Update and rename modular_exponentiation to modular_exponentiation.cpp

math/modular_exponentiation.cpp

@@ -0,0 +1,65 @@
+/**
+ * @file
+ * @brief C++ Program for Modular Exponentiation Iteratively.
+ * Calculate the value of an integer a raised to an integer exponent b 
+ * under modulo c. 
+ * @note The time complexity of this approach is O(log b).
+ *
+ * Example:
+ * (4^3) % 5 (where ^ stands for exponentiation and % for modulo)
+ * (4*4*4) % 5
+ * (4 % 5) * ( (4*4) % 5 )
+ * 4 * (16 % 5)
+ * 4 * 1
+ * 4
+ * We can also verify the result as 4^3 is 64 and 64 modulo 5 is 4
+ */
+
+#include <iostream> 
+
+
+/// Iterative Function to calculate a raised to exponent b 
+/// under modulo c in O(log b) using modular exponentiation.
+int power(int a, unsigned int b, int c) 
+{ 
+	int ans = 1;	    	/// Initialize the answer to be returned 
+
+	a = a % c;          	/// Update a if it is more than or 
+				/// equal to c 
+
+	if (a == 0) 
+	{
+	    return 0;       	/// In case a is divisible by c; 
+	}
+	
+	while (b > 0) 
+	{ 
+		/// If b is odd, multiply a with answer 
+		if (b & 1) 
+		{
+			ans = (ans*a) % c; 
+		}
+		
+		/// b must be even now 
+		b = b>>1;       // b = b/2 
+		a = (a*a) % c; 
+	}
+	
+	return ans; 
+} 
+
+/// Main Function
+int main() 
+{ 	
+	/// Give two numbers num1, num2 and modulo m
+	int num1 = 2; 
+	int num2 = 5; 
+	int m = 13; 
+	
+	std::cout << "The value of "<<num1<<" raised to exponent "<<num2<<" 
+	under modulo "<<m<<" is " << power(num1, num2, m); 
+	/// std::cout << "The value of "<<num1<<" raised to exponent "<<num2<<" 
+	/// under modulo "<<m<<" is " << power(num1, num2, m); 
+ 
+} 
+