documetnation for positive divisors

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Krishna Vedala 2020-05-27 16:22:10 -04:00
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@ -1,34 +1,39 @@
/// C++ Program to calculate number of divisors.
#include<iostream>
#include<vector>
/**
* @file
* @brief C++ Program to calculate number of divisors
*
* This algorithm use the prime factorization approach.
* Any number can be written in multiplication of its prime factors.
* Let N = P1^E1 * P2^E2 ... Pk^Ek
* Therefore. number-of-divisors(N) = (E1+1) * (E2+1) ... (Ek+1).
* Where P1, P2 ... Pk are prime factors and E1, E2 ... Ek are exponents respectively.
* <br/>Let N = P1^E1 * P2^E2 ... Pk^Ek
* <br/>Therefore. number-of-divisors(N) = (E1+1) * (E2+1) ... (Ek+1).
* <br/>Where P1, P2 ... Pk are prime factors and E1, E2 ... Ek are exponents
respectively.
*
* Example:-
* N = 36
* 36 = (3^2 * 2^2)
* number_of_positive_divisors(36) = (2+1) * (2+1) = 9.
* list of positive divisors of 36 = 1, 2, 3, 4, 6, 9, 12, 18, 36.
* <br/>N = 36
* <br/>36 = (3^2 * 2^2)
* <br/>number_of_positive_divisors(36) = (2+1) * (2+1) = 9.
* <br/>list of positive divisors of 36 = 1, 2, 3, 4, 6, 9, 12, 18, 36.
*
* Similarly if N is -36 at that time number of positive divisors remain same.
*
* Example:-
* N = -36
* -36 = -1 * (3^2 * 2^2)
* number_of_positive_divisors(-36) = (2+1) * (2+1) = 9.
* list of positive divisors of -36 = 1, 2, 3, 4, 6, 9, 12, 18, 36.
* <br/>N = -36
* <br/>-36 = -1 * (3^2 * 2^2)
* <br/>number_of_positive_divisors(-36) = (2+1) * (2+1) = 9.
* <br/>list of positive divisors of -36 = 1, 2, 3, 4, 6, 9, 12, 18, 36.
*
**/
#include <iostream>
#include <vector>
/**
* Algorithm
*/
int number_of_positive_divisors(int n) {
std::vector<int> prime_exponent_count;
for (int i=2; i*i <= n; i++) {
for (int i = 2; i * i <= n; i++) {
int prime_count = 0;
while (n % i == 0) {
prime_count += 1;
@ -44,13 +49,16 @@ int number_of_positive_divisors(int n) {
int divisors_count = 1;
for (int i=0; i < prime_exponent_count.size(); i++) {
divisors_count = divisors_count * (prime_exponent_count[i]+1);
for (int i = 0; i < prime_exponent_count.size(); i++) {
divisors_count = divisors_count * (prime_exponent_count[i] + 1);
}
return divisors_count;
}
/**
* Main function
*/
int main() {
int n;
std::cin >> n;