给定一个由N 个整数组成的数组arr[] ，任务是计算按位异或为奇数的对的数量，可以将其删除并替换为它们的按位或值，直到数组中不存在这样的对。
例子：
Input: arr[] = {5, 4, 7, 2}
Output: 2
Explanation:
Pair (5, 4): Bitwise XOR of 5 and 4 is 1. Remove this pair and add their Bitwise OR (= 5) into the array. Therefore, the modified array is {5, 7, 2}.
Pair (5, 2): Bitwise XOR of 5 and 2 is 7. Remove this pair and add their Bitwise OR (= 7) into the array. Therefore, the modified array is {7, 7}.
Therefore, the count of such pairs that can be removed is 2.
Input: arr[] = {2, 4, 6}
Output: 0
方法：根据以下观察可以解决给定的问题：
- 一对按位XOR是奇数仅当一个元件是奇数，而另一个是偶数。
- 因此，从数组中删除这样的一对并将它们的按位或添加到数组中不会影响数组中奇数元素的总数。但是偶数数组元素的数量减少了1 。
因此，这个想法是找到给定数组中存在的偶数元素的计数。如果偶数元素的计数为N ，则需要0 次移动。否则，打印count的值作为需要删除的对的结果计数。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to count the number of pairs
// required to be removed from the array
// and replaced by their Bitwise OR values
void countPairs(int arr[], int N)
{
// Stores the count of
// even array elements
int even = 0;
// Traverse the given array
for (int i = 0; i < N; i++) {
// Increment the count
// of even array elements
if (arr[i] % 2 == 0)
even++;
}
// If the array contains at
// least one odd array element
if (N - even >= 1) {
cout << even;
return;
}
// Otherwise, print 0
cout << 0;
}
// Driver Code
int main()
{
int arr[] = { 5, 4, 7, 2 };
int N = sizeof(arr) / sizeof(arr[0]);
countPairs(arr, N);
return 0;
} Java
// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
class GFG{
// Function to count the number of pairs
// required to be removed from the array
// and replaced by their Bitwise OR values
static void countPairs(int arr[], int N)
{
// Stores the count of
// even array elements
int even = 0;
// Traverse the given array
for(int i = 0; i < N; i++)
{
// Increment the count
// of even array elements
if (arr[i] % 2 == 0)
even++;
}
// If the array contains at
// least one odd array element
if (N - even >= 1)
{
System.out.println(even);
return;
}
// Otherwise, print 0
System.out.println(0);
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 5, 4, 7, 2 };
int N = arr.length;
countPairs(arr, N);
}
}
// This code is contributed by KingashPython3
# Python3 program for the above approach
# Function to count the number of pairs
# required to be removed from the array
# and replaced by their Bitwise OR values
def countPairs(arr, N):
# Stores the count of
# even array elements
even = 0
# Traverse the given array
for i in range(N):
# Increment the count
# of even array elements
if (arr[i] % 2 == 0):
even += 1
# If the array contains at
# least one odd array element
if (N - even >= 1):
print(even)
return
# Otherwise, print 0
print(0)
# Driver Code
if __name__ == "__main__":
arr = [ 5, 4, 7, 2 ]
N = len(arr)
countPairs(arr, N)
# This code is contributed by AnkThonC#
// C# program for the above approach
using System;
class GFG{
// Function to count the number of pairs
// required to be removed from the array
// and replaced by their Bitwise OR values
static void countPairs(int[] arr, int N)
{
// Stores the count of
// even array elements
int even = 0;
// Traverse the given array
for(int i = 0; i < N; i++)
{
// Increment the count
// of even array elements
if (arr[i] % 2 == 0)
even++;
}
// If the array contains at
// least one odd array element
if (N - even >= 1)
{
Console.WriteLine(even);
return;
}
// Otherwise, print 0
Console.WriteLine(0);
}
// Driver code
static void Main()
{
int[] arr = { 5, 4, 7, 2 };
int N = arr.Length;
countPairs(arr, N);
}
}
// This code is contributed by sanjoy_62Javascript
输出：
2时间复杂度： O(N)
辅助空间： O(1)
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