OctaveFunctions/ProjectEuler/C++/Problem12.cpp

//ProjectEuler/C++/Problem12.cpp
//Matthew Ellison
// Created: 9-27-18
//Modified: 9-28-18
//This file contains the program to calculate the answer to Problem 12 on ProjectEuler.net


#include <iostream>
#include <cmath>	//For fmod
#include <chrono>	//For the timer


//Counter how many divisors number has
unsigned long countDivisors(unsigned long number);


int main(){
	bool found = false;			//To flag whether the number has been found
	unsigned long sum = 1;		//The sum of the numbers up to counter
	unsigned long counter = 2;	//The next number to be added to sum
	const unsigned long goalDivisors = 500;	//The number of divisors that is being sought

	std::chrono::high_resolution_clock::time_point startTime = std::chrono::high_resolution_clock::now();
	while(!found){
		//If the number of divisors is correct set the flag
		if(countDivisors(sum) > goalDivisors){
			found = true;
		}
		//Otherwise add to the sum and increase the next numeber
		else{
			sum += counter;
			++counter;
		}
	}
	std::chrono::high_resolution_clock::time_point endTime = std::chrono::high_resolution_clock::now();

	//Print the results
	std::cout << "The triangular number " << sum << " is made with all number >= " << counter - 1 << " and has " << countDivisors << " divisors" << std::endl;
	std::cout << "The problem took " << std::chrono::duration_cast<std::chrono::milliseconds>(std::chrono::high_resolution_clock::duration(endTime - startTime)).count() << " milliseconds" << std::endl;

	std::cin.get();

	return 0;
}


unsigned long countDivisors(unsigned long number){
	unsigned long numDivisors = 0;	//Holds the number of divisors
	//You only need to go to sqrt(number). cnt * cnt is faster than sqrt()
	for(int cnt = 1;cnt * cnt < number;++cnt){
		//Check if the counter evenly divides the number
		//If it does the counter and the other number are both divisors
		if(fmod((double)number, (double)cnt) == 0){
			numDivisors += 2;
		}
	}
	return numDivisors;
}