結果
| 問題 | No.3101 Range Eratosthenes Query |
| コンテスト | |
| ユーザー |
 Astral__
|
| 提出日時 | 2025-04-11 21:51:07 |
| 言語 | C++23 (gcc 13.3.0 + boost 1.87.0) |
| 結果 |
AC
|
| 実行時間 | 571 ms / 3,000 ms |
| コード長 | 10,235 bytes |
| コンパイル時間 | 5,343 ms |
| コンパイル使用メモリ | 334,116 KB |
| 実行使用メモリ | 53,936 KB |
| 最終ジャッジ日時 | 2025-04-11 21:51:29 |
| 合計ジャッジ時間 | 20,424 ms |
|
ジャッジサーバーID （参考情報） |
judge3 / judge5 |
（要ログイン）
| ファイルパターン | 結果 |
|---|---|
| sample | AC * 2 |
| other | AC * 24 |
ソースコード
#include <bits/stdc++.h>
#if __has_include(<atcoder/all>)
#include <atcoder/all>
std::istream &operator>>(std::istream &is, atcoder::modint &v) {
long long value;
is >> value;
v = value;
return is;
}
std::ostream &operator<<(std::ostream &os, const atcoder::modint &v) {
os << v.val();
return os;
}
std::ostream &operator<<(std::ostream &os, const atcoder::modint998244353 &v) {
os << v.val();
return os;
}
std::istream &operator>>(std::istream &is, atcoder::modint998244353 &v) {
long long x;
is >> x;
v = x;
return is;
}
std::ostream &operator<<(std::ostream &os, const atcoder::modint1000000007 &v) {
os << v.val();
return os;
}
std::istream &operator>>(std::istream &is, atcoder::modint1000000007 &v) {
long long x;
is >> x;
v = x;
return is;
}
#endif
using namespace std;
using ll = long long;
using lint = __int128_t;
using pll = pair<ll, ll>;
#define newl '\n';
#define rep(i, s, t) for (ll i = s; i < (ll)(t); i++)
#define rrep(i, s, t) for (ll i = (ll)(t) - 1; i >= (ll)(s); i--)
#define all(x) begin(x), end(x)
#define SZ(x) ll(x.size())
#define eb emplace_back
#define pb push_back
#define TT template <typename T>
TT using vec = vector<T>;
TT using vvec = vec<vec<T>>;
TT using vvvec = vec<vvec<T>>;
TT using minheap = priority_queue<T, vector<T>, greater<T>>;
TT using maxheap = priority_queue<T>;
TT bool chmin(T &x, T y) {
return x > y ? (x = y, true) : false;
}
TT bool chmax(T &x, T y) {
return x < y ? (x = y, true) : false;
}
TT T smod(T x, T mod) {
x %= mod;
if (x < 0)
x += mod;
return x;
}
TT bool rng(T l, T x, T r) {
return l <= x && x < r;
}
TT T flr(T a, T b) {
if (b < 0)
a = -a, b = -b;
return a >= 0 ? a / b : (a + 1) / b - 1;
}
TT T cil(T a, T b) {
if (b < 0)
a = -a, b = -b;
return a > 0 ? (a - 1) / b + 1 : a / b;
}
TT T sqr(T x) {
return x * x;
}
//{0, 1, ... } -> {p[0], p[1], ...}
template <typename T, typename S>
void rearrange(vector<T> &A, vector<S> const &p) {
assert(p.size() == A.size());
vector<T> a = A;
for (int i = 0; i < ssize(A); ++i) {
a[i] = A[p[i]];
}
swap(a, A);
}
template <typename T, typename S, typename... Ts>
void rearrange(vector<T> &A, vector<S> p, vector<Ts> &...rest) {
rearrange(A, p);
(rearrange(rest, p), ...);
}
template <typename T, typename Compare, typename... Ts>
