Given two integers A and B, the task is to count the number of pronic numbers that are present in the range [A, B].
Examples:
Input: A = 3, B = 20
Output: 3
Explanation: The pronic numbers from the range [3, 20] are 6, 12, 20Input: A = 5000, B = 990000000
Output: 31393
Naive Approach: Follow the given steps to solve the problem:
- Initialize the count of pronic numbers to 0.
- For every number in the range [A, B], check whether it is a pronic integer or not
- If found to be true, increment the count.
- Finally, print the count
Below is the implementation of the above approach:
C++14
// C++ program for// the above approach#include <bits/stdc++.h>using namespace std;// Function to check if x// is a Pronic Number or notbool checkPronic(int x){ for (int i = 0; i <= (int)(sqrt(x)); i++) { // Check for Pronic Number by // multiplying consecutive // numbers if (x == i * (i + 1)) { return true; } } return false;}// Function to count pronic// numbers in the range [A, B]void countPronic(int A, int B){ // Initialise count int count = 0; // Iterate from A to B for (int i = A; i <= B; i++) { // If i is pronic if (checkPronic(i)) { // Increment count count++; } } // Print count cout << count;}// Driver Codeint main(){ int A = 3, B = 20; // Function call to count pronic // numbers in the range [A, B] countPronic(A, B); return 0;} |
Java
// Java program to implement // the above approach import java.util.*;class GFG{// Function to check if x// is a Pronic Number or notstatic boolean checkPronic(int x){ for (int i = 0; i <= (int)(Math.sqrt(x)); i++) { // Check for Pronic Number by // multiplying consecutive // numbers if ((x == i * (i + 1)) != false) { return true; } } return false;}// Function to count pronic// numbers in the range [A, B]static void countPronic(int A, int B){ // Initialise count int count = 0; // Iterate from A to B for (int i = A; i <= B; i++) { // If i is pronic if (checkPronic(i) != false) { // Increment count count++; } } // Print count System.out.print(count);}// Driver Codepublic static void main(String args[]){ int A = 3, B = 20; // Function call to count pronic // numbers in the range [A, B] countPronic(A, B);}}// This code is contributed by sanjoy_62. |
Python3
# Python3 program for the above approachimport math# Function to check if x# is a Pronic Number or notdef checkPronic(x) : for i in range(int(math.sqrt(x)) + 1): # Check for Pronic Number by # multiplying consecutive # numbers if (x == i * (i + 1)) : return True return False # Function to count pronic# numbers in the range [A, B]def countPronic(A, B) : # Initialise count count = 0 # Iterate from A to B for i in range(A, B + 1): # If i is pronic if (checkPronic(i) != 0) : # Increment count count += 1 # Print count print(count)# Driver CodeA = 3B = 20# Function call to count pronic# numbers in the range [A, B]countPronic(A, B)# This code is contributed by susmitakundugoaldanga. |
C#
// C# program for the above approachusing System;public class GFG{// Function to check if x// is a Pronic Number or notstatic bool checkPronic(int x){ for (int i = 0; i <= (int)(Math.Sqrt(x)); i++) { // Check for Pronic Number by // multiplying consecutive // numbers if ((x == i * (i + 1)) != false) { return true; } } return false;}// Function to count pronic// numbers in the range [A, B]static void countPronic(int A, int B){ // Initialise count int count = 0; // Iterate from A to B for (int i = A; i <= B; i++) { // If i is pronic if (checkPronic(i) != false) { // Increment count count++; } } // Print count Console.Write(count);}// Driver Codepublic static void Main(String[] args){ int A = 3, B = 20; // Function call to count pronic // numbers in the range [A, B] countPronic(A, B);}}// This code is contributed by code_hunt. |
Javascript
<script>// Javascript program for// the above approach// Function to check if x// is a Pronic Number or notfunction checkPronic(x){ for (let i = 0; i <= parseInt(Math.sqrt(x)); i++) { // Check for Pronic Number by // multiplying consecutive // numbers if (x == i * (i + 1)) { return true; } } return false;}// Function to count pronic// numbers in the range [A, B]function countPronic(A, B){ // Initialise count let count = 0; // Iterate from A to B for (let i = A; i <= B; i++) { // If i is pronic if (checkPronic(i)) { // Increment count count++; } } // Print count document.write(count);}// Driver Codelet A = 3, B = 20;// Function call to count pronic// numbers in the range [A, B]countPronic(A, B);</script> |
Output:
3
Time Complexity: O((B – A) * ?(B – A))
Auxiliary Space: O(1)
Efficient Approach: Follow the below steps to solve the problem:
- Define a function pronic(N) to find the count of pronic integers which are ? N.
