通过分别从两个给定范围中选择两个整数来计算可能的偶数和对

给定两个正整数X和Y ，任务是通过选择范围1到X 中的任何整数和范围1到Y 中的另一个整数来计算具有偶数和可能的对。

例子：

Input : X = 2, Y = 3 
Output : 3 
Explanation : All such possible pairs are {1, 1}, {1, 3}, {2, 2}.

处理方法：按照以下步骤解决问题：

• 初始化一个变量，比如cntXEvenNums ，以存储通过将X除以2获得的范围[1, X]中偶数的计数。
• 初始化一个变量，比如cntXOddNums ，以存储通过将(X + 1)除以2获得的范围[1, X] 中的奇数计数。
• 初始化一个变量，比如cntYEvenNums ，通过将Y除以2来存储1到Y之间偶数的计数。
• 初始化一个变量，比如cntYOddNums ，通过将(Y + 1)除以2来存储1到Y之间的奇数计数。
• 初始化变量，说cntPairs，用cntYEvenNums与cntYOddNums乘以cntXEvenNums和乘以cntXOddNums存储甚至和对的数量，发现两者的总和。
• 最后，打印得到的cntPairs的值。

下面是上述方法的实现：

C++

// C++ program to implement
// the above approach
 
#include 
using namespace std;
 
// Function to count even
// sum pairs in the given range
long long cntEvenSumPairs(long long X, long long Y)
{
    // Stores the count of even
    // numbers between 1 to X
    long long cntXEvenNums = X / 2;
 
    // Stores the count of odd
    // numbers between 1 to X
    long long cntXOddNums = (X + 1) / 2;
 
    // Stores the count of even
    // numbers between 1 to Y
    long long cntYEvenNums = Y / 2;
 
    // Stores the count of odd
    // numbers between 1 to Y
    long long cntYOddNums = (Y + 1) / 2;
 
    // Stores the count of
    // pairs having even sum
    long long cntPairs
        = (cntXEvenNums * 1LL * cntYEvenNums)
          + (cntXOddNums * 1LL * cntYOddNums);
 
    // Retuens the count of pairs
    // having even sum
    return cntPairs;
}
 
// Driver Code
int main()
{
    long long X = 2;
    long long Y = 3;
 
    cout << cntEvenSumPairs(X, Y);
    return 0;
}

Java

// Java program to implement
// the above approach
 
import java.io.*;
 
class GFG {
 
    // Function to count maximum  even
    // sum pairs in the given range
    static long cntEvenSumPairs(long X, long Y)
    {
 
        // Stores the count of even
        // numbers between 1 to X
        long cntXEvenNums = X / 2;
 
        // Stores the count of odd
        // numbers between 1 to X
        long cntXOddNums = (X + 1) / 2;
 
        // Stores the count of even
        // numbers between 1 to Y
        long cntYEvenNums = Y / 2;
 
        // Stores the count of odd
        // numbers between 1 to Y
        long cntYOddNums = (Y + 1) / 2;
 
        // Stores the count of
        // pairs having even sum
        long cntPairs = (cntXEvenNums * cntYEvenNums)
                        + (cntXOddNums * cntYOddNums);
 
        // Retuens the count of pairs
        // having even sum
        return cntPairs;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        long X = 2;
        long Y = 3;
 
        System.out.println(cntEvenSumPairs(X, Y));
    }
}

Python

# Python program to implement
# the above approach
 
# Function to count even
# sum pairs in the given range
def cntEvenSumPairs(X, Y):
     
    # Stores the count of even
    # numbers between 1 to X
    cntXEvenNums = X / 2
 
    # Stores the count of odd
    # numbers between 1 to X
    cntXOddNums = (X + 1) / 2
 
    # Stores the count of even
    # numbers between 1 to Y
    cntYEvenNums = Y / 2
 
    # Stores the count of odd
    # numbers between 1 to Y
    cntYOddNums = (Y + 1) / 2
 
    # Stores the count of
    # pairs having even sum
    cntPairs = ((cntXEvenNums * cntYEvenNums) +
                 (cntXOddNums * cntYOddNums))
  
    # Returns the count of pairs
    # having even sum
    return cntPairs
 
# Driver code
X = 2
Y = 3
 
print(cntEvenSumPairs(X, Y))
 
# This code is contributed by hemanth gadarla

C#

// C# program to implement
// the above approach
using System;
 
class GFG{
 
// Function to count maximum  even
// sum pairs in the given range
static long cntEvenSumPairs(long X, long Y)
{
     
    // Stores the count of even
    // numbers between 1 to X
    long cntXEvenNums = X / 2;
 
    // Stores the count of odd
    // numbers between 1 to X
    long cntXOddNums = (X + 1) / 2;
 
    // Stores the count of even
    // numbers between 1 to Y
    long cntYEvenNums = Y / 2;
 
    // Stores the count of odd
    // numbers between 1 to Y
    long cntYOddNums = (Y + 1) / 2;
 
    // Stores the count of
    // pairs having even sum
    long cntPairs = (cntXEvenNums * cntYEvenNums) +
                     (cntXOddNums * cntYOddNums);
 
    // Retuens the count of pairs
    // having even sum
    return cntPairs;
}
 
// Driver Code
public static void Main(string[] args)
{
    long X = 2;
    long Y = 3;
 
    Console.WriteLine(cntEvenSumPairs(X, Y));
}
}
 
// This code is contributed by chitranayal

Javascript

 
// Javascript program to implement
// the above approach   
 
     // Function to count maximum even
    // sum pairs in the given range
    function cntEvenSumPairs(X , Y) {
 
        // Stores the count of even
        // numbers between 1 to X
        var cntXEvenNums = parseInt(X / 2);
 
        // Stores the count of odd
        // numbers between 1 to X
        var cntXOddNums = parseInt((X + 1) / 2);
 
        // Stores the count of even
        // numbers between 1 to Y
        var cntYEvenNums = parseInt(Y / 2);
 
        // Stores the count of odd
        // numbers between 1 to Y
        var cntYOddNums =parseInt( (Y + 1) / 2);
 
        // Stores the count of
        // pairs having even sum
        var cntPairs = (cntXEvenNums * cntYEvenNums) +
        (cntXOddNums * cntYOddNums);
 
        // Retuens the count of pairs
        // having even sum
        return cntPairs;
    }
 
    // Driver Code
     
        var X = 2;
        var Y = 3;
 
        document.write(cntEvenSumPairs(X, Y));
 
// This code contributed by Rajput-Ji
 

输出：

3

时间复杂度：O(1)
辅助空间：O(1)