void
sequential()
{
int i;
for (i = 0; i < JOBS; i++)
calc_primes(1000);
}
int
main(int argc, char *argv[])
{
uint64_t max = argc > 1 ? atol(argv[1]) : 1000000000ull;
char *bitmap = calloc(max/8 + 1, 1);
calc_primes(bitmap, max);
printf("%ld\n", count_primes(bitmap, max));
free(bitmap);
return EXIT_SUCCESS;
}
int
main(int argc, char *argv[])
{
uint64_t max = argc > 1 ? atol(argv[1]) : 1000000000ull;
char *bitmap = calloc(max/30 + sqrtl(max) + 16*8, 1);
uint64_t adj_max;
int a_i;
/* adjust max to a multiple of 2,3,5 */
for (adj_max = (max - 1) / 30 * 30 + 1, a_i = 0;
adj_max + diff[a_i%8] <= max;
adj_max += diff[a_i++%8])
;
calc_primes(bitmap, adj_max);
printf("%ld\n", count_primes(bitmap, adj_max / 30 * 8 + num2bit(adj_max)) + 3);
free(bitmap);
return EXIT_SUCCESS;
}
/*!
* Calculates all the prime factors of the given value. Does this in a
* simplistic manner, using trial division on all values below the sqrt(n).
*
* \param n Value to factorize
*
* \return List of factors
*/
static struct factor *calc_prime_factors(long n, long max_factor, size_t *num_factors)
{
long i;
struct factor *ret = NULL;
struct factor *tail = NULL;
long *primes;
size_t num_primes;
primes = calc_primes(max_factor, &num_primes);
*num_factors = 0;
if(max_factor == 0) {
return ret;
}
for(i = 0; i < num_primes; i++) {
if(n % primes[i] == 0) {
*num_factors = *num_factors + 1;
if(!tail) {
ret = malloc(sizeof(struct factor));
tail = ret;
} else {
tail->next = malloc(sizeof(struct factor));
tail = tail->next;
}
tail->value = primes[i];
tail->next = NULL;
}
}
if(!ret && n <= max_factor) {
*num_factors = 1;
ret = malloc(sizeof(struct factor));
ret->value = n;
ret->next = NULL;
}
return ret;
}
void *
consumer(void *data)
{
struct job *job;
int *cnt;
int *item;
int total_primes = 0;
assert(cnt = malloc(sizeof(*cnt)));
*cnt = 0;
job = (struct job *) data;
while (NULL != (item = (int*) ds_queue_pop(job->job_queue))) {
(*cnt)++;
total_primes += calc_primes(1000);
free(item);
}
printf("Thread %s consumed %d items (%d primes).\n",
job->id, *cnt, total_primes);
ds_queue_push(job->results_queue, (void*) cnt);
return NULL;
}
int
problem35(int argc, char** argv)
{
int i;
int cnt1 = 1; // Counts for 2
int tmp;
int order;
const int end = 1e6; // 1e7;
calc_primes(end);
for (i = 3; i < end; i += 2){
if (!isPrime(i))
continue;
tmp = i;
order = 0;
while (tmp > 0) {
if (tmp % 10 % 2 == 0)
break;
tmp /= 10;
++order;
}
if (tmp > 0)
continue;
if (test_rotations(i, order)) {
// printf("Rotations are all primes: %d\n", i);
cnt1++;
}
}
printf("Circular Primes under %d: %d\n", end, cnt1);
return(0);
}
int problem27(int argc, char** argv) {
int a;
int b;
int n;
int cnt;
int max_num = MAX_NUM;
int max_prime_cnt_n = 0;
int max_prime_cnt_a = 0;
int max_prime_cnt_b = 0;
if (argc == 2) {
max_num = atoi(argv[1]);
if (max_num < 2) {
printf("Invalid Max: \"%s\" (%d)\n", argv[1], max_num);
return(1);
}
printf("Setting max number: %d\n", max_num);
}
cnt = calc_primes(max_num);
printf("%d primes under %d\n", cnt, max_num);
/* Prime numbers are now determined (any 0 bits in the bitmap) */
for (b = 2; b < max_num; ++b) {
/* Non primes are set to 1 in the BM */
/* When n = 0 => b must be a prime alreayd */
if (BM_IS_SET(b, bm_primes)) {
continue;
}
/* If a < -b, then when n = 1, the result would be negative => no prime */
/* a +=2 because when n = 1, (1 + a + b) needs to be an odd number */
for (a = -b; a < max_num; a +=2) {
for (n = 1; n < 100; ++n) {
if (!isPrime(((n*n) + a*n + b))) {
break;
}
}
if (n >= 100) {
printf("ERROR: n = %d, (a = %d, b = %d): I don't think it should "
"ever get this high\n", n, a, b);
}
if (n > max_prime_cnt_n) {
max_prime_cnt_n = n;
max_prime_cnt_a = a;
max_prime_cnt_b = b;
}
}
}
printf("Largest n = %d when a = %d, b = %d\n",
max_prime_cnt_n, max_prime_cnt_a, max_prime_cnt_b);
printf("Product of coefficients: %d\n", max_prime_cnt_a * max_prime_cnt_b);
if (bm_primes) {
free(bm_primes);
bm_primes = NULL;
}
return(0);
}