给定数组的索引范围[L，R]中的按位OR查询

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📜 给定数组的索引范围[L，R]中的按位OR查询

📅 最后修改于: 2021-05-25 04:26:31 🧑 作者: Mango

给定由范围[L，R]组成的N和Q个查询的数组arr [ ] 。任务是找到该索引范围内所有元素的按位“或”。

例子：

Input: arr[] = {1, 3, 1, 2, 3, 4}, q[] = {{0, 1}, {3, 5}}
Output:
3
7
1 OR 3 = 3
2 OR 3 OR 4 = 7

Input: arr[] = {1, 2, 3, 4, 5}, q[] = {{0, 4}, {1, 3}}
Output:
7
7

为什么编程需要懂一点英语

天真的方法：遍历该范围并找到该范围内所有数字的按位“或”。每个查询将花费O(n)时间。

有效的方法：如果我们看一下整数作为二进制数，我们可以很容易地看到，条件我我们的答案是一套^次的一点是，我的任何整数的范围[L，R]^个位应设置。
因此，我们将为每个位计算前缀计数。我们将使用它来查找设置了^第i位的范围内的整数数量。如果其大于0，那么还将设置我们答案的^第i位。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
#define MAX 100000
#define bitscount 32
using namespace std;
  
// Array to store bit-wise
// prefix count
int prefix_count[bitscount][MAX];
  
// Function to find the prefix sum
void findPrefixCount(int arr[], int n)
{
  
    // Loop for each bit
    for (int i = 0; i < bitscount; i++) {
        // Loop to find prefix count
        prefix_count[i][0] = ((arr[0] >> i) & 1);
        for (int j = 1; j < n; j++) {
            prefix_count[i][j] = ((arr[j] >> i) & 1);
            prefix_count[i][j] += prefix_count[i][j - 1];
        }
    }
}
  
// Function to answer query
int rangeOr(int l, int r)
{
  
    // To store the answer
    int ans = 0;
  
    // Loop for each bit
    for (int i = 0; i < bitscount; i++) {
        // To store the number of variables
        // with ith bit set
        int x;
        if (l == 0)
            x = prefix_count[i][r];
        else
            x = prefix_count[i][r]
                - prefix_count[i][l - 1];
  
        // Condition for ith bit
        // of answer to be set
        if (x != 0)
            ans = (ans | (1 << i));
    }
  
    return ans;
}
  
// Driver code
int main()
{
    int arr[] = { 7, 5, 3, 5, 2, 3 };
    int n = sizeof(arr) / sizeof(int);
  
    findPrefixCount(arr, n);
  
    int queries[][2] = { { 1, 3 }, { 4, 5 } };
    int q = sizeof(queries) / sizeof(queries[0]);
  
    for (int i = 0; i < q; i++)
        cout << rangeOr(queries[i][0],
                        queries[i][1])
             << endl;
  
    return 0;
}

Java

// Java implementation of the approach 
import java.io.*;
  
class GFG 
{
      
static int MAX = 100000;
static int bitscount = 32;
  
// Array to store bit-wise
// prefix count
static int [][]prefix_count = new int [bitscount][MAX];
  
// Function to find the prefix sum
static void findPrefixCount(int arr[], int n)
{
  
    // Loop for each bit
    for (int i = 0; i < bitscount; i++)
    {
        // Loop to find prefix count
        prefix_count[i][0] = ((arr[0] >> i) & 1);
        for (int j = 1; j < n; j++) 
        {
            prefix_count[i][j] = ((arr[j] >> i) & 1);
            prefix_count[i][j] += prefix_count[i][j - 1];
        }
    }
}
  
// Function to answer query
static int rangeOr(int l, int r)
{
  
    // To store the answer
    int ans = 0;
  
    // Loop for each bit
    for (int i = 0; i < bitscount; i++)
    {
        // To store the number of variables
        // with ith bit set
        int x;
        if (l == 0)
            x = prefix_count[i][r];
        else
            x = prefix_count[i][r]
                - prefix_count[i][l - 1];
  
        // Condition for ith bit
        // of answer to be set
        if (x != 0)
            ans = (ans | (1 << i));
    }
  
    return ans;
}
  
// Driver code
public static void main (String[] args) 
{
      
    int arr[] = { 7, 5, 3, 5, 2, 3 };
    int n = arr.length;
    findPrefixCount(arr, n);
    int queries[][] = { { 1, 3 }, { 4, 5 } };
    int q = queries.length;
    for (int i = 0; i < q; i++)
            System.out.println (rangeOr(queries[i][0],queries[i][1]));
  
}
}
  
// This code is contributed by Tushil.

