对按位或为偶数的对进行计数

给定大小为N的数组A []。任务是查找存在多少对(i，j)，以使A [i]或A [j]为偶数。
例子：

Input : N = 4
        A[] = { 5, 6, 2, 8 }
Output :3
Explanation :
Since pair of A[] = ( 5, 6 ), ( 5, 2 ), ( 5, 8 ),
( 6, 2 ), ( 6, 8 ), ( 2, 8 )
5 OR 6 = 7, 5 OR 2 = 7, 5 OR 8 = 13
6 OR 2 = 6, 6 OR 8 = 14, 2 OR 8 = 10
Total pair A( i, j ) = 6 and Even = 3

Input : N = 7
        A[] = {8, 6, 2, 7, 3, 4, 9}
Output :6

一个简单的解决方案是检查每对。

C++

// C++ program to count pairs with even OR
#include 
using namespace std;
 
int findEvenPair(int A[], int N)
{
    int evenPair = 0;
    for (int i = 0; i < N; i++) {
        for (int j = i + 1; j < N; j++) {
 
            // find OR operation
            // check odd or even
            if ((A[i] | A[j]) % 2 == 0)
                evenPair++;
        }
    }
    // return count of even pair
    return evenPair;
}
// Driver main
int main()
{
    int A[] = { 5, 6, 2, 8 };
    int N = sizeof(A) / sizeof(A[0]);
    cout << findEvenPair(A, N) << endl;
    return 0;
}

Java

// Java program to count
// pairs with even OR
import java.io.*;
 
class GFG
{
 
static int findEvenPair(int A[],
                        int N)
{
    int evenPair = 0;
    for (int i = 0; i < N; i++)
    {
        for (int j = i + 1; j < N; j++)
        {
 
            // find OR operation
            // check odd or even
            if ((A[i] | A[j]) % 2 == 0)
                evenPair++;
        }
    }
     
    // return count of even pair
    return evenPair;
}
 
// Driver Code
public static void main (String[] args)
{
    int A[] = { 5, 6, 2, 8 };
    int N = A.length;
    System.out.println(findEvenPair(A, N));
}
}
 
// This code is contributed
// by inder_verma.

Python 3

# Python 3 program to count pairs
# with even OR
 
def findEvenPair(A, N) :
    evenPair = 0
 
    for i in range(N) :
        for j in range(i+1, N):
 
            # find OR operation
            # check odd or even
            if (A[i] | A[j]) % 2 == 0 :
                evenPair += 1
 
    # return count of even pair
    return evenPair
 
# Driver Code
if __name__ == "__main__" :
 
    A = [ 5, 6, 2, 8]
    N = len(A)
 
    # function calling
    print(findEvenPair(A, N))
 
# This code is contributed by ANKITRAI1

C#

// C# program to count
// pairs with even OR
using System;
 
class GFG
{
 
static int findEvenPair(int []A,
                        int N)
{
    int evenPair = 0;
    for (int i = 0; i < N; i++)
    {
        for (int j = i + 1; j < N; j++)
        {
 
            // find OR operation
            // check odd or even
            if ((A[i] | A[j]) % 2 == 0)
                evenPair++;
        }
    }
     
    // return count of even pair
    return evenPair;
}
 
// Driver Code
public static void Main ()
{
    int []A = { 5, 6, 2, 8 };
    int N = A.Length;
    Console.WriteLine(findEvenPair(A, N));
}
}
 
// This code is contributed
// by inder_verma.

Javascript

// JavaScript program to count pairs with even OR
 
function findEvenPair(A, N)
{
    let evenPair = 0;
    for (let i = 0; i < N; i++) {
        for (let j = i + 1; j < N; j++) {
 
            // find OR operation
            // check odd or even
            if ((A[i] | A[j]) % 2 == 0)
                evenPair++;
        }
    }
    // return count of even pair
    return evenPair;
}
// Driver main
 
    let A = [5, 6, 2, 8 ];
    let N = A.length;
    document.write(findEvenPair(A, N));
     
// This code contributed by Rajput-Ji
 

输出：

3

时间复杂度：O(N ^ 2)
一个有效的解决方案是对最后一位为0的数字进行计数。然后返回count *(count – 1)/ 2。请注意，两个数字的OR只能在其最后一位为0的情况下才是偶数。

C++

// C++ program to count pairs with even OR
#include 
using namespace std;
 
int findEvenPair(int A[], int N)
{
    // Count total even numbers in
    // array.
    int count = 0;
    for (int i = 0; i < N; i++)
       if (!(A[i] & 1))
            count++;
             
    // return count of even pair
    return count * (count - 1) / 2;
}
 
// Driver main
int main()
{
    int A[] = { 5, 6, 2, 8 };
    int N = sizeof(A) / sizeof(A[0]);
    cout << findEvenPair(A, N) << endl;
    return 0;
}

Java

// Java program to count
// pairs with even OR
import java.io.*;
 
class GFG
{
static int findEvenPair(int A[], int N)
{
    // Count total even numbers in
    // array.
    int count = 0;
    for (int i = 0; i < N; i++)
    if ((!((A[i] & 1) > 0)))
            count++;
             
    // return count of even pair
    return count * (count - 1) / 2;
}
 
// Driver Code
public static void main (String[] args)
{
    int A[] = { 5, 6, 2, 8 };
    int N = A.length;
    System.out.println(findEvenPair(A, N));
}
}
 
// This code is contributed
// by inder_verma.

的Python 3

# Python 3 program to count
# pairs with even OR
 
def findEvenPair(A, N):
 
    # Count total even numbers
    # in array.
    count = 0
    for i in range(N):
        if (not (A[i] & 1)):
            count += 1
           
    # return count of even pair
    return count * (count - 1) // 2
 
# Driver Code
if __name__ == "__main__":
    A = [ 5, 6, 2, 8 ]
    N = len(A)
    print(findEvenPair(A, N))
 
# This code is contributed
# by ChitraNayal

C＃

// C# program to count
// pairs with even OR
using System;
 
class GFG
{
static int findEvenPair(int []A, int N)
{
    // Count total even numbers
    // in array.
    int count = 0;
    for (int i = 0; i < N; i++)
    if ((!((A[i] & 1) > 0)))
            count++;
             
    // return count of even pair
    return count * (count - 1) / 2;
}
 
// Driver Code
public static void Main (String[] args)
{
    int []A = { 5, 6, 2, 8 };
    int N = A.Length;
    Console.WriteLine(findEvenPair(A, N));
}
}
 
// This code is contributed
// by Kirti_Mangal

输出：

3

时间复杂度：O(N)