使用 Fenwick Tree 对给定范围内的元素进行 XOR 更新

给定一个整数数组A[]和由以下两种类型的查询组成的数组Q ：

• (1, L, R) ：返回索引L和R之间存在的所有元素的异或。
• (2, I, val) ：将 A[I]更新为A[I] XOR val 。

任务是使用 Fenwick 树解决每个查询并为第一种类型的每个查询打印 XOR。
例子：

Input: A[] = {2, 1, 1, 3, 2, 3, 4, 5, 6, 7, 8, 9} 
Q = {{ 1, 3, 8}, {2, 4, 6}, {1, 3, 8}} 
Output : 
XOR of elements in range 3 to 8 is 5 
XOR of elements in range 3 to 8 is 3 
Explanation: 
XOR of subarray { 3, 2, 3, 4, 5, 6 } is 5. 
After 2nd query arr[4] gets replaced by 4. 
Xor of subarray { 3, 4, 3, 4, 5, 6 } is 3.
Input :A[] = {2, 1, 1, 3, 2, 3, 4, 5, 6, 7, 8, 9} 
Q = {{1, 0, 9}, {2, 3, 6}, {2, 5, 5}, {2, 8, 1}, {1, 0, 9}} 
Output : 
XOR of elements in range 0 to 9 is 0 
XOR of elements in range 0 to 9 is 2

方法：

• 对于类型 1 的查询，使用 getXor() 返回范围[1, R]和范围[1, L-1]中元素的异或。
• 在 getXor() 中，对于i从索引到其所有祖先直到 1，继续使用BITree[i]计算异或。为了在getXor()视图中获得第 i 个索引的祖先，我们只需要通过i = i – i&(-i)从 i 中减去 LSB(最低有效位) 。最后返回最终的 XOR 值。
• 对于类型 2 的查询，将 A[index] 更新为A[index] ^ val 。通过调用updateBIT()更新 BITree[] 中包含此元素的所有范围。
• 在 updateBIT() 中，对于从索引到其所有祖先直到N 的每个i ，将 BITree[i] 更新为 BITree[i] ^ val 。为了在updateBit()视图中获得第 i 个索引的祖先，我们只需要通过i = i + i&(-i)从 i 添加 LSB(最低有效位) 。

下面是上述方法的实现：

C++

// C++ Program to find XOR of
// elements in given range [L, R].
 
#include 
using namespace std;
 
// Returns XOR of arr[0..index].
// This function assumes that the
// array is preprocessed and partial
// XORs of array elements are stored
// in BITree[].
int getXOR(int BITree[], int index)
{
    int ans = 0;
    index += 1;
 
    // Traverse ancestors
    // of BITree[index]
    while (index > 0) {
 
        // XOR current element
        // of BIT to ans
        ans ^= BITree[index];
 
        // Update index to that
        // of the parent node in
        // getXor() view by
        // subtracting LSB(Least
        // Significant Bit)
        index -= index & (-index);
    }
    return ans;
}
 
// Updates the Binary Index Tree by
// replacing all ancestores of index
// by their respective XOR with val
void updateBIT(int BITree[], int n,
               int index, int val)
{
    index = index + 1;
 
    // Traverse all ancestors
    // and XOR with 'val'.
    while (index <= n) {
 
        // XOR 'val' to current
        // node of BIT
        BITree[index] ^= val;
 
        // Update index to that
        // of the parent node in
        // updateBit() view by
        // adding LSB(Least
        // Significant Bit)
        index += index & (-index);
    }
}
 
// Constructs and returns a Binary
// Indexed Tree for the given array
int* constructBITree(int arr[], int n)
{
    // Create and initialize
    // the Binary Indexed Tree
    int* BITree = new int[n + 1];
    for (int i = 1; i <= n; i++)
        BITree[i] = 0;
 
    // Store the actual values in
    // BITree[] using update()
    for (int i = 0; i < n; i++)
        updateBIT(BITree, n, i, arr[i]);
 
    return BITree;
}
 
int rangeXor(int BITree[], int l, int r)
{
    return getXOR(BITree, r)
           ^ getXOR(BITree, l - 1);
}
 