Python3

# Python3 implementation of the approach 
  
import numpy as np
  
MAX = 100000
bitscount = 32
  
# Array to store bit-wise 
# prefix count 
prefix_count = np.zeros((bitscount,MAX)); 
  
# Function to find the prefix sum 
def findPrefixCount(arr, n) :
  
    # Loop for each bit 
    for i in range(0, bitscount) :
        # Loop to find prefix count 
        prefix_count[i][0] = ((arr[0] >> i) & 1); 
          
        for j in range(1, n) :
            prefix_count[i][j] = ((arr[j] >> i) & 1); 
            prefix_count[i][j] += prefix_count[i][j - 1]; 
  
# Function to answer query 
def rangeOr(l, r) : 
  
    # To store the answer 
    ans = 0; 
  
    # Loop for each bit 
    for i in range(bitscount) :
          
        # To store the number of variables 
        # with ith bit set 
        x = 0; 
          
        if (l == 0) :
            x = prefix_count[i][r]; 
        else :
            x = prefix_count[i][r] - prefix_count[i][l - 1]; 
  
        # Condition for ith bit 
        # of answer to be set 
        if (x != 0) :
            ans = (ans | (1 << i)); 
              
    return ans; 
  
  
# Driver code 
if __name__ == "__main__" : 
  
    arr = [ 7, 5, 3, 5, 2, 3 ]; 
    n = len(arr);
  
    findPrefixCount(arr, n); 
  
    queries = [ [ 1, 3 ], [ 4, 5 ] ]; 
      
    q = len(queries); 
  
    for i in range(q) :
        print(rangeOr(queries[i][0], queries[i][1]));
  
# This code is contributed by AnkitRai01
\

C#

// C# implementation of the approach 
using System;
  
class GFG 
{
      
static int MAX = 100000;
static int bitscount = 32;
  
// Array to store bit-wise
// prefix count
static int [,]prefix_count = new int [bitscount,MAX];
  
// Function to find the prefix sum
static void findPrefixCount(int []arr, int n)
{
  
    // Loop for each bit
    for (int i = 0; i < bitscount; i++)
    {
        // Loop to find prefix count
        prefix_count[i,0] = ((arr[0] >> i) & 1);
        for (int j = 1; j < n; j++) 
        {
            prefix_count[i,j] = ((arr[j] >> i) & 1);
            prefix_count[i,j] += prefix_count[i,j - 1];
        }
    }
}
  
// Function to answer query
static int rangeOr(int l, int r)
{
  
    // To store the answer
    int ans = 0;
  
    // Loop for each bit
    for (int i = 0; i < bitscount; i++)
    {
        // To store the number of variables
        // with ith bit set
        int x;
        if (l == 0)
            x = prefix_count[i,r];
        else
            x = prefix_count[i,r]
                - prefix_count[i,l - 1];
  
        // Condition for ith bit
        // of answer to be set
        if (x != 0)
            ans = (ans | (1 << i));
    }
  
    return ans;
}
  
// Driver code
public static void Main (String[] args) 
{
      
    int []arr = { 7, 5, 3, 5, 2, 3 };
    int n = arr.Length;
    findPrefixCount(arr, n);
    int [,]queries = { { 1, 3 }, { 4, 5 } };
    int q = queries.GetLength(0);
    for (int i = 0; i < q; i++)
            Console.WriteLine(rangeOr(queries[i,0],queries[i,1]));
  
}
}
  
// This code is contributed by 29AjayKumar

PHP

> $i) & 1);
        for ($j = 1; $j < $n; $j++)
        {
            $prefix_count[$i][$j] = (($arr[$j] >> $i) & 1);
            $prefix_count[$i][$j] += $prefix_count[$i][$j - 1];
        }
    }
}
  
// Function to answer query
function rangeOr($l, $r)
{
    global $MAX,$bitscount,$prefix_count;
      
    // To store the answer
    $ans = 0;
  
    // Loop for each bit
    for ($i = 0; $i < $bitscount; $i++) 
    {
        // To store the number of variables
        // with ith bit set
          
        if ($l == 0)
            $x = $prefix_count[$i][$r];
        else
            $x = $prefix_count[$i][$r]
                - $prefix_count[$i][l - 1];
  
        // Condition for ith bit
        // of answer to be set
        if ($x != 0)
            $ans = ($ans | (1 << $i));
    }
  
    return $ans;
}
  
    // Driver code
    $arr =array( 7, 5, 3, 5, 2, 3 );
    $n = sizeof($arr) / sizeof($arr[0]);
  
    findPrefixCount($arr, $n);
  
    $queries = array(array( 1, 3 ), array( 4, 5 ));
    $q = sizeof($queries) / sizeof($queries[0]);
  
    for ($i = 0; $i < $q; $i++)
        echo rangeOr($queries[$i][0],
                        $queries[$i][1])."\n";
  
    return 0;
      
    // This code is contributed by ChitraNayal
?>

输出：

7
3

预计算的时间复杂度为O(n)，每个查询都可以在O(1)中回答