40 lines
996 B
C
40 lines
996 B
C
/* If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.
|
|
*
|
|
* Find the sum of all the multiples of 3 or 5 below 1000.*/
|
|
|
|
#define _POSIX_C_SOURCE 199309L
|
|
|
|
#include <stdio.h>
|
|
#include <stdlib.h>
|
|
#include <time.h>
|
|
|
|
int main(int argc, char **argv)
|
|
{
|
|
int i, sum = 0;
|
|
double elapsed;
|
|
struct timespec start, end;
|
|
|
|
clock_gettime(CLOCK_MONOTONIC, &start);
|
|
|
|
/* Simple brute-force approach: try every number between 3 and 999,
|
|
* check if it's a multiple of 3 or 5, if yes add it to the total.*/
|
|
for(i = 3; i < 1000; i++)
|
|
{
|
|
if (i % 3 == 0 || i % 5 == 0)
|
|
{
|
|
sum += i;
|
|
}
|
|
}
|
|
|
|
clock_gettime(CLOCK_MONOTONIC, &end);
|
|
|
|
elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
|
|
|
|
printf("Project Euler, Problem 1\n");
|
|
printf("Answer: %d\n", sum);
|
|
|
|
printf("Elapsed time: %.9lf seconds\n", elapsed);
|
|
|
|
return 0;
|
|
}
|