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Segmented sieve of Eratosthenes implementation in C++
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#include <bits/stdc++.h> | |
using namespace std; | |
typedef long long ll; | |
vector<ll> sieve(ll n) { | |
vector<ll> is_prime(n + 1, true), res; | |
for (ll i = 2; i <= n; i++) { | |
if (is_prime[i]) { | |
res.push_back(i); | |
for (ll j = i * i; j <= n; j += i) is_prime[j] = false; | |
} | |
} | |
return res; | |
} | |
vector<bool> segmented_sieve(ll l, ll r) { | |
vector<ll> primes = sieve(sqrt(r)); | |
vector<bool> is_prime(r - l + 1, true); | |
for(ll i : primes){ | |
ll min_mul; | |
if(i > l) min_mul = i; | |
else { | |
min_mul = l; | |
if(l % i) min_mul += i - l % i; | |
} | |
for(ll j = min_mul; j - l < is_prime.size(); j += i){ | |
if(j == i) continue; | |
is_prime[j - l] = false; | |
} | |
} | |
if (l == 1) is_prime[0] = false; | |
return is_prime; | |
} | |
int main() { | |
ll l, r; | |
cin >> l >> r; | |
vector<bool> is_prime_segmented = segmented_sieve(l, r); | |
for(ll i = l; i <= r; i++){ | |
if(is_prime_segmented[i - l]) cout << i << '\n'; | |
} | |
} |
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