使用位掩码将2的幂乘以所需的总和

给定整数N ，任务是要找到数字，将其乘以2的幂后，将得到整数N。

例子：

Input: N = 12345
Output: 0, 3, 4, 5, 12, 13
Explanation:
12345 = 2^0 + 2^3 + 2^4 + 2^5 + 2^12 + 2^13

Input: N = 10000
Output: 4, 8, 9, 10, 13
Explanation:
10000 = 2^4 + 2^8 + 2^9 + 2^10 + 2^13

方法：

• 由于每个数字都可以表示为2的幂的和，因此当在N的二进制表示形式中设置^第i^个位时，任务是打印每个i 。

• Illustration:
  (29)₁₀ = (11101)₂
  Thus, in 29, {0, 2, 3, 4} are the indices of the set bits.
  2⁰ + 2² + 2³ + 2⁴
  = 1 + 4 + 8 + 16
  = 29

• 为了检查^第i位是否已设置，我们只需要检查：

  if( N & (1 << i) ) == 1
  Illustration:
  (29)₁₀ = (11101)₂
  0^th bit: 11101 & 1 = 1
  1^st bit: 11101 & 10 = 0
  2^nd bit: 11101 & 100 = 1
  3^rd bit: 11101 & 1000 = 1
  4^th bit: 11101 & 10000 = 1

下面是上述方法的实现。

C++

// C++ Program to split N
// into numbers which when
// added after being raised
// to the power of 2 gives N
  
#include 
using namespace std;
  
// Function to print the numbers
// which raised to power of 2
// adds up to N
void PowerOfTwo(long long N)
{
    for (int i = 0; i < 64; i++) {
        long long x = 1;
  
        // Checking if i-th bit is
        // set in N or not by
        // shifting 1 i bits to
        // the left and performing
        // AND with N
        if (N & (x << i))
            cout << i << " ";
    }
}
  
// Driver Code
int main()
{
  
    long long N = 12345;
    PowerOfTwo(N);
  
    return 0;
}

Java

// Java Program to split N
// into numbers which when
// added after being raised
// to the power of 2 gives N
class GFG{
  
// Function to print the numbers
// which raised to power of 2
// adds up to N
static void PowerOfTwo(long N)
{
    for (int i = 0; i < 64; i++) 
    {
        long x = 1;
  
        // Checking if i-th bit is
        // set in N or not by
        // shifting 1 i bits to
        // the left and performing
        // AND with N
        if ((N & (x << i)) > 0)
            System.out.print(i + " ");
    }
}
  
// Driver Code
public static void main(String[] args)
{
    long N = 12345;
    PowerOfTwo(N);
}
}
  
// This code is contributed by Amit Katiyar

Python3

# Python3 program to split N
# into numbers which when
# added after being raised
# to the power of 2 gives N
  
# Function to print the numbers
# which raised to power of 2
# adds up to N
def PowerOfTwo(N):
  
    for i in range(0, 64): 
        x = 1
          
        # Checking if i-th bit is
        # set in N or not by
        # shifting 1 i bits to
        # the left and performing
        # AND with N
        if (N & (x << i)) > 0:
            print(i, end = " ")
          
# Driver Code 
if __name__ == "__main__": 
      
    # Given number
    N = 12345;
  
    # Function call
    PowerOfTwo(N)
  
# This code is contributed by rock_cool

C#

// C# Program to split N
// into numbers which when
// added after being raised
// to the power of 2 gives N
using System;
class GFG{
  
// Function to print the numbers
// which raised to power of 2
// adds up to N
static void PowerOfTwo(long N)
{
    for (int i = 0; i < 64; i++) 
    {
        long x = 1;
  
        // Checking if i-th bit is
        // set in N or not by
        // shifting 1 i bits to
        // the left and performing
        // AND with N
        if ((N & (x << i)) > 0)
            Console.Write(i + " ");
    }
}
  
// Driver Code
public static void Main()
{
    long N = 12345;
    PowerOfTwo(N);
}
}
  
// This code is contributed by Nidhi_biet

输出：

0 3 4 5 12 13

时间复杂度： O(log N)
空间复杂度： O(1)

有关替代方法，请参考2的幂以求和