int main() {
int i;
int prime;
primegen pg;
primegen_init(&pg);
for(i = 0; i < 10001; i++)
prime = primegen_next(&pg);
printf("%d\n", prime);
return 0;
}
/****************************************************************************
* Break down the number into prime factors using DJB's sieve code, which
* is about 5 to 10 times faster than the Seive of Eratosthenes.
*
* @param number
* The integer that we are factoring. It can be any value up to 64 bits
* in size.
* @param factors
* The list of all the prime factors, zero terminated.
* @param non_factors
* A list of smallest numbers that aren't prime factors. We return
* this because we are going to use prime non-factors for finding
* interesting numbers.
****************************************************************************/
unsigned
sieve_prime_factors(uint64_t number, PRIMEFACTORS factors, PRIMEFACTORS non_factors, double *elapsed)
{
primegen pg;
clock_t start;
clock_t stop;
uint64_t prime;
uint64_t max;
unsigned factor_count = 0;
unsigned non_factor_count = 0;
/*
* We only need to seive up to the square-root of the target number. Only
* one prime factor can be bigger than the square root, so once we find
* all the other primes, the square root is the only one left.
* Note: you have to link to the 'm' math library for some gcc platforms.
*/
max = (uint64_t)sqrt(number + 1.0);
/*
* Init the DJB primegen library.
*/
primegen_init(&pg);
/*
* Enumerate all the primes starting with 2
*/
start = clock();
for (;;) {
/* Seive the next prime */
prime = primegen_next(&pg);
/* If we've reached the square root, then that's as far as we need
* to go */
if (prime > max)
break;
/* If this prime is not a factor (evenly divisible with no remainder)
* then loop back and get the next prime */
if ((number % prime) != 0) {
if (non_factor_count < 12)
non_factors[non_factor_count++] = prime;
continue;
}
/* Else we've found a prime factor, so add this to the list of primes */
factors[factor_count++] = prime;
/* At the end, we may have one prime factor left that's bigger than the
* sqrt. Therefore, as we go along, divide the original number
* (possibly several times) by the prime factor so that this large
* remaining factor will be the only one left */
while ((number % prime) == 0)
number /= prime;
/* exit early if we've found all prime factors. comment out this
* code if you want to benchmark it */
if (number == 1 && non_factor_count > 10)
break;
}
/*
* See if there is one last prime that's bigger than the square root.
* Note: This is the only number that can be larger than 32-bits in the
* way this code is written.
*/
if (number != 1)
factors[factor_count++] = number;
/*
* Zero terminate the results.
*/
factors[factor_count] = 0;
non_factors[non_factor_count] = 0;
/*
* Since prime factorization takes a long time, especially on slow
* CPUs, we benchmark it to keep track of performance.
*/
stop = clock();
if (elapsed)
*elapsed = ((double)stop - (double)start)/(double)CLOCKS_PER_SEC;
/* should always be at least 1, because if the number itself is prime,
* then that's it's only prime factor */
return factor_count;
}