제출 #222257

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
222257		mode149256	말 (IOI15_horses)	C++14	17 / 100	868 ms	130592 KiB

horses

/*input

*/
#include "horses.h"
#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 ll 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';
}

vl x, y;
int n;
struct maxNode {
	int l, r;
	pl did;
	maxNode *left = nullptr;
	maxNode *right = nullptr;

	maxNode(int a, int b): l(a), r(b) {
		if (l != r) {
			left = new maxNode(l, (l + r) / 2);
			right = new maxNode((l + r) / 2 + 1, r);
			did = max(left->did, right->did);
		} else
			did = {y[l], l};
	}

	void upd(int pos, int val) {
		if (l == r) {
			did.x = val;
			return;
		} else if (pos <= (l + r) / 2)
			left->upd(pos, val);
		else
			right->upd(pos, val);

		did = max(left->did, right->did);
	}

	pl gauk(int a, int b) {
		if (r < a or b < l) return pi(-MOD, -MOD);
		else if (a <= l and r <= b) return did;
		else
			return max(left->gauk(a, b), right->gauk(a, b));
	}
};

struct prodNode {
	int l, r;
	ll prod;
	prodNode *left = nullptr;
	prodNode *right = nullptr;

	prodNode(int a, int b): l(a), r(b) {
		if (l != r) {
			left = new prodNode(l, (l + r) / 2);
			right = new prodNode((l + r) / 2 + 1, r);
			prod = (left->prod * right->prod) % MOD;
		} else
			prod = x[l];
	}

	void upd(int pos, ll val) {
		if (l == r) {
			prod = val;
			return;
		} else if (pos <= (l + r) / 2)
			left->upd(pos, val);
		else
			right->upd(pos, val);

		prod = (left->prod * right->prod) % MOD;
	}

	ll gauk(int a, int b) {
		if (r < a or b < l) return 1;
		else if (a <= l and r <= b) return prod;
		else
			return (left->gauk(a, b) * right->gauk(a, b)) % MOD;
	}
};

maxNode *yg;
prodNode *xs;
set<int, greater<int> > indes;

int compute() {
	if (indes.empty()) {
		pi ans = yg->did;
		return ans.x;
	}
	int last = n;

	ll currProd = 1;
	int currBest = -1;

	int cn = 0;
	for (auto it = indes.begin(); it != indes.end() and cn < 60; it++, cn++) {
		pi dabY = yg->gauk(*it, last - 1);

		// printf("ind = %d, dabY = %d %d\n", *it, dabY.x, dabY.y);
		if (currBest == -1 or dabY.x > currProd * y[currBest]) {
			currBest = dabY.y;
			currProd = x[*it];
		} else {
			currProd *= x[*it];
			if (currProd > MOD) break;
		}
		// printf("currBest = %lld, currProd  = %lld\n", currBest, currProd);

		last = *it;
	}
	if (last != 0 and currProd <= MOD) {
		pi dabY = yg->gauk(0, last - 1);
		if (dabY.x > currProd * y[currBest]) {
			currBest = dabY.y;
			currProd = x[dabY.y];
		}
	}

	return (int)((ll)xs->gauk(0, currBest) * y[currBest]) % MOD;
}

int init(int N, int X[], int Y[]) {
	n = N;
	for (int i = 0; i < N; ++i)
	{
		x.emplace_back(X[i]);
		y.emplace_back(Y[i]);
		if (X[i] != 1) indes.insert(i);
	}
	yg = new maxNode(0, N - 1);
	xs = new prodNode(0, N - 1);

	return compute();
}


int updateX(int pos, int val) {
	if (val != 1) indes.insert(pos);
	else if (val == 1) indes.erase(pos);
	xs->upd(pos, val);
	return compute();
}

int updateY(int pos, int val) {
	yg->upd(pos, val);
	return compute();
}

• Subtask 1: 17 / 17
• Subtask 2: 0 / 17
• Subtask 3: 0 / 20
• Subtask 4: 0 / 23
• Subtask 5: 0 / 23