Submission #917172

#	Time	Username	Problem	Language	Result	Execution time	Memory
917172		GrindMachine	Xylophone (JOI18_xylophone)	C++17	47 / 100	51 ms	1968 KiB

xylophone

#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);
	};

	int pos1 = -1;
	rev(i,n-1,1){
		if(query(i,n) == n-1){
			pos1 = i;
			break;
		}
	}

	vector<int> a(n+5);
	a[pos1] = 1;
	int sign = 1;

	for(int i = pos1+1; i <= n; ++i){
		int d1 = f(i-1,i);
		a[i] = a[i-1]+d1*sign;
		if(i < n){
			int d2 = f(i,i+1);
			if(d1+d2 != f(i-1,i+1)){
				sign *= -1;
			}
		}
	}

	sign = 1;

	rev(i,pos1-1,1){
		int d1 = f(i,i+1);
		a[i] = a[i+1]+d1*sign;
		if(i > 1){
			int d2 = f(i-1,i);
			if(d1+d2 != f(i-1,i+1)){
				sign *= -1;
			}
		}
	}

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

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