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 <iostream>
Include dependency graph for number_of_positive_divisors.cpp:

Go to the source code of this file.

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 81 of file number_of_positive_divisors.cpp.

81 {
82 tests();
83 int n;
84 std::cin >> n;
85 if (n == 0) {
86 std::cout << "All non-zero numbers are divisors of 0 !" << std::endl;
87 } else {
88 std::cout << "Number of positive divisors is : ";
89 std::cout << number_of_positive_divisors(n) << std::endl;
90 }
91 return 0;
92}
int number_of_positive_divisors(int n)

◆ 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 33 of file number_of_positive_divisors.cpp.

33 {
34 if (n < 0) {
35 n = -n; // take the absolute value of n
36 }
37
38 int number_of_divisors = 1;
39
40 for (int i = 2; i * i <= n; i++) {
41 // This part is doing the prime factorization.
42 // Note that we cannot find a composite divisor of n unless we would
43 // already previously find the corresponding prime divisor and dvided
44 // n by that prime. Therefore, all the divisors found here will
45 // actually be primes.
46 // The loop terminates early when it is left with a number n which
47 // does not have a divisor smaller or equal to sqrt(n) - that means
48 // the remaining number is a prime itself.
49 int prime_exponent = 0;
50 while (n % i == 0) {
51 // Repeatedly divide n by the prime divisor n to compute
52 // the exponent (e_i in the algorithm description).
53 prime_exponent++;
54 n /= i;
55 }
56 number_of_divisors *= prime_exponent + 1;
57 }
58 if (n > 1) {
59 // In case the remaining number n is a prime number itself
60 // (essentially p_k^1) the final answer is also multiplied by (e_k+1).
61 number_of_divisors *= 2;
62 }
63
64 return number_of_divisors;
65}

◆ tests()

void tests ( )

Test implementations

Definition at line 70 of file number_of_positive_divisors.cpp.

70 {
71 assert(number_of_positive_divisors(36) == 9);
72 assert(number_of_positive_divisors(-36) == 9);
73 assert(number_of_positive_divisors(1) == 1);
74 assert(number_of_positive_divisors(2011) == 2); // 2011 is a prime
75 assert(number_of_positive_divisors(756) == 24); // 756 = 2^2 * 3^3 * 7
76}