#include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include using namespace std; using lint = long long; using pint = pair; using plint = pair; struct fast_ios { fast_ios(){ cin.tie(nullptr), ios::sync_with_stdio(false), cout << fixed << setprecision(20); }; } fast_ios_; #define ALL(x) (x).begin(), (x).end() #define FOR(i, begin, end) for(int i=(begin),i##_end_=(end);i=i##_begin_;i--) #define REP(i, n) FOR(i,0,n) #define IREP(i, n) IFOR(i,0,n) template void ndarray(vector& vec, const V& val, int len) { vec.assign(len, val); } template void ndarray(vector& vec, const V& val, int len, Args... args) { vec.resize(len), for_each(begin(vec), end(vec), [&](T& v) { ndarray(v, val, args...); }); } template bool chmax(T &m, const T q) { return m < q ? (m = q, true) : false; } template bool chmin(T &m, const T q) { return m > q ? (m = q, true) : false; } int floor_lg(long long x) { return x <= 0 ? -1 : 63 - __builtin_clzll(x); } template pair operator+(const pair &l, const pair &r) { return make_pair(l.first + r.first, l.second + r.second); } template pair operator-(const pair &l, const pair &r) { return make_pair(l.first - r.first, l.second - r.second); } template vector sort_unique(vector vec) { sort(vec.begin(), vec.end()), vec.erase(unique(vec.begin(), vec.end()), vec.end()); return vec; } template int arglb(const std::vector &v, const T &x) { return std::distance(v.begin(), std::lower_bound(v.begin(), v.end(), x)); } template int argub(const std::vector &v, const T &x) { return std::distance(v.begin(), std::upper_bound(v.begin(), v.end(), x)); } template istream &operator>>(istream &is, vector &vec) { for (auto &v : vec) is >> v; return is; } template ostream &operator<<(ostream &os, const vector &vec) { os << '['; for (auto v : vec) os << v << ','; os << ']'; return os; } template ostream &operator<<(ostream &os, const array &arr) { os << '['; for (auto v : arr) os << v << ','; os << ']'; return os; } #if __cplusplus >= 201703L template istream &operator>>(istream &is, tuple &tpl) { std::apply([&is](auto &&... args) { ((is >> args), ...);}, tpl); return is; } template ostream &operator<<(ostream &os, const tuple &tpl) { os << '('; std::apply([&os](auto &&... args) { ((os << args << ','), ...);}, tpl); return os << ')'; } #endif template ostream &operator<<(ostream &os, const deque &vec) { os << "deq["; for (auto v : vec) os << v << ','; os << ']'; return os; } template ostream &operator<<(ostream &os, const set &vec) { os << '{'; for (auto v : vec) os << v << ','; os << '}'; return os; } template ostream &operator<<(ostream &os, const unordered_set &vec) { os << '{'; for (auto v : vec) os << v << ','; os << '}'; return os; } template ostream &operator<<(ostream &os, const multiset &vec) { os << '{'; for (auto v : vec) os << v << ','; os << '}'; return os; } template ostream &operator<<(ostream &os, const unordered_multiset &vec) { os << '{'; for (auto v : vec) os << v << ','; os << '}'; return os; } template ostream &operator<<(ostream &os, const pair &pa) { os << '(' << pa.first << ',' << pa.second << ')'; return os; } template ostream &operator<<(ostream &os, const map &mp) { os << '{'; for (auto v : mp) os << v.first << "=>" << v.second << ','; os << '}'; return os; } template ostream &operator<<(ostream &os, const unordered_map &mp) { os << '{'; for (auto v : mp) os << v.first << "=>" << v.second << ','; os << '}'; return os; } #ifdef HITONANODE_LOCAL const string COLOR_RESET = "\033[0m", BRIGHT_GREEN = "\033[1;32m", BRIGHT_RED = "\033[1;31m", BRIGHT_CYAN = "\033[1;36m", NORMAL_CROSSED = "\033[0;9;37m", RED_BACKGROUND = "\033[1;41m", NORMAL_FAINT = "\033[0;2m"; #define dbg(x) cerr << BRIGHT_CYAN << #x << COLOR_RESET << " = " << (x) << NORMAL_FAINT << " (L" << __LINE__ << ") " << __FILE__ << COLOR_RESET << endl #define dbgif(cond, x) ((cond) ? cerr << BRIGHT_CYAN << #x << COLOR_RESET << " = " << (x) << NORMAL_FAINT << " (L" << __LINE__ << ") " << __FILE__ << COLOR_RESET << endl : cerr) #else #define dbg(x) (x) #define dbgif(cond, x) 