将按位与的对数作为偶数进行计数

给定一个数组$N$整数。任务是找到对(i，j)的对数，以使A [i]和A [j]为偶数。

例子：

Input: N = 4, A[] = { 5, 1, 3, 2 }
Output: 3
Since pair of A[] are:
( 5, 1 ), ( 5, 3 ), ( 5, 2 ), ( 1, 3 ), ( 1, 2 ), ( 3, 2 )
5 AND 1 = 1, 5 AND 3 = 1, 5 AND 2 = 0,
1 AND 3 = 1, 1 AND 2 = 0,
3 AND 2 = 2
Total even pair A( i, j ) = 3

Input : N = 6, A[] = { 5, 9, 0, 6, 7, 3 }
Output : 9
Since pair of A[] =
( 5, 9 ) = 1, ( 5, 0 ) = 0, ( 5, 6 ) = 4, ( 5, 7 ) = 5, ( 5, 3 ) = 1,
( 9, 0 ) = 0, ( 9, 6 ) = 0, ( 9, 7 ) = 1, ( 9, 3 ) = 1,
( 0, 6 ) = 0, ( 0, 7 ) = 0, ( 0, 3 ) = 0,
( 6, 7 ) = 6, ( 6, 3 ) = 2,
( 7, 3 ) = 3

幼稚的方法是检查每一对，并打印偶数对。

下面是上述方法的实现：

C++

// C++ program to count pair with
// bitwise-AND as even number
 
#include 
using namespace std;
 
// Function to count number of pairs EVEN bitwise AND
int findevenPair(int A[], int N)
{
    int i, j;
 
    // variable for counting even pairs
    int evenPair = 0;
 
    // find all pairs
    for (i = 0; i < N; i++) {
        for (j = i + 1; j < N; j++) {
 
            // find AND operation
            // to check evenpair
            if ((A[i] & A[j]) % 2 == 0)
                evenPair++;
        }
    }
 
    // return number of even pair
    return evenPair;
}
 
// Driver Code
int main()
{
 
    int a[] = { 5, 1, 3, 2 };
    int n = sizeof(a) / sizeof(a[0]);
 
    cout << findevenPair(a, n) << endl;
 
    return 0;
}

Java

// Java program to count pair with
// bitwise-AND as even number
import java.io.*;
 
class GFG
{
 
// Function to count number of
// pairs EVEN bitwise AND
static int findevenPair(int []A,
                        int N)
{
    int i, j;
 
    // variable for counting even pairs
    int evenPair = 0;
 
    // find all pairs
    for (i = 0; i < N; i++)
    {
        for (j = i + 1; j < N; j++)
        {
 
            // find AND operation
            // to check evenpair
            if ((A[i] & A[j]) % 2 == 0)
                evenPair++;
        }
    }
 
    // return number of even pair
    return evenPair;
}
 
// Driver Code
public static void main (String[] args)
{
    int []a = { 5, 1, 3, 2 };
    int n = a.length;
     
    System.out.println(findevenPair(a, n));
}
}
 
// This code is contributed by anuj_67..

Python3

# Python 3 program to count pair with
# bitwise-AND as even number
 
# Function to count number of
# pairs EVEN bitwise AND
def findevenPair(A, N):
 
    # variable for counting even pairs
    evenPair = 0
 
    # find all pairs
    for i in range(0, N):
        for j in range(i + 1, N):
             
            # find AND operation to
            # check evenpair
            if ((A[i] & A[j]) % 2 == 0):
                evenPair += 1
 
    # return number of even pair
    return evenPair
 
# Driver Code
a = [ 5, 1, 3, 2 ]
n = len(a)
 
print(findevenPair(a, n))
 
# This code is contributed
# by PrinciRaj1992

C#

// C# program to count pair with
// bitwise-AND as even number
using System;
 
class GFG
{
 
// Function to count number of
// pairs EVEN bitwise AND
static int findevenPair(int []A,
                        int N)
{
    int i, j;
 
    // variable for counting even pairs
    int evenPair = 0;
 
    // find all pairs
    for (i = 0; i < N; i++)
    {
        for (j = i + 1; j < N; j++)
        {
 
            // find AND operation
            // to check evenpair
            if ((A[i] & A[j]) % 2 == 0)
                evenPair++;
        }
    }
 
    // return number of even pair
    return evenPair;
}
 
// Driver Code
public static void Main ()
{
    int []a = { 5, 1, 3, 2 };
    int n = a.Length;
     
    Console.WriteLine(findevenPair(a, n));
}
}
 
// This code is contributed by anuj_67..

