Submission #205421

#	Time	Username	Problem	Language	Result	Execution time	Memory
205421		mode149256	Rice Hub (IOI11_ricehub)	C++14	100 / 100	21 ms	1528 KiB

ricehub

/*input

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


int besthub(int R, int L, int X[], ll B)
{
	vl pref(R, 0);
	for (int i = 0; i < R; ++i)
		pref[i] = (i ? pref[i - 1] : 0) + X[i];

	auto sum = [&](int a, int b) -> ll {
		return pref[b] - (a ? pref[a - 1] : 0);
	};

	auto galima = [&](int K) -> bool {
		// printf("in galima, K = %d\n", K);
		for (int i = K - 1; i < R; ++i)
		{
			int l = i - K + 1;
			int r = i;
			int m = (l + r) / 2;

			if (K & 1) {
				ll mid = X[m];

				ll reik = mid * (K / 2) - sum(l, m - 1);
				reik += sum(m + 1, r) - mid * (K / 2);

				// printf("i = %d, mid = %lld, reik = %lld\n", i, mid, reik);
				if (reik <= B) return true;
			} else {
				// K = 4
				// K/2 = 2
				// 0 1 2 3
				ll mid = (X[m] + X[m + 1]);

				ll reik = mid * (K / 2) - 2 * sum(l, m);
				reik += 2 * sum(m + 1, r) - mid * (K / 2);

				if (reik <= 2 * B) return true;
			}
		}
		return false;
	};

	int l = 1;
	int h = R;
	int m;

	// for (int i = 0; i < R; ++i)
	// 	printf("%d ", X[i]);
	// printf("\n");
	// printf("B = %lld\n", B);
	while (l < h) {
		m = (l + h + 1) / 2;
		// printf("l = %d, m = %d, r = %d, galima = %d\n",
		//        l, m, h, int(galima(m)));
		if (galima(m))
			l = m;
		else
			h = m - 1;
	}

	return l;
}


/* 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: 17 / 17
• Subtask 2: 25 / 25
• Subtask 3: 26 / 26
• Subtask 4: 32 / 32