查询用更新的给定值用其XOR替换每个数组元素

给定一个数组，该数组最初由0作为唯一的存在元素，并且由以下两种类型的Q操作组成：

• Add(X)：将X插入数组。
• Update(X)：将每个数组元素A _i替换为A _i ^ X ，其中^是XOR操作。

例子：

Input: Q = 2
Add(5)
Update(2)
Output: 2 7
Explanation:
Initially, Array = {0}
Query 1: Array = {0, 5}
Query 2: Array = {0^2, 5^2} => {2, 7}

Input: Q = 2
Add(3)
Add(5)
Output: 0 3 5 

方法：这个想法是存储一个变量，该变量存储将与每个数组元素进行XOR运算的值，并且在将元素插入数组中以消除XOR值的影响时，请应用XOR操作本身。请按照以下步骤解决问题：

• 维护一个变量，例如effect ，以存储最终的XOR值。
• 对于操作Add(X)：在列表末尾添加X并使用来更新其值 X ^效应。
• 对于操作更新(X)：更新“效果”的值 具有效果^X。这是为了获得最新的XOR效果。
• 完成上述所有操作后，使用效果的XOR更新所有数组元素，即，对于所有A _i ，更新A _i = A _i ^ effect 。

下面是上述方法的实现：

C++

// C++ program of the above approach
 
#include 
using namespace std;
 
#define int long long
 
// Function to insert an element
// into the array
void add(vector& arr, int x)
{
    arr.push_back(x);
}
 
// Function to update every array
// element a[i] by a[i] ^ x
void update(int& effect, int x)
{
    effect = effect ^ x;
}
 
// Function to compute the final
// results after the operations
void computeResults(vector& arr,
                    int effect)
{
    for (int i = 0; i < arr.size(); i++) {
        arr[i] = arr[i] ^ effect;
        cout << arr[i] << " ";
    }
}
 
// Driver Code
signed main()
{
    vector arr = { 0 };
    int effect = 0;
 
    // Queries
    add(arr, 5);
    update(effect, 2);
 
    // Function Call
    computeResults(arr, effect);
}

Java

// Java program of
// the above approach
import java.util.*;
class GFG{
   
static Vector arr =
              new Vector();
static int effect;
   
// Function to insert an element
// into the array
static void add(int x)
{
  arr.add(x);
}
 
// Function to update every array
// element a[i] by a[i] ^ x
static void update(int x)
{
  effect = effect ^ x;
}
 
// Function to compute the final
// results after the operations
static void computeResults()
{
  for (int i = 0; i < arr.size(); i++)
  {
    arr.set(i, arr.get(i) ^ effect);
    System.out.print(arr.get(i) + " ");
  }
}
 
// Driver Code
public static void main(String[] args)
{
  arr = new Vector();
  arr.add(0);
  effect = 0;
 
  // Queries
  add(5);
  update(2);
 
  // Function Call
  computeResults();
}
}
 
// This code is contributed by Rajput-Ji

Python3

# Python3 program of the above approach
 
# Function to insert an element
# into the array
def add(arr, x):
     
    arr.append(x)
 
# Function to update every array
# element a[i] by a[i] ^ x
def update(effect, x):
 
    effect = effect ^ x
    return effect
 
# Function to compute the final
# results after the operations
def computeResults(arr, effect):
 
    for i in range(len(arr)):
        arr[i] = arr[i] ^ effect
        print(arr[i], end = " ")
 
# Driver Code
if __name__ == '__main__':
 
    arr = [0]
    effect = 0
 
    # Queries
    add(arr, 5)
    effect = update(effect, 2)
 
    # Function call
    computeResults(arr, effect)
 
# This code is contributed by mohit kumar 29

C#

// C# program of
// the above approach
using System;
using System.Collections.Generic;
class GFG{
   
static List arr = new List();
static int effect;
   
// Function to insert an element
// into the array
static void add(int x)
{
  arr.Add(x);
}
 
// Function to update every array
// element a[i] by a[i] ^ x
static void update(int x)
{
  effect = effect ^ x;
}
 
// Function to compute the final
// results after the operations
static void computeResults()
{
  for (int i = 0; i < arr.Count; i++)
  {
    arr[i] = arr[i] ^ effect;
    Console.Write(arr[i] + " ");
  }
}
 
// Driver Code
public static void Main(String[] args)
{
  arr = new List();
  arr.Add(0);
  effect = 0;
 
  // Queries
  add(5);
  update(2);
 
  // Function Call
  computeResults();
}
}
 
// This code is contributed by shikhasingrajput

输出：

2 7

时间复杂度： O(Q)
辅助空间： O(1)