计算带偶数和奇数乘积的有序对的数量

给定一个由n个正数组成的数组，任务是计算带有偶数和奇数乘积的有序对的数量。有序对意味着(a，b)和(b，a)是不同的。
例子：

Input: n = 3, arr[] = {1, 2, 7} 
Output: Even product Pairs = 4, Odd product Pairs = 2 
The ordered pairs are (1, 2), (1, 7), (2, 1), (7, 1), (2, 7), (7, 2) 
Pairs with Odd product: (1, 7), (7, 1) 
Pairs with Even product: (1, 2), (2, 7), (2, 1), (7, 2)
Input: n = 6, arr[] = {2, 4, 5, 9, 1, 8} 
Output: Even product Pairs = 24, Odd product Pairs = 6 
 

方法：
仅当两个数字均为奇数时，两个数字的乘积才为奇数。所以：

Number of odd product pairs = (count of odd numbers) * (count of odd numbers – 1)

并且偶数产品对的数量将是奇数产品对的数量的倒数。所以：

Number of even product pairs = Total Number of pairs – Number of odd product pairs

下面是上述方法的实现：

C++

// C++ implementation of the above approach
#include 
using namespace std;
 
// function to count odd product pair
int count_odd_pair(int n, int a[])
{
    int odd = 0, even = 0;
 
    for (int i = 0; i < n; i++) {
 
        // if number is even
        if (a[i] % 2 == 0)
            even++;
 
        // if number is odd
        else
            odd++;
    }
 
    // count of ordered pairs
    int ans = odd * (odd - 1);
 
    return ans;
}
 
// function to count even product pair
int count_even_pair(int odd_product_pairs, int n)
{
    int total_pairs = (n * (n - 1));
    int ans = total_pairs - odd_product_pairs;
    return ans ;
}
 
// Driver code
int main()
{
 
    int n = 6;
    int a[] = { 2, 4, 5, 9, 1, 8 };
 
    int odd_product_pairs = count_odd_pair(n, a);
 
    int even_product_pairs = count_even_pair(
        odd_product_pairs, n);
 
    cout << "Even Product Pairs = "
         << even_product_pairs
         << endl;
    cout << "Odd Product Pairs= "
         << odd_product_pairs
         << endl;
 
    return 0;
}

Java

// Java  implementation of the above approach
import java.io.*;
 
class GFG {
     
     
// function to count odd product pair
static int count_odd_pair(int n, int a[])
{
    int odd = 0, even = 0;
 
    for (int i = 0; i < n; i++) {
 
        // if number is even
        if (a[i] % 2 == 0)
            even++;
 
        // if number is odd
        else
            odd++;
    }
 
    // count of ordered pairs
    int ans = odd * (odd - 1);
 
    return ans;
}
 
// function to count even product pair
static int count_even_pair(int odd_product_pairs, int n)
{
    int total_pairs = (n * (n - 1));
    int ans = total_pairs - odd_product_pairs;
    return ans;
}
 
// Driver code
    public static void main (String[] args) {
 
        int n = 6;
        int []a = { 2, 4, 5, 9, 1, 8 };
 
        int odd_product_pairs = count_odd_pair(n, a);
 
        int even_product_pairs = count_even_pair(
            odd_product_pairs, n);
 
        System.out.println( "Even Product Pairs = "+
            even_product_pairs );
          
        System.out.println("Odd Product Pairs= "+
             odd_product_pairs );
     
    }
}
//This Code is Contributed by ajit

Python3

# Python3 implementation of
# above approach
 
# function to count odd product pair
def count_odd_pair(n, a):
    odd = 0
    even = 0
    for i in range(0,n):
         
        # if number is even
        if a[i] % 2==0:
            even=even+1
        # if number is odd
        else:
            odd=odd+1
     
    # count of ordered pairs
    ans = odd * (odd - 1)
    return ans
 
# function to count even product pair
def count_even_pair(odd_product_pairs, n):
    total_pairs = (n * (n - 1))
    ans = total_pairs - odd_product_pairs
    return ans
 
#Driver code
if __name__=='__main__':
    n = 6
    a = [2, 4, 5, 9, 1 ,8]
 
    odd_product_pairs = count_odd_pair(n, a)
    even_product_pairs = (count_even_pair
                       (odd_product_pairs, n))
 
    print("Even Product Pairs = "
          ,even_product_pairs)
    print("Odd Product Pairs= "
          ,odd_product_pairs)
 
# This code is contributed by
# Shashank_Sharma

C#

// C#  implementation of the above approach
using System;
 
public class GFG{
     
         
// function to count odd product pair
static int count_odd_pair(int n, int []a)
{
    int odd = 0, even = 0;
 
    for (int i = 0; i < n; i++) {
 
        // if number is even
        if (a[i] % 2 == 0)
            even++;
 
        // if number is odd
        else
            odd++;
    }
 
    // count of ordered pairs
    int ans = odd * (odd - 1);
 
    return ans;
}
 
// function to count even product pair
static int count_even_pair(int odd_product_pairs, int n)
{
    int total_pairs = (n * (n - 1));
    int ans = total_pairs - odd_product_pairs;
    return ans;
}
 
// Driver code
     
static public void Main (){
        int n = 6;
        int []a = { 2, 4, 5, 9, 1, 8 };
 
        int odd_product_pairs = count_odd_pair(n, a);
 
        int even_product_pairs = count_even_pair(
            odd_product_pairs, n);
 
        Console.WriteLine( "Even Product Pairs = "+
            even_product_pairs );
         
        Console.WriteLine("Odd Product Pairs= "+
            odd_product_pairs );
    }
}

Javascript

 
// JavaScript implementation of the above approach
 
// function to count odd product pair
function count_odd_pair(n, a)
{
    let odd = 0, even = 0;
 
    for (let i = 0; i < n; i++) {
 
        // if number is even
        if (a[i] % 2 == 0)
            even++;
 
        // if number is odd
        else
            odd++;
    }
 
    // count of ordered pairs
    let ans = odd * (odd - 1);
 
    return ans;
}
 
// function to count even product pair
function count_even_pair(odd_product_pairs, n)
{
    let total_pairs = (n * (n - 1));
    let ans = total_pairs - odd_product_pairs;
    return ans ;
}
 
// Driver code
    let n = 6;
    let a = [ 2, 4, 5, 9, 1, 8 ];
 
    let odd_product_pairs = count_odd_pair(n, a);
 
    let even_product_pairs = count_even_pair(
        odd_product_pairs, n);
 
    document.write("Even Product Pairs = "
        + even_product_pairs
        + "<br>");
    document.write("Odd Product Pairs= "
        + odd_product_pairs
        + "<br>");
 
// This code is contributed by Surbhi Tyagi.
 

输出：

Even Product Pairs = 24
Odd Product Pairs= 6