제출 #1302321

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1302321		nicolo_010	Horses (IOI15_horses)	C++20	0 / 100	151 ms	43476 KiB

horses

#include <bits/stdc++.h>
#include "horses.h"
using namespace std;
using ll = long long;
using pii = pair<ll, ll>;
const int MOD = 1e9+7;

const ll INF = 1e18;

struct SegTree {
	vector<ll> tree;
	SegTree() {
	}

	void init(int m) {
		tree.assign(4*m, 0);
	}

	void query(int p, int l, int r, int i, int x) {
		if (l > i || r < i) return;
		if (l==r) {
			tree[p] = x;
		}
		else {
			int m = (l+r)/2;
			query(2*p, l, m, i, x);
			query(2*p+1, m+1, r, i, x);
			ll i1 = tree[2*p];
			ll i2 = tree[2*p+1];
			tree[p] = (i1*i2) % MOD;
		}
	}
	ll rsq(int p, int l, int r, int i, int j) {
		if (l > j || r < i) return 1;
		if (i <= l && r <= j) {
			return tree[p];
		}
		else {
			int m = (l+r)/2;
			ll i1 = rsq(2*p, l, m, i, j);
			ll i2 = rsq(2*p+1, m+1, r, i, j);
			return (i1*i2)%MOD;
		}
	}

	void querymx(int p, int l, int r, int i, int x) {
		if (l > i || r < i) return;
		if (l == r) {
			tree[p] = x;
		}
		else {
			int m = (l+r)/2;
			querymx(2*p, l, m, i, x);
			querymx(2*p+1, m+1, r, i, x);
			int i1 = tree[2*p];
			int i2 = tree[2*p+1];
			tree[p] = max(i1, i2);
		}
	}

	int rmq(int p, int l, int r, int i, int j) {
		if (l > j || r < i) return 0;
		if (i <= l && r <= j) {
			return tree[p];
		}
		else {
			int m = (l+r)/2;
			int i1 = rmq(2*p, l, m, i, j);
			int i2 = rmq(2*p+1, m+1, r, i, j);
			return max(i1, i2);
		}
	}


};
SegTree st, mxtr;
vector<ll> x, y;
set<int> s;
int n;

ll compute() {
	auto it = s.end();
	for (int i=0; i<min(61, (int)s.size()); i++) {
		it--;
	}
	ll prod=1;
	bool moduled = false;
	int idx = *it;
	it++;
	//cout << *it << endl;
	ll y0 = (x[idx] == 1 ? mxtr.rmq(1, 0, n-1, idx, x[*it]-1) : y[idx]);
	for (it; it != s.end(); it++) {

		if (*it == n) break;
		auto nxt = next(it);
		ll y1 = (x[*it] == 1 ? mxtr.rmq(1, 0, n-1, *it, x[*nxt]-1) : y[*it]);
		prod *= x[*it];
		//cout << *it << " " << y0 << " " << idx << " " << y1 << " " << *nxt << endl;
		if (prod >= MOD) {
			prod %= MOD;
			moduled = true;
		}
		if (moduled) {
			idx = *it;
			y0 = y1;
			prod = 1;
			moduled = false;
		}
		else {
			if (prod*y1 >= y0) {
				idx = *it;
				prod = 1;
				y0 = y1;
				moduled = false;
			}
		}
	}
	prod = st.rsq(1, 0, n-1, 0, idx);
	return (prod*y0)%MOD;
}

int init(int N, int* a, int* b) {
	n = N;
	x.assign(n, 0);
	y.assign(n, 0);
	for (int i=0; i<n; i++) {
		x[i] = a[i];
		y[i] = b[i];
	}
	st.init(n);
	mxtr.init(n);
	for (int i=0; i<n; i++) {
		st.query(1, 0, n-1, i, x[i]);
		mxtr.querymx(1, 0, n-1, i, y[i]);
	}
	s.insert(0);
	s.insert(n);
	for (int i=1; i<n; i++) {
		if (x[i] == 1 && x[i-1] != 1) {
			s.insert(i);
		}
	}
	return compute();
}

int updateX(int pos, int val) {
	st.query(1, 0, n-1, pos, val);
	if (pos != 0 && x[pos] == 1 && val != 1) {
		auto it = s.upper_bound(pos);
		it--;
		s.insert(pos);
	}
	else if (x[pos] != 1 && val >= 2) {
		if (pos == 0 || x[pos-1] != 1) {
			s.insert(pos);
		}
	}
	x[pos] = val;
	return compute();
}

int updateY(int pos, int val) {
	y[pos] = val;
	mxtr.querymx(1, 0, n-1, pos, val);
	return compute();
}

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