C++ Program to calculate the number of positive divisors. More...

#include <cassert>
#include <iostream>

Include dependency graph for number_of_positive_divisors.cpp:

Functions
int	number_of_positive_divisors (int n)

void	tests ()

int	main ()

Detailed Description

C++ Program to calculate the number of positive divisors.

This algorithm uses the prime factorization approach. Any positive integer can be written as a product of its prime factors.
Let \(N = p_1^{e_1} \times p_2^{e_2} \times\cdots\times p_k^{e_k}\) where \(p_1,\, p_2,\, \dots,\, p_k\) are distinct prime factors of \(N\) and \(e_1,\, e_2,\, \dots,\, e_k\) are respective positive integer exponents.
Each positive divisor of \(N\) is in the form \(p_1^{g_1}\times p_2^{g_2}\times\cdots\times p_k^{g_k}\) where \(0\le g_i\le e_i\) are integers for all \(1\le i\le k\).
Finally, there are \((e_1+1) \times (e_2+1)\times\cdots\times (e_k+1)\) positive divisors of \(N\) since we can choose every \(g_i\) independently.

Example:
\(N = 36 = (3^2 \cdot 2^2)\)
\(\mbox{number_of_positive_divisors}(36) = (2+1) \cdot (2+1) = 9\).
list of positive divisors of 36 = 1, 2, 3, 4, 6, 9, 12, 18, 36.

Similarly, for N = -36 the number of positive divisors remain same.

Function Documentation

◆ main()

int main ( void )

Main function

           {
    tests();
    int n;
    std::cin >> n;
    if (n == 0) {
        std::cout << "All non-zero numbers are divisors of 0 !" << std::endl;
    } else {
        std::cout << "Number of positive divisors is : ";
        std::cout << number_of_positive_divisors(n) << std::endl;
    }
    return 0;
}

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◆ number_of_positive_divisors()

int number_of_positive_divisors ( int n )

Function to compute the number of positive divisors.

Parameters

n	number to compute divisors for

Returns: number of positive divisors of n (or 1 if n = 0)

                                       {
    if (n < 0) {
        n = -n; // take the absolute value of n
    }
 
    int number_of_divisors = 1;
 
    for (int i = 2; i * i <= n; i++) {
        // This part is doing the prime factorization.
        // Note that we cannot find a composite divisor of n unless we would
        // already previously find the corresponding prime divisor and dvided
        // n by that prime. Therefore, all the divisors found here will
        // actually be primes.
        // The loop terminates early when it is left with a number n which
        // does not have a divisor smaller or equal to sqrt(n) - that means
        // the remaining number is a prime itself.
        int prime_exponent = 0;
        while (n % i == 0) {
            // Repeatedly divide n by the prime divisor n to compute
            // the exponent (e_i in the algorithm description).
            prime_exponent++;
            n /= i;
        }
        number_of_divisors *= prime_exponent + 1;
    }
    if (n > 1) {
        // In case the remaining number n is a prime number itself
        // (essentially p_k^1) the final answer is also multiplied by (e_k+1).
        number_of_divisors *= 2;
    }
 
    return number_of_divisors;
}

◆ tests()

void tests ( )

Test implementations