44 lines
1.1 KiB
C
44 lines
1.1 KiB
C
/* By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6th prime is 13.
|
|
*
|
|
* What is the 10 001st prime number?*/
|
|
|
|
#define _POSIX_C_SOURCE 199309L
|
|
|
|
#include <stdio.h>
|
|
#include <stdlib.h>
|
|
#include <time.h>
|
|
#include "projecteuler.h"
|
|
|
|
int main(int argc, char **argv)
|
|
{
|
|
int count = 1, n = 1, target = 10001;
|
|
double elapsed;
|
|
struct timespec start, end;
|
|
|
|
clock_gettime(CLOCK_MONOTONIC, &start);
|
|
|
|
/* Brute force approach: start with count=1 and check every odd number
|
|
* (2 is the only even prime), if it's prime increment count, until the
|
|
* target prime is reached.*/
|
|
while(count != target)
|
|
{
|
|
n += 2;
|
|
/* Use the function in projecteuler.c to check if a number is prime.*/
|
|
if(is_prime(n))
|
|
{
|
|
count++;
|
|
}
|
|
}
|
|
|
|
clock_gettime(CLOCK_MONOTONIC, &end);
|
|
|
|
elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
|
|
|
|
printf("Project Euler, Problem 7\n");
|
|
printf("Answer: %d\n", n);
|
|
|
|
printf("Elapsed time: %.9lf seconds\n", elapsed);
|
|
|
|
return 0;
|
|
}
|