Submission #669798

#	Time	Username	Problem	Language	Result	Execution time	Memory
669798		ghostwriter	Semiexpress (JOI17_semiexpress)	C++14	100 / 100	15 ms	8792 KiB

semiexpress

#include <bits/stdc++.h>
using namespace std;
#ifdef LOCAL
#include <debug.h>
#else
#define debug(...)
#endif
#define ft front
#define bk back
#define st first
#define nd second
#define ins insert
#define ers erase
#define pb push_back
#define pf push_front
#define _pb pop_back
#define _pf pop_front
#define lb lower_bound
#define ub upper_bound
#define mtp make_tuple
#define bg begin
#define ed end
#define all(x) (x).bg(), (x).ed()
#define sz(x) (int)(x).size()
typedef long long ll; typedef unsigned long long ull;
typedef double db; typedef long double ldb;
typedef pair<int, int> pi; typedef pair<ll, ll> pll;
typedef vector<int> vi; typedef vector<ll> vll; typedef vector<pi> vpi; typedef vector<pll> vpll;
typedef string str;
#define FOR(i, l, r) for (int i = (l); i <= (r); ++i)
#define FOS(i, r, l) for (int i = (r); i >= (l); --i)
#define FRN(i, n) for (int i = 0; i < (n); ++i)
#define FSN(i, n) for (int i = (n) - 1; i >= 0; --i)
#define EACH(i, x) for (auto &i : (x))
#define WHILE while
template<typename T> T gcd(T a, T b) { WHILE(b) { a = a % b; swap(a, b); } return a; }
template<typename T> T lcm(T a, T b) { return a / gcd(a, b) * b; }
#define file "TEST"
mt19937 rd(chrono::steady_clock::now().time_since_epoch().count());
ll rand(ll l, ll r) { return uniform_int_distribution<ll>(l, r)(rd); }
/*
----------------------------------------------------------------
    END OF TEMPLATE
----------------------------------------------------------------
    Tran The Bao - ghostwriter
    Training for VOI23 gold medal
----------------------------------------------------------------
    GOAT
----------------------------------------------------------------
*/
const int oo = 2e9 + 5;
const int M = 3005;
int n, m, k, a, b, c, s[M], d[M][M], pos[M], rs = 0;
ll t;
void input() {
	cin >> n >> m >> k;
    cin >> a >> b >> c;
    cin >> t;
    FOR(i, 1, m) cin >> s[i];
}
void solve() {
	s[m + 1] = s[m] + 1;
	FOR(i, 1, m) {
		int cur = s[i];
		ll minn = 1LL * (s[i] - 1) * b;
		FOR(j, 1, k - m + 2) {
			if (minn > t || cur == s[i + 1]) break;
			ll pos = min((t - minn) / a + cur + 1, 1LL * s[i + 1]);
			d[i][j] = pos - cur;
			minn += 1LL * (pos - cur) * c;
			cur = pos;
		}
		rs += d[i][1];
		pos[i] = 0;
	}
	FOR(i, 1, k - m) {
		pi maxx = {0, 0};
		FOR(j, 1, m) {
			if (1LL * (s[j] - 1) * b > t || !d[j][pos[j] + 2]) continue;
			maxx = max(maxx, {d[j][pos[j] + 2], j});
		}
		if (!maxx.st) break;
		rs += maxx.st;
		++pos[maxx.nd];
	}
	cout << rs - 1 << '\n';
}
signed main() {
    ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    // freopen(file".inp", "r", stdin);
    // freopen(file".out", "w", stdout);
    input();
    solve();
    return 0;
}
/*
----------------------------------------------------------------
From Benq:
    stuff you should look for
        * int overflow, array bounds
        * special cases (n=1?)
        * do smth instead of nothing and stay organized
        * WRITE STUFF DOWN
        * DON'T GET STUCK ON ONE APPROACH
----------------------------------------------------------------
*/

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