Added a cpp for problem 1

ProjectEuler/C++/Problem1.cpp

@@ -0,0 +1,45 @@
+//ProjectEuler/C++/Problem1.cpp
+//Matthew Ellison
+// Created: 9-28-18
+//Modified: 9-28-18
+//What is the sum of all the multiples of 3 or 5 that are less than 1000
+
+
+#include <iostream>
+#include <vector>
+#include <chrono>
+
+
+int main(){
+	unsigned long fullSum = 0;		//For the sum of all the numbers
+	std::vector<unsigned long> numbers;	//Holds all the numbers
+
+	std::chrono::high_resolution_clock::time_point startTime = std::chrono::high_resolution_clock::now();
+	//Step through every number < 1000 and see if either 3 or 5 divide it evenly
+	for(int cnt = 0;cnt < 1000;++cnt){
+		//If either divides it then add it to the vector
+		if((cnt % 3) == 0){
+			numbers.push_back(cnt);
+		}
+		else if((cnt % 5) == 0){
+			numbers.push_back(cnt);
+		}
+	}
+
+	//Get the sum of all numbers
+	for(unsigned long num : numbers){
+		fullSum += num;
+	}
+
+	//Calculate the time needed to run the algorithm
+	std::chrono::high_resolution_clock::time_point endTime = std::chrono::high_resolution_clock::now();
+	std::chrono::high_resolution_clock::duration dur = std::chrono::duration_cast<std::chrono::nanoseconds>(endTime - startTime);
+
+	//Print the output
+	std::cout << "The sum of all the numbers < 1000 that are divisible by 3 or 5 is " << fullSum
+			<< "\nIt took " << dur.count() << " nanoseconds to run this algorithm" << std::endl;
+
+	std::cin.get();
+
+	return 0;
+}