0 #endif struct CountPrimes { // Count Primes less than or equal to x (\pi(x)) for each x = N / i (i = 1, ..., N) in O(N^(2/3)) time // Learned this algorihtm from https://old.yosupo.jp/submission/14650 // Reference: https://min-25.hatenablog.com/entry/2018/11/11/172216 using Int = long long; Int n, n2, n3, n6; std::vector is_prime; // [0, 0, 1, 1, 0, 1, 0, 1, ...] std::vector primes; // primes up to O(N^(1/2)), [2, 3, 5, 7, ...] int s; // size of vs std::vector vs; // [N, ..., n2, n2 - 1, n2 - 2, ..., 3, 2, 1] std::vector pi; // pi[i] = (# of primes s.t. <= vs[i]) is finally obtained std::vector _fenwick; int getidx(Int a) const { return a <= n2 ? s - a : n / a - 1; } // vs[i] >= a を満たす最大の i を返す void _fenwick_rec_update(int i, Int cur, bool first) { // pi[n3:] に対して cur * (primes[i] 以上の素因数) の数の寄与を減じる if (!first) { for (int x = getidx(cur) - n3; x >= 0; x -= (x + 1) & (-x - 1)) _fenwick[x]--; } for (int j = i; cur * primes[j] <= vs[n3]; j++) _fenwick_rec_update(j, cur * primes[j], false); } CountPrimes(Int n_) : n(n_), n2((Int)sqrtl(n)), n3((Int)cbrtl(n)), n6((Int)sqrtl(n3)) { is_prime.assign(n2 + 300, 1), is_prime[0] = is_prime[1] = 0; // `+ 300`: for (size_t p = 2; p < is_prime.size(); p++) { if (is_prime[p]) { primes.push_back(p); for (size_t j = p * 2; j < is_prime.size(); j += p) is_prime[j] = 0; } } for (Int now = n; now; now = n / (n / now + 1)) vs.push_back(now); // [N, N / 2, ..., 1], Relevant integers (decreasing) length ~= 2sqrt(N) s = vs.size(); // pi[i] = (# of integers x s.t. x <= vs[i], (x is prime or all factors of x >= p)) // pre = (# of primes less than p) // 最小の素因数 p = 2, ..., について篩っていく pi.resize(s); for (int i = 0; i < s; i++) pi[i] = vs[i] - 1; int pre = 0; auto trans = [&](int i, Int p) { pi[i] -= pi[getidx(vs[i] / p)] - pre; }; size_t ip = 0; // [Sieve Part 1] For each prime p satisfying p <= N^(1/6) (Only O(N^(1/6) / log N) such primes exist), // O(sqrt(N)) simple operation is conducted. // - Complexity of this part: O(N^(2/3) / logN) for (; primes[ip] <= n6; ip++, pre++) { const auto &p = primes[ip]; for (int i = 0; p * p <= vs[i]; i++) trans(i, p); } // [Sieve Part 2] For each prime p satisfying N^(1/6) < p <= N^(1/3), // point-wise & Fenwick tree-based hybrid update is used // - first N^(1/3) elements are simply updated by quadratic algorithm. // - Updates of latter segments are managed by Fenwick tree. // - Complexity of this part: O(N^(2/3)) (O(N^(2/3)/log N) operations for Fenwick tree (O(logN) per query)) _fenwick.assign(s - n3, 0); // Fenwick tree, inversed order (summation for upper region) auto trans2 = [&](int i, Int p) { int j = getidx(vs[i] / p); auto z = pi[j]; if (j >= n3) { for (j -= n3; j < int(_fenwick.size()); j += (j + 1) & (-j - 1)) z += _fenwick[j]; } pi[i] -= z - pre; }; for (; primes[ip] <= n3; ip++, pre++) { const auto &p = primes[ip]; for (int i = 0; i < n3 and p * p <= vs[i]; i++) trans2(i, p); // upto n3, total trans2 times: O(N^(2/3) / logN) _fenwick_rec_update(ip, primes[ip], true); // total update times: O(N^(2/3) / logN) } for (int i = s - n3 - 1; i >= 0; i--) { int j = i + ((i + 1) & (-i - 1)); if (j < s - n3) _fenwick[i] += _fenwick[j]; } for (int i = 0; i < s - n3; i++) pi[i + n3] += _fenwick[i]; // [Sieve Part 3] For each prime p satisfying N^(1/3) < p <= N^(1/2), use only simple updates. // - Complexity of this part: O(N^(2/3) / logN) // \sum_i (# of factors of vs[i] of the form p^2, p >= N^(1/3)) = \sum_{i=1}^{N^(1/3)} \pi(\sqrt(vs[i]))) // = sqrt(N) \sum_i^{N^(1/3)} i^{-1/2} / logN = O(N^(2/3) / logN) // (Note: \sum_{i=1}^{N} i^{-1/2} = O(sqrt N) https://math.stackexchange.com/questions/2600796/finding-summation-of-inverse-of-square-roots ) for (; primes[ip] <= n2; ip++, pre++) { const auto &p = primes[ip]; for (int i = 0; p * p <= vs[i]; i++) trans(i, p); } } }; int main() { long long L, R; cin >> L >> R; long long ret = CountPrimes(R).pi[0]; if (L > 1) ret -= CountPrimes(L - 1).pi[0]; ret += CountPrimes(R * 2).pi[0]; ret -= CountPrimes(L * 2).pi[0]; cout << ret << '\n'; }