TheAlgorithms/C++ 1.0.0
All the algorithms implemented in C++
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number_of_positive_divisors.cpp File Reference

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

#include <cassert>
Include dependency graph for number_of_positive_divisors.cpp:

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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.

Definition in file number_of_positive_divisors.cpp.

Function Documentation

◆ main()

int main ( void )

Main function

Definition at line 80 of file number_of_positive_divisors.cpp.

80 {
81 tests();
82 return 0;
83}

◆ number_of_positive_divisors()

int number_of_positive_divisors ( int n)

Function to compute the number of positive divisors.

Parameters
nnumber to compute divisors for
Returns
number of positive divisors of n (or 1 if n = 0)

Definition at line 32 of file number_of_positive_divisors.cpp.

32 {
33 if (n < 0) {
34 n = -n; // take the absolute value of n
35 }
36
37 int number_of_divisors = 1;
38
39 for (int i = 2; i * i <= n; i++) {
40 // This part is doing the prime factorization.
41 // Note that we cannot find a composite divisor of n unless we would
42 // already previously find the corresponding prime divisor and dvided
43 // n by that prime. Therefore, all the divisors found here will
44 // actually be primes.
45 // The loop terminates early when it is left with a number n which
46 // does not have a divisor smaller or equal to sqrt(n) - that means
47 // the remaining number is a prime itself.
48 int prime_exponent = 0;
49 while (n % i == 0) {
50 // Repeatedly divide n by the prime divisor n to compute
51 // the exponent (e_i in the algorithm description).
52 prime_exponent++;
53 n /= i;
54 }
55 number_of_divisors *= prime_exponent + 1;
56 }
57 if (n > 1) {
58 // In case the remaining number n is a prime number itself
59 // (essentially p_k^1) the final answer is also multiplied by (e_k+1).
60 number_of_divisors *= 2;
61 }
62
63 return number_of_divisors;
64}

◆ tests()

void tests ( )

Test implementations

Definition at line 69 of file number_of_positive_divisors.cpp.

69 {
70 assert(number_of_positive_divisors(36) == 9);
71 assert(number_of_positive_divisors(-36) == 9);
72 assert(number_of_positive_divisors(1) == 1);
73 assert(number_of_positive_divisors(2011) == 2); // 2011 is a prime
74 assert(number_of_positive_divisors(756) == 24); // 756 = 2^2 * 3^3 * 7
75}
int number_of_positive_divisors(int n)