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📜  范围的按位或(或|)

📅  最后修改于: 2021-05-25 04:14:02             🧑  作者: Mango

给定两个整数L和R。确定[L，R](包括两个端点)范围内所有整数的按位“或”。

例子： 

Input: L = 3, R = 8
Output: 15
3 | 4 | 5 | 6 | 7 | 8 = 15

Input: L = 12, R = 18
Output: 31
12 | 13 | 14 | 15 | 16 | 17 | 18 = 31

天真的方法是遍历L和R之间的所有整数，并对所有数字进行按位或运算。

一种有效的方法是遵循以下步骤：

  1. 在两个数字(L和R)中找到最高有效位(MSB)的位置
  2. 如果两个MSB的位置都不同，则将从max(MSB1，MSB2)开始的所有位(包括此不同的位)设置为Oth位，即，将所有0≤i≤max(MSB1，MSB2)的值相加(1 << i)在答案中。
  3. 如果两个MSB的位置相同，则
    • 将对应于MSB的该位置1或在答案中添加值(1 << MSB)。
    • 从两个数字(L和R)中减去值(1 << MSB)。
    • 重复步骤1，2和3。

下面给出的是L = 18和R = 21时上述算法的工作原理。 

L = 18, R = 21
The result is initially 0.
The position of Most Significant Bit in L = 4
Position of Most Significant Bit in R = 4
Since positions are same, add value (1 << 4) i.e. 16 to the result.

Subtract (1 << 4) from L, L becomes 2.
Subtract (1 << 4) from R, R becomes 5.

Now, Position of MSB in L is 1
Position of MSB in R is 2
Since positions are different all value (1 << i) for all 
0 ≤ i ≤ max(MSB1, MSB2)
i.e. Add ((1 << 2) + (1 << 1) + (1 << 0)) = 7
Hence, final result is 16 + 7 = 23.

下面是上述方法的实现。

C++
// C++ Program to find the bitwise
// OR of all the intgers in range L-R
#include 
using namespace std;
  
// Returns the Most Significant Bit
// Position (MSB)
int MSBPosition(long long int N)
{
    int msb_p = -1;
    while (N) {
        N = N >> 1;
        msb_p++;
    }
    return msb_p;
}
  
// Returns the Bitwise OR of all
// integers between L and R
long long int findBitwiseOR(long long int L,
                            long long int R)
{
    long long int res = 0;
  
    // Find the MSB position in L
    int msb_p1 = MSBPosition(L);
  
    // Find the MSB position in R
    int msb_p2 = MSBPosition(R);
  
    while (msb_p1 == msb_p2) {
        long long int res_val = (1 << msb_p1);
  
        // Add this value until msb_p1 and
        // msb_p2 are same;
        res += res_val;
  
        L -= res_val;
        R -= res_val;
  
        // Calculate msb_p1 and msb_p2
        msb_p1 = MSBPosition(L);
        msb_p2 = MSBPosition(R);
    }
    // Find the max of msb_p1 and msb_p2
    msb_p1 = max(msb_p1, msb_p2);
  
    // Set all the bits from msb_p1 upto
    // 0th bit in the result
    for (int i = msb_p1; i >= 0; i--) {
        long long int res_val = (1 << i);
        res += res_val;
    }
    return res;
}
  
// Driver Code
int main()
{
    int L = 12, R = 18;
    cout << findBitwiseOR(L, R) << endl;
    return 0;
}

Java
// Java Program to find 
// the bitwise OR of all 
// the intgers in range L-R
import java.io.*;
  
class GFG
{
  
// Returns the Most Significant 
// Bit Position (MSB)
static int MSBPosition(long N)
{
    int msb_p = -1;
    while (N > 0)
    {
        N = N >> 1;
        msb_p++;
    }
    return msb_p;
}
  
// Returns the Bitwise 
// OR of all integers 
// between L and R
static long findBitwiseOR(long L,
                          long R)
{
    long res = 0;
  
    // Find the MSB 
    // position in L
    int msb_p1 = MSBPosition(L);
  
    // Find the MSB 
    // position in R
    int msb_p2 = MSBPosition(R);
  
    while (msb_p1 == msb_p2)
    {
        long res_val = (1 << msb_p1);
  
        // Add this value until 
        // msb_p1 and msb_p2 are same;
        res += res_val;
  
        L -= res_val;
        R -= res_val;
  
        // Calculate msb_p1 
        // and msb_p2
        msb_p1 = MSBPosition(L);
        msb_p2 = MSBPosition(R);
    }
      
    // Find the max of 
    // msb_p1 and msb_p2
    msb_p1 = Math.max(msb_p1, 
                      msb_p2);
  
    // Set all the bits 
    // from msb_p1 upto
    // 0th bit in the result
    for (int i = msb_p1; i >= 0; i--) 
    {
        long res_val = (1 << i);
        res += res_val;
    }
    return res;
}
  
// Driver Code
public static void main (String[] args) 
{
    int L = 12, R = 18;
    System.out.println(findBitwiseOR(L, R));
}
}
  
// This code is contributed
// by anuj_67.