void rearrange(vector<T> &A, Compare cmp, vector<Ts> &...rest) {
vector<int> p(ssize(A));
iota(p.begin(), p.end(), 0);
sort(p.begin(), p.end(), cmp);
rearrange(A, p);
(rearrange(rest, p), ...);
}
struct io_setup {
io_setup() {
ios::sync_with_stdio(false);
std::cin.tie(nullptr);
cout << fixed << setprecision(15);
}
} io_setup;
template <class T1, class T2>
ostream &operator<<(ostream &os, const pair<T1, T2> &p) {
os << p.first << " " << p.second;
return os;
}
TT ostream &operator<<(ostream &os, const vector<T> &v) {
for (size_t i = 0; i < v.size(); i++) {
os << v[i] << (i + 1 != v.size() ? " " : "");
}
return os;
}
template <typename T, size_t n>
ostream &operator<<(ostream &os, const array<T, n> &v) {
for (size_t i = 0; i < n; i++) {
os << v[i] << (i + 1 != n ? " " : "");
}
return os;
}
template <typename T> ostream &operator<<(ostream &os, const vvec<T> &v) {
for (size_t i = 0; i < v.size(); i++) {
os << v[i] << (i + 1 != v.size() ? "\n" : "");
}
return os;
}
TT istream &operator>>(istream &is, vector<T> &v) {
for (size_t i = 0; i < v.size(); i++) {
is >> v[i];
}
return is;
}
#if __has_include(<debug/debug.hpp>)
#include <debug/debug.hpp>
#else
#define dbg(...) true
#define DBG(...) true
#define OUT(...) true
#endif
struct notlinear_sieve {
int n;
vector<int> sm;
notlinear_sieve(int max_n) : n(max_n), sm(max_n + 1) {
assert(1 <= n);
iota(sm.begin(), sm.end(), 0);
if (n >= 2)
sm[2] = 2;
for (int j = 4; j <= n; j += 2)
sm[j] = 2;
for (int i = 3; i * i <= n; i += 2) {
if (sm[i] != i)
continue;
for (int j = i * 2; j <= n; j += i) {
if (sm[j] == j)
sm[j] = i;
}
}
}
bool is_prime(int v) const noexcept {
assert(v <= n);
if (v <= 1)
return false;
return sm[v] == v;
}
vector<int> primes(int max_n) const noexcept {
assert(1 <= max_n && max_n <= n);
vector<int> ret;
for (int i = 2; i <= max_n; i++)
if (is_prime(i))
ret.push_back(i);
return ret;
}
// sorted
vector<pair<int, int>> factorize(int v) const noexcept {
assert(1 <= v && v <= n);
vector<pair<int, int>> ret;
while (sm[v] != v) {
int tmp = v;
int c = 0;
while (tmp % sm[v] == 0)
c++, tmp /= sm[v];
ret.emplace_back(sm[v], c);
v = tmp;
}
if (v != 1)
ret.emplace_back(v, 1);
return ret;
}
int divcnt(int v) const noexcept {
assert(1 <= v && v <= n);
auto ps = factorize(v);
int ret = 1;
for (auto [p, c] : ps)
ret *= (c + 1);
return ret;
}
// not sorted
vector<int> divs(int v) const noexcept {
assert(1 <= v && v <= n);
auto ps = factorize(v);
int sz = 1;
for (auto [p, c] : ps)
sz *= (c + 1);
vector<int> ret(sz);
ret[0] = 1;
int r = 1;
for (auto [p, c] : ps) {
int nr = r;
for (int j = 0; j < c; j++) {
for (int k = 0; k < r; k++) {
ret[nr] = p * ret[nr - r];
nr++;
}
}
r = nr;
}
return ret;
}
// 偶数...+1 奇数...-1 p^2...0
template <typename T> vector<T> mobius(int N) const {
assert(N <= n);
