给定两个数组，计算所有总和为奇数的对

给定N和M个整数的两个数组。任务是从两个数组中查找元素组成的无序对的数量，以使它们的总和为奇数。
注意：一个元素只能是一对。
例子：

Input: a[] = {9, 14, 6, 2, 11}, b[] = {8, 4, 7, 20} 
Output: 3 
{9, 20}, {14, 7} and {11, 8}
Input: a[] = {2, 4, 6}, b[] = {8, 10, 12} 
Output: 0 
 

方法：计算两个数组中的奇数和偶数的数目，对数的答案将是min(odd1，even2)+ min(odd2，even1)，因为奇数+偶数只是奇数。
下面是上述方法的实现：

C++

// C++ program to implement
// the above approach
#include 
using namespace std;
 
// Function that returns the number of pairs
int count_pairs(int a[], int b[], int n, int m)
{
 
    // Count of odd and even numbers
    int odd1 = 0, even1 = 0;
    int odd2 = 0, even2 = 0;
 
    // Traverse in the first array
    // and count the number of odd
    // and evene numbers in them
    for (int i = 0; i < n; i++) {
        if (a[i] % 2)
            odd1++;
        else
            even1++;
    }
 
    // Traverse in the second array
    // and count the number of odd
    // and evene numbers in them
    for (int i = 0; i < m; i++) {
        if (b[i] % 2)
            odd2++;
        else
            even2++;
    }
 
    // Count the number of pairs
    int pairs = min(odd1, even2) + min(odd2, even1);
 
    // Return the number of pairs
    return pairs;
}
 
// Driver code
int main()
{
    int a[] = { 9, 14, 6, 2, 11 };
    int b[] = { 8, 4, 7, 20 };
    int n = sizeof(a) / sizeof(a[0]);
    int m = sizeof(b) / sizeof(b[0]);
    cout << count_pairs(a, b, n, m);
 
    return 0;
}

Java

// Java program to implement
// the above approach
 
class GFG {
 
    // Function that returns the number of pairs
    static int count_pairs(int a[], int b[], int n, int m)
    {
 
        // Count of odd and even numbers
        int odd1 = 0, even1 = 0;
        int odd2 = 0, even2 = 0;
 
        // Traverse in the first array
        // and count the number of odd
        // and evene numbers in them
        for (int i = 0; i < n; i++) {
            if (a[i] % 2 == 1) {
                odd1++;
            }
            else {
                even1++;
            }
        }
 
        // Traverse in the second array
        // and count the number of odd
        // and evene numbers in them
        for (int i = 0; i < m; i++) {
            if (b[i] % 2 == 1) {
                odd2++;
            }
            else {
                even2++;
            }
        }
 
        // Count the number of pairs
        int pairs = Math.min(odd1, even2) + Math.min(odd2, even1);
 
        // Return the number of pairs
        return pairs;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int a[] = { 9, 14, 6, 2, 11 };
        int b[] = { 8, 4, 7, 20 };
        int n = a.length;
        int m = b.length;
        System.out.println(count_pairs(a, b, n, m));
    }
}
 
// This code contributed by Rajput-Ji

Python3

# Python 3 program to implement
# the above approach
 
# Function that returns
# the number of pairs
def count_pairs(a, b, n, m):
     
    # Count of odd and even numbers
    odd1 = 0
    even1 = 0
    odd2 = 0
    even2 = 0
 
    # Traverse in the first array
    # and count the number of odd
    # and evene numbers in them
    for i in range(n):
        if (a[i] % 2):
            odd1 += 1
        else:
            even1 += 1
 
    # Traverse in the second array
    # and count the number of odd
    # and evene numbers in them
    for i in range(m):
        if (b[i] % 2):
            odd2 += 1
        else:
            even2 += 1
 
    # Count the number of pairs
    pairs = (min(odd1, even2) +
             min(odd2, even1))
 
    # Return the number of pairs
    return pairs
 
# Driver code
if __name__ == '__main__':
    a = [9, 14, 6, 2, 11]
    b = [8, 4, 7, 20]
    n = len(a)
    m = len(b)
    print(count_pairs(a, b, n, m))
 
# This code is contributed by
# Surendra_Gangwar

C#

// C# program to implement
// the above approach
using System;
 
class GFG {
 
    // Function that returns the number of pairs
    static int count_pairs(int[] a, int[] b, int n, int m)
    {
 
        // Count of odd and even numbers
        int odd1 = 0, even1 = 0;
        int odd2 = 0, even2 = 0;
 
        // Traverse in the first array
        // and count the number of odd
        // and evene numbers in them
        for (int i = 0; i < n; i++) {
            if (a[i] % 2 == 1) {
                odd1++;
            }
            else {
                even1++;
            }
        }
 
        // Traverse in the second array
        // and count the number of odd
        // and evene numbers in them
        for (int i = 0; i < m; i++) {
            if (b[i] % 2 == 1) {
                odd2++;
            }
            else {
                even2++;
            }
        }
 
        // Count the number of pairs
        int pairs = Math.Min(odd1, even2) + Math.Min(odd2, even1);
 
        // Return the number of pairs
        return pairs;
    }
 
    // Driver code
    static public void Main()
    {
        int[] a = { 9, 14, 6, 2, 11 };
        int[] b = { 8, 4, 7, 20 };
        int n = a.Length;
        int m = b.Length;
        Console.WriteLine(count_pairs(a, b, n, m));
    }
}
 
// This code contributed by ajit.

Javascript

 
// JavaScript program to implement
// the above approach
 
// Function that returns the number of pairs
function count_pairs(a, b, n, m)
{
 
    // Count of odd and even numbers
    let odd1 = 0, even1 = 0;
    let odd2 = 0, even2 = 0;
 
    // Traverse in the first array
    // and count the number of odd
    // and even numbers in them
    for (let i = 0; i < n; i++) {
        if (a[i] % 2)
            odd1++;
        else
            even1++;
    }
 
    // Traverse in the second array
    // and count the number of odd
    // and even numbers in them
    for (let i = 0; i < m; i++) {
        if (b[i] % 2)
            odd2++;
        else
            even2++;
    }
 
    // Count the number of pairs
    let pairs = Math.min(odd1, even2) + Math.min(odd2, even1);
 
    // Return the number of pairs
    return pairs;
}
 
// Driver code
    let a = [ 9, 14, 6, 2, 11 ];
    let b = [ 8, 4, 7, 20 ];
    let n = a.length;
    let m = b.length;
    document.write(count_pairs(a, b, n, m));
 
 
// This code is contributed by Surbhi Tyagi.
 

输出：

3

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