给定一个由N 个正整数组成的数组arr[]使得arr[i]表示第i个包包含arr[i]钻石和一个正整数K ，任务是找到可以在其中获得的最大钻石数正好K分钟，如果掉落一个袋子需要1分钟，那么如果一个有P颗钻石的袋子掉下来，那么它会变成[P/2]颗钻石，并且获得P颗钻石。
例子：
Input: arr[] = {2, 1, 7, 4, 2}, K = 3
Output: 14
Explanation:
The initial state of bags is {2, 1, 7, 4, 2}.
Operation 1: Take all diamonds from third bag i.e., arr[2](= 7), the state of bags becomes: {2, 1, 3, 4, 2}.
Operation 2: Take all diamonds from fourth bag i.e., arr[3](= 4), the state of bags becomes: {2, 1, 3, 2, 2}.
Operation 3: Take all diamonds from Third bag i.e., arr[2](= 3), the state of bags becomes{2, 1, 1, 2, 2}.
Therefore, the total diamonds gains is 7 + 4 + 3 = 14.
Input: arr[] = {7, 1, 2}, K = 2
Output: 10
方法：可以在最大堆的帮助下使用贪心方法解决给定的问题。请按照以下步骤解决问题：
- 初始化一个优先级队列，比如PQ ，并将给定数组的所有元素插入PQ 。
- 初始化一个变量，比如ans为0来存储最终获得的最大钻石。
- 迭代一个循环，直到优先级队列PQ不为空且K > 0的值：
- 弹出优先级队列的顶部元素并将弹出的元素添加到变量ans 。
- 将弹出的元素除以2并将其插入优先级队列PQ 。
- 将K的值减少1 。
- 完成以上步骤后，打印ans的值作为结果。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to find the maximum number
// of diamonds that can be gained in
// exactly K minutes
void maxDiamonds(int A[], int N, int K)
{
// Stores all the array elements
priority_queue pq;
// Push all the elements to the
// priority queue
for (int i = 0; i < N; i++) {
pq.push(A[i]);
}
// Stores the required result
int ans = 0;
// Loop while the queue is not
// empty and K is positive
while (!pq.empty() && K--) {
// Store the top element
// from the pq
int top = pq.top();
// Pop it from the pq
pq.pop();
// Add it to the answer
ans += top;
// Divide it by 2 and push it
// back to the pq
top = top / 2;
pq.push(top);
}
// Print the answer
cout << ans;
}
// Driver Code
int main()
{
int A[] = { 2, 1, 7, 4, 2 };
int K = 3;
int N = sizeof(A) / sizeof(A[0]);
maxDiamonds(A, N, K);
return 0;
} Java
// Java program for the above approach
import java.util.*;
class GFG{
// Function to find the maximum number
// of diamonds that can be gained in
// exactly K minutes
static void maxDiamonds(int A[], int N, int K)
{
// Stores all the array elements
PriorityQueue pq = new PriorityQueue<>(
(a, b) -> b - a);
// Push all the elements to the
// priority queue
for(int i = 0; i < N; i++)
{
pq.add(A[i]);
}
// Stores the required result
int ans = 0;
// Loop while the queue is not
// empty and K is positive
while (!pq.isEmpty() && K-- > 0)
{
// Store the top element
// from the pq
int top = pq.peek();
// Pop it from the pq
pq.remove();
// Add it to the answer
ans += top;
// Divide it by 2 and push it
// back to the pq
top = top / 2;
pq.add(top);
}
// Print the answer
System.out.print(ans);
}
// Driver Code
public static void main(String[] args)
{
int A[] = { 2, 1, 7, 4, 2 };
int K = 3;
int N = A.length;
maxDiamonds(A, N, K);
}
}
// This code is contributed by 29AjayKumar Python3
# Python3 program for the above approach
# Function to find the maximum number
# of diamonds that can be gained in
# exactly K minutes
def maxDiamonds(A, N, K):
# Stores all the array elements
pq = []
# Push all the elements to the
# priority queue
for i in range(N):
pq.append(A[i])
pq.sort()
# Stores the required result
ans = 0
# Loop while the queue is not
# empty and K is positive
while (len(pq) > 0 and K > 0):
pq.sort()
# Store the top element
# from the pq
top = pq[len(pq) - 1]
# Pop it from the pq
pq = pq[0:len(pq) - 1]
# Add it to the answer
ans += top
# Divide it by 2 and push it
# back to the pq
top = top // 2;
pq.append(top)
K -= 1
# Print the answer
print(ans)
# Driver Code
if __name__ == '__main__':
A = [ 2, 1, 7, 4, 2 ]
K = 3
N = len(A)
maxDiamonds(A, N, K)
# This code is contributed by SURENDRA_GANGWAR14时间复杂度： O((N + K)*log N)
辅助空间： O(N)
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