46 lines
1.1 KiB
C
46 lines
1.1 KiB
C
/* The decimal number, 585 = 1001001001_2 (binary), is palindromic in both bases.
|
|
*
|
|
* Find the sum of all numbers, less than one million, which are palindromic in base 10 and base 2.
|
|
*
|
|
* (Please note that the palindromic number, in either base, may not include leading zeros.)*/
|
|
|
|
#define _POSIX_C_SOURCE 199309L
|
|
|
|
#include <stdio.h>
|
|
#include <stdlib.h>
|
|
#include <time.h>
|
|
#include "projecteuler.h"
|
|
|
|
#define N 1000000
|
|
|
|
int main(int argc, char **argv)
|
|
{
|
|
int i, sum = 0;
|
|
double elapsed;
|
|
struct timespec start, end;
|
|
|
|
clock_gettime(CLOCK_MONOTONIC, &start);
|
|
|
|
/* Brute force approach. For every number below 1 million,
|
|
* check if they're palindrome in base 2 and 10 using the
|
|
* function implemented in projecteuler.c.*/
|
|
for(i = 1; i < N; i += 2)
|
|
{
|
|
if(is_palindrome(i, 10) && is_palindrome(i, 2))
|
|
{
|
|
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 36\n");
|
|
printf("Answer: %d\n", sum);
|
|
|
|
printf("Elapsed time: %.9lf seconds\n", elapsed);
|
|
|
|
return 0;
|
|
}
|