计算具有偶数和的整数对

给定两个整数N和M ，任务是计算所有可能的整数对(i, j) (1 ≤ i ≤ N, 1 ≤ j ≤ M)，使得i + j是偶数。

例子：

Input: N = 6, M = 4 
Output: 12 
Explanation: The pairs (1, 1), (1, 3), (2, 2), (2, 4), (3, 1), (3, 3), (4, 2), (4, 4), (5, 1), (5, 3), (6, 2), (6, 4) satisfy the required condition. Therefore, the count is 12.

Input: N = 2 and M = 8 
Output: 8 
 

朴素的方法：解决这个问题的最简单的方法是遍历范围[1, M]中的每个数字，遍历范围[1, N]并生成所有可能的对。对于每个可能的对，检查其总和是否为偶数。如果发现为真，则增加计数。最后，打印获得的计数。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to count pairs with even sum
int countEvenPairs(int N, int M)
{
     
    // Stores the count of pairs
    // with even sum
    int count = 0;
 
    // Traverse the range 1 to N
    for(int i = 1; i <= N; i++)
    {
         
        // Traverse the range 1 to M
        for(int j = 1; j <= M; j++)
        {
             
            // Check if the sum of the
            // pair (i, j) is even or not
            if ((i + j) % 2 == 0)
            {
                 
                // Update count
                count++;
            }
        }
    }
 
    // Return the count
    return count;
}
 
// Driver Code
int main()
{
    int N = 4;
    int M = 6;
     
    cout << countEvenPairs(N, M) << endl;
     
    return 0;
}
 
// This code is contributed by akhilsaini

Java

// Java program for the above approach
import java.io.*;
 
class GFG {
 
    // Function to count pairs with even sum
    public static int countEvenPairs(
        int N, int M)
    {
 
        // Stores the count of pairs
        // with even sum
        int count = 0;
 
        // Traverse the range 1 to N
        for (int i = 1; i <= N; i++) {
 
            // Traverse the range 1 to M
            for (int j = 1; j <= M; j++) {
 
                // Check if the sum of the
                // pair (i, j) is even or not
                if ((i + j) % 2 == 0) {
 
                    // Update count
                    count++;
                }
            }
        }
 
        // Return the count
        return count;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int N = 4;
        int M = 6;
        System.out.print(
            countEvenPairs(N, M));
    }
}

Python3

# Python3 program for the above approach
 
# Function to count pairs with even sum
def countEvenPairs(N, M):
     
    # Stores the count of pairs
    # with even sum
    count = 0
     
    # Traverse the range 1 to N
    for i in range(1, N + 1):
     
        # Traverse the range 1 to M
        for j in range(1, M + 1):
         
            # Check if the sum of the
            # pair (i, j) is even or not
            if ((i + j) % 2 == 0):
               
                # Update count
                count += 1
     
    # Return the count
    return count
 
# Driver Code
if __name__ == '__main__':
     
  N = 4
  M = 6
   
  print(countEvenPairs(N, M))
   
# This code is contributed by akhilsaini

C#

// C# program for the above approach
using System;
 
class GFG{
 
// Function to count pairs with even sum
public static int countEvenPairs(int N, int M)
{
     
    // Stores the count of pairs
    // with even sum
    int count = 0;
 
    // Traverse the range 1 to N
    for(int i = 1; i <= N; i++)
    {
         
        // Traverse the range 1 to M
        for(int j = 1; j <= M; j++)
        {
             
            // Check if the sum of the
            // pair (i, j) is even or not
            if ((i + j) % 2 == 0)
            {
                 
                // Update count
                count++;
            }
        }
    }
     
    // Return the count
    return count;
}
 
// Driver Code
public static void Main()
{
    int N = 4;
    int M = 6;
     
    Console.WriteLine(
        countEvenPairs(N, M));
}
}
 
// This code is contributed by akhilsaini

Javascript

 
// Javascript program for the above approach
 
// Function to count pairs with even sum
function countEvenPairs(N, M)
{
     
    // Stores the count of pairs
    // with even sum
    var count = 0;
 
    // Traverse the range 1 to N
    for(var i = 1; i <= N; i++)
    {
         
        // Traverse the range 1 to M
        for(var j = 1; j <= M; j++)
        {
             
            // Check if the sum of the
            // pair (i, j) is even or not
            if ((i + j) % 2 == 0)
            {
                 
                // Update count
                count++;
            }
        }
    }
 
    // Return the count
    return count;
}
 
// Driver code
var N = 4;
var M = 6;
 
document.write(countEvenPairs(N, M));
 
// This code is contributed by Kirti
 

输出：

12

时间复杂度： O(N ² )
空间复杂度： O(1)

