Submission #58356

#	Time	Username	Problem	Language	Result	Execution time	Memory
58356		Benq	Xylophone (JOI18_xylophone)	C++14	100 / 100	129 ms	840 KiB

xylophone

#include "xylophone.h"

#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>

using namespace std;
using namespace __gnu_pbds;
 
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<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<cd> vcd;

template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;

#define FOR(i, a, b) for (int i=a; i<(b); i++)
#define F0R(i, a) for (int i=0; i<(a); i++)
#define FORd(i,a,b) for (int i = (b)-1; i >= a; i--)
#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)

#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()

const int MOD = 1000000007;
const ll INF = 1e18;
const int MX = 5001;

int d1[MX], d2[MX], res[MX];

/*
vi A = {0,5,3,4,1,2};

void answer(int a, int b) {
    cout << a << " " << b << "\n";
}

int query(int l, int r) {
    int mx = l; FOR(i,l,r+1) if (A[i] > A[mx]) mx = i;
    int mn = l; FOR(i,l,r+1) if (A[i] < A[mn]) mn = i;
    return A[mx]-A[mn];
}*/

void solve(int N) {
	FOR(i,1,N) d1[i] = query(i,i+1);
	FOR(i,1,N-1) d2[i] = query(i,i+2);
	FOR(i,1,N-1) if (max(0,max(d1[i],d1[i]+d1[i+1]))-min(0,min(d1[i],d1[i]+d1[i+1])) != d2[i]) d1[i+1] *= -1;
    /*FOR(i,1,N) cout << d1[i] << " ";
    cout << "\n";
    FOR(i,1,N-1) cout << d2[i] << " ";
    cout << "\n";*/
    
	FOR(i,2,N+1) res[i] = res[i-1]+d1[i-1];
	int lo = 1, hi = 1;
	FOR(i,2,N+1) {
		if (res[i] < res[lo]) lo = i;
		if (res[i] > res[hi]) hi = i;
	}
	if (lo > hi) {
		FOR(i,1,N+1) res[i] *= -1;
		swap(lo,hi);
	}

	int t = 1-res[lo];
	FOR(i,1,N+1) res[i] += t;

	for(int i = 1; i <= N; i++) {
		answer(i, res[i]);
	}

}

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