【发布时间】：2014-11-03 04:23:57
【问题描述】：
在网上偶然发现了这种高效的分段素筛，请帮我理解工作原理，尤其是next向量的使用
段大小的具体选择如何影响性能？
const int L1D_CACHE_SIZE = 32768;
void segmented_sieve(int64_t limit, int segment_size = L1D_CACHE_SIZE)
{
int sqrt = (int) std::sqrt((double) limit);
int64_t count = (limit < 2) ? 0 : 1;
int64_t s = 2;
int64_t n = 3;
// vector used for sieving
std::vector<char> sieve(segment_size);
// generate small primes <= sqrt
std::vector<char> is_prime(sqrt + 1, 1);
for (int i = 2; i * i <= sqrt; i++)
if (is_prime[i])
for (int j = i * i; j <= sqrt; j += i)
is_prime[j] = 0;
std::vector<int> primes;
std::vector<int> next;
for (int64_t low = 0; low <= limit; low += segment_size)
{
std::fill(sieve.begin(), sieve.end(), 1);
// current segment = interval [low, high]
int64_t high = std::min(low + segment_size - 1, limit);
// store small primes needed to cross off multiples
for (; s * s <= high; s++)
{
if (is_prime[s])
{
primes.push_back((int) s);
next.push_back((int)(s * s - low));
}
}
// sieve the current segment
for (std::size_t i = 1; i < primes.size(); i++)
{
int j = next[i];
for (int k = primes[i] * 2; j < segment_size; j += k)
sieve[j] = 0;
next[i] = j - segment_size;
}
for (; n <= high; n += 2)
if (sieve[n - low]) // n is a prime
count++;
}
std::cout << count << " primes found." << std::endl;
}
【问题讨论】：
-
保留您 push_back 的向量。
标签： c++ algorithm primes sieve-of-eratosthenes