50 lines
1.2 KiB
C
50 lines
1.2 KiB
C
/* Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be:
|
|
*
|
|
* 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...
|
|
*
|
|
* By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms.*/
|
|
|
|
#define _POSIX_C_SOURCE 199309L
|
|
|
|
#include <stdio.h>
|
|
#include <stdlib.h>
|
|
#include <time.h>
|
|
|
|
#define N 4000000
|
|
|
|
int main(int argc, char **argv)
|
|
{
|
|
int fib0 = 1, fib1 = 2, fib2, sum = 2;
|
|
double elapsed;
|
|
struct timespec start, end;
|
|
|
|
clock_gettime(CLOCK_MONOTONIC, &start);
|
|
|
|
fib2 = fib0 + fib1;
|
|
|
|
/* Simple brute-force approach: generate every value in the Fibonacci
|
|
* sequence smaller than 4 million and if it's even add it to the total.*/
|
|
while(fib2 <= N)
|
|
{
|
|
if(fib2 % 2 == 0)
|
|
{
|
|
sum += fib2;
|
|
}
|
|
|
|
fib0 = fib1;
|
|
fib1 = fib2;
|
|
fib2 = fib0 + fib1;
|
|
}
|
|
|
|
clock_gettime(CLOCK_MONOTONIC, &end);
|
|
|
|
elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
|
|
|
|
printf("Project Euler, Problem 2\n");
|
|
printf("Answer: %d\n", sum);
|
|
|
|
printf("Elapsed time: %.9lf seconds\n", elapsed);
|
|
|
|
return 0;
|
|
}
|