给定一个由N个正整数组成的数组arr [] ，任务是计算其元素的按位或为偶数的子数组的数量。
例子：
Input: arr[] = {1, 5, 4, 2, 6 }
Output: 6
Explanation:
The subarrays with even Bitwise OR are {4}, {2}, {6}, {2, 6}, {4, 2}, {4, 2, 6}.
Therefore, the number of subarrays having even Bitwise OR are 6.
Input: arr[] ={2, 5, 6, 8}
Output: 4
天真的方法：解决问题的最简单方法是生成所有子数组，如果任何子数组的按位或为偶数，则增加此类子数组的数量。检查所有子数组后，打印获得的计数作为结果。
时间复杂度： O(N 3 )
辅助空间： O(1)
高效方法：可以通过观察以下事实来优化上述方法：如果子数组中的任何元素为奇数，则必定会使子数组的按位或为奇数。因此，该想法是找到数组中连续段的长度，该长度是偶数，并将其对总计数的贡献相加。
请按照以下步骤解决问题：
- 初始化一个变量，例如count ，以存储具有按位或为偶数的子数组的总数。
- 初始化一个变量L ，以存储偶数个相邻元素的计数。
- 遍历给定数组arr []并执行以下步骤：
- 如果当前元素是偶数，则将L的值增加1 。
- 否则，将值L *(L +1)/ 2添加到变量count并将L更新为0 。
- 如果L的值不为零，则将L *(L + 1)/ 2加到变量count上。
- 完成上述步骤后，打印count的值作为结果。
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
// subarrays having even Bitwise OR
int bitOr(int arr[], int N)
{
// Store number of subarrays
// having even bitwise OR
int count = 0;
// Store the length of the current
// subarray having even numbers
int length = 0;
// Traverse the array
for (int i = 0; i < N; i++) {
// If the element is even
if (arr[i] % 2 == 0) {
// Increment size of the
// current continuous sequence
// of even array elements
length++;
}
// If arr[i] is odd
else {
// If length is non zero
if (length != 0) {
// Adding contribution of
// subarrays consisting
// only of even numbers
count += ((length)
* (length + 1))
/ 2;
}
// Make length of subarray as 0
length = 0;
}
}
// Add contribution of previous subarray
count += ((length) * (length + 1)) / 2;
// Return total count of subarrays
return count;
}
// Driver Code
int main()
{
int arr[] = { 1, 5, 4, 2, 6 };
int N = sizeof(arr) / sizeof(arr[0]);
// Function Call
cout << bitOr(arr, N);
return 0;
} Java
// Java program for the above approach
import java.util.*;
class GFG{
// Function to count the number of
// subarrays having even Bitwise OR
static int bitOr(int arr[], int N)
{
// Store number of subarrays
// having even bitwise OR
int count = 0;
// Store the length of the current
// subarray having even numbers
int length = 0;
// Traverse the array
for(int i = 0; i < N; i++)
{
// If the element is even
if (arr[i] % 2 == 0)
{
// Increment size of the
// current continuous sequence
// of even array elements
length++;
}
// If arr[i] is odd
else
{
// If length is non zero
if (length != 0)
{
// Adding contribution of
// subarrays consisting
// only of even numbers
count += ((length) * (length + 1)) / 2;
}
// Make length of subarray as 0
length = 0;
}
}
// Add contribution of previous subarray
count += ((length) * (length + 1)) / 2;
// Return total count of subarrays
return count;
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 1, 5, 4, 2, 6 };
int N = arr.length;
// Function Call
System.out.print(bitOr(arr, N));
}
}
// This code is contributed by splevel62Python3
# Python3 program for the above approach
# Function to count the number of
# subarrays having even Bitwise OR
def bitOr(arr, N):
# Store number of subarrays
# having even bitwise OR
count = 0
# Store the length of the current
# subarray having even numbers
length = 0
# Traverse the array
for i in range(N):
# If the element is even
if (arr[i] % 2 == 0):
# Increment size of the
# current continuous sequence
# of even array elements
length += 1
# If arr[i] is odd
else:
# If length is non zero
if (length != 0):
# Adding contribution of
# subarrays consisting
# only of even numbers
count += ((length) * (length + 1)) // 2
# Make length of subarray as 0
length = 0
# Add contribution of previous subarray
count += ((length) * (length + 1)) // 2
# Return total count of subarrays
return count
# Driver Code
if __name__ == '__main__':
arr = [ 1, 5, 4, 2, 6 ]
N = len(arr)
# Function Call
print (bitOr(arr, N))
# This code is contributed by mohit kumar 29C#
// C# program for the above approach
using System;
class GFG
{
// Function to count the number of
// subarrays having even Bitwise OR
static int bitOr(int[] arr, int N)
{
// Store number of subarrays
// having even bitwise OR
int count = 0;
// Store the length of the current
// subarray having even numbers
int length = 0;
// Traverse the array
for(int i = 0; i < N; i++)
{
// If the element is even
if (arr[i] % 2 == 0)
{
// Increment size of the
// current continuous sequence
// of even array elements
length++;
}
// If arr[i] is odd
else
{
// If length is non zero
if (length != 0)
{
// Adding contribution of
// subarrays consisting
// only of even numbers
count += ((length) * (length + 1)) / 2;
}
// Make length of subarray as 0
length = 0;
}
}
// Add contribution of previous subarray
count += ((length) * (length + 1)) / 2;
// Return total count of subarrays
return count;
}
// Driver code
static void Main()
{
int[] arr = { 1, 5, 4, 2, 6 };
int N = arr.Length;
// Function Call
Console.Write(bitOr(arr, N));
}
}
// This code is contributed by sanjoy_62.输出：
6时间复杂度： O(N)
辅助空间： O(1)