Submission #204757

#	Time	Username	Problem	Language	Result	Execution time	Memory
204757		mode149256	Xylophone (JOI18_xylophone)	C++14	100 / 100	125 ms	456 KiB

xylophone

/*input
1
4
2
1
4
4
2

2, 1, 5, 3, 4
*/
#include <bits/stdc++.h>
#include "xylophone.h"
using namespace std;

typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;

typedef pair<int, int> pi;
typedef pair<ll, ll> pl;
typedef pair<ld, ld> pd;

typedef vector<int> vi;
typedef vector<vi> vii;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<vl> vll;
typedef vector<pi> vpi;
typedef vector<vpi> vpii;
typedef vector<pl> vpl;
typedef vector<cd> vcd;
typedef vector<pd> vpd;
typedef vector<bool> vb;
typedef vector<vb> vbb;
typedef std::string str;
typedef std::vector<str> vs;

#define x first
#define y second
#define debug(...) cout<<"["<<#__VA_ARGS__<<": "<<__VA_ARGS__<<"]\n"

const int MOD = 1000000007;
const ll INF = std::numeric_limits<ll>::max();
const int MX = 100101;
const ld PI = 3.14159265358979323846264338327950288419716939937510582097494L;

template<typename T>
pair<T, T> operator+(const pair<T, T> &a, const pair<T, T> &b) { return pair<T, T>(a.x + b.x, a.y + b.y); }
template<typename T>
pair<T, T> operator-(const pair<T, T> &a, const pair<T, T> &b) { return pair<T, T>(a.x - b.x, a.y - b.y); }
template<typename T>
T operator*(const pair<T, T> &a, const pair<T, T> &b) { return (a.x * b.x + a.y * b.y); }
template<typename T>
T operator^(const pair<T, T> &a, const pair<T, T> &b) { return (a.x * b.y - a.y * b.x); }

template<typename T>
void print(vector<T> vec, string name = "") {
	cout << name;
	for (auto u : vec)
		cout << u << ' ';
	cout << '\n';
}


// int query(int s, int t) {
// 	int ret;
// 	cin >> ret;
// 	return ret;
// }

// vi final(MX);
// void answer(int i, int a) {
// 	final[i - 1] = a;
// }

void solve(int N) {
	vi kiek(N - 1);
	for (int i = 0; i < N - 1; ++i)
		kiek[i] = query(i + 1, i + 2);

	vi ats(N);

	vi tri(N);
	ats[0] = 0;
	ats[1] = kiek[0];

	bool inc = true;

	for (int i = 0; i < N - 2; ++i)
	{
		tri[i] = query(i + 1, i + 3);
		if (tri[i] == kiek[i] + kiek[i + 1]) {
			if (inc) {
				// ats[i + 1] = ats[i] + kiek[i];
				ats[i + 2] = ats[i + 1] + kiek[i + 1];
			} else {
				// ats[i + 1] = ats[i] - kiek[i];
				ats[i + 2] = ats[i + 1] - kiek[i + 1];
			}
		} else {
			if (inc) {
				// ats[i + 1] = ats[i] + kiek[i];
				ats[i + 2] = ats[i + 1] - kiek[i + 1];
			} else {
				// ats[i + 1] = ats[i] - kiek[i];
				ats[i + 2] = ats[i + 1] + kiek[i + 1];
			}
			inc = !inc;
		}
	}

	int diff = 1 - *min_element(ats.begin(), ats.end());
	for (auto &u : ats)
		u += diff;

	int maz = int(min_element(ats.begin(), ats.end()) - ats.begin());
	int did = int(max_element(ats.begin(), ats.end()) - ats.begin());

	if (did < maz)
	{
		for (auto &u : ats)
			u = N - u + 1;
	}

	for (int i = 0; i < N; ++i)
		answer(i + 1, ats[i]);
	// print(ats);
}


/* Look for:
* special cases (n=1?)
* overflow (ll vs int?)
* the exact constraints (multiple sets are too slow for n=10^6 :( )
* array bounds
*/

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