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:heavy_check_mark: $n$ 以下の素数の数え上げ
(math/counting_primes.hpp)

$n$ 以下の素数の数え上げ

使い方

counting_primes(ll n) : $n$ 以下の素数の数を返す。 $O(\frac{n^{3/4}}{\log n})$

参考

Depends on

Verified with

Code

#pragma once

#include "iroot.hpp"

#include<vector>

//https://judge.yosupo.jp/submission/61551
//https://rsk0315.hatenablog.com/entry/2021/05/18/015511
long long counting_primes(const long long N) {
	if (N <= 1) return 0;
	if (N == 2) return 1;
	const int v = iroot(N);
	int s = (v + 1) / 2;
	std::vector<int> smalls(s);
	for (int i = 1; i < s; i++) smalls[i] = i;
	std::vector<int> roughs(s);
	for (int i = 0; i < s; i++) roughs[i] = 2 * i + 1;
	std::vector<long long> larges(s);
	for (int i = 0; i < s; i++) larges[i] = (N / (2 * i + 1) - 1) / 2;
	std::vector<bool> skip(v + 1);
	const auto divide = [](long long n, long long d) -> int { return (double)n / d;};
	const auto half = [](int n) -> int { return (n - 1) >> 1;};
	int pc = 0;
	for (int p = 3; p <= v; p += 2) if (!skip[p]) {
		int q = p * p;
		if ((long long)q * q > N) break;
		skip[p] = true;
		for (int i = q; i <= v; i += 2 * p) skip[i] = true;
		int ns = 0;
		for (int k = 0; k < s; k++) {
			int i = roughs[k];
			if (skip[i]) continue;
			long long d = (long long)i * p;
			larges[ns] = larges[k] - (d <= v ? larges[smalls[d >> 1] - pc] : smalls[half(divide(N, d))]) + pc;
			roughs[ns++] = i;
		}
		s = ns;
		for (int i = half(v), j = ((v / p) - 1) | 1; j >= p; j -= 2) {
			int c = smalls[j >> 1] - pc;
			for (int e = (j * p) >> 1; i >= e; i--) smalls[i] -= c;
		}
		pc++;
	}
	larges[0] += (long long)(s + 2 * (pc - 1)) * (s - 1) / 2;
	for (int k = 1; k < s; k++) larges[0] -= larges[k];
	for (int l = 1; l < s; l++) {
		long long q = roughs[l];
		long long M = N / q;
		int e = smalls[half(M / q)] - pc;
		if (e < l + 1) break;
		long long t = 0;
		for (int k = l + 1; k <= e; k++)
			t += smalls[half(divide(M, roughs[k]))];
		larges[0] += t - (long long)(e - l) * (pc + l - 1);
	}
	return larges[0] + 1;
}
#line 2 "math/counting_primes.hpp"

#line 2 "math/iroot.hpp"

#include<cmath>
#include<limits>

unsigned long long iroot(unsigned long long n, int k=2){
	constexpr unsigned long long LIM = std::numeric_limits<unsigned long long>::max();
	if(n <= 1 || k == 1){
		return n;
	}
	if(k >= 64){
		return 1;
	}
	if(k == 2){
		return sqrtl(n);
	}

	if(n == LIM)n--;

	auto safe_mul = [&](unsigned long long &x, unsigned long long &y) -> void {
		if(x <= LIM / y){
			x *= y;
		}else{
			x = LIM;
		}
	};

	auto power = [&](unsigned long long a, int b) -> unsigned long long {
		unsigned long long ret = 1;
		while(b){
			if(b & 1)safe_mul(ret, a);
			safe_mul(a, a);
			b >>= 1;
		}
		return ret;
	};

	unsigned long long ret = (k == 3 ? cbrt(n)-1 : std::pow(n, std::nextafter(1.0/double(k), 0.0)));
	while(power(ret+1, k) <= n)ret++;
	return ret;
}
#line 4 "math/counting_primes.hpp"

#include<vector>

//https://judge.yosupo.jp/submission/61551
//https://rsk0315.hatenablog.com/entry/2021/05/18/015511
long long counting_primes(const long long N) {
	if (N <= 1) return 0;
	if (N == 2) return 1;
	const int v = iroot(N);
	int s = (v + 1) / 2;
	std::vector<int> smalls(s);
	for (int i = 1; i < s; i++) smalls[i] = i;
	std::vector<int> roughs(s);
	for (int i = 0; i < s; i++) roughs[i] = 2 * i + 1;
	std::vector<long long> larges(s);
	for (int i = 0; i < s; i++) larges[i] = (N / (2 * i + 1) - 1) / 2;
	std::vector<bool> skip(v + 1);
	const auto divide = [](long long n, long long d) -> int { return (double)n / d;};
	const auto half = [](int n) -> int { return (n - 1) >> 1;};
	int pc = 0;
	for (int p = 3; p <= v; p += 2) if (!skip[p]) {
		int q = p * p;
		if ((long long)q * q > N) break;
		skip[p] = true;
		for (int i = q; i <= v; i += 2 * p) skip[i] = true;
		int ns = 0;
		for (int k = 0; k < s; k++) {
			int i = roughs[k];
			if (skip[i]) continue;
			long long d = (long long)i * p;
			larges[ns] = larges[k] - (d <= v ? larges[smalls[d >> 1] - pc] : smalls[half(divide(N, d))]) + pc;
			roughs[ns++] = i;
		}
		s = ns;
		for (int i = half(v), j = ((v / p) - 1) | 1; j >= p; j -= 2) {
			int c = smalls[j >> 1] - pc;
			for (int e = (j * p) >> 1; i >= e; i--) smalls[i] -= c;
		}
		pc++;
	}
	larges[0] += (long long)(s + 2 * (pc - 1)) * (s - 1) / 2;
	for (int k = 1; k < s; k++) larges[0] -= larges[k];
	for (int l = 1; l < s; l++) {
		long long q = roughs[l];
		long long M = N / q;
		int e = smalls[half(M / q)] - pc;
		if (e < l + 1) break;
		long long t = 0;
		for (int k = l + 1; k <= e; k++)
			t += smalls[half(divide(M, roughs[k]))];
		larges[0] += t - (long long)(e - l) * (pc + l - 1);
	}
	return larges[0] + 1;
}
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