답안 #562953

# 제출 시각 아이디 문제 언어 결과 실행 시간 메모리
562953 2022-05-15T16:59:46 Z Bungmint 비밀 (JOI14_secret) C++17
100 / 100
485 ms 4420 KB
// Copyright © 2022 Youngmin Park. All rights reserved.
//#pragma GCC optimize("O3")
//#pragma GCC target("avx2")
#include <bits/stdc++.h>
#include "secret.h"	
using namespace std;

using ll = long long;
using vi = vector<int>;
using pii = pair<int, int>;
using vpi = vector<pii>;
using pll = pair<ll, ll>;
using vl = vector<ll>;
using vpl = vector<pll>;
using ld = long double;
template <typename T, size_t SZ>
using ar = array<T, SZ>;
template <typename T>
using pqg = priority_queue<T, vector<T>, greater<T>>;

#define all(v) (v).begin(), (v).end()
#define pb push_back
#define sz(x) (int)(x).size()
#define fi first
#define se second
#define lb lower_bound
#define ub upper_bound

constexpr int INF = 1e9;
constexpr ll LINF = 1e18;
const ld PI = acos((ld)-1.0);
constexpr int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
template <typename T>
constexpr bool ckmin(T &a, const T &b) { return b < a ? a = b, 1 : 0; }
template <typename T>
constexpr bool ckmax(T &a, const T &b) { return b > a ? a = b, 1 : 0; }

template <typename A, typename B>
ostream &operator<<(ostream &os, const pair<A, B> &p)
{
	return os << '(' << p.first << ", " << p.second << ')';
}
template <typename T_container, typename T = typename enable_if<!is_same<T_container, string>::value, typename T_container::value_type>::type>
ostream &operator<<(ostream &os, const T_container &v)
{
	os << '{';
	string sep;
	for (const T &x : v)
		os << sep << x, sep = ", ";
	return os << '}';
}
template <typename T>
ostream &operator<<(ostream &os, const deque<T> &v) {
	os << vector<T>(all(v));
	return os;
}
template <typename T, typename S, typename C>
ostream &operator<<(ostream &os, priority_queue<T, S, C> pq) {
	vector<T> v;
	while (sz(pq)) {
		v.pb(pq.top());
		pq.pop();
	}
	os << v;
	return os;
}
void dbg_out()
{
	cerr << "\033[0m" << endl;
}
template <typename Head, typename... Tail>
void dbg_out(Head H, Tail... T)
{
	cerr << ' ' << H;
	dbg_out(T...);
}
#ifdef LOCAL
#define dbg(...) cerr << "\033[1;35m(" << #__VA_ARGS__ << "):\033[33m", dbg_out(__VA_ARGS__)
#else
#define dbg(...) 42
#endif

inline namespace RecursiveLambda
{
	template <typename Fun>
	struct y_combinator_result
	{
		Fun fun_;
		template <typename T>
		explicit y_combinator_result(T &&fun) : fun_(forward<T>(fun)) {}
		template <typename... Args>
		decltype(auto) operator()(Args &&...args)
		{
			return fun_(ref(*this), forward<Args>(args)...);
		}
	};
	template <typename Fun>
	decltype(auto) y_combinator(Fun &&fun)
	{
		return y_combinator_result<decay_t<Fun>>(forward<Fun>(fun));
	}
};

inline namespace Range {
	class ForwardRange {
		int src, dst;

	  public:
	  	explicit constexpr ForwardRange(const int l, const int r) : src(l), dst(r) {}
		explicit constexpr ForwardRange(const int n) : src(0), dst(n) {}
		constexpr ForwardRange begin() const { return *this; }
		constexpr monostate end() const { return {}; }
		constexpr bool operator!=(monostate) const { return src < dst; }
		constexpr void operator++() const {}
		constexpr int operator*() { return src++; }
	};
	class BackwardRange {
		int src, dst;

	  public:
	  	explicit constexpr BackwardRange(const int l, const int r) : src(r), dst(l) {}
		explicit constexpr BackwardRange(const int n) : src(n), dst(0) {}
		constexpr BackwardRange begin() const { return *this; }
		constexpr monostate end() const { return {}; }
		constexpr bool operator!=(monostate) const { return src > dst; }
		constexpr void operator++() const {}
		constexpr int operator*() { return --src; }
	};
	using rep = ForwardRange;
	using per = BackwardRange;
};

constexpr int K = 10;
constexpr int MX = 1024;
int layer[K][MX];
int a[MX];
int n;
int NN;

void DnC(int L, int R, int lev) {
	if (lev <= 0) return;
	int M = (L + R) >> 1;
	int prod = -1;
	for (int i : rep(M, R)) {
		if (i >= NN) break;
		int nxt = prod == -1 ? a[i] : Secret(prod, a[i]);
		layer[lev][i] = prod = nxt;
	}
	prod = -1;
	for (int i : per(L, M)) {
		if (i >= NN) break;
		int nxt = prod == -1 ? a[i] : Secret(a[i], prod);
		layer[lev][i] = prod = nxt;
	}
	DnC(L, M, lev - 1), DnC(M, R, lev - 1);
}

void Init(int N, int A[]) {
	NN = N;
	for (int i : rep(N)) a[i] = A[i];
	n = 1;
	while ((1 << n) < N) n++;
	DnC(0, (1 << n), n - 1);
}
int Query(int L, int R) {
	if (L == R) return a[L];
	if (L == R - 1) return Secret(a[L], a[R]);
	int z = 31 - __builtin_clz(L ^ R);
	int x = layer[z][L], y = layer[z][R];
	return Secret(x, y);
}
# 결과 실행 시간 메모리 Grader output
1 Correct 124 ms 2388 KB Output is correct - number of calls to Secret by Init = 3578, maximum number of calls to Secret by Query = 1
2 Correct 128 ms 2400 KB Output is correct - number of calls to Secret by Init = 3586, maximum number of calls to Secret by Query = 1
3 Correct 135 ms 2448 KB Output is correct - number of calls to Secret by Init = 4097, maximum number of calls to Secret by Query = 1
4 Correct 451 ms 4420 KB Output is correct - number of calls to Secret by Init = 7979, maximum number of calls to Secret by Query = 1
5 Correct 469 ms 4288 KB Output is correct - number of calls to Secret by Init = 7993, maximum number of calls to Secret by Query = 1
6 Correct 467 ms 4284 KB Output is correct - number of calls to Secret by Init = 7993, maximum number of calls to Secret by Query = 1
7 Correct 465 ms 4320 KB Output is correct - number of calls to Secret by Init = 7993, maximum number of calls to Secret by Query = 1
8 Correct 479 ms 4356 KB Output is correct - number of calls to Secret by Init = 7993, maximum number of calls to Secret by Query = 1
9 Correct 485 ms 4388 KB Output is correct - number of calls to Secret by Init = 7993, maximum number of calls to Secret by Query = 1
10 Correct 472 ms 4312 KB Output is correct - number of calls to Secret by Init = 7993, maximum number of calls to Secret by Query = 1