以0到N的二进制表示形式翻转/翻转的总位数

给定一个整数N ，任务是查找被翻转以按顺序获得从0到N的所有数字的位数。

例子：

Input: N = 5 
Output: 8 
Explanation: 
Let’s represent numbers from 0 to 5 in binary: 
000 -> 001 : 1 bit toggled 
001 -> 010 : 2 bits toggled 
010 -> 011 : 1 bit toggled 
011 -> 100 : 3 bits toggled 
100 -> 101 : 1 bit toggled 
Hence, 8 bits toggled

Input: N = 1 
Output: 1

方法：
请按照以下步骤解决问题：

• 需要解决以下问题：

 
 The rightmost bit toggles every time. 
(000) -> (001) -> (010) -> (011) 
So, the contribution of this bit to the count will be N. 
The next bit toggles after every 2 numbers. 
(000) -> (001) -> (010) -> (011) -> (100) 
Hence, the contribution of this bit to the count will be N/2.

 

• 因此，我们可以得出结论，第i^个最低有效位将对触发计数贡献N /(2 ⁱ ) 。
• 因此，其中i在[0，log ₂ N]范围内的N /(2 ⁱ )之和给出了所需的答案。
• 因此，初始化变量ans 。将N加到ans并将N更新为N / 2 。重复此过程，直到N变为0，以获得最终结果。

下面是上述方法的实现：

C++

// C++ program to count
// the number of toggles
// required to generate
// all numbers from 0 to N
 
#include 
using namespace std;
 
typedef long long int ll;
 
// Function to count and print
// the required number of
// toggles
void solve(ll N)
{
    // Store the count
    // of toggles
    ll ans = 0;
 
    while (N != 0) {
 
        // Add the contribution
        // of the current LSB
        ans += N;
 
        // Update N
        N /= 2;
    }
    // Print the result
    cout << ans << endl;
}
// Driver code
int main()
{
    ll N = 5;
    solve(N);
    return 0;
}

Java

// Java program to count the
// number of toggles required
// to generate all numbers
// from 0 to N
class GFG{
 
// Function to count and print
// the required number of
// toggles
static void solve(int N)
{
     
    // Store the count
    // of toggles
    int ans = 0;
 
    while (N != 0)
    {
         
        // Add the contribution
        // of the current LSB
        ans += N;
 
        // Update N
        N /= 2;
    }
     
    // Print the result
    System.out.println(ans);
}
 
// Driver code
public static void main(String []args)
{
    int N = 5;
     
    solve(N);
}
}
 
// This code is contributed by Ritik Bansal

Python3

# Python3 program to count
# the number of toggles
# required to generate
# all numbers from 0 to N
 
# Function to count and pr
# the required number of
# toggles
def solve(N):
     
    # Store the count
    # of toggles
    ans = 0
 
    while (N != 0):
 
        # Add the contribution
        # of the current LSB
        ans += N
 
        # Update N
        N //= 2
     
    # Print the result
    print(ans)
 
# Driver code
N = 5
 
solve(N)
 
# This code is contributed by code_hunt

C#

// C# program to count the
// number of toggles required
// to generate all numbers
// from 0 to N
using System;
 
class GFG{
 
// Function to count and print
// the required number of
// toggles
static void solve(int N)
{
     
    // Store the count
    // of toggles
    int ans = 0;
 
    while (N != 0)
    {
         
        // Add the contribution
        // of the current LSB
        ans += N;
 
        // Update N
        N /= 2;
    }
     
    // Print the result
    Console.Write(ans);
}
 
// Driver code
public static void Main(string []args)
{
    int N = 5;
     
    solve(N);
}
}
 
// This code is contributed by rock_cool

Javascript

 
// Javascript program to count the
// number of toggles required to
// generate all numbers from 0 to N
 
// Function to count and print
// the required number of
// toggles
function solve(N)
{
     
    // Store the count
    // of toggles
    let ans = 0;
 
    while (N != 0)
    {
         
        // Add the contribution
        // of the current LSB
        ans += N;
 
        // Update N
        N = parseInt(N / 2, 10);
    }
     
    // Print the result
    document.write(ans);
}
     
// Driver code   
let N = 5;
 
solve(N);
     
// This code is contributed by divyeshrabadiya07
 

输出：

8

时间复杂度： O(log N)
辅助空间： O(1)