Count the number of pairs (a, b) such that 1 ≤ a ≤ x, 1 ≤ b ≤ y and a+b is even

Given two positive integers. Count the number of pairs (a, b) such that 1 ≤ a ≤ x, 1 ≤ b ≤ y and a+b is even.

Examples:

Input: 2 7
Output: Count of pairs is 7

Input: 1 1
Output: Count of pairs is 1 

Input: 4 9
Output: Count of pairs is 18

Input: 3 5
Output: Count of pairs is 8

Simple Approach :
A simple approach will be to run two nested loops to count all possible pair of elements such that 1 ≤ a ≤  x, 1 ≤ b ≤ y and a+b is even.
If yes then we increment count and return the count.

Implementation:

// A O(n ^ 2) time and O(1) extra space C++ program to
// count the number of pairs (a, b) 
// such that 1 ≤ a ≤  x, 1 ≤ b ≤ y and a+b  is even  
#include <bits/stdc++.h>

using namespace std;

// Returns the number of pairs  (a, b) 
// such that 1 ≤ a ≤  x, 1 ≤ b ≤ y and a+b  is even  
int getPairsCount(int x, int y) {
  int count = 0;
  for (int i = 1; i <= x; i++) {

    for (int j = 1; j <= y; j++) {
      if ((i + j) % 2 == 0) // check pair is even or not 

        count++;
    }

  }
  return count;

}

// Driver function to test the above function
int main() {

  int x = 2, y = 7;
  cout << "Count of pairs is " << getPairsCount(x, y);
  return 0;

}

Output:

Count of pairs is 7

Time Complexity: O(n^2)  
Auxiliary Space: O(1)

Efficient Approach :
If a is even and b is even then sum of pair (a,b) is also even but if a is odd and b is odd then sum of pair (a,b) will be even.
Total even numbers between 1 to x will be x/2
Total odd numbers between 1 to x  will be (x+1)/2
Total even numbers between 1 to y will be y/2
Total odd numbers between 1 to y will be (y+1)/2

Therefore, the number of pairs (a,b) where a+b is even :  

(x/2) * (y/2)+((x+1)/2)*((y+1)/2) 

Implementation:

// A O(1) time and O(1) extra space C++ program to
// count the number of pairs (a, b) 
// such that 1 ≤ a ≤  x, 1 ≤ b ≤ y and a+b  is even  
#include <bits/stdc++.h>

using namespace std;

// Returns the number of pairs  (a, b) 
// such that 1 ≤ a ≤  x, 1 ≤ b ≤ y and a+b  is even  
int getPairsCount(int x, int y) {

  int even_x = x / 2; // store the half of x
  int even_y = y / 2; // store the half of y 

  int odd_x = (x + 1) / 2;
  int odd_y = (y + 1) / 2;

  int res = even_x * even_y + odd_x * odd_y;

  return res;

}

// Driver function to test the above function
int main() {
  int x = 2, y = 7;
  cout << "Count of pairs is " << getPairsCount(x, y);
  return 0;

}

Output:

Count of pairs is 7

Time Complexity: O(1)
Auxiliary Space: O(1)