Submission #307098

#	Time	Username	Problem	Language	Result	Execution time	Memory
307098		syy	Xylophone (JOI18_xylophone)	C++17	0 / 100	1 ms	388 KiB

xylophone

#include "xylophone.h"
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
#define FOR(i, a, b) for(int i = (int)a; i <= (int)b; i++)
#define DEC(i, a, b) for(int i = (int)a; i >= (int)b; i--)
typedef pair<int, int> pi;
typedef pair<pi, int> pii;
typedef pair<pi, pi> pipi;
#define f first
#define s second
typedef vector<int> vi;
typedef vector<pi> vpi;
typedef vector<pii> vpii;
#define pb push_back
#define pf push_front
#define all(v) v.begin(), v.end()
#define disc(v) sort(all(v)); v.resize(unique(all(v)) - v.begin());
#define INF (int) 1e9 + 100
#define LLINF (ll) 1e18
#define fastio ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define sandybridge __attribute__((optimize("Ofast"), target("arch=sandybridge")))
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());    //can be used by calling rng() or shuffle(A, A+n, rng)
inline ll rand(ll x, ll y) { ++y; return (rng() % (y-x)) + x; } //inclusivesss

int n, arr[5005], adj[5005], tri[5005];

void solve(int N) {
	n = N;
	FOR(i, 1, n-1) adj[i] = query(i, i+1);
	FOR(i, 1, n-2) tri[i] = query(i, i+2);
	arr[1] = 1, arr[2] = 1 + adj[1];
	FOR(i, 1, n-2) {
		if (tri[i] == adj[i] + adj[i+1]) {
			if (arr[i+1] > arr[i]) arr[i+2] = arr[i] - tri[i];
			else arr[i+2] = arr[i] + tri[i];
		} else {
			if (arr[i+1] > arr[i]) arr[i+2] = arr[i+1] - adj[i+1];
			else arr[i+2] = arr[i+1] + adj[i+1];
		}
	}
	pi s = pi(INF, 0), b = pi(0, 0);
	FOR(i, 1, n) s = min(s, pi(arr[i], i)), b = max(b, pi(arr[i], i));
	if (s.s > b.s) {
		arr[1] = 1, arr[2] = 1 - adj[1];
		FOR(i, 1, n-2) {
			if (tri[i] == adj[i] + adj[i+1]) {
				if (arr[i+1] > arr[i]) arr[i+2] = arr[i] - tri[i];
				else arr[i+2] = arr[i] + tri[i];
			} else {
				if (arr[i+1] > arr[i]) arr[i+2] = arr[i+1] - adj[i+1];
				else arr[i+2] = arr[i+1] + adj[i+1];
			}
		}
		s = pi(INF, 0), b = pi(0, 0);
		FOR(i, 1, n) s = min(s, pi(arr[i], i)), b = max(b, pi(arr[i], i));
	}
	FOR(i, 1, n) answer(i, arr[i] + n-b.f);
}

• Subtask 1: 0 / 11
• Subtask 2: 0 / 36
• Subtask 3: 0 / 53