给定大小为N的数组arr []和整数K ，该任务是通过最多将K个元素重复除以2来找到可以减少为1的最大数组元素数的任务。
注意：对于奇数数组元素，采用其ceil的除法值。
例子：
Input: arr[] = {5, 8, 4, 7}, K = 5
Output: 2
Explanation:
5 needs 3 operations(5→3→2→1).
8 needs 3 operations(8→4→2→1).
4 needs 2 operations(4→2→1).
7 needs 3 operations(7→4→2→1)
Therefore, in 5 operations, the maximum number of array elements that can be reduced to 1 are 2, either (4, 5), (4, 8) or (4, 7).
Input: arr[] = {5, 8, 5, 7}, K = 5
Output: 1
方法：为了最大程度地增加元素数量，我们的想法是按升序对数组进行排序，并从第一个索引开始减少元素，并通过减少第i个元素所需的操作数来减少K。请按照以下步骤解决问题：
- 初始化一个变量，例如cnt ，以存储所需数量的元素。
- 以递增顺序对数组arr []进行排序。
- 使用变量i遍历数组arr []并执行以下步骤：
- 存储将arr [i]减少到1所需的操作数为opr = ceil( log2(arr [i]) ) 。
- 由opr递减K。
- 如果K的值小于0 ，则跳出循环。否则，将cnt增加1 。
- 完成上述步骤后，打印cnt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 count the maximum number of
// array elements that can be reduced to 1
// by repeatedly dividing array elements by 2
void findMaxNumbers(int arr[], int n, int k)
{
// Sort the array in ascending order
sort(arr, arr + n);
// Store the count of array elements
int cnt = 0;
// Traverse the array
for (int i = 0; i < n; i++) {
// Store the number of operations
// required to reduce arr[i] to 1
int opr = ceil(log2(arr[i]));
// Decrement k by opr
k -= opr;
// If k becomes less than 0,
// then break out of the loop
if (k < 0) {
break;
}
// Increment cnt by 1
cnt++;
}
// Print the answer
cout << cnt;
}
// Driver Code
int main()
{
int arr[] = { 5, 8, 4, 7 };
int N = sizeof(arr) / sizeof(arr[0]);
int K = 5;
findMaxNumbers(arr, N, K);
return 0;
} Java
// Java program for the above approach
import java.util.*;
public class GFG
{
// Function to count the maximum number of
// array elements that can be reduced to 1
// by repeatedly dividing array elements by 2
static void findMaxNumbers(int arr[], int n, int k)
{
// Sort the array in ascending order
Arrays.sort(arr);
// Store the count of array elements
int cnt = 0;
// Traverse the array
for (int i = 0; i < n; i++)
{
// Store the number of operations
// required to reduce arr[i] to 1
int opr = (int)Math.ceil((Math.log(arr[i]) / Math.log(2)));
// Decrement k by opr
k -= opr;
// If k becomes less than 0,
// then break out of the loop
if (k < 0) {
break;
}
// Increment cnt by 1
cnt++;
}
// Print the answer
System.out.println(cnt);
}
// Driver Code
public static void main(String args[])
{
int arr[] = { 5, 8, 4, 7 };
int N = arr.length;
int K = 5;
findMaxNumbers(arr, N, K);
}
}
// This code is contributed by jana_sayantan.Python3
# Python3 program to implement
# the above approach
import math
# Function to count the maximum number of
# array elements that can be reduced to 1
# by repeatedly dividing array elements by 2
def findMaxNumbers(arr, n, k) :
# Sort the array in ascending order
arr.sort()
# Store the count of array elements
cnt = 0
# Traverse the array
for i in range(n):
# Store the number of operations
# required to reduce arr[i] to 1
opr = math.ceil(math.log2(arr[i]))
# Decrement k by opr
k -= opr
# If k becomes less than 0,
# then break out of the loop
if (k < 0) :
break
# Increment cnt by 1
cnt += 1
# Prthe answer
print(cnt)
# Driver Code
arr = [ 5, 8, 4, 7 ]
N = len(arr)
K = 5
findMaxNumbers(arr, N, K)
# This code is contributed by splevel62.C#
// C# program for the above approach
using System;
public class GFG
{
// Function to count the maximum number of
// array elements that can be reduced to 1
// by repeatedly dividing array elements by 2
static void findMaxNumbers(int[] arr, int n, int k)
{
// Sort the array in ascending order
Array.Sort(arr);
// Store the count of array elements
int cnt = 0;
// Traverse the array
for (int i = 0; i < n; i++)
{
// Store the number of operations
// required to reduce arr[i] to 1
int opr = (int)Math.Ceiling((Math.Log(arr[i]) / Math.Log(2)));
// Decrement k by opr
k -= opr;
// If k becomes less than 0,
// then break out of the loop
if (k < 0) {
break;
}
// Increment cnt by 1
cnt++;
}
// Print the answer
Console.Write(cnt);
}
// Driver Code
public static void Main(String[] args)
{
int[] arr = { 5, 8, 4, 7 };
int N = arr.Length;
int K = 5;
findMaxNumbers(arr, N, K);
}
}
// This code is contributed by susmitakundugoaldanga.输出：
2时间复杂度： O(N * log N)
辅助空间： O(1)