Example 1:
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Input: [1,12,-5,-6,50,3], k = 4
Output: 12.75
Explanation: Maximum average is (12-5-6+50)/4 = 51/4 = 12.75
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Note:
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1 <= k <= n <= 30,000.
Elements of the given array will be in the range [-10,000, 10,000].

class Solution {
public:
    double findMaxAverage(vector<int>& nums, int k) {
        int sum=accumulate(nums.begin(),nums.begin()+k,0);
        int maxSum=sum;
        for(int i=k;i<nums.size();i++)
        {
            maxSum=max(maxSum, sum+=nums[i]-nums[i-k]);
        }
        return maxSum/double(k);
    }
};