Problem Explanation:

You're given several test cases, each consisting of two integers n and k. For each test case, you need to determine the number of values x that satisfy certain conditions.

Based on the provided code, it seems the problem involves splitting the number n into two parts (r and n - r) and checking if their product is less than or equal to n * k. The goal is to find the largest valid split and output twice the largest valid r.

Key Concepts:

• Objective: For each test case, you are given n and k. You need to:

  1. Find how many values of r can split n into two parts such that the product of the two parts is less than or equal to n * k.
  2. Output twice the largest valid r that satisfies this condition.

• Mathematical Formulation: The problem checks the condition r * (n - r) <= n * k, where r is an integer between 1 and n/2.

Steps in the Solution:

1. Input:

   • You receive multiple test cases where each test case contains two integers n and k.
   • The number n represents an upper limit, and k is a factor that impacts how large the product of the split can be.

2. Initial Check:

   • First, check whether the product of n/2 * (n - n/2) is less than or equal to n * k.
     • This is because splitting n evenly into two parts (n/2 and n - n/2) generally maximizes the product of the two parts.
   • If this condition is satisfied, or if n == 1 (since splitting doesn't make sense for n = 1), print n - 1 and stop further calculation.
     • For n == 1, there's only one valid number x (which is 0), so the result is n - 1 = 0.

3. Binary Search:

   • If the initial condition fails, use binary search to find the largest r such that r * (n - r) <= n * k.
   • Binary search works because:
     • The function r * (n - r) behaves in a predictable way as r increases: it starts small, peaks at n/2, and then decreases.
   • The idea is to efficiently find the largest value of r that satisfies the condition.
   • The search is conducted within the range [1, n/2] because the split becomes redundant after n/2 (e.g., r > n/2 would be the same as having r <= n/2 due to symmetry).

4. Binary Search Logic:

   • You calculate the midpoint mid of the current search range [l, r].
   • If mid * (n - mid) <= n * k, it means this split is valid, and you try to move the left pointer l to explore larger values of r.
   • If the condition is not satisfied, move the right pointer r to explore smaller values of r.
   • The goal is to find the largest r for which the condition holds.

5. Final Result:

   • Once the binary search is complete, the largest valid r is found, and the result is 2 * r (as required by the problem).

Example Walkthrough:

Test Case 1:

• Input: n = 5, k = 1

  • First, check if n/2 * (n - n/2) <= n * k.
  • 5 / 2 * (5 - 5 / 2) = 2 * 3 = 6
  • 5 * 1 = 5

  • Since 6 > 5, we move to the binary search.

  • Perform binary search over the range [1, 5/2 = 2]:

  • Try mid = 1: 1 * (5 - 1) = 1 * 4 = 4, which is less than or equal to 5 * 1 = 5. So, move l = 2.

  • Try mid = 2: 2 * (5 - 2) = 2 * 3 = 6, which is greater than 5. So, move r = 1.

  • The largest valid r is 1, and the result is 2 * r = 2.

Test Case 2:

• Input: n = 5, k = 2

  • First, check if n/2 * (n - n/2) <= n * k.
  • 5 / 2 * (5 - 5 / 2) = 2 * 3 = 6
  • 5 * 2 = 10
  • Since 6 <= 10, print n - 1 = 4.

Summary of Solution Steps:

1. For each test case:
   • Compute whether the product of two halves of n (n/2 * (n - n/2)) is less than or equal to n * k.
   • If yes, print n - 1.
   • If no, use binary search to find the largest value of r such that r * (n - r) <= n * k.
   • Print 2 * r as the result.

This process ensures that you efficiently find the correct number of valid splits for each test case based on the given condition.