LSB的按位与，按位OR和按位XOR为1的数组中的总对

给定大小为N的数组arr [] 。任务是找到对数(arr [i]，arr [j])为cntAND ， cntOR和cntXOR ，以便：

1. cntAND：最低有效位的按位与为1的对的计数。
2. cntOR：最低有效位的按位或为1的对数。
3. cntXOR：最低有效位的按位XOR为1的对的计数。

例子：

Input: arr[] = {1, 2, 3} 
Output: 
cntXOR = 2 
cntAND = 1 
cntOR = 3 
Array elements in binary are {01, 10, 11} 
Total XOR pairs: 2 i.e., (1, 2) and (2, 3) 
Total AND pairs: 1 i.e., (1, 3) 
Total OR pairs: 3 i.e., (1, 2), (2, 3) and (1, 3)
Input: arr[] = {1, 3, 4, 2} 
Output: 
cntXOR = 4 
cntAND = 1 
cntOR = 5 
 

方法：

1. 要获得数组元素的LSB ，首先我们要计算总的偶数和奇数元素。偶数元素的LSB为0，奇数元素的LSB为1。
2. 为了
   • XOR为1时，两个元素的LSB必须不同。
   • AND为1时，两个元素的LSB必须为1。
   • OR为1时，至少一个元素的LSB应为1。
3. 因此，所需对的总数
   • 对于XOR： cntXOR = cntOdd * cntEven
   • 对于AND： cntAND = cntOdd *(cntOdd – 1)/ 2
   • 对于OR： cntOR =(cntOdd * cntEven)+ cntOdd *(cntOdd – 1)/ 2

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to find the count of required pairs
void CalculatePairs(int a[], int n)
{
 
    // To store the count of elements which
    // give remainder 0 i.e. even values
    int cnt_zero = 0;
 
    // To store the count of elements which
    // give remainder 1 i.e. odd values
    int cnt_one = 0;
 
    for (int i = 0; i < n; i++) {
 
        if (a[i] % 2 == 0)
            cnt_zero += 1;
        else
            cnt_one += 1;
    }
 
    long int total_XOR_pairs = cnt_zero * cnt_one;
    long int total_AND_pairs = (cnt_one) * (cnt_one - 1) / 2;
    long int total_OR_pairs = cnt_zero * cnt_one
                              + (cnt_one) * (cnt_one - 1) / 2;
 
    cout << "cntXOR = " << total_XOR_pairs << endl;
    cout << "cntAND = " << total_AND_pairs << endl;
    cout << "cntOR = " << total_OR_pairs << endl;
}
 
// Driver code
int main()
{
    int a[] = { 1, 3, 4, 2 };
    int n = sizeof(a) / sizeof(a[0]);
 
    CalculatePairs(a, n);
 
    return 0;
}

Java

// Java implementation of the approach
class GFG {
 
    // Function to find the count of required pairs
    static void CalculatePairs(int a[], int n)
    {
 
        // To store the count of elements which
        // give remainder 0 i.e. even values
        int cnt_zero = 0;
 
        // To store the count of elements which
        // give remainder 1 i.e. odd values
        int cnt_one = 0;
 
        for (int i = 0; i < n; i++) {
 
            if (a[i] % 2 == 0)
                cnt_zero += 1;
            else
                cnt_one += 1;
        }
 
        int total_XOR_pairs = cnt_zero * cnt_one;
        int total_AND_pairs = (cnt_one) * (cnt_one - 1) / 2;
        int total_OR_pairs = cnt_zero * cnt_one
                             + (cnt_one) * (cnt_one - 1) / 2;
 
        System.out.println("cntXOR = " + total_XOR_pairs);
        System.out.println("cntAND = " + total_AND_pairs);
        System.out.println("cntOR = " + total_OR_pairs);
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int a[] = { 1, 3, 4, 2 };
        int n = a.length;
 
        CalculatePairs(a, n);
    }
}

Python3

# Python3 program to find number of pairs
 
# Function to find the count of required pairs
def CalculatePairs(a, n):
 
    # To store the count of elements which
    # give remainder 0 i.e. even values
    cnt_zero = 0
 
    # To store the count of elements which
    # give remainder 1 i.e. odd values
    cnt_one = 0
 
    for i in range(0, n):
        if (a[i] % 2 == 0):
            cnt_zero += 1
        else:
            cnt_one += 1
     
    total_XOR_pairs = cnt_zero * cnt_one
    total_AND_pairs = (cnt_one) * (cnt_one - 1) / 2
    total_OR_pairs = cnt_zero * cnt_one + (cnt_one) * (cnt_one - 1) / 2
 
    print("cntXOR = ", int(total_XOR_pairs))
    print("cntAND = ", int(total_AND_pairs))
    print("cntOR = ", int(total_OR_pairs))
     
 
# Driver code
if __name__ == '__main__':
     
    a = [1, 3, 4, 2]
    n = len(a)
     
    # Print the count
    CalculatePairs(a, n)

C#

// C# implementation of the approach
using System;
 
class GFG {
 
    // Function to find the count of required pairs
    static void CalculatePairs(int[] a, int n)
    {
 
        // To store the count of elements which
        // give remainder 0 i.e. even values
        int cnt_zero = 0;
 
        // To store the count of elements which
        // give remainder 1 i.e. odd values
        int cnt_one = 0;
 
        for (int i = 0; i < n; i++) {
 
            if (a[i] % 2 == 0)
                cnt_zero += 1;
            else
                cnt_one += 1;
        }
 
        int total_XOR_pairs = cnt_zero * cnt_one;
        int total_AND_pairs = (cnt_one) * (cnt_one - 1) / 2;
        int total_OR_pairs = cnt_zero * cnt_one
                             + (cnt_one) * (cnt_one - 1) / 2;
 
        Console.WriteLine("cntXOR = " + total_XOR_pairs);
        Console.WriteLine("cntAND = " + total_AND_pairs);
        Console.WriteLine("cntOR = " + total_OR_pairs);
    }
 
    // Driver code
    public static void Main()
    {
        int[] a = { 1, 3, 4, 2 };
        int n = a.Length;
 
        // Print the count
        CalculatePairs(a, n);
    }
}

Javascript

 
// JavaScript implementation of the approach
 
// Function to find the count of required pairs
function CalculatePairs(a, n)
{
 
    // To store the count of elements which
    // give remainder 0 i.e. even values
    let cnt_zero = 0;
 
    // To store the count of elements which
    // give remainder 1 i.e. odd values
    let cnt_one = 0;
 
    for (let i = 0; i < n; i++) {
 
        if (a[i] % 2 == 0)
            cnt_zero += 1;
        else
            cnt_one += 1;
    }
 
    let total_XOR_pairs = cnt_zero * cnt_one;
    let total_AND_pairs = (cnt_one) * (cnt_one - 1) / 2;
    let total_OR_pairs = cnt_zero * cnt_one
                            + (cnt_one) * (cnt_one - 1) / 2;
 
    document.write("cntXOR = " + total_XOR_pairs + "<br>");
    document.write("cntAND = " + total_AND_pairs + "<br>");
    document.write("cntOR = " + total_OR_pairs + "<br>");
}
 
// Driver code
 
    let a = [ 1, 3, 4, 2 ];
    let n = a.length;
 
    CalculatePairs(a, n);
 
// This code is contributed by Surbhi Tyagi
 

输出：

cntXOR = 4
cntAND = 1
cntOR = 5

时间复杂度： O(N)