Find The Triangular Number With More Than N Divisors

Question

My code is a general solution to the following problem from Project Euler #12:

What is the value of the first triangle number to have over five hundred divisors?

MY CODE

#include <iostream>
int generate_triangular(int n);
int enumerate_divisors(int n);
int main()
{
    std::cout<<"You want to find the first triangular number with less than or more than how many divisors? "<<std::endl;
    int user_input;
    std::cin>>user_input;
    bool result_found = false;
    int test_number = 1;
    int target;
    while(!result_found){
        int factor_count = enumerate_divisors(generate_triangular(test_number));
        if(factor_count < user_input){
            test_number++;
        }else{
            result_found = true;
            target = generate_triangular(test_number);
        }
    }
    std::cout<<"The "<<test_number<<" th triangular number"<<std::endl;
    std::cout<<target<<std::endl;
}
int generate_triangular(int n){
    return (n*(n+1))/2;
}
int enumerate_divisors(int n){
    int divisor_count = 0;
    for(int i = 1;i*i <= n;i++){
        if(n%i == 0){
            divisor_count++;
        }
    }
    return 2*divisor_count;
}

I found my solution to be quite fast for the purposes of the question (it found the first triangular number with over 500 divisors in about .2 seconds). Wondering how I can make it faster, and about any general code suggestions to be made.

Answer 1

Your naming of variables is hard to reason about. When I think about what your program is doing you, I come up with the following sequence of steps:

1. Take in a target divisor count.
2. Index through each triangular number n and check if it satisfies the condition.
3. If it doesn't increment n and repeat.
4. Once it does print out the results.

Here is how I would restructure your main loop:

std::cout << "You want to find the first triangular number with less than or more than how many divisors? " << std::endl;
int target;
std::cin >> target;
// Indexes the triangle numbers.
int n = 0;
// Keeps track of how many divisors.
int divisors = 0;
do {
    n++;
    divisors = enumerate_divisors(generate_triangular(n));
} while (divisors < target);
std::cout << "The " << n <<" th triangular number" << std::endl;
std::cout << generate_triangular(n) << std::endl;

Also, add some spaces in between your function declarations, it's hard to read when everything is clumped together. But don't over do it (in particular, don't double space everything).

Answer 2

Your enumerate_divisors(n) gives a wrong result if n is a perfect square.

Note that the question asks for a triangular number that has over 500 divisors, so you would have to enter 501 to perform the desired calculation.

I agree with @Dair that your main loop is convoluted. In particular, result_found is a flag variable, and you should structure your loops to eliminate flag variables. I would write it this way:

int divisor_threshold = …;
int i;
long t;
for (i = 1; divisor_count(t = triangular(i)) <= divisor_threshold; i++);
std::cout << "The " << i << "th triangular number " << t
          << " has over " << divisor_threshold << " divisors.\n";