高效的方法：上述方法可以基于以下观察进行优化：

• Even number + Even number = Even number
• Odd number + Odd number = Even number

请按照以下步骤解决问题：

• 初始化两个变量，比如nEven和nOdd ，以存储最多N的奇数和偶数整数。
• 初始化两个变量，比如mEven和mOdd ，以存储最多M的偶数和奇数整数。
• 最后，使用公式计算所需的对数：

count = nEven * mEven + nOdd * mOdd

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to count even pairs
int countEvenPairs(int N, int M)
{
 
    // Stores count of pairs having even sum
    int count = 0;
 
    // Stores count of even numbers up to N
    int nEven = floor(N / 2);
 
    // Stores count of odd numbers up to N
    int nOdd = ceil(N / 2);
 
    // Stores count of even numbers up to M
    int mEven = floor(M / 2);
 
    // Stores count of odd numbers up to M
    int mOdd = ceil(M / 2);
    count = nEven * mEven + nOdd * mOdd;
 
    // Return the count
    return count;
}
 
// Driver Code
int main()
{
    int N = 4;
    int M = 6;
    cout << countEvenPairs(N, M);
    return 0;
}
 
// This code is contributed by Dharanendra L V

Java

// Java program for the above approach
import java.io.*;
 
class GFG {
 
    // Function to count even pairs
    public static int countEvenPairs(
        int N, int M)
    {
 
        // Stores count of pairs having even sum
        int count = 0;
 
        // Stores count of even numbers up to N
        int nEven = (int)Math.floor((double)N / 2);
 
        // Stores count of odd numbers up to N
        int nOdd = (int)Math.ceil((double)N / 2);
 
        // Stores count of even numbers up to M
        int mEven = (int)Math.floor((double)M / 2);
 
        // Stores count of odd numbers up to M
        int mOdd = (int)Math.ceil((double)M / 2);
 
        count = nEven * mEven + nOdd * mOdd;
 
        // Return the count
        return count;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int N = 4;
        int M = 6;
        System.out.print(countEvenPairs(N, M));
    }
}

蟒蛇3

# Python3 program for the above approach
import math
 
# Function to count even pairs
def countEvenPairs(N, M):
   
    # Stores count of pairs having even sum
    count = 0;
 
    # Stores count of even numbers up to N
    nEven = int(math.floor(N / 2));
 
    # Stores count of odd numbers up to N
    nOdd = int(math.ceil(N / 2));
 
    # Stores count of even numbers up to M
    mEven = int(math.floor(M / 2));
 
    # Stores count of odd numbers up to M
    mOdd = int(math.ceil(M / 2));
 
    count = nEven * mEven + nOdd * mOdd;
 
    # Return the count
    return count;
 
# Driver Code
if __name__ == '__main__':
    N = 4;
    M = 6;
    print(countEvenPairs(N, M));
 
# This code is contributed by 29AjayKumar

C＃

// C# program for the above approach
using System;
 
class GFG
{
 
  // Function to count even pairs
  public static int countEvenPairs(int N, int M)
  {
 
    // Stores count of pairs having even sum
    int count = 0;
 
    // Stores count of even numbers up to N
    int nEven = (int)Math.Floor((double)N / 2);
 
    // Stores count of odd numbers up to N
    int nOdd = (int)Math.Ceiling((double)N / 2);
 
    // Stores count of even numbers up to M
    int mEven = (int)Math.Floor((double)M / 2);
 
    // Stores count of odd numbers up to M
    int mOdd = (int)Math.Ceiling((double)M / 2);
    count = nEven * mEven + nOdd * mOdd;
 
    // Return the count
    return count;
  }
 
  // Driver Code
  public static void Main(String[] args)
  {
    int N = 4;
    int M = 6;
    Console.Write(countEvenPairs(N, M));
  }
}
 
// This code is contributed by shikhasingrajput

Javascript

 
// Javascript program for the above approach
 
// Function to count even pairs
function countEvenPairs(N, M)
{
     
    // Stores count of pairs having even sum
    let count = 0;
 
    // Stores count of even numbers up to N
    nEven = parseInt(Math.floor(N / 2));
 
    // Stores count of odd numbers up to N
    nOdd = parseInt(Math.ceil(N / 2));
 
    // Stores count of even numbers up to M
    mEven = parseInt(Math.floor(M / 2));
 
    // Stores count of odd numbers up to M
    mOdd = parseInt(Math.ceil(M / 2));
 
    count = nEven * mEven + nOdd * mOdd;
 
    // Return the count
    return count;
}
 
// Driver Code
let N = 4;
let M = 6;
 
document.write(countEvenPairs(N, M));
 
// This code is contributed by mohan1240760
 

输出：

12

时间复杂度： O(1)
辅助空间： O(1)