Updated problem 1 to faster solution

headers/Problems/Problem1.hpp

@@ -38,6 +38,7 @@ private:
 	static uint64_t MAX_NUMBER;	//The highest number to be tested
 	//Instance variables
 	uint64_t fullSum;		//For the sum of all the numbers
+	uint64_t sumOfProgression(uint64_t multiple);	//Gets the sum of the progression of the multiple
 public:
 	//Constructor
 	Problem1();
@@ -51,7 +52,7 @@ public:
 
 /* Results:
 The sum of all the numbers < 1000 that are divisible by 3 or 5 is 233168
-It took an average of 957.000 nanoseconds to run this problem over 100 iterations
+It took an average of 50.000 nanoseconds to run this problem over 100 iterations
 */
 
 #endif //PROBLEM1_HPP

src/Problems/Problem1.cpp

@@ -26,6 +26,7 @@
 #include <sstream>
 #include <cinttypes>
 #include <vector>
+#include <cmath>
 #include "Problems/Problem1.hpp"
 
 
@@ -36,6 +37,12 @@ uint64_t Problem1::MAX_NUMBER = 999;
 Problem1::Problem1() : Problem("What is the sum of all the multiples of 3 or 5 that are less than 1000?"), fullSum(0){
 }
 
+uint64_t Problem1::sumOfProgression(uint64_t multiple){
+	uint64_t numTerms = std::floor(MAX_NUMBER / multiple);	//This gets the number of multiples of a particular number that is < MAX_NUMBER
+	//The sum of progression formula is (n / 2)(a + l). n = number of terms, a = multiple, l = last term
+	return ((numTerms / 2.0) * (multiple + (numTerms * multiple)));
+}
+
 //Operational functions
 //Solve the problem
 void Problem1::solve(){
@@ -46,16 +53,8 @@ void Problem1::solve(){
 	//Start the timer
 	timer.start();
 
-	//Step through every number < 1000 and see if either 3 or 5 divide it evenly
-	for(uint64_t cnt = 1;cnt <= MAX_NUMBER;++cnt){
-		//If either 3 or 5 divides it evenly then add it to the vector
-		if((cnt % 3) == 0){
-			fullSum += cnt;
-		}
-		else if((cnt % 5) == 0){
-			fullSum += cnt;
-		}
-	}
+	//Get the sum of the progressions of 3 and 5 and remove the sum of progressions of the overlap
+	fullSum = sumOfProgression(3) + sumOfProgression(5) - sumOfProgression(3 * 5);
 
 	//Stop the timer
 	timer.stop();