Submission #362009

#	Time	Username	Problem	Language	Result	Execution time	Memory
362009		Galebickosikasa	Semiexpress (JOI17_semiexpress)	C++17	100 / 100	1 ms	492 KiB

semiexpress

// #pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
// #pragma GCC target("sse,sse2,sse3,ssse3,sse4,sse4.1,sse4.2,popcnt,abm,mmx,avx")
 
#include <iostream>
#include <vector>
#include <cmath>
#include <algorithm>
#include <unordered_map>
#include <set>
#include <map>
#include <queue>
#include <random>
#include <chrono>

#define fi first
#define se second
#define pb push_back
#define ll long long
#define ld long double
#define hm unordered_map 
#define pii pair<int, int>
#define sz(a) (int)a.size()
#define all(a) a.begin(), a.end()
#define cinv(v) for (auto& x: v) cin >> x
#define fr(i, n) for (int i = 0; i < n; ++i)
#define fl(i, l, n) for (int i = l; i < n; ++i)

#define int ll

template <typename T1, typename T2> inline bool chkmin(T1 &x, const T2 &y) {if (x > y) {x = y; return 1;} return 0;}
template <typename T1, typename T2> inline bool chkmax(T1 &x, const T2 &y) {if (x < y) {x = y; return 1;} return 0;}

using namespace std;

#ifdef LOCAL
	#define dbg(x) cerr << #x << " : " << x << '\n'
#else 
	#define dbg(x)
#endif

//tg: @runningcherry
 
template <typename T1, typename T2> ostream& operator << (ostream& out, const pair<T1, T2>& v) {
	out << v.fi << ", " << v.se;
	return out;
}

template<typename T> ostream& operator << (ostream& out, const vector<T>& v) {
	for (auto& x: v) out << x << " ";
	return out;
}

template <typename T1, typename T2> istream& operator >> (istream& in, pair<T1, T2>& a) {
	in >> a.fi >> a.se;
	return in;
}

const ll inf = (ll) 2e9;
const ld pi = asin (1) * 2;
const ld eps = 1e-8;
const ll mod = (ll)1e9 + 7;
const ll ns = 97;

const int maxn = 2e5 + 20;

mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());

struct gay {
	int i, t, x;

	inline bool operator < (const gay& other) const {
		return x > other.x;
	}
};

signed main () {
	ios_base::sync_with_stdio(false);
	cin.tie(nullptr);
	cout.tie(nullptr);
	int n, m, k, a, b, c, t;
	cin >> n >> m >> k >> a >> b >> c >> t;
	vector<int> goo (m);
	cinv (goo);
	multiset <gay> kek;
	for (auto& x: goo) --x;
	int res = b * goo.back () <= t;
	fr (i, m - 1) {
		int l = goo[i], r = min (goo[i + 1] - 1, goo[i] + (t - b * goo[i]) / a);
		if (b * goo[i] <= t) {
			dbg (l);
			dbg (r);
			dbg (goo[i]);
			dbg (r - l + 1);
			res += r - l + 1;
		}
		if (r + 1 != goo[i + 1] && (t - b * goo[i] - (r + 1 - goo[i]) * c) >= 0) 
			kek.insert ({r + 1, b * goo[i] + (r + 1 - goo[i]) * c, min (goo[i + 1] - 1, r + 1 + (t - b * goo[i] - (r + 1 - goo[i]) * c) / a) - r});
	}
	fr (i, k - m) {
		if (kek.empty ()) break;
		auto ptt = *kek.begin ();
		kek.erase (kek.begin ());
		dbg (ptt.i);
		res += ptt.x;
		int r = ptt.i + ptt.x;
		auto it = lower_bound (all (goo), r) - goo.begin ();
		if (ptt.t + (r - ptt.i) * c <= t && r < goo[it]) 
			kek.insert ({r, ptt.t + (r - ptt.i) * c, min (goo[it] - 1, r + (t - ptt.t - (r - ptt.i) * c) / a) - r + 1});
	}
	cout << res - 1 << '\n';




}

• Subtask 1: 18 / 18
• Subtask 2: 30 / 30
• Subtask 3: 52 / 52