用按位异或作为奇数对进行计数

给定一个由N个整数组成的数组，任务是找到对(i，j)的对数，以使A [i] ^ A [j]为奇数。
例子：

Input : N = 5
        A[] =  { 5, 4, 7, 2, 1}
Output :6
Since pair of A[] =
( 5, 4 ) = 1( 5, 7 ) = 2( 5, 2 ) = 7( 5, 1 ) = 4
( 4, 7 ) = 3( 4, 2 ) = 6( 4, 1 ) = 5
( 7, 2 ) = 5( 7, 1 ) = 6
( 2, 1 ) = 3
Total XOR ODD pair  = 6

Input : N = 7
        A[] = { 7, 2, 8, 1, 0, 5, 11 }
Output :12
Since pair of A[] =
( 7, 2 ) = 5( 7, 8 ) = 15( 7, 1 ) = 6( 7, 0 ) = 7( 7, 5 ) = 2( 7, 11 ) = 12
( 2, 8 ) = 10( 2, 1 ) = 3( 2, 0 ) = 2( 2, 5 ) = 7( 2, 11 ) = 9
( 8, 1 ) = 9( 8, 0 ) = 8( 8, 5 ) = 13( 8, 11 ) = 3
( 1, 0 ) = 1( 1, 5 ) = 4( 1, 11 ) = 10
( 0, 5 ) = 5( 0, 11 ) = 11
( 5, 11 ) = 14

天真的方法是检查每对并打印对数。
下面是上述方法的实现：

C++

// C++ program to count pairs
// with XOR giving a odd number
#include 
using namespace std;
 
// Function to count number of odd pairs
int findOddPair(int A[], int N)
{
    int i, j;
 
    // variable for counting odd pairs
    int oddPair = 0;
 
    // find all pairs
    for (i = 0; i < N; i++) {
        for (j = i + 1; j < N; j++) {
 
            // find XOR operation
            // check odd or even
            if ((A[i] ^ A[j]) % 2 != 0)
                oddPair++;
        }
        cout << endl;
    }
 
    // return number of odd pair
    return oddPair;
}
 
// Driver Code
int main()
{
 
    int A[] = { 5, 4, 7, 2, 1 };
    int N = sizeof(A) / sizeof(A[0]);
 
    // calling function findOddPair
    // and print number of odd pair
    cout << findOddPair(A, N) << endl;
 
    return 0;
}

Java

// Java program to count pairs
// with XOR giving a odd number
 
class GFG
{
     
// Function to count
// number of odd pairs
static int findOddPair(int A[],
                       int N)
{
    int i, j;
 
    // variable for counting
    // odd pairs
    int oddPair = 0;
 
    // find all pairs
    for (i = 0; i < N; i++)
    {
        for (j = i + 1; j < N; j++)
        {
 
            // find XOR operation
            // check odd or even
            if ((A[i] ^ A[j]) % 2 != 0)
                oddPair++;
        }
     
    }
 
    // return number
    // of odd pair
    return oddPair;
}
 
// Driver Code
public static void main(String args[])
{
    int A[] = { 5, 4, 7, 2, 1 };
    int N = A.length;
 
    // calling function findOddPair
    // and print number of odd pair
    System.out.println(findOddPair(A, N));
}
}
 
// This code is contributed
// by Kirti_Mangal

Python3

# Python3 program to count pairs
# with XOR giving a odd number
 
# Function to count number of odd pairs
def findOddPair(A, N) :
 
    # variable for counting odd pairs
    oddPair = 0
 
    # find all pairs
    for i in range(0, N) :
        for j in range(i+1, N) :
 
            # find XOR operation
            # check odd or even
            if ((A[i] ^ A[j]) % 2 != 0):
                oddPair+=1
 
    # return number of odd pair
    return oddPair
 
# Driver Code
if __name__=='__main__':
    A = [5, 4, 7, 2, 1 ]
    N = len(A)
 
# calling function findOddPair
# and print number of odd pair
    print(findOddPair(A, N))
 
# This code is contributed by Smitha Dinesh Semwal

C#

// C# program to count pairs
// with XOR giving a odd number
using System;
 
class GFG
{
     
// Function to count
// number of odd pairs
static int findOddPair(int[] A,
                    int N)
{
    int i, j;
 
    // variable for counting
    // odd pairs
    int oddPair = 0;
 
