Added number-of-divisors.cpp in math directory

math/number-of-divisors.cpp

@@ -0,0 +1,47 @@
+/// C++ Program to calculate number of divisors.
+
+#include<iostream>
+#include<vector>
+
+/**
+ * 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.
+ *
+ * Example:- N=36
+ * 36 = (3^2 * 2^2)
+ * numbe-of-divisors(36) = (2+1) * (2+1) = 9.
+ * list of divisors of 36 = 1, 2, 3, 4, 6, 9, 12, 18, 36.
+**/
+
+int number_of_divisors(int n) {
+    std::vector<int> prime_exponent_count;
+    for (int i=2; i*i <= n; i++) {
+        int prime_count = 0;
+        while (n % i == 0) {
+            prime_count += 1;
+            n /= i;
+        }
+        if (prime_count != 0) {
+            prime_exponent_count.push_back(prime_count);
+        }
+    }
+    int divisors_count = 1;
+    // If n is prime at that time vector prime_exponent_count will remain empty.
+    for (int i=0; i<prime_exponent_count.size(); i++) {
+        divisors_count = divisors_count * (prime_exponent_count[i]+1);
+    }
+    if(divisors_count == 1 and n != 1) {
+        // Prime number has exactly 2 divisors 1 and itself.
+        divisors_count = 2;
+    }
+    return divisors_count;
+}
+
+int main() {
+    int n;
+    std::cin >> n;
+    std::cout << number_of_divisors(n) << std::endl;
+}