=nums[0];
          for(int i=1;i<n;i++)
            pre[i]=pre[i-1]+nums[i];
        return f(0,1,k,nums,pre);
    }
};
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