Leetcode 1962. Remove stones to minimize the total amount (should be rounded up)

Give you an array of integers piles , Array Subscript from 0 Start , among piles[i] It means the first one i The number of stones in the pile . I'll give you another integer k , Please do the following just k Time ：

Choose any pile of stones piles[i] , And from it remove floor(piles[i] / 2) Stone .
Be careful ： You can be right The same pile The stone performs this operation multiple times .

Return execution k After the first operation , The remaining stones Minimum total .

floor(x) by Less than or be equal to x Of Maximum Integers .（ namely , Yes x Rounding down ）.（ According to the example , The error is described here ）

Example 1：

 Input ：piles = [5,4,9], k = 2
 Output ：12
 explain ： Possible execution scenarios are as follows ：
-  Right.  2  Remove the rockfill , The distribution of stones becomes  [5,4,5] .
-  Right.  0  Remove the rockfill , The distribution of stones becomes  [3,4,5] .
 The total number of stones left is  12 .

Example 2：

 Input ：piles = [4,3,6,7], k = 3
 Output ：12
 explain ： Possible execution scenarios are as follows ：
-  Right.  2  Remove the rockfill , The distribution of stones becomes  [4,3,3,7] .
-  Right.  3  Remove the rockfill , The distribution of stones becomes  [4,3,3,4] .
-  Right.  0  Remove the rockfill , The distribution of stones becomes  [2,3,3,4] .
 The total number of stones left is  12 .

Tips ：

• 1 <= piles.length <= 105
• 1 <= piles[i] <= 104
• 1 <= k <= 105

Code：

class Solution {
    
public:
    int minStoneSum(vector<int>& piles, int k) {
    
        priority_queue<int>q;

        for(int i=0;i<piles.size();i++)    
        {
    
            q.push(piles[i]);
        }
        for(int i=0;i<k;i++)
        {
    
            int num;

            if(q.top()%2)
            {
    
                num=q.top()/2+1;
            }
            else
                num=q.top()/2;
            
            
            q.pop();
            q.push(num);
        }
        int res=0;
        while (q.size()) {
    
            res+=q.top();
            q.pop();
        }
        return res;
    }
};