    // find all pairs
    for (i = 0; i < N; i++)
    {
        for (j = i + 1; j < N; j++)
        {
 
            // find XOR operation
            // check odd or even
            if ((A[i] ^ A[j]) % 2 != 0)
                oddPair++;
        }
     
    }
 
    // return number
    // of odd pair
    return oddPair;
}
 
// Driver Code
public static void Main()
{
    int[] A = { 5, 4, 7, 2, 1 };
    int N = A.Length;
 
    // calling function findOddPair
    // and print number of odd pair
    Console.WriteLine(findOddPair(A, N));
}
}
 
// This code is contributed
// by Akanksha Rai(Abby_akku)
 
 
# calling function findOddPair
# and print number of odd pair
print(findOddPair(a, n))

Javascript

// JavaScript program to count pairs
// with XOR giving a odd number 
 
// Function to count number of odd pairs
function findOddPair(A, N)
{
    let i, j;
 
    // variable for counting odd pairs
    let oddPair = 0;
 
    // find all pairs
    for (i = 0; i < N; i++) {
        for (j = i + 1; j < N; j++) {
 
            // find XOR operation
            // check odd or even
            if ((A[i] ^ A[j]) % 2 != 0)
                oddPair++;
        }
    }
 
    // return number of odd pair
    return oddPair;
}
 
// Driver Code
 
 
    let A = [ 5, 4, 7, 2, 1 ];
    let N = A.length;
 
    // calling function findOddPair
    // and print number of odd pair
    document.write(findOddPair(A, N) + "<br>");
 
// This code is contributed by Surbhi Tyagi.
 

输出：

6

时间复杂度：O(N ^ 2)
一个有效的解决方案是计算偶数。然后返回count *(N –计数)。

C++

// C++ program to count pairs
// with XOR giving a odd number
#include 
using namespace std;
 
// Function to count number of odd pairs
int findOddPair(int A[], int N)
{
    int i, count = 0;
 
    // find all pairs
    for (i = 0; i < N; i++) {
        if (A[i] % 2 == 0)
            count++;
    }
 
    // return number of odd pair
    return count * (N - count);
}
 
// Driver Code
int main()
{
    int a[] = { 5, 4, 7, 2, 1 };
    int n = sizeof(a) / sizeof(a[0]);
 
    // calling function findOddPair
    // and print number of odd pair
    cout << findOddPair(a, n) << endl;
 
    return 0;
}

Java

// Java program to count pairs
// with XOR giving a odd number
class GFG
{
// Function to count
// number of odd pairs
static int findOddPair(int A[],
                       int N)
{
    int i, count = 0;
 
    // find all pairs
    for (i = 0; i < N; i++)
    {
        if (A[i] % 2 == 0)
            count++;
    }
 
    // return number of odd pair
    return count * (N - count);
}
 
// Driver Code
public static void main(String[] arg)
{
    int a[] = { 5, 4, 7, 2, 1 };
    int n = a.length ;
 
    // calling function findOddPair
    // and print number of odd pair
    System.out.println(findOddPair(a, n));
}
}
 
// This code is contributed
// by Smitha

Python3

# Python3 program to count pairs
# with XOR giving a odd number
 
# Function to count number of odd pairs
def findOddPair(A, N) :
 
    count = 0
 
    # find all pairs
    for i in range(0 , N) :
        if (A[i] % 2 == 0) :
            count+=1
     
    # return number of odd pair
    return count * (N - count)
 
# Driver Code
if __name__=='__main__':
    a = [5, 4, 7, 2, 1]
    n = len(a)
    print(findOddPair(a,n))
 
# this code is contributed by Smitha Dinesh Semwal

C＃

// C# program to count pairs
// with XOR giving a odd number
using System;
 
class GFG
{
// Function to count
// number of odd pairs
static int findOddPair(int []A,
                       int N)
{
    int i, count = 0;
 
    // find all pairs
    for (i = 0; i < N; i++)
    {
        if (A[i] % 2 == 0)
            count++;
    }
 
    // return number of odd pair
    return count * (N - count);
}
 
// Driver Code
public static void Main()
{
    int []a = { 5, 4, 7, 2, 1 };
    int n = a.Length ;
 
    // calling function findOddPair
    // and print number of odd pair
    Console.Write(findOddPair(a, n));
}
}
 
// This code is contributed
// by Smitha

输出：

6

时间复杂度：O(N)