oj.uz

답안 #250556

# 제출 시각 아이디 문제 언어 결과 실행 시간 메모리
250556 2020-07-18T10:55:12 Z mode149256 Candies (JOI18_candies) C++14
0 / 100
 3 ms 512 KB 
/*input
5
3
5
1
7
6
*/
#include <bits/stdc++.h>
using namespace std;

namespace my_template {
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 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';
}
}
using namespace my_template;

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

using lii = list<pl>::iterator;


struct comp {
	bool operator()(const lii &a, const lii &b) {
		return (*a).x > (*b).x or ((*a).x == (*b).x and (*a).y < (*b).y);
	}
};

int main() {
	ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	int N;
	cin >> N;
	list<pl> sk;
	set<lii, comp > did;
	for (int i = 0; i < N; ++i)
	{
		int a; cin >> a;
		sk.push_back({a, i});
		did.insert((--sk.end()));
	}
	// printf("%d\n", (*did.begin()).y);

	ll suma = 0;
	for (int i = 0; i < (N + 1) / 2; ++i)
	{
		auto it = (*did.begin());
		ll val = (*it).x;
		suma += (ll)val;
		did.erase(did.begin());

		ll kitas = -val;
		if (it != sk.begin()) {
			auto prv = prev(it);
			kitas += (*prv).x;
			did.erase(prv);
			sk.erase(prv);
		}

		if (next(it) != sk.end()) {
			auto nxt = next(it);
			kitas += (*nxt).x;
			did.erase(nxt);
			sk.erase(nxt);
		}

		(*it).x = kitas;
		did.insert(it);

		printf("%lld\n", suma);
	}

}

/* 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
0 / 8.0

# 결과 실행 시간 메모리 Grader output
1 Incorrect 3 ms 512 KB Output isn't correct
2 Halted 0 ms 0 KB -

Subtask #2
0 / 92.0

# 결과 실행 시간 메모리 Grader output
1 Incorrect 3 ms 512 KB Output isn't correct
2 Halted 0 ms 0 KB -