제출 #212412

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
212412		mode149256	Hiring (IOI09_hiring)	C++14	7 / 100	209 ms	18572 KiB

hiring

/*input
3 40
10 1
10 2
10 3 
*/
#include <bits/stdc++.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';
}

struct ppl {
	ll S, Q;
	int i;
};

int main() {
	ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	int N;
	ll W;
	cin >> N >> W;
	vector<ppl> sk(N);

	for (int i = 0; i < N; ++i)
	{
		int a, b; cin >> a >> b;
		sk[i] = {a, b, i + 1};
	}

	sort(sk.begin(), sk.end(), [&](const ppl & a, const ppl & b) {
		return a.Q * b.S < b.Q * a.S;
	});

	ll sumQ = 0;

	int kiek = 0;
	int l = 1, r = 0;
	ll sumokejau = MOD;

	int j = 1;
	sumQ = sk[0].Q;

	for (int i = 0; i < N; ++i)
	{
		while (j < N and (sumQ + sk[j].Q)*sk[i].S <= W * sk[i].Q) {
			sumQ += sk[j].Q;
			j++;
		}

		if (sumQ * sk[i].S <= W * sk[i].Q) {
			if (j - i > kiek or (j - i == kiek and sumQ * sk[i].S * sk[l].Q < sumokejau * sk[i].Q)) {
				l = i;
				r = j - 1;
				kiek = j - i;
				sumokejau = sumQ * sk[i].S;
			}
		}

		sumQ -= sk[i].Q;
	}

	printf("%d\n", kiek);
	for (int i = l; i <= r; ++i)
	{
		printf("%d\n", sk[i].i);
	}
}

/* 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: 5 / 30
• Subtask 2: 2 / 28
• Subtask 3: 0 / 6
• Subtask 4: 0 / 6
• Subtask 5: 0 / 6
• Subtask 6: 0 / 6
• Subtask 7: 0 / 6
• Subtask 8: 0 / 6
• Subtask 9: 0 / 6