查找范围 [2, N] 中的所有 M 使得直到 M 的按位或等于直到 M-1 的值

给定一个整数N ，任务是在[2, N]范围内找到所有可能的整数M ，使得直到 M 的所有正值的按位或与直到M-1的所有正值的按位或相同。

例子：

Input: N = 4
Output: 1
Explanation: Bitwise OR till 3 = 1 | 2 | 3 = 3.
Bitwse OR till 2 = 1 | 2 = 3.

Input: N = 7
Output: 4

方法：解决此问题的方法基于以下观察：

Consider p(x) to the bitwise OR till x. So p(x) = 1 | 2 | 3 | . . . | (x-1) | x
Given p(x) = 1 | 2 | 3 | . . . | x – 1 | x. Therefore, p(x + 1) will be different from p(x) if there is a new “1” bit in (x + 1) that isn’t present in the binary sequence of p(x).

Now, let us observe the pattern:

Decimal Number	Binary Number
1	1
2	10
3	11
4	100
5	101
6	110
7	111
8	1000
9	1001

We can see that a new “1” bit that hasn’t previously appeared in the range [1, x] appears at every power of 2.
As such, p(x) = 1 | 2 | 3 | . . . | x – 1 | x
= 2^a+1 – 1, where a = log₂x.
This implies that, for a given a, there will be ( 2^{a + 1} – 2^a – 1 ) values of x where p(x) = p(x – 1).

编程需要懂一点英语

请按照以下步骤解决此问题：

计算a = log ₂ N。
迭代从1到a的 2 的幂（例如使用变量exp ），并将ans （最初为 0）增加(2 ^{exp + 1} – 2 ^exp – 1) 。
最后，计算N和2 ^a之间的对通过将(n – 2 ^a )添加到ans 。

C++

// C++ code to implement the approach
  
#include 
using namespace std;
  
// Function to calculate
// total matches in the range
int checkXORrange(int n)
{
    int ans = 0;
    int a = log2(n);
  
    for (int exp = 1; exp <= a; exp++)
        ans += pow(2, exp) - pow(2, exp - 1) - 1;
  
    ans += n - pow(2, a);
    return ans;
}
  
// Driver code
int main()
{
    int N = 7;
  
    // Function call
    cout << checkXORrange(N) << endl;
    return 0;
}

C

// C code to implement the approach
  
#include 
#include 
  
// Function to calculate
// total matches in the range
int checkXORrange(int n)
{
    int ans = 0;
    int a = log2(n);
    for (int exp = 1; exp <= a; exp++)
        ans += pow(2, exp) - pow(2, exp - 1) - 1;
    ans += n - pow(2, a);
    return ans;
}
  
// Driver code
int main()
{
    int N = 7;
  
    // Function call
    printf("%d\n", checkXORrange(N));
    return 0;
}

Python3

# Python3 code to implement the approach
  
import math
  
# Function to calculate 
# total matches in the range
def checkXORrange(n):
    ans = 0
    a = int(math.log2(n))
    for exp in range(1, a + 1):
        ans += 2 ** exp - 2 ** (exp - 1) - 1
    ans += n - 2 ** a
    return ans
  
# Driver Code
if __name__ == "__main__":
    N = 7
    print(checkXORrange(N))

输出

时间复杂度： O(log ₂ N)
辅助空间： O(1)