sol1.c File Reference

Problem 12 solution More...

#include <math.h>
#include <stdio.h>
#include <stdlib.h>

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Functions
long	count_divisors (long long n)
int	main (int argc, char **argv)

Detailed Description

Problem 12 solution

Author

Krishna Vedala

Function Documentation

count_divisors()

long count_divisors

(

long long

n

)

Get number of divisors of a given number

If \(x = a \times b\), then both \(a\) and \(b\) are divisors of \(x\). Since multiplication is commutative, we only need to search till a maximum of \(a=b = a^2\) i.e., till \(\sqrt{x}\). At every integer till then, there are eaxctly 2 divisors and at \(a=b\), there is only one divisor.

{
    long num_divisors = 0;
 
    for (long long i = 1; i < sqrtl(n) + 1; i++)
        if (n % i == 0)
            num_divisors += 2;
        else if (i * i == n)
            num_divisors += 1;
 
    return num_divisors;
}

main()

int main	(	int	argc,
		char **	argv
	)

Main function

{
    int MAX_DIVISORS = 500;
    long i = 1, num_divisors;
    long long triangle_number = 1;
 
    if (argc == 2)
        MAX_DIVISORS = atoi(argv[1]);
 
    while (1)
    {
        i++;
        triangle_number += i;
        num_divisors = count_divisors(triangle_number);
        if (num_divisors > MAX_DIVISORS)
            break;
    }
 
    printf("First Triangle number with more than %d divisors: %lld\n",
           MAX_DIVISORS, triangle_number);
 
    return 0;
}

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