結果

問題 No.1657 Sum is Prime (Easy Version)
ユーザー MMRZ
提出日時 2025-01-08 00:04:22
言語 C++23
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 2 ms / 2,000 ms
コード長 4,587 bytes
コンパイル時間 4,228 ms
コンパイル使用メモリ 283,720 KB
実行使用メモリ 5,248 KB
最終ジャッジ日時 2025-01-08 00:04:28
合計ジャッジ時間 5,702 ms
ジャッジサーバーID
（参考情報） 		judge3 / judge2
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ファイルパターン 結果
sample AC * 3
other AC * 21
権限があれば一括ダウンロードができます

ソースコード

diff # 

# include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ull = unsigned long long;
const double pi = acos(-1);
template<class T>constexpr T inf() { return ::std::numeric_limits<T>::max(); }
template<class T>constexpr T hinf() { return inf<T>() / 2; }
template <typename T_char>T_char TL(T_char cX) { return tolower(cX); }
template <typename T_char>T_char TU(T_char cX) { return toupper(cX); }
template<class T> bool chmin(T& a,T b) { if(a > b){a = b; return true;} return false; }
template<class T> bool chmax(T& a,T b) { if(a < b){a = b; return true;} return false; }
int popcnt(unsigned long long n) { int cnt = 0; for (int i = 0; i < 64; i++)if ((n >> i) & 1)cnt++; return cnt; }
int d_sum(ll n) { int ret = 0; while (n > 0) { ret += n % 10; n /= 10; }return ret; }
int d_cnt(ll n) { int ret = 0; while (n > 0) { ret++; n /= 10; }return ret; }
ll gcd(ll a, ll b) { if (b == 0)return a; return gcd(b, a%b); };
ll lcm(ll a, ll b) { ll g = gcd(a, b); return a / g*b; };
ll MOD(ll x, ll m){return (x%m+m)%m; }
ll FLOOR(ll x, ll m) {ll r = (x%m+m)%m; return (x-r)/m; }
template<class T> using dijk = priority_queue<T, vector<T>, greater<T>>;
# define all(qpqpq)           (qpqpq).begin(),(qpqpq).end()
# define UNIQUE(wpwpw)        (wpwpw).erase(unique(all((wpwpw))),(wpwpw).end())
# define LOWER(epepe)         transform(all((epepe)),(epepe).begin(),TL<char>)
# define UPPER(rprpr)         transform(all((rprpr)),(rprpr).begin(),TU<char>)
# define rep(i,upupu)         for(ll i = 0, i##_len = (upupu);(i) < (i##_len);(i)++)
# define reps(i,opopo)        for(ll i = 1, i##_len = (opopo);(i) <= (i##_len);(i)++)
# define len(x)                ((ll)(x).size())
# define bit(n)               (1LL << (n))
# define pb push_back
# define exists(c, e)         ((c).find(e) != (c).end())

struct INIT{
	INIT(){
		std::ios::sync_with_stdio(false);
		std::cin.tie(0);
		cout << fixed << setprecision(20);
	}
}INIT;

namespace mmrz {
	void solve();
}

int main(){
	mmrz::solve();
}
#define debug(...) (static_cast<void>(0))

using namespace mmrz;



unsigned long long iroot(unsigned long long n, int k=2){
	constexpr unsigned long long LIM = -1;
	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 : pow(n, nextafter(1.0/double(k), 0.0)));
	while(power(ret+1, k) <= n)ret++;
	return ret;
}

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

void SOLVE(){
	ll l, r;
	cin >> l >> r;
	cout << counting_primes(r) - counting_primes(l-1) + counting_primes(2*r) - counting_primes(2*l) << endl;
}

void mmrz::solve(){
	int t = 1;
	//cin >> t;
	while(t--)SOLVE();
}
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