제출 #917183

#제출 시각아이디문제언어결과실행 시간메모리
917183GrindMachineXylophone (JOI18_xylophone)C++17
100 / 100
68 ms2440 KiB
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace std;
using namespace __gnu_pbds;

template<typename T> using Tree = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
typedef long long int ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;

#define fastio ios_base::sync_with_stdio(false); cin.tie(NULL)
#define pb push_back
#define endl '\n'
#define sz(a) (int)a.size()
#define setbits(x) __builtin_popcountll(x)
#define ff first
#define ss second
#define conts continue
#define ceil2(x,y) ((x+y-1)/(y))
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define yes cout << "Yes" << endl
#define no cout << "No" << endl

#define rep(i,n) for(int i = 0; i < n; ++i)
#define rep1(i,n) for(int i = 1; i <= n; ++i)
#define rev(i,s,e) for(int i = s; i >= e; --i)
#define trav(i,a) for(auto &i : a)

template<typename T>
void amin(T &a, T b) {
    a = min(a,b);
}

template<typename T>
void amax(T &a, T b) {
    a = max(a,b);
}

#ifdef LOCAL
#include "debug.h"
#else
#define debug(x) 42
#endif

/*

read some solutions a long time ago, remember some key ideas from there 

*/

const int MOD = 1e9 + 7;
const int N = 1e5 + 5;
const int inf1 = int(1e9) + 5;
const ll inf2 = ll(1e18) + 5;

#include "xylophone.h"

void solve(int n) {
	map<pii,int> mp;
	auto f = [&](int l, int r){
		pii px = {l,r};
		if(mp.count(px)) return mp[px];
		return mp[px] = query(l,r);
	};

	vector<int> d(n+5);
	d[1] = f(1,2);
	int sign = 1;

	for(int i = 2; i < n; ++i){
		d[i] = sign*f(i,i+1);
		if(abs(d[i-1])+abs(d[i]) != f(i-1,i+1)){
			sign *= -1;
			d[i] *= -1;
		}
	}

	sign = 1;

	rep1(iter,2){
		if(iter == 2){
			sign = -1;
		}

		rep1(i,n-1) d[i] *= sign;

		int mn_pref = inf1;
		int pref = 0;
		rep1(i,n){
			pref += d[i-1];
			amin(mn_pref,pref);
		}

		vector<int> a(n+5);
		a[1] = -mn_pref+1;
		for(int i = 2; i <= n; ++i){
			a[i] = a[i-1]+d[i-1];
		}

		bool ok = true;
		int pos1 = -1, posn = -1;

		rep1(i,n){
			if(a[i] < 1 or a[i] > n){
				ok = false;
				break;
			}

			if(a[i] == 1) pos1 = i;
			else if(a[i] == n) posn = i;
		}

		if(pos1 > posn) ok = false;
		if(!ok) conts;

		rep1(i,n){
			answer(i,a[i]);
		}

		return;
	}

	assert(0);
}
#Verdict Execution timeMemoryGrader output
Fetching results...
#Verdict Execution timeMemoryGrader output
Fetching results...
#Verdict Execution timeMemoryGrader output
Fetching results...