結果

問題 No.811 約数の個数の最大化
ユーザー stoqstoq
提出日時 2021-08-12 02:34:42
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 160 ms / 2,000 ms
コード長 5,210 bytes
コンパイル時間 4,753 ms
コンパイル使用メモリ 267,680 KB
最終ジャッジ日時 2025-01-23 17:40:49
ジャッジサーバーID
(参考情報)
judge3 / judge2
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 3
other AC * 12
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 #
プレゼンテーションモードにする

#define MOD_TYPE 2
#pragma region Macros
#include <bits/stdc++.h>
using namespace std;
#include <atcoder/all>
using namespace atcoder;
#if 0
#include <boost/multiprecision/cpp_dec_float.hpp>
#include <boost/multiprecision/cpp_int.hpp>
using Int = boost::multiprecision::cpp_int;
using lld = boost::multiprecision::cpp_dec_float_100;
#endif
#if 1
#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#endif
using ll = long long int;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using pld = pair<ld, ld>;
template <typename Q_type>
using smaller_queue = priority_queue<Q_type, vector<Q_type>, greater<Q_type>>;
#if MOD_TYPE == 1
constexpr ll MOD = ll(1e9 + 7);
#else
#if MOD_TYPE == 2
constexpr ll MOD = 998244353;
#else
constexpr ll MOD = 0;
#endif
#endif
using mint = static_modint<MOD>;
constexpr int INF = (int)1e9 + 10;
constexpr ll LINF = (ll)4e18;
constexpr double PI = acos(-1.0);
constexpr double EPS = 1e-11;
constexpr int Dx[] = {0, 0, -1, 1, -1, 1, -1, 1, 0};
constexpr int Dy[] = {1, -1, 0, 0, -1, -1, 1, 1, 0};
#define REP(i, m, n) for (ll i = m; i < (ll)(n); ++i)
#define rep(i, n) REP(i, 0, n)
#define REPI(i, m, n) for (int i = m; i < (int)(n); ++i)
#define repi(i, n) REPI(i, 0, n)
#define MP make_pair
#define MT make_tuple
#define YES(n) cout << ((n) ? "YES" : "NO") << "\n"
#define Yes(n) cout << ((n) ? "Yes" : "No") << "\n"
#define possible(n) cout << ((n) ? "possible" : "impossible") << "\n"
#define Possible(n) cout << ((n) ? "Possible" : "Impossible") << "\n"
#define Yay(n) cout << ((n) ? "Yay!" : ":(") << "\n"
#define all(v) v.begin(), v.end()
#define NP(v) next_permutation(all(v))
#define dbg(x) cerr << #x << ":" << x << "\n";
struct io_init {
io_init() {
cin.tie(0);
ios::sync_with_stdio(false);
cout << setprecision(30) << setiosflags(ios::fixed);
};
} io_init;
template <typename T>
inline bool chmin(T &a, T b) {
if (a > b) {
a = b;
return true;
}
return false;
}
template <typename T>
inline bool chmax(T &a, T b) {
if (a < b) {
a = b;
return true;
}
return false;
}
inline ll CEIL(ll a, ll b) { return (a + b - 1) / b; }
template <typename A, size_t N, typename T>
inline void Fill(A (&array)[N], const T &val) {
fill((T *)array, (T *)(array + N), val);
}
template <typename T>
vector<T> compress(vector<T> &v) {
vector<T> val = v;
sort(all(val)), val.erase(unique(all(val)), val.end());
for (auto &&vi : v) vi = lower_bound(all(val), vi) - val.begin();
return val;
}
template <typename T, typename U>
constexpr istream &operator>>(istream &is, pair<T, U> &p) noexcept {
is >> p.first >> p.second;
return is;
}
template <typename T, typename U>
constexpr ostream &operator<<(ostream &os, pair<T, U> p) noexcept {
os << p.first << " " << p.second;
return os;
}
ostream &operator<<(ostream &os, mint m) {
os << m.val();
return os;
}
random_device seed_gen;
mt19937_64 engine(seed_gen());
struct BiCoef {
vector<mint> fact_, inv_, finv_;
BiCoef(int n) noexcept : fact_(n, 1), inv_(n, 1), finv_(n, 1) {
fact_.assign(n, 1), inv_.assign(n, 1), finv_.assign(n, 1);
for (int i = 2; i < n; i++) {
fact_[i] = fact_[i - 1] * i;
inv_[i] = -inv_[MOD % i] * (MOD / i);
finv_[i] = finv_[i - 1] * inv_[i];
}
}
mint C(ll n, ll k) const noexcept {
if (n < k || n < 0 || k < 0) return 0;
return fact_[n] * finv_[k] * finv_[n - k];
}
mint P(ll n, ll k) const noexcept { return C(n, k) * fact_[k]; }
mint H(ll n, ll k) const noexcept { return C(n + k - 1, k); }
mint Ch1(ll n, ll k) const noexcept {
if (n < 0 || k < 0) return 0;
mint res = 0;
for (int i = 0; i < n; i++)
res += C(n, i) * mint(n - i).pow(k) * (i & 1 ? -1 : 1);
return res;
}
mint fact(ll n) const noexcept {
if (n < 0) return 0;
return fact_[n];
}
mint inv(ll n) const noexcept {
if (n < 0) return 0;
return inv_[n];
}
mint finv(ll n) const noexcept {
if (n < 0) return 0;
return finv_[n];
}
};
BiCoef bc(2000010);
#pragma endregion
const int MAX_N = 1e6;
ll can_div[MAX_N] = {};
void init_prime() {
can_div[1] = -1;
for (ll i = 2; i < MAX_N; i++) {
if (can_div[i] != 0) continue;
for (ll j = i + i; j < MAX_N; j += i) can_div[j] = i;
}
}
struct init_prime_ {
init_prime_() { init_prime(); };
} init_prime_;
inline bool is_prime(ll n) {
if (n <= 1) return false;
return !can_div[n];
}
void factorization(int n, unordered_map<ll, int> &res) {
if (n <= 1) return;
if (!can_div[n]) {
++res[n];
return;
}
++res[can_div[n]];
factorization(n / can_div[n], res);
}
int common(int n, int m) {
unordered_map<ll, int> A, B;
factorization(n, A);
factorization(m, B);
int sum = 0;
for (auto [p, e] : A) {
sum += min(e, B[p]);
}
return sum;
}
int divisor(int n) {
unordered_map<ll, int> mp;
factorization(n, mp);
int cnt = 1;
for (auto [p, e] : mp) cnt *= e + 1;
return cnt;
}
void solve() {
int n, k;
cin >> n >> k;
pii p{INF, -1};
for (int i = 1; i < n; i++) {
if (common(n, i) >= k) {
chmin(p, pii{-divisor(i), i});
}
}
cout << p.second << "\n";
}
int main() { solve(); }
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