Javascript

// Javascript program to count pair with
// bitwise-AND as even number
 
// Function to count number of pairs EVEN bitwise AND
function findevenPair(A, N)
{
    let i, j;
 
    // variable for counting even pairs
    let evenPair = 0;
 
    // find all pairs
    for (i = 0; i < N; i++) {
        for (j = i + 1; j < N; j++) {
 
            // find AND operation
            // to check evenpair
            if ((A[i] & A[j]) % 2 == 0)
                evenPair++;
        }
    }
 
    // return number of even pair
    return evenPair;
}
 
// Driver Code
 
let a = [ 5, 1, 3, 2 ];
let n = a.length;
 
document.write(findevenPair(a, n));
 
// This code is contributed by souravmahato348.

输出：

3

一种有效的方法是观察两个数字的按位与，即使两个数字中的至少一个是偶数也将是偶数。因此，计算数组中所有的奇数，例如count 。现在，具有两个奇数元素的对的数量将为count *(count-1)/ 2 。

因此，具有至少一个偶数元素的元素数将是：

Total Pairs - Count of pair with both odd elements

下面是上述方法的实现：

C++

// C++ program to count pair with
// bitwise-AND as even number
 
#include 
using namespace std;
 
// Function to count number of pairs
// with EVEN bitwise AND
int findevenPair(int A[], int N)
{
    int count = 0;
 
    // count odd numbers
    for (int i = 0; i < N; i++)
        if (A[i] % 2 != 0)
            count++;
 
    // count odd pairs
    int oddCount = count * (count - 1) / 2;
 
    // return number of even pair
    return (N * (N - 1) / 2) - oddCount;
}
 
// Driver Code
int main()
{
 
    int a[] = { 5, 1, 3, 2 };
    int n = sizeof(a) / sizeof(a[0]);
 
    cout << findevenPair(a, n) << endl;
 
    return 0;
}

Java

// Java program to count pair with
// bitwise-AND as even number
 
 
import java.io.*;
 
class GFG {
  
 
 
// Function to count number of pairs
// with EVEN bitwise AND
static int findevenPair(int A[], int N)
{
    int count = 0;
 
    // count odd numbers
    for (int i = 0; i < N; i++)
        if (A[i] % 2 != 0)
            count++;
 
    // count odd pairs
    int oddCount = count * (count - 1) / 2;
 
    // return number of even pair
    return (N * (N - 1) / 2) - oddCount;
}
 
// Driver Code
 
    public static void main (String[] args) {
            int a[] = { 5, 1, 3, 2 };
    int n =a.length;
 
    System.out.print( findevenPair(a, n));
    }
}
// This code is contributed by anuj_67..

Python3

# Python 3 program to count pair with
# bitwise-AND as even number
 
# Function to count number of pairs
# with EVEN bitwise AND
def findevenPair(A, N):
    count = 0
 
    # count odd numbers
    for i in range(0, N):
        if (A[i] % 2 != 0):
            count += 1
 
    # count odd pairs
    oddCount = count * (count - 1) / 2
 
    # return number of even pair
    return (int)((N * (N - 1) / 2) - oddCount)
 
# Driver Code
a = [5, 1, 3, 2 ]
n = len(a)
 
print(findevenPair(a, n))
 
# This code is contributed
# by PrinciRaj1992

C＃

// C# program to count pair with
// bitwise-AND as even number
 
using System;
 
public class GFG{
     
// Function to count number of pairs
// with EVEN bitwise AND
static int findevenPair(int []A, int N)
{
    int count = 0;
 
    // count odd numbers
    for (int i = 0; i < N; i++)
        if (A[i] % 2 != 0)
            count++;
 
    // count odd pairs
    int oddCount = count * (count - 1) / 2;
 
    // return number of even pair
    return (N * (N - 1) / 2) - oddCount;
}
 
// Driver Code
 
    static public void Main (){
    int []a = { 5, 1, 3, 2 };
    int n =a.Length;
    Console.WriteLine( findevenPair(a, n));
    }
}
// This code is contributed by ajit

Java脚本

 
// Javascript program to count pair with
// bitwise-AND as even number
 
// Function to count number of pairs
// with EVEN bitwise AND
function findevenPair(A, N)
{
    let count = 0;
 
    // Count odd numbers
    for(let i = 0; i < N; i++)
        if (A[i] % 2 != 0)
            count++;
 
    // Count odd pairs
    let oddCount = parseInt((count *
                            (count - 1)) / 2);
 
    // Return number of even pair
    return parseInt((N * (N - 1)) / 2) - oddCount;
}
 
// Driver Code
let a = [ 5, 1, 3, 2 ];
let n = a.length;
 
document.write(findevenPair(a, n));
 
// This code is contributed by subhammahato348
 

输出：

3

时间复杂度：O(N)

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