给定一个整数数组，我们需要获得所有子数组XOR的总XOR，其中子数组XOR可以通过对其所有元素进行XOR运算获得。
例子 ：
Input : arr[] = [3, 5, 2, 4, 6]
Output : 7
Total XOR of all subarray XORs is,
(3) ^ (5) ^ (2) ^ (4) ^ (6)
(3^5) ^ (5^2) ^ (2^4) ^ (4^6)
(3^5^2) ^ (5^2^4) ^ (2^4^6)
(3^5^2^4) ^ (5^2^4^6) ^
(3^5^2^4^6) = 7
Input : arr[] = {1, 2, 3, 4}
Output : 0
Total XOR of all subarray XORs is,
(1) ^ (2) ^ (3) ^ (4) ^
(1^2) ^ (2^3) ^ (3^4) ^
(1^2^3) ^ (2^3^4) ^
(1^2^3^4) = 0我们在下面的文章中讨论了O(n)解决方案。
所有子数组XOR的XOR |套装1
如上文所述，第i个索引处的元素频率由(i + 1)*(Ni)给出，其中N是数组的大小
可能有4种情况：
情况1：我是奇数，N是奇数
令i = 2k + 1，N = 2m + 1
freq [i] =(((2k + 1)+1)*((2m + 1)-(2k + 1))= 4(mk)(k + 1)=偶数
情况2：我是奇数，N是偶数
令i = 2k + 1，N = 2m
freq [i] =(((2k + 1)+1)*((2m)-(2k + 1))= 2(k + 1)(2m-2k-1)=偶数
情况3：我是偶数，N是奇数
令i = 2k，N = 2m + 1
freq [i] =(((2k)+1)*((2m + 1)-(2k))= 2k(2m-2k + 1)+(2m-2k)+1 =奇数
情况4：我是偶数，N是偶数
令i = 2k，N = 2m
freq [i] =((2k)+1)*((2m)-(2k))= 2(mk)(2k + 1)=偶数
由此可以得出结论，如果数组中元素的总数为偶数，则在任何位置的元素的频率都是偶数。因此，总异或将为0。元素的个数为奇数，则偶数位置的元素的频率为奇数，加法位置为偶数。因此，我们只需要查找偶数位置元素的XOR。
以下是上述想法的实现：
C++
// C++ program to get total xor of all subarray xors
#include
using namespace std;
// Returns XOR of all subarray xors
int getTotalXorOfSubarrayXors(int arr[], int N)
{
// if even number of terms are there, all
// numbers will appear even number of times.
// So result is 0.
if (N % 2 == 0)
return 0;
// else initialize result by 0 as (a xor 0 = a)
int res = 0;
for (int i = 0; i Java
// Java program to get total
// xor of all subarray xors
import java.io.*;
class GFG
{
// Returns XOR of all
// subarray xors
static int getTotalXorOfSubarrayXors(int arr[],
int N)
{
// if even number of terms are
// there, all numbers will appear
// even number of times. So result is 0.
if (N % 2 == 0)
return 0;
// else initialize result
// by 0 as (a xor 0 = a)
int res = 0;
for (int i = 0; i < N; i += 2)
res ^= arr[i];
return res;
}
// Driver Code
public static void main (String[] args)
{
int arr[] = {3, 5, 2, 4, 6};
int N = arr.length;
System.out.println(
getTotalXorOfSubarrayXors(arr, N));
}
}
// This code is contributed by ajitPython 3
# Python 3 program to get total xor
# of all subarray xors
# Returns XOR of all subarray xors
def getTotalXorOfSubarrayXors(arr, N):
# if even number of terms are there,
# all numbers will appear even number
# of times. So result is 0.
if (N % 2 == 0):
return 0
# else initialize result by 0
# as (a xor 0 = a)
res = 0
for i in range(0, N, 2):
res ^= arr[i]
return res
# Driver code
if __name__ == "__main__":
arr = [3, 5, 2, 4, 6]
N = len(arr)
print(getTotalXorOfSubarrayXors(arr, N))
# This code is contributed by ita_cC#
// C# program to get total
// xor of all subarray xors
using System;
class GFG
{
// Returns XOR of all
// subarray xors
static int getTotalXorOfSubarrayXors(int []arr,
int N)
{
// if even number of terms
// are there, all numbers
// will appear even number
// of times. So result is 0.
if (N % 2 == 0)
return 0;
// else initialize result
// by 0 as (a xor 0 = a)
int res = 0;
for (int i = 0; i < N; i += 2)
res ^= arr[i];
return res;
}
// Driver Code
static void Main()
{
int []arr = {3, 5, 2, 4, 6};
int N = arr.Length;
Console.Write(getTotalXorOfSubarrayXors(arr, N));
}
}
// This code is contributed by aj_36PHP
Javascript
输出：
7