前N个非负整数的连续位差之和

给定正整数N ，任务是找出从0到N的所有连续位差之和。
注意：如果两个(例如(3，4))的位长度不同，则在开头(011，100)后面附加0。
例子：

Input: N = 3 
Output: 4 
Explanation: 
Bit differences of (0, 1) + (1, 2) + (2, 3) = 1 + 2 + 1 = 4.
Input: N = 7 
Output: 11

天真的方法：
最简单的方法是按位比较范围内的两个连续值，并找出这两个数字相差多少位。将此差值加到总和上。这样获得的最终总和是所需的解决方案。
时间复杂度： O(N)
方法：
为了优化上述解决方案，需要进行以下观察：

• 连续的数字位差遵循一个模式，即每个等于(2 ⁱ )的值X与先前的数字以及(2 ⁱ – 1)数在X和(2 ⁱ之上)的位差为(i +1)。 – 1)X以下的数字遵循相同的模式。
• 对于X = 4(2 ² )，i = 2的位差为(2 +1)，数字(1、2、3)和(5、6、7)遵循相同的位差模式。

For X = 4, the pattern is as follows:
NUM   BIT Diff
1     1(0, 1)
2     2(1, 2)
3     1(2, 3)
4     3(3, 4)
5     1(4, 5)
6     2(5, 6)
7     1(6, 7)

请按照以下步骤解决问题：

• 对于每个N ，找到小于或等于N的最接近的数字，即2的幂。假设该数字为M。
• 对于所有小于M的数字，可以使用以下递归方法找出连续位差的总和。

 
 

Count(i) = (i + 1) + 2 * Count(i – 1) 
where i is the exponent of the nearest power of 2.

 

• 初始化大小为65(基于0的索引)的数组，以存储使用递归函数Count()时获得的值，以便将来，如果需要相同的Count()值，则可以直接获取它们而无需递归调用Count()函数以节省时间。
• 使用以下公式，对大于M的剩余数重复相同的过程。

 
 

Sum = Sum + (i+1) + Count(i-1)

 

例如：
对于N = 10，使用Count(3)计算最接近的2的幂，即M = 8，然后对剩余的大于8的数重复该过程。
下面是上述方法的实现：

C++

// C++ program for the above problem
#include 
using namespace std;
 
long long a[65] = { 0 };
 
// Recursive function to count
// the sum of bit differences
// of numbers from 1 to
// pow(2, (i+1)) - 1
long long Count(int i)
{
    // base cases
    if (i == 0)
        return 1;
 
    else if (i < 0)
        return 0;
 
    // Recursion call if the sum
    // of bit difference of numbers
    // around i are not calculated
    if (a[i] == 0) {
        a[i] = (i + 1) + 2 * Count(i - 1);
        return a[i];
    }
 
    // return the sum of bit
    // differences if already
    // calculated
    else
        return a[i];
}
 
// Function to calculate the
// sum of bit differences up to N
long long solve(long long n)
{
    long long i, sum = 0;
 
    while (n > 0) {
 
        // nearest smaller power
        // of 2
        i = log2(n);
 
        // remaining numbers
        n = n - pow(2, i);
 
        // calculate the count
        // of bit diff
        sum = sum + (i + 1) + Count(i - 1);
    }
    return sum;
}
 
// Driver code
int main()
{
    long long n = 7;
    cout << solve(n) << endl;
    return 0;
}

Java

// Java program for the above problem
import java.util.*;
class GFG{
   
static int a[] = new int[65];
   
// Recursive function to count
// the sum of bit differences
// of numbers from 1 to
// pow(2, (i+1)) - 1
static int Count(int i)
{
     
    // base cases
    if (i == 0)
        return 1;
  
    else if (i < 0)
        return 0;
  
    // Recursion call if the sum
    // of bit difference of numbers
    // around i are not calculated
    if (a[i] == 0)
    {
        a[i] = (i + 1) + 2 * Count(i - 1);
        return a[i];
    }
  
    // return the sum of bit
    // differences if already
    // calculated
    else
        return a[i];
}
  
// Function to calculate the
// sum of bit differences up to N
static int solve(int n)
{
    int i, sum = 0;
  
    while (n > 0)
    {
  
        // nearest smaller power
        // of 2
        i = (int)(Math.log(n) / Math.log(2));
 