// Driver Code
int main()
{
    int A[] = { 2, 1, 1, 3, 2, 3,
                4, 5, 6, 7, 8, 9 };
    int n = sizeof(A) / sizeof(A[0]);
 
    vector > q
        = { { 1, 0, 9 },
            { 2, 3, 6 },
            { 2, 5, 5 },
            { 2, 8, 1 },
            { 1, 0, 9 } };
 
    // Create the Binary Indexed Tree
    int* BITree = constructBITree(A, n);
 
    // Solve each query in Q
    for (int i = 0; i < q.size(); i++) {
        int id = q[i][0];
 
        if (id == 1) {
            int L = q[i][1];
            int R = q[i][2];
            cout << "XOR of elements "
                 << "in given range is "
                 << rangeXor(BITree, L, R)
                 << "\n";
        }
        else {
            int idx = q[i][1];
            int val = q[i][2];
            A[idx] ^= val;
 
            // Update the values of all
            // ancestors of idx
            updateBIT(BITree, n, idx, val);
        }
    }
 
    return 0;
}

Java

// Java Program to find XOR of
// elements in given range [L, R].
import java.util.*;
 
class GFG{
 
// Returns XOR of arr[0..index].
// This function assumes that the
// array is preprocessed and partial
// XORs of array elements are stored
// in BITree[].
static int getXOR(int BITree[], int index)
{
    int ans = 0;
    index += 1;
 
    // Traverse ancestors
    // of BITree[index]
    while (index > 0)
    {
 
        // XOR current element
        // of BIT to ans
        ans ^= BITree[index];
 
        // Update index to that
        // of the parent node in
        // getXor() view by
        // subtracting LSB(Least
        // Significant Bit)
        index -= index & (-index);
    }
    return ans;
}
 
// Updates the Binary Index Tree by
// replacing all ancestores of index
// by their respective XOR with val
static void updateBIT(int BITree[], int n,
                      int index, int val)
{
    index = index + 1;
 
    // Traverse all ancestors
    // and XOR with 'val'.
    while (index <= n)
    {
 
        // XOR 'val' to current
        // node of BIT
        BITree[index] ^= val;
 
        // Update index to that
        // of the parent node in
        // updateBit() view by
        // adding LSB(Least
        // Significant Bit)
        index += index & (-index);
    }
}
 
// Constructs and returns a Binary
// Indexed Tree for the given array
static int[] constructBITree(int arr[], int n)
{
    // Create and initialize
    // the Binary Indexed Tree
    int []BITree = new int[n + 1];
    for (int i = 1; i <= n; i++)
        BITree[i] = 0;
 
    // Store the actual values in
    // BITree[] using update()
    for (int i = 0; i < n; i++)
        updateBIT(BITree, n, i, arr[i]);
 
    return BITree;
}
 
static int rangeXor(int BITree[], int l, int r)
{
    return getXOR(BITree, r) ^
           getXOR(BITree, l - 1);
}
 
// Driver Code
public static void main(String[] args)
{
    int A[] = { 2, 1, 1, 3, 2, 3,
                4, 5, 6, 7, 8, 9 };
    int n = A.length;
 
    int [][]q = { { 1, 0, 9 },
                  { 2, 3, 6 },
                  { 2, 5, 5 },
                  { 2, 8, 1 },
                  { 1, 0, 9 } };
 
    // Create the Binary Indexed Tree
    int []BITree = constructBITree(A, n);
 
    // Solve each query in Q
    for (int i = 0; i < q.length; i++)
    {
        int id = q[i][0];
 
        if (id == 1)
        {
            int L = q[i][1];
            int R = q[i][2];
            System.out.print("XOR of elements " +
                           "in given range is " +
                         rangeXor(BITree, L, R) + "\n");
        }
        else
        {
            int idx = q[i][1];
            int val = q[i][2];
            A[idx] ^= val;
 
            // Update the values of all
            // ancestors of idx
            updateBIT(BITree, n, idx, val);
        }
    }
}
}
 