- Calculate the square root of N, say X.
- If product of X and X – 1 is less than or equal to N, return N.
- Otherwise, return N – 1.
- Print pronic(B) – pronic(A – 1) to get the count of pronic integers in the range [A, B]
Below is the implementation of the above approach:
C++14
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to count pronic// numbers in the range [A, B]int pronic(int num){ // Check upto sqrt N int N = (int)sqrt(num); // If product of consecutive // numbers are less than equal to num if (N * (N + 1) <= num) { return N; } // Return N - 1 return N - 1;}// Function to count pronic// numbers in the range [A, B]int countPronic(int A, int B){ // Subtract the count of pronic numbers // which are <= (A - 1) from the count // f pronic numbers which are <= B return pronic(B) - pronic(A - 1);}// Driver Codeint main(){ int A = 3; int B = 20; // Function call to count pronic // numbers in the range [A, B] cout << countPronic(A, B); return 0;} |
Java
// Java program for the above approachimport java.util.*;class GFG{// Function to count pronic// numbers in the range [A, B]static int pronic(int num){ // Check upto sqrt N int N = (int)Math.sqrt(num); // If product of consecutive // numbers are less than equal to num if (N * (N + 1) <= num) { return N; } // Return N - 1 return N - 1;}// Function to count pronic// numbers in the range [A, B]static int countPronic(int A, int B){ // Subtract the count of pronic numbers // which are <= (A - 1) from the count // f pronic numbers which are <= B return pronic(B) - pronic(A - 1);}// Driver Codepublic static void main(String[] args){ int A = 3; int B = 20; // Function call to count pronic // numbers in the range [A, B] System.out.print(countPronic(A, B));}}// This code is contributed by shikhasingrajput |
Python3
# Python3 program for the above approach# Function to count pronic# numbers in the range [A, B]def pronic(num) : # Check upto sqrt N N = int(num ** (1/2)); # If product of consecutive # numbers are less than equal to num if (N * (N + 1) <= num) : return N; # Return N - 1 return N - 1;# Function to count pronic# numbers in the range [A, B]def countPronic(A, B) : # Subtract the count of pronic numbers # which are <= (A - 1) from the count # f pronic numbers which are <= B return pronic(B) - pronic(A - 1);# Driver Codeif __name__ == "__main__" : A = 3; B = 20; # Function call to count pronic # numbers in the range [A, B] print(countPronic(A, B)); # This code is contributed by AnkThon |
C#
// C# program for the above approachusing System;public class GFG{// Function to count pronic// numbers in the range [A, B]static int pronic(int num){ // Check upto sqrt N int N = (int)Math.Sqrt(num); // If product of consecutive // numbers are less than equal to num if (N * (N + 1) <= num) { return N; } // Return N - 1 return N - 1;}// Function to count pronic// numbers in the range [A, B]static int countPronic(int A, int B){ // Subtract the count of pronic numbers // which are <= (A - 1) from the count // f pronic numbers which are <= B return pronic(B) - pronic(A - 1);}// Driver Codepublic static void Main(String[] args){ int A = 3; int B = 20; // Function call to count pronic // numbers in the range [A, B] Console.Write(countPronic(A, B));}}// This code is contributed by 29AjayKumar |
Javascript
<script>// Javascript program for the above approach// Function to count pronic// numbers in the range [A, B]function pronic(num){ // Check upto sqrt N var N = parseInt(Math.sqrt(num)); // If product of consecutive // numbers are less than equal to num if (N * (N + 1) <= num) { return N; } // Return N - 1 return N - 1;}// Function to count pronic// numbers in the range [A, B]function countPronic(A, B){ // Subtract the count of pronic numbers // which are <= (A - 1) from the count // f pronic numbers which are <= B return pronic(B) - pronic(A - 1);}// Driver Codevar A = 3;var B = 20;// Function call to count pronic// numbers in the range [A, B]document.write(countPronic(A, B));// This code is contributed by noob2000.</script> |
Output:
3
Time Complexity: O(log2(B))
Auxiliary Space: O(1)
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