Python3
# Python3 Program to find the bitwise 
# OR of all the intgers in range L-R 
  
# Returns the Most Significant Bit 
# Position (MSB) 
def MSBPosition(N) : 
   
    msb_p = -1
    while (N) :
        N = N >> 1 
        msb_p += 1 
   
    return msb_p 
   
  
# Returns the Bitwise OR of all 
# integers between L and R 
def findBitwiseOR(L, R) :
  
    res = 0
  
    # Find the MSB position in L 
    msb_p1 = MSBPosition(L)
  
    # Find the MSB position in R 
    msb_p2 = MSBPosition(R) 
  
    while (msb_p1 == msb_p2) :
        res_val = (1 << msb_p1)
  
        # Add this value until msb_p1 and 
        # msb_p2 are same; 
        res += res_val
  
        L -= res_val
        R -= res_val
  
        # Calculate msb_p1 and msb_p2 
        msb_p1 = MSBPosition(L) 
        msb_p2 = MSBPosition(R) 
       
    # Find the max of msb_p1 and msb_p2 
    msb_p1 = max(msb_p1, msb_p2) 
  
    # Set all the bits from msb_p1 upto 
    # 0th bit in the result 
    for i in range(msb_p1, -1, -1) : 
        res_val = (1 << i)
        res += res_val
      
    return res 
   
  
# Driver Code 
if __name__ == "__main__" :
   
    L , R= 12 ,18 
    print(findBitwiseOR(L, R)) 
  
# This code is contributed by Ryuga

C#
// C# Program to find 
// the bitwise OR of all 
// the intgers in range L-R
using System;
  
class GFG
{
  
// Returns the Most Significant 
// Bit Position (MSB)
static int MSBPosition(long N)
{
    int msb_p = -1;
    while (N > 0)
    {
        N = N >> 1;
        msb_p++;
    }
    return msb_p;
}
  
// Returns the Bitwise 
// OR of all integers 
// between L and R
static long findBitwiseOR(long L,
                          long R)
{
    long res = 0;
  
    // Find the MSB 
    // position in L
    int msb_p1 = MSBPosition(L);
  
    // Find the MSB 
    // position in R
    int msb_p2 = MSBPosition(R);
  
    while (msb_p1 == msb_p2)
    {
        long res_val = (1 << msb_p1);
  
        // Add this value until 
        // msb_p1 and msb_p2 are same;
        res += res_val;
  
        L -= res_val;
        R -= res_val;
  
        // Calculate msb_p1 
        // and msb_p2
        msb_p1 = MSBPosition(L);
        msb_p2 = MSBPosition(R);
    }
      
    // Find the max of 
    // msb_p1 and msb_p2
    msb_p1 = Math.Max(msb_p1, 
                      msb_p2);
  
    // Set all the bits 
    // from msb_p1 upto
    // 0th bit in the result
    for (int i = msb_p1; i >= 0; i--) 
    {
        long res_val = (1 << i);
        res += res_val;
    }
    return res;
}
  
// Driver Code
public static void Main () 
{
    int L = 12, R = 18;
    Console.WriteLine(findBitwiseOR(L, R));
}
}
  
// This code is contributed
// by anuj_67.

PHP
> 1;
        $msb_p++;
    }
    return $msb_p;
}
  
// Returns the Bitwise 
// OR of all integers 
// between L and R
function findBitwiseOR($L, $R)
{
    $res = 0;
  
    // Find the MSB
    // position in L
    $msb_p1 = MSBPosition($L);
  
    // Find the MSB
    // position in R
    $msb_p2 = MSBPosition($R);
  
    while ($msb_p1 == $msb_p2) 
    {
        $res_val = (1 << $msb_p1);
  
        // Add this value until 
        // msb_p1 and msb_p2 are same;
        $res += $res_val;
  
        $L -= $res_val;
        $R -= $res_val;
  
        // Calculate msb_p1 
        // and msb_p2
        $msb_p1 = MSBPosition($L);
        $msb_p2 = MSBPosition($R);
    }
      
    // Find the max of
    // msb_p1 and msb_p2
    $msb_p1 = max($msb_p1, 
                  $msb_p2);
  
    // Set all the bits from msb_p1 
    // upto 0th bit in the result
    for ($i = $msb_p1; $i >= 0; $i--) 
    {
        $res_val = (1 << $i);
        $res += $res_val;
    }
    return $res;
}
  
// Driver Code
$L = 12; $R = 18;
echo findBitwiseOR($L, $R);
  
// This code is contributed
// by anuj_67.
?>

输出： 
31

时间复杂度： O(N)，其中N是最高有效位。