给定一个数组nrr [] ,该数组arr []由N个不同的整数组成,并且查询类型为{X,Y}的查询Q [] [] ,则每个查询的任务是在将X替换为Y后,找到所有数组元素的按位XOR 。大批。
例子:
Input: arr[] = {1, 2, 3, 4, 5} Q = {(1, 4) (3, 6) (2, 3)}
Output:4 1 0
Explanation:
Query 1: The array modifies to {4, 2, 3, 4, 5} and XOR = 4
Query 2: The array modifies to {4, 2, 6, 4, 5} and XOR = 1
Query 3: The array modifies to {4, 3, 6, 4, 5} and XOR = 0
Input: arr[] = {5, 7, 8, 9, } Q = {(5, 6) (8, 1)}
Output: 0 9
Explanation:
Query 1: The array modifies to {6, 7, 8, 9} and XOR = 0
Query 2: The array modifies to {6, 7, 1, 9} and XOR = 9
方法:
方法是使用按位XOR属性:
- 假设,有三个元件A,B,和与X,和它们的异或是,total_xor = A ^ B ^ X。
- 要从total_xor中除去X的贡献,请执行total_xor ^ X。可以从以下事实验证: A ^ B ^ X ^ X = A ^ B (元素与自身的异或为0)。
- 要在total_xor中添加Y的贡献,只需执行total_xor ^ Y即可。
下面是上述方法的实现:
C++
// C++ Program to implement
// the above approach
#include
using namespace std;
// Stores the bitwise XOR
// of array elements
int total_xor;
// Function to find the total xor
void initialize_xor(int arr[], int n)
{
// Loop to find the xor
// of all the elements
for (int i = 0; i < n; i++) {
total_xor = total_xor ^ arr[i];
}
}
// Function to find the XOR
// after replacing all occurrences
// of X by Y for Q queries
void find_xor(int X, int Y)
{
// Remove contribution of
// X from total_xor
total_xor = total_xor ^ X;
// Adding contribution of
// Y to total_xor
total_xor = total_xor ^ Y;
// Print total_xor
cout << total_xor << "\n";
}
// Driver Code
int main()
{
int arr[] = { 5, 7, 8, 9 };
int n = sizeof(arr) / sizeof(arr[0]);
initialize_xor(arr, n);
vector > Q = { { 5, 6 }, { 8, 1 } };
for (int i = 0; i < Q.size(); i++) {
find_xor(Q[i][0], Q[i][1]);
}
return 0;
}
Java
// Java Program to implement
// the above approach
import java.util.*;
class GFG{
// Stores the bitwise XOR
// of array elements
static int total_xor;
// Function to find the total xor
static void initialize_xor(int arr[],
int n)
{
// Loop to find the xor
// of all the elements
for (int i = 0; i < n; i++)
{
total_xor = total_xor ^ arr[i];
}
}
// Function to find the XOR
// after replacing all occurrences
// of X by Y for Q queries
static void find_xor(int X, int Y)
{
// Remove contribution of
// X from total_xor
total_xor = total_xor ^ X;
// Adding contribution of
// Y to total_xor
total_xor = total_xor ^ Y;
// Print total_xor
System.out.print(total_xor + "\n");
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 5, 7, 8, 9 };
int n = arr.length;
initialize_xor(arr, n);
int [][]Q = { { 5, 6 }, { 8, 1 } };
for (int i = 0; i < Q.length; i++)
{
find_xor(Q[i][0], Q[i][1]);
}
}
}
// This code is contributed by Rohit_ranjan
Python3
# Python3 program to implement
# the above approach
# Stores the bitwise XOR
# of array elements
global total_xor
total_xor = 0
# Function to find the total xor
def initialize_xor(arr, n):
global total_xor
# Loop to find the xor
# of all the elements
for i in range(n):
total_xor = total_xor ^ arr[i]
# Function to find the XOR
# after replacing all occurrences
# of X by Y for Q queries
def find_xor(X, Y):
global total_xor
# Remove contribution of
# X from total_xor
total_xor = total_xor ^ X
# Adding contribution of
# Y to total_xor
total_xor = total_xor ^ Y
# Print total_xor
print(total_xor)
# Driver Code
if __name__ == '__main__':
arr = [ 5, 7, 8, 9 ]
n = len(arr)
initialize_xor(arr, n)
Q = [ [ 5, 6 ], [ 8, 1 ] ]
# Function call
for i in range(len(Q)):
find_xor(Q[i][0], Q[i][1])
# This code is contributed by Shivam Singh
C#
// C# program to implement
// the above approach
using System;
class GFG{
// Stores the bitwise XOR
// of array elements
static int total_xor;
// Function to find the total xor
static void initialize_xor(int []arr,
int n)
{
// Loop to find the xor
// of all the elements
for(int i = 0; i < n; i++)
{
total_xor = total_xor ^ arr[i];
}
}
// Function to find the XOR
// after replacing all occurrences
// of X by Y for Q queries
static void find_xor(int X, int Y)
{
// Remove contribution of
// X from total_xor
total_xor = total_xor ^ X;
// Adding contribution of
// Y to total_xor
total_xor = total_xor ^ Y;
// Print total_xor
Console.Write(total_xor + "\n");
}
// Driver Code
public static void Main(String[] args)
{
int []arr = { 5, 7, 8, 9 };
int n = arr.Length;
initialize_xor(arr, n);
int [,]Q = { { 5, 6 }, { 8, 1 } };
for(int i = 0; i < Q.GetLength(0); i++)
{
find_xor(Q[i, 0], Q[i, 1]);
}
}
}
// This code is contributed by 29AjayKumar
输出:
0
9
时间复杂度: O(N + sizeof(Q))
辅助空间: O(1)