2530. Maximal Score After Applying K Operations 

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class Solution {
    public long maxKelements(int[] nums, int k) {
        long score = 0;

        Queue<Integer> maxHeap = new PriorityQueue<>((o1, o2) -> Integer.compare(o2, o1));
        for (int num : nums) {
            maxHeap.offer(num);
        }
        for (int i = 0; i < k && !maxHeap.isEmpty(); i++) {
            int num = maxHeap.poll();
            score += num;
            maxHeap.offer((num + 2) / 3); // 注意此处向上取整的实现
        }

        return score;
    }
}

References

2530. Maximal Score After Applying K Operations