NumberTheory/Math/CheckPrime.h

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• Last update: 2025-05-25 00:26:18+07:00
• Link: https://codeforces.com/blog/entry/91632

Depends on

• template.h

Verified with

• NumberTheory/Counting_primes.test.cpp
• NumberTheory/Enumerate_primes.test.cpp

Code

#include "../../template.h"


bool BruteForce(ll n) {
    if (n == 2 || n == 3) return true;
    if (n <= 1 || n % 2 == 0 || n % 3 == 0) return false;
    for (ll i = 5; i * i <= n; i += 6)
        if (n % i == 0 || n % (i+2) == 0)
            return false;
    return true;
}

// https://codeforces.com/blog/entry/91632
ll Meissel(ll n) {
    vector<ll> v;
    for (ll i = 1; i*i <= n; i++) {
        v.push_back(i);
        v.push_back(n / i);
    }
    sort(all(v));
    unique(v);
    ll sq = sqrt(n);
    auto geti = [&](ll x) {
        if (x <= sq) return x-1;
        return (int)v.size() - n / x;
    };
    vector<ll> dp = v;
    ll a = 0;
    for (ll p = 2; p*p <= n; p++) {
        if (dp[geti(p)] == dp[geti(p-1)]) continue;
        a++;
        for (int i = (int)v.size()-1; i >= 0; i--) {
            if (v[i] < p * p) break;
            dp[i] -= dp[geti(v[i] / p)] - a;
        }
    }
    return dp[geti(n)] - 1;
}

#line 2 "template.h"

#include <bits/stdc++.h>
using namespace std;
 
#define ll long long
#define MOD (ll)(1e9+7)
#define all(x) (x).begin(),(x).end()
#define unique(x) x.erase(unique(all(x)), x.end())
#define INF32 ((1ull<<31)-1)
#define INF64 ((1ull<<63)-1)
#define inf (ll)1e18

#define vi vector<int>
#define pii pair<int, int>
#define pll pair<ll, ll>
#define fi first
#define se second

const int mod = 998244353;

void solve();

int main(){
    ios_base::sync_with_stdio(false);cin.tie(NULL);
    // cin.exceptions(cin.failbit);
    // int t; cin >> t;
    // while(t--)
        solve();
    cerr << "\nTime run: " << 1000 * clock() / CLOCKS_PER_SEC << "ms" << '\n';
    return 0;
}
#line 2 "NumberTheory/Math/CheckPrime.h"


bool BruteForce(ll n) {
    if (n == 2 || n == 3) return true;
    if (n <= 1 || n % 2 == 0 || n % 3 == 0) return false;
    for (ll i = 5; i * i <= n; i += 6)
        if (n % i == 0 || n % (i+2) == 0)
            return false;
    return true;
}

// https://codeforces.com/blog/entry/91632
ll Meissel(ll n) {
    vector<ll> v;
    for (ll i = 1; i*i <= n; i++) {
        v.push_back(i);
        v.push_back(n / i);
    }
    sort(all(v));
    unique(v);
    ll sq = sqrt(n);
    auto geti = [&](ll x) {
        if (x <= sq) return x-1;
        return (int)v.size() - n / x;
    };
    vector<ll> dp = v;
    ll a = 0;
    for (ll p = 2; p*p <= n; p++) {
        if (dp[geti(p)] == dp[geti(p-1)]) continue;
        a++;
        for (int i = (int)v.size()-1; i >= 0; i--) {
            if (v[i] < p * p) break;
            dp[i] -= dp[geti(v[i] / p)] - a;
        }
    }
    return dp[geti(n)] - 1;
}

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