  
        // remaining numbers
        n = n - (int)Math.pow(2, i);
  
        // calculate the count
        // of bit diff
        sum = sum + (i + 1) + Count(i - 1);
    }
    return sum;
}
  
// Driver code
public static void main(String[] args)
{
    int  n = 7;
    System.out.println(solve(n));
}
}
 
// This code is contributed by rock_cool

Python3

# Python3 program for the above problem
import math
 
a = [0] * 65
 
# Recursive function to count
# the sum of bit differences
# of numbers from 1 to
# pow(2, (i+1)) - 1
def Count(i):
     
    # Base cases
    if (i == 0):
        return 1
 
    elif (i < 0):
        return 0
 
    # Recursion call if the sum
    # of bit difference of numbers
    # around i are not calculated
    if (a[i] == 0):
        a[i] = (i + 1) + 2 * Count(i - 1)
        return a[i]
 
    # Return the sum of bit
    # differences if already
    # calculated
    else:
        return a[i]
 
# Function to calculate the
# sum of bit differences up to N
def solve(n):
     
    sum = 0
 
    while (n > 0):
 
        # Nearest smaller power
        # of 2
        i = int(math.log2(n))
 
        # Remaining numbers
        n = n - pow(2, i)
 
        # Calculate the count
        # of bit diff
        sum = sum + (i + 1) + Count(i - 1)
     
    return sum
 
# Driver code
n = 7
 
print(solve(n))
 
# This code is contributed by sanjoy_62

C#

// C# program for the above problem
using System;
class GFG{
 
static int []a = new int[65];
 
// Recursive function to count
// the sum of bit differences
// of numbers from 1 to
// pow(2, (i+1)) - 1
static int Count(int i)
{
     
    // base cases
    if (i == 0)
        return 1;
 
    else if (i < 0)
        return 0;
 
    // Recursion call if the sum
    // of bit difference of numbers
    // around i are not calculated
    if (a[i] == 0)
    {
        a[i] = (i + 1) + 2 * Count(i - 1);
        return a[i];
    }
 
    // return the sum of bit
    // differences if already
    // calculated
    else
        return a[i];
}
 
// Function to calculate the
// sum of bit differences up to N
static int solve(int n)
{
    int i, sum = 0;
 
    while (n > 0)
    {
 
        // nearest smaller power
        // of 2
        i = (int)(Math.Log(n) / Math.Log(2));
 
 
        // remaining numbers
        n = n - (int)Math.Pow(2, i);
 
        // calculate the count
        // of bit diff
        sum = sum + (i + 1) + Count(i - 1);
    }
    return sum;
}
 
// Driver code
public static void Main(String[] args)
{
    int n = 7;
    Console.Write(solve(n));
}
}
 
// This code is contributed by shivanisinghss2110

Javascript

 
    // Javascript program for the above problem
     
    let a = new Array(65);
    a.fill(0);
   
    // Recursive function to count
    // the sum of bit differences
    // of numbers from 1 to
    // pow(2, (i+1)) - 1
    function Count(i)
    {
        // base cases
        if (i == 0)
            return 1;
 
        else if (i < 0)
            return 0;
 
        // Recursion call if the sum
        // of bit difference of numbers
        // around i are not calculated
        if (a[i] == 0) {
            a[i] = (i + 1) + 2 * Count(i - 1);
            return a[i];
        }
 
        // return the sum of bit
        // differences if already
        // calculated
        else
            return a[i];
    }
 
    // Function to calculate the
    // sum of bit differences up to N
    function solve(n)
    {
        let i, sum = 0;
 
        while (n > 0) {
 
            // nearest smaller power
            // of 2
            i = parseInt(Math.log(n) / Math.log(2), 10);
 
            // remaining numbers
            n = n - parseInt(Math.pow(2, i), 10);
 
            // calculate the count
            // of bit diff
            sum = sum + (i + 1) + Count(i - 1);
        }
        return sum;
    }
 
    let n = 7;
    document.write(solve(n));
     

输出：

11

时间复杂度： O(log(N))
辅助空间： O(1)