Given two integers X and Y, the task is to count the number of pairs (m, n), such that m / n = m % n and 1 ? m ? x and 1 ? n ? y.
Examples:
Input: X = 4, Y = 5
Output: 2
Explanation: The pairs (3, 2) and (4, 3) satisfy the condition.Input: X = 3, Y = 1
Output : 0
Approach: The given problem can be solved based on the following observations:
- For the condition to be satisfied, the numerator must be of the form (kn + k). Therefore, (kn + k) / n = (kn + k) % n = k.
- It also implies that k < n. Therefore, k * k < k * n + k <= x. Hence, k < sqrt(x).
- Therefore, iterating from 1 to sqrt(x) for the numerator is sufficient.
- Rewriting k * n + k ? x gives us n <= (x / k – 1) . Also, n > k and n <= y from the constraints.
- For each possible numerator value, count the possible denominator values and update the total count.
Below is the implementation of the above approach.
C++
// C++ Program for the above approach#include <bits/stdc++.h>using namespace std;// Function to calculate the number// of pairs satisfying (m / n = m % n)void countOfPairs(int x, int y){ int count = 0; // Iterate from 1 to sqrt(x) for (int k = 1; k * k <= x; ++k) { // Combining the conditions - // 1) n > k // 2) n <= y // 3) n <= (x/ k -1) count += max(0, min(y, x / k - 1) - k); } cout << count << "\n";}// Driver codeint main(){ int x = 4; int y = 5; countOfPairs(x, y); return 0;} |
Java
// Java Program for the above approachimport java.io.*;class GFG { // Function to calculate the number // of pairs satisfying (m / n = m % n) static void countOfPairs(int x, int y) { int count = 0; // Iterate from 1 to sqrt(x) for (int k = 1; k * k <= x; ++k) { // Combining the conditions - // 1) n > k // 2) n <= y // 3) n <= (x/ k -1) count += Math.max( 0, Math.min(y, x / k - 1) - k); } System.out.print(count); } // Driver code public static void main(String[] args) { int x = 4; int y = 5; countOfPairs(x, y); }} |
Python3
# python 3 Program for the above approachfrom math import sqrt# Function to calculate the number# of pairs satisfying (m / n = m % n)def countOfPairs(x, y): count = 0 # Iterate from 1 to sqrt(x) for k in range(1,int(sqrt(x)) + 1, 1): # Combining the conditions - # 1) n > k # 2) n <= y # 3) n <= (x/ k -1) count += max(0, min(y, x / k - 1) - k) print(int(count))# Driver codeif __name__ == '__main__': x = 4 y = 5 countOfPairs(x, y) # This code is contributed by bgangwar59. |
C#
// C# Program for the above approachusing System;public class GFG { // Function to calculate the number // of pairs satisfying (m / n = m % n) static void countOfPairs(int x, int y) { int count = 0; // Iterate from 1 to sqrt(x) for (int k = 1; k * k <= x; ++k) { // Combining the conditions - // 1) n > k // 2) n <= y // 3) n <= (x/ k -1) count += Math.Max( 0, Math.Min(y, x / k - 1) - k); } Console.Write(count); } // Driver Code static public void Main() { int x = 4; int y = 5; countOfPairs(x, y); }} |
Javascript
<script>// JavaScript Program for the above approach// Function to calculate the number// of pairs satisfying (m / n = m % n)function countOfPairs(x, y){ var count = 0; var k; // Iterate from 1 to sqrt(x) for (k = 1; k * k <= x; ++k) { // Combining the conditions - // 1) n > k // 2) n <= y // 3) n <= (x/ k -1) count += Math.max(0, Math.min(y, x / k - 1) - k); } document.write(count + "<br>");}// Driver code var x = 4; var y = 5; countOfPairs(x, y);</script> |
Output:
2
Time Complexity: O(?X)
Auxiliary Space: O(1)
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