OctaveFunctions/ProjectEuler/C++/Problem1.cpp

//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;
}