给定一个由n个整数组成的数组A。任务是计算将给定数组元素分为两个不相交的组的方法的数量，以使每个组的元素的XOR相等。
例子：
Input : A[] = { 1, 2, 3 }
Output : 3
{(1), (2, 3)}, {(2), (1, 3)}, {(3), (1, 2)}
are three ways with equal XOR value of two
groups.
Input : A[] = { 5, 2, 3, 2 }
Output : 0让我们将第一组中所有元素之间的XOR表示为G 1 ，将第二组中所有元素之间的XOR表示为G 2 。现在，下面的关系是总是正确的：克1-⊕克2- = A 1⊕A 2 …⊕。 ⊕A n 。
因此，对于G 1 = G 2 ，数组A的所有元素之间的xor等于0。因此，在这种情况下，答案将是(2 n – 2)/ 2 =(2 n-1 – 1)。在第二种情况下，当所有元素之间的XOR不为0时，我们将无法拆分数组。答案将为0。
C++
// CPP Program to count number of ways to split
// array into two groups such that each group
// has equal XOR value
#include
using namespace std;
// Return the count number of ways to split
// array into two groups such that each group
// has equal XOR value.
int countgroup(int a[], int n)
{
int xs = 0;
for (int i = 0; i < n; i++)
xs = xs ^ a[i];
// We can split only if XOR is 0. Since
// XOR of all is 0, we can consider all
// subsets as one group.
if (xs == 0)
return (1 << (n-1)) - 1;
return 0;
}
// Driver Program
int main()
{
int a[] = { 1, 2, 3 };
int n = sizeof(a)/sizeof(a[0]);
cout << countgroup(a, n) << endl;
return 0;
} Java
// Java Program to count number of ways
// to split array into two groups such
// that each group has equal XOR value
import java.io.*;
import java.util.*;
class GFG {
// Return the count number of ways to split
// array into two groups such that each group
// has equal XOR value.
static int countgroup(int a[], int n) {
int xs = 0;
for (int i = 0; i < n; i++)
xs = xs ^ a[i];
// We can split only if XOR is 0. Since
// XOR of all is 0, we can consider all
// subsets as one group.
if (xs == 0)
return (1 << (n - 1)) - 1;
return 0;
}
// Driver program
public static void main(String args[]) {
int a[] = {1, 2, 3};
int n = a.length;
System.out.println(countgroup(a, n));
}
}
// This code is contributed by Nikita Tiwari.Python3
# Python3 code to count number of ways
# to split array into two groups such
# that each group has equal XOR value
# Return the count of number of ways
# to split array into two groups such
# that each group has equal XOR value.
def countgroup(a, n):
xs = 0
for i in range(n):
xs = xs ^ a[i]
# We can split only if XOR is 0.
# Since XOR of all is 0, we can
# consider all subsets as one group.
if xs == 0:
return (1 << (n-1)) - 1
return 0
# Driver Program
a = [1, 2, 3]
n = len(a)
print(countgroup(a, n))
# This code is contributed by "Sharad_Bhardwaj".C#
// C# Program to count number of ways
// to split array into two groups such
// that each group has equal XOR value
using System;
class GFG {
// Return the count number of ways to split
// array into two groups such that each group
// has equal XOR value.
static int countgroup(int[] a, int n)
{
int xs = 0;
for (int i = 0; i < n; i++)
xs = xs ^ a[i];
// We can split only if XOR is 0. Since
// XOR of all is 0, we can consider all
// subsets as one group.
if (xs == 0)
return (1 << (n - 1)) - 1;
return 0;
}
// Driver program
public static void Main()
{
int[] a = { 1, 2, 3 };
int n = a.Length;
Console.WriteLine(countgroup(a, n));
}
}
// This code is contributed by vt_m.PHP
Javascript
输出：
3