从0到N的数字的位差总和|套装2

给定数字N ，任务是为从0到N的每个连续数字计算二进制表示形式中相应不同位的总数。

例子：

Input: N = 5 
Output: 8 
Explanation: 
Binary Representation of numbers are: 
0 -> 000, 
1 -> 001, 
2 -> 010, 
3 -> 011, 
4 -> 100, 
5 -> 101 
Between 1 and 0 -> 1 bit is different 
Between 2 and 1 -> 2 bits are different 
Between 3 and 2 -> 1 bit is different 
Between 4 and 3 -> 3 bits are different 
Between 5 and 4 -> 1 bit is different 
Total = 1 + 2 + 1 + 3 + 1 = 8
Input: N = 11 
Output: 19  

对于朴素和有效的方法，请参阅本文的上一篇文章。
更有效的方法：为了优化上述方法，我们可以使用递归。要解决此问题，需要进行以下观察

Number:     0 1 2 3 4  5  6  7
Difference: 1 2 1 3 1  2  1  4
Sum:        1 3 4 7 8 10 11 15

我们可以观察到，对于N = [1,2,3,4，…] ，连续元素中不同位的总和形成序列[1、3、4、7、8，……] 。因此，本系列的^第N^个术语将是我们所需的答案，可以将其计算为：

a(n) = a(n / 2) + n; with base case as a(1) = 1  

以下是上述递归方法的实现：

C++

// C++ program to find the sum
// of bit differences between
// consecutive numbers
// from 0 to N using recursion
 
#include 
using namespace std;
 
// Recursive function to find sum
// of different bits between
// consecutive numbers from 0 to N
int totalCountDifference(int n)
{
 
    // Base case
    if (n == 1)
        return 1;
 
    // Calculate the Nth term
    return n
           + totalCountDifference(n / 2);
}
 
// Driver Code
int main()
{
    // Given Number
    int N = 5;
 
    // Function Call
    cout << totalCountDifference(N);
    return 0;
}

Java

// Java program to find the sum
// of bit differences between
// consecutive numbers from
// 0 to N using recursion
class GFG{
 
// Recursive function to find sum
// of different bits between
// consecutive numbers from 0 to N
static int totalCountDifference(int n)
{
 
    // Base case
    if (n == 1)
        return 1;
 
    // Calculate the Nth term
    return n + totalCountDifference(n / 2);
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given number
    int N = 5;
 
    // Function call
    System.out.println(totalCountDifference(N));
}
}
 
// This code is contributed by himanshu77

Python3

# Python3 program to find the sum
# of bit differences between
# consecutive numbers from
# 0 to N using recursion
 
# Recursive function to find sum
# of different bits between
# consecutive numbers from 0 to N
def totalCountDifference (n):
 
    # Base case
    if (n == 1):
        return 1
 
    # Calculate the Nth term
    return n + totalCountDifference(n // 2)
 
# Driver code
 
# Given number
N = 5
 
# Function call
print(totalCountDifference(N))
 
# This code is contributed by himanshu77

C#

// C# program to find the sum
// of bit differences between
// consecutive numbers from
// 0 to N using recursion
using System;
 
class GFG{
 
// Recursive function to find sum
// of different bits between
// consecutive numbers from 0 to N
static int totalCountDifference(int n)
{
 
    // Base case
    if (n == 1)
        return 1;
 
    // Calculate the Nth term
    return n + totalCountDifference(n / 2);
}
 
// Driver Code
public static void Main()
{
     
    // Given number
    int N = 5;
 
    // Function call
    Console.WriteLine(totalCountDifference(N));
}
}
 
// This code is contributed by himanshu77

Javascript

// JavaScript program to find the sum
// of bit differences between
// consecutive numbers from
// 0 to N using recursion
 
// Recursive function to find sum
// of different bits between
// consecutive numbers from 0 to N
function totalCountDifference(n)
{
   
    // Base case
    if (n == 1)
        return 1;
   
    // Calculate the Nth term
    return n + totalCountDifference(Math.floor(n / 2));
}
     
// Driver Code
     
           // Given number
    let N = 5;
   
    // Function call
    document.write(totalCountDifference(N));
              

输出：

8

时间复杂度： O(log ₂ N)
辅助空间： O(1)

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