📌  相关文章
📜  表示一个数字所需的给定幂的最小数量2

📅  最后修改于: 2021-04-22 02:12:59             🧑  作者: Mango

给定一个整数x和一个数组arr []，每个元素的幂为2。该任务是从数组中找到2的最小整数次幂，相加后得出x 。如果不可能用给定的数组元素表示x ，则打印-1 。

例子：

方法：对于2的乘方，让我们计算给定数组中元素的数量，其值等于此值。我们称它为cnt 。显然，我们可以贪婪地获得值x(因为元素的所有较少值都是元素的所有较大值的除数)。

现在让我们将2的所有幂从30迭代到0 。 deg为当前度。我们可以取min(x / 2 deg ，cnt deg )个元素，其值等于2 deg 。任其治愈。将cur加到答案中，并从x减去2度* cur 。重复该过程，直到无法再减小x为止。如果在遍历所有幂之后，x仍然不为零，则打印-1 。否则，打印答案。

下面是上述方法的实现：

C++
// C++ implementation of the approach
#include 
using namespace std;
  
// Function to return the minimum number
// of given integer powers of 2 required
// to represent a number as sum of these powers
int power_of_two(int n, int a[], int x)
{
  
    // To store the count of powers of two
    vector cnt(31);
  
    for (int i = 0; i < n; ++i) {
  
        // __builtin_ctz(a[i]) returns the count 
        // of trailing 0s in a[i]
        ++cnt[__builtin_ctz(a[i])];
    }
  
    int ans = 0;
    for (int i = 30; i >= 0 && x > 0; --i) {
  
        // If current power is available
        // in the array and can be used
        int need = min(x >> i, cnt[i]);
  
        // Update the answer
        ans += need;
  
        // Reduce the number
        x -= (1 << i) * need;
    }
  
    // If the original number is not reduced to 0
    // It cannot be represented as the sum
    // of the given powers of 2
    if (x > 0)
        ans = -1;
  
    return ans;
}
  
// Driver code
int main()
{
    int arr[] = { 2, 2, 4, 4, 8 }, x = 6;
    int n = sizeof(arr) / sizeof(arr[0]);
  
    cout << power_of_two(n, arr, x);
  
    return 0;
}

Java
// Java implementation of the approach 
import java.util.*;
  
class GFG
{
      
// __builtin_ctz(a[i]) returns the count 
// of trailing 0s in a[i] 
static int __builtin_ctz(int a)
{
    int count = 0;
    for(int i = 0; i < 40; i++)
    if(((a >> i) & 1) == 0)
    {
        count++;
    }
    else 
        break;
    return count;
}
  
// Function to return the minimum number 
// of given integer powers of 2 required 
// to represent a number as sum of these powers 
static int power_of_two(int n, int a[], int x) 
{ 
  
    // To store the count of powers of two 
    Vector cnt = new Vector(); 
      
    for (int i = 0; i < 31; ++i)
        cnt.add(0);
  
    for (int i = 0; i < n; ++i) 
    { 
  
        // __builtin_ctz(a[i]) returns the count 
        // of trailing 0s in a[i] 
          
        cnt.set(__builtin_ctz(a[i]), 
        (cnt.get(__builtin_ctz(a[i]))==null) ?
        1 : cnt.get(__builtin_ctz(a[i]))+1); 
    } 
  
    int ans = 0; 
    for (int i = 30; i >= 0 && x > 0; --i) 
    { 
  
        // If current power is available 
        // in the array and can be used 
        int need = Math.min(x >> i, cnt.get(i)); 
  
        // Update the answer 
        ans += need; 
  
        // Reduce the number 
        x -= (1 << i) * need; 
    } 
  
    // If the original number is not reduced to 0 
    // It cannot be represented as the sum 
    // of the given powers of 2 
    if (x > 0) 
        ans = -1; 
  
    return ans; 
} 
  
// Driver code 
public static void main(String args[])
{ 
    int arr[] = { 2, 2, 4, 4, 8 }, x = 6; 
    int n = arr.length; 
    System.out.println(power_of_two(n, arr, x)); 
} 
}
  
// This code is contributed by Arnab Kundu

python
# Python3 implementation of the approach
  
# Function to return the minimum number
# of given eger powers of 2 required
# to represent a number as sum of these powers
def power_of_two( n, a, x):
  
  
    # To store the count of powers of two
    cnt=[0 for i in range(31)]
  
    for i in range(n):
        # __builtin_ctz(a[i]) returns the count
        # of trailing 0s in a[i]
        count = 0
        xx = a[i]
        while ((xx & 1) == 0):
            xx = xx >> 1
            count += 1
  
        cnt[count]+=1
  
    ans = 0
    for i in range(30,-1,-1):
        if x<=0:
            continue
  
        # If current power is available
        # in the array and can be used
        need = min(x >> i, cnt[i])
  
        # Update the answer
        ans += need
  
        # Reduce the number
        x -= (1 << i) * need
  
  
    # If the original number is not reduced to 0
    # It cannot be represented as the sum
    # of the given powers of 2
    if (x > 0):
        ans = -1
  
    return ans
  
  
# Driver code
  
arr=[2, 2, 4, 4, 8 ]
x = 6
n = len(arr)
  
print(power_of_two(n, arr, x))
  
# This code is contributed by mohit kumar 29

C#
// C# implementation of the approach 
using System;
  
class GFG
{
      
// __builtin_ctz(a[i]) returns the count 
// of trailing 0s in a[i] 
static int __builtin_ctz(int a)
{
    int count = 0;
    for(int i = 0; i < 40; i++)
    if(((a >> i) & 1) == 0)
    {
        count++;
    }
    else
        break;
    return count;
}
  
// Function to return the minimum number 
// of given integer powers of 2 required 
// to represent a number as sum of these powers 
static int power_of_two(int n, int []a, int x) 
{ 
  
    // To store the count of powers of two 
    int[] cnt = new int[32]; 
  
    for (int i = 0; i < n; ++i) 
    { 
  
        // __builtin_ctz(a[i]) returns the count 
        // of trailing 0s in a[i] 
          
        cnt[__builtin_ctz(a[i])] = 
        cnt[__builtin_ctz(a[i])] == 
        0?1 : cnt[__builtin_ctz(a[i])] + 1; 
    } 
  
    int ans = 0; 
    for (int i = 30; i >= 0 && x > 0; --i) 
    { 
  
        // If current power is available 
        // in the array and can be used 
        int need = Math.Min(x >> i, cnt[i]); 
  
        // Update the answer 
        ans += need; 
  
        // Reduce the number 
        x -= (1 << i) * need; 
    } 
  
    // If the original number is not reduced to 0 
    // It cannot be represented as the sum 
    // of the given powers of 2 
    if (x > 0) 
        ans = -1; 
  
    return ans; 
} 
  
// Driver code 
static void Main()
{ 
    int []arr = { 2, 2, 4, 4, 8 };
    int x = 6; 
    int n = arr.Length; 
    Console.WriteLine(power_of_two(n, arr, x)); 
} 
}
  
// This code is contributed by mits

输出： 
2