Submission #219578

#	Time	Username	Problem	Language	Result	Execution time	Memory
219578		mode149256	Boxes with souvenirs (IOI15_boxes)	C++14	10 / 100	7 ms	384 KiB

boxes

/*input

*/
#include <bits/stdc++.h>
#include "boxes.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';
}


long long delivery(int N, int K, int L, int p[]) {
	vl pos(N);
	for (int i = 0; i < N; ++i)
		pos[i] = p[i];

	vl isKaires(N);
	vl isDesines(N);

	// isKaires[0] = 0;
	// isDesines[N] = 0;

	auto disK = [&](int i) {
		return pos[i] + min(pos[i], L - pos[i]);
	};

	auto disD = [&](int i) {
		return L - pos[i] + min(pos[i], L - pos[i]);
	};

	{
		for (int i = 0; i < N; ++i)
			isKaires[i] = (i - K < 0 ? 0 : isKaires[i - K]) + disK(i);

		for (int i = N - 1; i >= 0; i--)
			isDesines[i] = (i + K >= N ? 0 : isDesines[i - K]) + disD(i);
	}

	ll ats = 1e17;
	for (int i = -1; i < N; ++i)
	{
		ats = min(ats, (i >= 0 ? isKaires[i] : 0) + (i + 1 < N ? isDesines[i + 1] : 0));
		// if (i) ats = min(ats, isKaires[i - 1] + isDesines[N - i]);
	}

	// printf("%lld\n", ats);
	return ats;
}

• Subtask 1: 0 / 10
• Subtask 2: 10 / 10
• Subtask 3: 0 / 15
• Subtask 4: 0 / 15
• Subtask 5: 0 / 20
• Subtask 6: 0 / 30