51 lines
1.5 KiB
C
51 lines
1.5 KiB
C
/* Let d(n) be defined as the sum of proper divisors of n (numbers less than n which divide evenly into n).
|
|
* If d(a) = b and d(b) = a, where a ≠ b, then a and b are an amicable pair and each of a and b are called amicable numbers.
|
|
*
|
|
* For example, the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55 and 110; therefore d(220) = 284.
|
|
* The proper divisors of 284 are 1, 2, 4, 71 and 142; so d(284) = 220.
|
|
*
|
|
* Evaluate the sum of all the amicable numbers under 10000.*/
|
|
|
|
#define _POSIX_C_SOURCE 199309L
|
|
|
|
#include <stdio.h>
|
|
#include <stdlib.h>
|
|
#include <math.h>
|
|
#include <time.h>
|
|
#include "projecteuler.h"
|
|
|
|
int main(int argc, char **argv)
|
|
{
|
|
int i, n, sum = 0;
|
|
double elapsed;
|
|
struct timespec start, end;
|
|
|
|
clock_gettime(CLOCK_MONOTONIC, &start);
|
|
|
|
for(i = 2; i < 10000; i++)
|
|
{
|
|
/* Calculate the sum of proper divisors with the function
|
|
* implemented in projecteuler.c.*/
|
|
n = sum_of_divisors(i, 1);
|
|
/* If i!=n and the sum of proper divisors of n=i,
|
|
* sum the pair of numbers and add it to the total.*/
|
|
if(i != n && sum_of_divisors(n, 1) == i)
|
|
{
|
|
sum += i + n;
|
|
}
|
|
}
|
|
|
|
sum /= 2;
|
|
|
|
clock_gettime(CLOCK_MONOTONIC, &end);
|
|
|
|
elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
|
|
|
|
printf("Project Euler, Problem 21\n");
|
|
printf("Answer: %d\n", sum);
|
|
|
|
printf("Elapsed time: %.9lf seconds\n", elapsed);
|
|
|
|
return 0;
|
|
}
|