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++ 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 = N
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
int main()
{
int N = 10;
cout << possibleTriplets(N);
return 0;
}
// Java program for the above approach
import 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.
# Python program for the above approach
import 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 = N
def 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 Code
if __name__ == "__main__":
N = 10
print(possibleTriplets(N))
# This code is contributed by Potta Lokesh
// C# program for the above approach
using 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 = 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()
{
int N = 10;
Console.Write(possibleTriplets(N));
}
}
// This code is contributed by SURENDRA_GANGWAR.
<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)