// This code is contributed by Princi Singh

Python3

# Python3 program to find XOR of
# elements in given range [L, R].
 
# Returns XOR of arr[0..index].
# This function assumes that the
# array is preprocessed and partial
# XORs of array elements are stored
# in BITree[].
def getXOR(BITree, index):
 
    ans = 0
    index += 1
 
    # Traverse ancestors
    # of BITree[index]
    while (index > 0):
     
        # XOR current element
        # of BIT to ans
        ans ^= BITree[index]
 
        # Update index to that
        # of the parent node in
        # getXor() view by
        # subtracting LSB(Least
        # Significant Bit)
        index -= index & (-index)
     
    return ans
 
# Updates the Binary Index Tree by
# replacing all ancestores of index
# by their respective XOR with val
def updateBIT(BITree, n, index, val):
 
    index = index + 1
 
    # Traverse all ancestors
    # and XOR with 'val'.
    while (index <= n):
     
        # XOR 'val' to current
        # node of BIT
        BITree[index] ^= val
 
        # Update index to that
        # of the parent node in
        # updateBit() view by
        # adding LSB(Least
        # Significant Bit)
        index += index & (-index)
     
# Constructs and returns a Binary
# Indexed Tree for the given array
def constructBITree(arr, n):
 
    # Create and initialize
    # the Binary Indexed Tree
    BITree = [0] * (n + 1)
     
    # Store the actual values in
    # BITree[] using update()
    for i in range(n):
        updateBIT(BITree, n, i, arr[i])
 
    return BITree
 
def rangeXor(BITree, l, r):
 
    return (getXOR(BITree, r) ^
            getXOR(BITree, l - 1))
            
# Driver Code
if __name__ == "__main__":
     
    A = [ 2, 1, 1, 3, 2, 3,
          4, 5, 6, 7, 8, 9 ]
    n = len(A)
 
    q = [ [ 1, 0, 9 ], [ 2, 3, 6 ],
          [ 2, 5, 5 ], [ 2, 8, 1 ],
          [ 1, 0, 9 ] ]
 
    # Create the Binary Indexed Tree
    BITree = constructBITree(A, n)
 
    # Solve each query in Q
    for i in range(len(q)):
        id = q[i][0]
 
        if (id == 1):
            L = q[i][1]
            R = q[i][2]
            print("XOR of elements in "
                  "given range is ",
                  rangeXor(BITree, L, R))
        else:
            idx = q[i][1]
            val = q[i][2]
            A[idx] ^= val
 
            # Update the values of all
            # ancestors of idx
            updateBIT(BITree, n, idx, val)
 
# This code is contributed by jana_sayantan

C#

// C# program to find XOR of
// elements in given range [L, R].
using System;
 
class GFG{
 
// Returns XOR of arr[0..index].
// This function assumes that the
// array is preprocessed and partial
// XORs of array elements are stored
// in BITree[].
static int getXOR(int []BITree, int index)
{
    int ans = 0;
    index += 1;
 
    // Traverse ancestors
    // of BITree[index]
    while (index > 0)
    {
 
        // XOR current element
        // of BIT to ans
        ans ^= BITree[index];
 
        // Update index to that
        // of the parent node in
        // getXor() view by
        // subtracting LSB(Least
        // Significant Bit)
        index -= index & (-index);
    }
    return ans;
}
 
// Updates the Binary Index Tree by
// replacing all ancestores of index
// by their respective XOR with val
static void updateBIT(int []BITree, int n,
                      int index, int val)
{
    index = index + 1;
 
    // Traverse all ancestors
    // and XOR with 'val'.
    while (index <= n)
    {
 
        // XOR 'val' to current
        // node of BIT
        BITree[index] ^= val;
 
        // Update index to that
        // of the parent node in
        // updateBit() view by
        // adding LSB(Least
        // Significant Bit)
        index += index & (-index);
    }
}
 
// Constructs and returns a Binary
// Indexed Tree for the given array
static int[] constructBITree(int []arr,
                             int n)
{
     
    // Create and initialize
    // the Binary Indexed Tree
    int []BITree = new int[n + 1];
    for(int i = 1; i <= n; i++)
       BITree[i] = 0;
 
    // Store the actual values in
    // BITree[] using update()
    for(int i = 0; i < n; i++)
       updateBIT(BITree, n, i, arr[i]);
 
    return BITree;
}
 
static int rangeXor(int []BITree, int l,
                                  int r)
{
    return getXOR(BITree, r) ^
           getXOR(BITree, l - 1);
}
 