vector<T> ret(N + 1, 1);
for (int p = 2; p <= N; p++)
if (is_prime(p)) {
for (int q = p; q <= N; q += p) {
if ((q / p) % p == 0)
ret[q] = 0;
else
ret[q] = -ret[q];
}
}
return ret;
}
// 以下4つは素因数ごとの累積和と思うと良い。計算量はO(nloglogn)
// zeta_transform... 結合則 + 交換則 ならなんでも乗る
// mobius_transform ... 結合 + 交換 + 逆元の存在 ならなんでも乗る
// f -> F 約数の添字をadd
template <typename T> vector<T> divisor_zeta_transform(vector<T> A) const {
int N = int(A.size()) - 1;
assert(N <= n);
for (int p = 2; p <= N; p++) {
if (is_prime(p)) {
for (int k = 1; k * p <= N; k++) {
A[k * p] += A[k];
}
}
}
return A;
}
// F -> f
template <typename T>
vector<T> divisor_mobius_transform(vector<T> A) const {
int N = int(A.size()) - 1;
assert(N <= n);
for (int p = 2; p <= N; p++) {
if (is_prime(p)) {
for (int k = N / p; k >= 1; k--) {
A[k * p] -= A[k];
}
}
}
return A;
}
// f -> F 倍数の添字をadd
template <typename T> vector<T> multiple_zeta_transform(vector<T> A) const {
int N = int(A.size()) - 1;
assert(N <= n);
for (int p = 2; p <= N; p++) {
if (is_prime(p)) {
for (int k = N / p; k >= 1; k--) {
A[k] += A[k * p];
}
}
}
return A;
}
// F -> f
template <typename T>
vector<T> multiple_mobius_transform(vector<T> A) const {
int N = int(A.size()) - 1;
assert(N <= n);
for (int p = 2; p <= N; p++) {
if (is_prime(p)) {
for (int k = 1; k <= N / p; k++) {
A[k] -= A[k * p];
}
}
}
return A;
}
};
template <class T> struct fenwick_tree {
public:
fenwick_tree() : n(0) {
}
explicit fenwick_tree(int n) : n(n), data(n), raw(n, T()), S(T()) {
}
void add(int p, T x) {
assert(0 <= p && p < n);
raw[p] += x;
S += x;
p++;
while (p <= n) {
data[p - 1] += x;
p += p & -p;
}
}
T sum(int r) const {
T s = 0;
while (r > 0) {
s += data[r - 1];
r -= r & -r;
}
return s;
}
T prod(int l, int r) const {
assert(0 <= l && l <= r && r <= n);
return sum(r) - sum(l);
}
T all_prod() const {
return S;
}
T get(int p) const {
return raw[p];
}
template <class F> int max_right(F f) const {
assert(f(0));
T s = 0;
int p = 0;
for (int i = 32 - __builtin_clz(n) - 1; i >= 0; i--) {
int k = p + (1 << i);
if (k <= n && f(s + data[k - 1])) {
s += data[k - 1];
p = k;
}
}
return p;
}
private:
int n;
vector<T> data;
vector<T> raw;
T S;
};
const ll M = 1'000'000;
notlinear_sieve sieve(M+10);
int main() {
ll q;
cin >> q;
fenwick_tree<ll> fen(M+10);
vector ls(M + 10, vector<pll>());
vector<ll> ans(q, -1);
rep(qi, 0, q) {
ll l, r;
cin >> l >> r;
r++;
ls[r].eb(l, qi);
}
rep(r, 1, M + 2) {
for (auto [l, i] : ls[r]) {
ans[i] = fen.prod(l, r);
}
fen.add(r, 1);
// rの最大の約数は？
if(r==1) continue;
auto ds = sieve.divs(r);
sort(all(ds));
reverse(all(ds));
fen.add(ds[1], -1);
}
for(auto v : ans) cout << v << newl;
}
/*
同じ議論を繰り返さない
do smth instead of nothing and stay organized
WRITE STUFF DOWN
DON'T GET STUCK ON ONE APPROACH
*/