从0到N的连续数字的汉明差总和|套装2

给定数字N ，任务是找到从0到N的连续数字的汉明差之和。

Hamming Distance between two integers is the number of bits which are different at the same position in both numbers. 
 

例子：

Input: 5 
Output: 8 
Explanation: 
Difference between (0, 1) = 1, (1, 2) = 2, 
(2, 3) = 1, (3, 4) = 3, (4, 5) = 1. 
So the total sum is 1 + 2 + 1 + 3 + 1 = 8
Input: 9 
Output: 16 
 

朴素和对数方法：朴素的方法是指从0到N的数字的位差之和，而对数方法是基于检查N是否为2的幂来计算和。

方法：在本文中，讨论了一种基于观察从LSB到MSB的所有位中发生的更改数量的方法。
请按照以下步骤解决问题：

• 可以看出，最低有效位将改变N次。第二最低有效位将改变下限(N / 2)次(即，如果N为奇数，则为(N – 1)/ 2)，如果N为偶数，则为N / 2。因此，第i^个比特将改变floor(N / 2 ⁱ )次。
  

• 因此，为了解决这个问题，我们可以简单地存储和

  N + floor(N/2) + … floor(N/2ⁱ)

• 直到该项不为零。

下面是上述方法的实现：

C++

// C++ Program to implement 
// the above approach 
  
#include  
using namespace std; 
  
// Function to calculate and 
// return the hamming distance 
// between all consecutive 
// numbers from 0 to N 
int TotalHammingDistance(int n) 
{ 
    int i = 1, sum = 0; 
    while (n / i > 0) { 
        sum = sum + n / i; 
        i = i * 2; 
    } 
    return sum; 
} 
  
// Driver Code 
int main() 
{ 
    int N = 9; 
    cout << TotalHammingDistance(N); 
    return 0; 
}

Java

// Java program to implement the
// above approach
import java.util.*;
  
class GFG{
  
// Function to calculate and
// return the hamming distance
// between all consecutive 
// numbers from 0 to N
static int TotalHammingDistance(int n)
{
    int i = 1, sum = 0;
    while (n / i > 0) 
    {
        sum = sum + n / i;
        i = i * 2;
    }
    return sum;
}
  
// Driver code
public static void main(String[] args)
{
    int N = 9;
      
    System.out.println(TotalHammingDistance(N));
}
}
  
// This code is contributed by offbeat

Python3

# Python3 program to implement 
# the above approach 
  
# Function to calculate and 
# return the hamming distance 
# between all consecutive 
# numbers from 0 to N 
def TotalHammingDistance(n): 
      
    i = 1
    sum = 0
      
    while (n // i > 0): 
        sum = sum + n // i 
        i = i * 2
          
    return sum
  
# Driver Code 
if __name__ == '__main__': 
      
    N = 9
      
    print(TotalHammingDistance(N)) 
  
# This code is contributed by mohit kumar 29

C#

// C# Program to implement 
// the above approach 
using System; 
class GFG{ 
  
// Function to calculate and 
// return the hamming distance 
// between all consecutive 
// numbers from 0 to N 
static int TotalHammingDistance(int n) 
{ 
    int i = 1, sum = 0; 
    while (n / i > 0) 
    { 
        sum = sum + n / i; 
        i = i * 2; 
    } 
    return sum; 
} 
  
// Driver Code 
public static void Main() 
{ 
    int N = 9; 
    Console.Write(TotalHammingDistance(N)); 
} 
} 
  
// This code is contributed by Code_Mech

输出：

16

时间复杂度： O(logN)
辅助空间： O(1)