// Driver Code
public static void Main(String[] args)
{
    int []A = { 2, 1, 1, 3, 2, 3,
                4, 5, 6, 7, 8, 9 };
    int n = A.Length;
     
    int [,]q = { { 1, 0, 9 },
                 { 2, 3, 6 },
                 { 2, 5, 5 },
                 { 2, 8, 1 },
                 { 1, 0, 9 } };
 
    // Create the Binary Indexed Tree
    int []BITree = constructBITree(A, n);
 
    // Solve each query in Q
    for(int i = 0; i < q.GetLength(0); i++)
    {
       int id = q[i, 0];
        
       if (id == 1)
       {
           int L = q[i, 1];
           int R = q[i, 2];
           Console.Write("XOR of elements " +
                       "in given range is " +
                     rangeXor(BITree, L, R) + "\n");
       }
       else
       {
           int idx = q[i, 1];
           int val = q[i, 2];
           A[idx] ^= val;
            
           // Update the values of
           // all ancestors of idx
           updateBIT(BITree, n, idx, val);
       }
    }
}
}
 
// This code is contributed by sapnasingh4991

Javascript

 
// Javascript Program to find XOR of
// elements in given range [L, R].
 
// Returns XOR of arr[0..index].
// This function assumes that the
// array is preprocessed and partial
// XORs of array elements are stored
// in BITree[].
function getXOR(BITree, index)
{
    let ans = 0;
    index += 1;
 
    // Traverse ancestors
    // of BITree[index]
    while (index > 0) {
 
        // XOR current element
        // of BIT to ans
        ans ^= BITree[index];
 
        // Update index to that
        // of the parent node in
        // getXor() view by
        // subtracting LSB(Least
        // Significant Bit)
        index -= index & (-index);
    }
    return ans;
}
 
// Updates the Binary Index Tree by
// replacing all ancestores of index
// by their respective XOR with val
function updateBIT(BITree, n, index, val)
{
    index = index + 1;
 
    // Traverse all ancestors
    // and XOR with 'val'.
    while (index <= n) {
 
        // XOR 'val' to current
        // node of BIT
        BITree[index] ^= val;
 
        // Update index to that
        // of the parent node in
        // updateBit() view by
        // adding LSB(Least
        // Significant Bit)
        index += index & (-index);
    }
}
 
// Constructs and returns a Binary
// Indexed Tree for the given array
function constructBITree(arr, n)
{
    // Create and initialize
    // the Binary Indexed Tree
    let BITree = new Array(n + 1);
    for (let i = 1; i <= n; i++)
        BITree[i] = 0;
 
    // Store the actual values in
    // BITree[] using update()
    for (let i = 0; i < n; i++)
        updateBIT(BITree, n, i, arr[i]);
 
    return BITree;
}
 
function rangeXor(BITree, l, r)
{
    return getXOR(BITree, r)
           ^ getXOR(BITree, l - 1);
}
 
// Driver Code
    let A = [ 2, 1, 1, 3, 2, 3,
                4, 5, 6, 7, 8, 9 ];
    let n = A.length;
 
    let q
        = [ [ 1, 0, 9 ],
            [ 2, 3, 6 ],
            [ 2, 5, 5 ],
            [ 2, 8, 1 ],
            [ 1, 0, 9 ] ];
 
    // Create the Binary Indexed Tree
    let BITree = constructBITree(A, n);
 
    // Solve each query in Q
    for (let i = 0; i < q.length; i++) {
        let id = q[i][0];
 
        if (id == 1) {
            let L = q[i][1];
            let R = q[i][2];
            document.write("XOR of elements "
                 + "in given range is "
                 + rangeXor(BITree, L, R)
                 + "<br>");
        }
        else {
            let idx = q[i][1];
            let val = q[i][2];
            A[idx] ^= val;
 
            // Update the values of all
            // ancestors of idx
            updateBIT(BITree, n, idx, val);
        }
    }
 

输出：

XOR of elements in given range is 0
XOR of elements in given range is 2

getXor() 的时间复杂度： O(log N)
updateBIT() 的时间复杂度： O(log N)
总体时间复杂度： O(M * log N)，其中 M 和 N 分别是给定数组的查询次数和大小。

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