按位或为奇数的对数

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]为奇数。
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 Odd = 3

Input : N = 7
            A[] = {8, 6, 2, 7, 3, 4, 9}
Output :15

一个简单的解决方案是检查每一对，并找到按位或，并用按位或将所有此类对计数为奇数。
下面是上述方法的实现：

C++

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

Java

// Java program to count pairs
// with odd OR
class GFG
{
// Function to count pairs with odd OR
static int findOddPair(int A[], int N)
{
    int oddPair = 0;
    for (int i = 0; i < N; i++)
    {
        for (int j = i + 1; j < N; j++)
        {
 
            // find OR operation
            // check odd or odd
            if ((A[i] | A[j]) % 2 != 0)
                oddPair++;
        }
    }
 
    // return count of odd pair
    return oddPair;
}
 
// Driver Code
public static void main(String []args)
{
    int A[] = { 5, 6, 2, 8 };
 
    int N = A.length;
 
    System.out.println(findOddPair(A, N));
}
}
 
// This code is contributed by ANKITRAI1

Python3

# Python3 program to count pairs with odd OR
 
  
# Function to count pairs with odd OR
def findOddPair(A, N):
     
    oddPair = 0
    for i in range(0, N):
        for j in range(i+1, N):
  
            # find OR operation
            # check odd or odd
            if ((A[i] | A[j]) % 2 != 0):
                oddPair+=1
 
    # return count of odd pair
    return oddPair
 
  
# Driver Code
def main():
     
    A = [ 5, 6, 2, 8 ]
  
    N = len(A)
  
    print(findOddPair(A, N))
 
if __name__ == '__main__':
    main()
# This code is contributed by PrinciRaj1992

C#

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

Javascript

// Javascript program to count pairs with odd OR
 
// Function to count pairs with odd OR
function findOddPair(A, N)
{
    let oddPair = 0;
    for (let i = 0; i < N; i++)
    {
        for (let j = i + 1; j < N; j++)
        {
 
            // find OR operation
            // check odd or odd
            if ((A[i] | A[j]) % 2 != 0)
                oddPair++;
        }
    }
 
    // return count of odd pair
    return oddPair;
}
 
// Driver Code
let A = [ 5, 6, 2, 8 ];
let N = A.length;
document.write(findOddPair(A, N));
 
// This code is contributed by souravmahato348.

输出：

3

时间复杂度：O(N ² )
一个有效的解决方案是对偶数OR的对进行计数，并用对总数减去它们，以得到奇数的逐位OR的对。为此，对最后一位为0的数字进行计数。然后，偶数按位或的对数= count *(count – 1)/ 2 ，对的总数为N *(N-1)/ 2。
因此，与ODD按位或的对将为：

Total Pairs - Pairs with EVEN Bitwise-OR

下面是上述方法的实现：

C++

// C++ program to count pairs with odd OR
#include 
using namespace std;
 
// Function to count pairs with odd OR
int countOddPair(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++;
 
    // Even pair count
    int evenPairCount = count * (count - 1) / 2;
 
    // Total pairs
    int totPairs = N * (N - 1) / 2;
 
    // Return Odd pair count
    return totPairs - evenPairCount;
}
 
// Driver main
int main()
{
    int A[] = { 5, 6, 2, 8 };
    int N = sizeof(A) / sizeof(A[0]);
 
    cout << countOddPair(A, N) << endl;
 
    return 0;
}

Java

// Java program to count pairs with odd OR
 
public class GFG {
 
// Function to count pairs with odd OR
    static int countOddPair(int A[], int N) {
        // Count total even numbers in
        // array
 
        int count = 0;
        for (int i = 0; i < N; i++) {
            if ((A[i] % 2 != 1)) {
                count++;
            }
        }
 
        // Even pair count
        int evenPairCount = count * (count - 1) / 2;
 
        // Total pairs
        int totPairs = N * (N - 1) / 2;
 
        // Return Odd pair count
        return totPairs - evenPairCount;
    }
 
// Driver main
    public static void main(String[] args) {
        int A[] = {5, 6, 2, 8};
        int N = A.length;
 
        System.out.println(countOddPair(A, N));
 
    }
}

Python3

# Python 3program to count pairs with odd OR
 
# Function to count pairs with odd OR
def countOddPair(A, N):
     
    # Count total even numbers in
    # array
    count = 0
    for i in range(0, N):
        if (A[i] % 2 != 1):
            count+=1
 
    # Even pair count
    evenPairCount = count * (count - 1) / 2
 
    # Total pairs
    totPairs = N * (N - 1) / 2
 
    # Return Odd pair count
    return (int)(totPairs - evenPairCount)
     
# Driver Code
A = [ 5, 6, 2, 8 ]
 
N = len(A)
 
print(countOddPair(A, N))
 
# This code is contributed by PrinciRaj1992

C＃

// C# program to count pairs with odd OR
using System;
 
public class GFG {
 
// Function to count pairs with odd OR
    static int countOddPair(int []A, int N) {
        // Count total even numbers in
        // array
  
        int count = 0;
        for (int i = 0; i < N; i++) {
            if ((A[i] % 2 != 1)) {
                count++;
            }
        }
  
        // Even pair count
        int evenPairCount = count * (count - 1) / 2;
  
        // Total pairs
        int totPairs = N * (N - 1) / 2;
  
        // Return Odd pair count
        return totPairs - evenPairCount;
    }
  
// Driver main
    public static void Main() {
        int []A = {5, 6, 2, 8};
        int N = A.Length;
  
        Console.WriteLine(countOddPair(A, N));
  
    }
}
/*This code is contributed by PrinciRaj1992*/

输出：

3

时间复杂度：O(N)