Given a positive integer N, the task is to find the number of triplets (A, B, C) where A, B, C are positive integers such that the product of two numbers added with the third number is N i.e., A * B + C = N.
Examples:
Input: N = 3
Output: 3
Explanation:
Following are the possible triplets satisfying the given criteria:
- (1, 1, 2): The value of 1*1 + 2 = 3.
- (1, 2, 1): The value of 1*2 + 1 = 3.
- (2, 1, 1): The value of 2*1 + 1 = 3.
Therefore, the total count of such triplets is 3.
Input: N = 5
Output: 8
Approach: The given problem can be solved by rearranging the equation A * B + C = N as A * B = N – C. Now, the only possible values A and B can have to satisfy the above equation is the divisors of N – C. For Example, if the value of N – C = 18, having 6 divisors that are 1, 2, 3, 6, 9, 18. So, values of A, B satisfying the above equation are: (1, 18), (2, 9), (3, 6), (6, 3), (9, 2), (18, 1). So, for the value of N – C = 18, possible values of A, B are 6, i.e., the number of divisors of N – C(= 18). Follow the steps below to solve the given problem:
- Initialize a variable, say count as 0 that stores the total count of possible triplets.
- Iterate a loop over the range [1, N – 1] using the variable i and for each value, i find the total count of divisors of (N – i) and add it to the variable count.
- After completing the above steps, print the value of count as the resultant number of triplets.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to find the divisors of// the number (N - i)int countDivisors(int n){ // Stores the resultant count of // divisors of (N - i) int divisors = 0; int i; // Iterate over range [1, sqrt(N)] for (i = 1; i * i < n; i++) { if (n % i == 0) { divisors++; } } if (i - (n / i) == 1) { i--; } for (; i >= 1; i--) { if (n % i == 0) { divisors++; } } // Return the total divisors return divisors;}// Function to find the number of triplets// such that A * B - C = Nint possibleTriplets(int N){ int count = 0; // Loop to fix the value of C for (int i = 1; i < N; i++) { // Adding the number of // divisors in count count += countDivisors(N - i); } // Return count of triplets return count;}// Driver Codeint main(){ int N = 10; cout << possibleTriplets(N); return 0;} |
Java
// Java program for the above approachimport java.io.*;class GFG{ // Function to find the divisors of // the number (N - i) static int countDivisors(int n) { // Stores the resultant count of // divisors of (N - i) int divisors = 0; int i; // Iterate over range [1, sqrt(N)] for (i = 1; i * i < n; i++) { if (n % i == 0) { divisors++; } } if (i - (n / i) == 1) { i--; } for (; i >= 1; i--) { if (n % i == 0) { divisors++; } } // Return the total divisors return divisors; } // Function to find the number of triplets // such that A * B - C = N static int possibleTriplets(int N) { int count = 0; // Loop to fix the value of C for (int i = 1; i < N; i++) { // Adding the number of // divisors in count count += countDivisors(N - i); } // Return count of triplets return count; } // Driver Code public static void main (String[] args) { int N = 10; System.out.println(possibleTriplets(N)); }}// This code is contributed by Dharanendra L V. |
Python3
# Python program for the above approachimport math# function to find the divisors of# the number (N - i)def countDivisors(n): # Stores the resultant count of # divisors of (N - i) divisors = 0 # Iterate over range [1, sqrt(N)] for i in range(1, math.ceil(math.sqrt(n))+1): if n % i == 0: divisors = divisors+1 if (i - (n / i) == 1): i = i-1 for i in range(math.ceil(math.sqrt(n))+1, 1, -1): if (n % i == 0): divisors = divisors+1 # Return the total divisors return divisors # def to find the number of triplets # such that A * B - C = Ndef possibleTriplets(N): count = 0 # Loop to fix the value of C for i in range(1, N): # Adding the number of # divisors in count count = count + countDivisors(N - i) # Return count of triplets return count # Driver Code# Driver Codeif __name__ == "__main__": N = 10 print(possibleTriplets(N))# This code is contributed by Potta Lokesh |
C#
// C# program for the above approachusing System;using System.Collections.Generic;class GFG{// Function to find the divisors of// the number (N - i)static int countDivisors(int n){ // Stores the resultant count of // divisors of (N - i) int divisors = 0; int i; // Iterate over range [1, sqrt(N)] for (i = 1; i * i < n; i++) { if (n % i == 0) { divisors++; } } if (i - (n / i) == 1) { i--; } for (; i >= 1; i--) { if (n % i == 0) { divisors++; } } // Return the total divisors return divisors;}// Function to find the number of triplets// such that A * B - C = Nstatic int possibleTriplets(int N){ int count = 0; // Loop to fix the value of C for (int i = 1; i < N; i++) { // Adding the number of // divisors in count count += countDivisors(N - i); } // Return count of triplets return count;}// Driver Codepublic static void Main(){ int N = 10; Console.Write(possibleTriplets(N));}}// This code is contributed by SURENDRA_GANGWAR. |
Javascript
<script>// javascript program for the above approach // Function to find the divisors of // the number (N - i) function countDivisors(n) { // Stores the resultant count of // divisors of (N - i) var divisors = 0; var i; // Iterate over range [1, sqrt(N)] for (i = 1; i * i < n; i++) { if (n % i == 0) { divisors++; } } if (i - (n / i) == 1) { i--; } for (; i >= 1; i--) { if (n % i == 0) { divisors++; } } // Return the total divisors return divisors; } // Function to find the number of triplets // such that A * B - C = N function possibleTriplets(N) { var count = 0; // Loop to fix the value of C for (var i = 1; i < N; i++) { // Adding the number of // divisors in count count += countDivisors(N - i); } // Return count of triplets return count; } // Driver Code var N = 10; document.write(possibleTriplets(N));// This code is contributed by shikhasingrajput </script> |
Output:
23
Time Complexity: O(N*sqrt(N))
Auxiliary Space: O(1)
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