在给定数组的索引范围 [L, R] 中查询按位 AND

给定一个由N和Q查询组成的数组arr[] ，其中包含范围[L, R] 。任务是找到该索引范围内所有元素的按位与。
例子：

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

天真的方法：遍历范围并找到该范围内所有数字的按位与。对于每个查询，这将花费 O(n) 时间。
有效的方法：如果我们将整数视为二进制数，我们可以很容易地看到，我们答案的^第i位要设置的条件是应该设置范围[L, R]内所有整数的^第i位。
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。如果它等于范围的大小，那么我们的答案的^第i位也将被设置。
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 rangeAnd(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 == r - l + 1)
            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 << rangeAnd(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 rangeAnd(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 == r - l + 1)
            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 (rangeAnd(queries[i][0],queries[i][1]));
         
 
}
}
 
// This code is contributed by ajit.

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 == r - l + 1) :
            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 rangeAnd(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 == r - l + 1)
            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(rangeAnd(queries[i,0],queries[i,1]));
         
}
}
 
// This code contributed by Rajput-Ji

Javascript

   
    // Javascript implementation of the approach
     
    let MAX = 100000;
    let bitscount =32;
 
    // Array to store bit-wise
    // prefix count
    let prefix_count = new Array(bitscount);
     
    for (let i = 0; i < bitscount; i++)
    {
        prefix_count[i] = new Array(MAX);
        for (let j = 0; j < MAX; j++)
        {
            prefix_count[i][j] = 0;
        }
    }
 
    // Function to find the prefix sum
    function findPrefixCount(arr, n)
    {
        // Loop for each bit
        for (let i = 0; i < bitscount; i++)
        {
 
            // Loop to find prefix count
            prefix_count[i][0] = ((arr[0] >> i) & 1);
            for (let 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 rangeAnd(l, r)
    {
 
        // To store the answer
        let ans = 0;
 
        // Loop for each bit
        for (let i = 0; i < bitscount; i++)
        {
            // To store the number of variables
            // with ith bit set
            let 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 == r - l + 1)
                ans = (ans | (1 << i));
        }
 
        return ans;
    }
 
    let arr = [ 7, 5, 3, 5, 2, 3 ];
    let n = arr.length;
       
    findPrefixCount(arr, n);
       
    let queries = [ [ 1, 3 ], [ 4, 5 ] ];
    let q = queries.length;
       
    for (let i = 0; i < q; i++)
          document.write(rangeAnd(queries[i][0],queries[i][1]) + "</br>");

输出：

1
2

预计算的时间复杂度为 O(n) 并且每个查询都可以在 O(1) 中回答

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