# 1962. Remove Stones to Minimize the Total

Medium
You are given a 0-indexed integer array `piles`, where `piles[i]` represents the number of stones in the `ith` pile, and an integer `k`. You should apply the following operation exactly `k` times:
• Choose any `piles[i]` and remove `floor(piles[i] / 2)` stones from it.
Notice that you can apply the operation on the same pile more than once.
Return the minimum possible total number of stones remaining after applying the `k` operations.
`floor(x)` is the greatest integer that is smaller than or equal to `x` (i.e., rounds `x` down).
Example 1:
Input: piles = [5,4,9], k = 2
Output: 12
Explanation: Steps of a possible scenario are:
- Apply the operation on pile 2. The resulting piles are [5,4,5].
- Apply the operation on pile 0. The resulting piles are [3,4,5].
The total number of stones in [3,4,5] is 12.
Example 2:
Input: piles = [4,3,6,7], k = 3
Output: 12
Explanation: Steps of a possible scenario are:
- Apply the operation on pile 2. The resulting piles are [4,3,3,7].
- Apply the operation on pile 3. The resulting piles are [4,3,3,4].
- Apply the operation on pile 0. The resulting piles are [2,3,3,4].
The total number of stones in [2,3,3,4] is 12.
Constraints:
• `1 <= piles.length <= 10^5`
• `1 <= piles[i] <= 10^4`
• `1 <= k <= 10^5`

### 解題

class Solution {
public:
int minStoneSum(vector<int>& piles, int k) {
priority_queue <int> pq(piles.begin(), piles.end());
int ans = accumulate(piles.begin(), piles.end(), 0);
while (k > 0) {
int t = pq.top();
pq.pop();
pq.push(t - t / 2);
ans -= t / 2;
k--;
}
return ans;
}
};