与 N 的按位异或大于 N 的 X 的所有可能值的计数

给定一个整数N计算X的值的数量，使得X ⊕ N > N ，  其中⊕表示按位异或运算

例子：

Input: N = 10
Output: 5 
Explanation: The five possible value satisfying the above condition are:
1⊕10 = 11, 4⊕10 = 14, 5⊕10 = 15, 6⊕10 = 12, 7⊕10 = 13

Input: N = 8
Output: 7
Explanation: The seven possible value satisfying the above condition are:
1⊕8 = 9, 2⊕8 = 10, 3⊕8 = 11, 4⊕8 = 12, 5⊕8 = 13, 6⊕8 = 14, 7⊕8 = 15

朴素方法：解决此问题的简单方法是从1 迭代到 N并在X XOR N >N for 0 < X < N时增加计数。最后，返回值的数量。

时间复杂度： O(N)
辅助空间： O(1)

有效方法：最有效的方法是找到最接近 2 的下一个幂的数字，该数字已设置所有位，最后从原始数字中减去它。下面给出了解决方案的数学观察：

If number N can be written by using k bits then number  X must use k-1 bits at most because:-

• If a number X has a same bits as number N Xoring both the number will result in a number less than N which wont satisfy our condition X xor N > N.
• If a number X is greater than number N then Xoring both the number  will always result X is greater than N which wont satisfy our inequality.
  So the problem reduces to find the count of the number that lies in the given range [x , 2^k – 1]. We are taking 2^k – 1 because this is the maximum number that can be form when all its n bits are set.
• The require number equals  2^k – 1 -N .

请按照以下步骤解决问题：

• 求 2 的最接近的下一个幂。
• 从最接近的 2 的幂中减去 1，以便给定数字的所有位都已设置
• 最后，从原始数字中减去并返回它

下面是上述方法的实现：

C++

// C++ program for find the count of
// number of value of X such that  N XOR X >N
 
#include 
using namespace std;
 
// Function for  calculating the total
// possible value satisfy the condition
// X⊕N > N and 0 < X < N
void maximumXor(long long N)
{
    long long num = N;
    long long int count;
 
    count = 0;
    while (N > 0) {
 
        // Count the total number of bits
        count++;
        N = N / 2;
    }
    long long next_power = ((long long)1 << count);
 
    // Finding the next power of 2
    // of the given number
    long long result = next_power - 1;
 
    // Finding the number nearest to
    // the nextpower of 2 which has all
    // its bits set
    cout << result - num;
}
 
// Driver Code
int main()
{
    int N = 10;
    maximumXor(N);
    return 0;
}

Java

// Java program for find the count of
// number of value of X such that  N XOR X >N
import java.util.*;
public class GFG {
 
  // Function for  calculating the total
  // possible value satisfy the condition
  // X⊕N > N and 0 < X < N
  static void maximumXor(long N)
  {
    long num = N;
    long count = 0;
 
    count = 0;
    while (N > 0) {
 
      // Count the total number of bits
      count++;
      N = N / 2;
    }
    long next_power = ((long)1 << count);
 
    // Finding the next power of 2
    // of the given number
    long result = next_power - 1;
 
    // Finding the number nearest to
    // the nextpower of 2 which has all
    // its bits set
    System.out.print(result - num);
  }
 
  // Driver Code
  public static void main(String args[])
  {
    int N = 10;
    maximumXor(N);
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Python3

# python3 program for find the count of
# number of value of X such that N XOR X >N
 
# Function for calculating the total
# possible value satisfy the condition
# X⊕N > N and 0 < X < N
def maximumXor(N):
 
    num = N
    count = 0
 
    while (N > 0):
 
        # Count the total number of bits
        count += 1
        N = N // 2
 
    next_power = (1 << count)
 
    # Finding the next power of 2
    # of the given number
    result = next_power - 1
 
    # Finding the number nearest to
    # the nextpower of 2 which has all
    # its bits set
    print(result - num)
 
# Driver Code
if __name__ == "__main__":
 
    N = 10
    maximumXor(N)
 
# This code is contributed by rakeshsahni

C#

// C# program for find the count of
// number of value of X such that  N XOR X >N
using System;
 
public class GFG
{
 
  // Function for  calculating the total
  // possible value satisfy the condition
  // X⊕N > N and 0 < X < N
  static void maximumXor(long N)
  {
    long num = N;
    long count = 0;
 
    count = 0;
    while (N > 0) {
 
      // Count the total number of bits
      count++;
      N = N / 2;
    }
    long next_power = (1 << (int)count);
 
    // Finding the next power of 2
    // of the given number
    long result = next_power - 1;
 
    // Finding the number nearest to
    // the nextpower of 2 which has all
    // its bits set
    Console.Write(result - num);
  }
 
  // Driver Code
  public static void Main(String []args)
  {
    int N = 10;
    maximumXor(N);
  }
}
 
// This code is contributed by 29AjayKumar

Javascript

       // JavaScript code for the above approach
 
       // Function for  calculating the total
       // possible value satisfy the condition
       // X⊕N > N and 0 < X < N
       function maximumXor(N) {
           let num = N;
           let count;
 
           count = 0;
           while (N > 0) {
 
               // Count the total number of bits
               count++;
               N = Math.floor(N / 2);
           }
           let next_power = (1 << count);
 
           // Finding the next power of 2
           // of the given number
           let result = next_power - 1;
 
           // Finding the number nearest to
           // the nextpower of 2 which has all
           // its bits set
           document.write(result - num);
       }
 
       // Driver Code
       let N = 10;
       maximumXor(N);
 
    // This code is contributed by Potta Lokesh
   

输出

5

时间复杂度：O(log(N))
辅助空间：O(1)