Submission #1138206

#	Time	Username	Problem	Language	Result	Execution time	Memory
1138206		Ekber_Ekber	Horses (IOI15_horses)	C++20	17 / 100	1597 ms	35400 KiB

horses

#include "horses.h"
#include <bits/stdc++.h>
#define ll long long
#define itn int
#define endl "\n"
#define ff first
#define ss second
#define pb push_back
#define ppb pop_back
#define ins insert
#define lb lower_bound
#define ub upper_bound
#define bs binary_search
#define count1 __builtin_popcount
#define all(v) v.begin(), v.end()
using namespace std;

struct point{
	ll ans, sum;
};

constexpr ll MAX = 5e+5 + 3, INF = 2e+9, MOD = 1e+9 + 7, K = log2(MAX);
ll n;
vector <ll> a, b;
vector <point> t(4 * MAX);

void build(ll v, ll tl, ll tr) {
	if (tl == tr) {
		t[v].ans = (a[tl] * b[tl]) % MOD;
		t[v].sum = a[tl];
		return;
	}
	ll tm = (tl + tr) / 2;
	build(v*2, tl, tm);
	build(v*2+1, tm+1, tr);
	t[v].sum = (t[v*2].sum * t[v*2+1].sum) % MOD;
	t[v].ans = max(t[v*2].ans, (t[v*2].sum * t[v*2+1].ans)) % MOD;
}

point find(ll v, ll tl, ll tr, ll l, ll r) {
	if (l > r) return {INF, INF};
	if (tl == l && tr == r) return t[v];
	ll tm = (tl + tr) / 2;
	point res1 = find(v*2, tl, tm, l, min(r, tm));
	point res2 = find(v*2+1, tm+1, tr, max(l, tm+1), r);
	point res;
	res.sum = (res1.sum * res2.sum) % MOD;
	res.ans = max(res1.ans, (res1.sum * res2.ans)) % MOD;
	return res;
}

void updatex(ll v, ll tl, ll tr, ll i, ll x) {
	if (tl == tr) {
		t[v].ans = (x * b[tl]) % MOD;
		t[v].sum = x;
		a[tl] = x;
		return;
	}
	ll tm = (tl + tr) / 2;
	if (i <= tm) {
		updatex(v*2, tl, tm, i, x);
	}
	else {
		updatex(v*2+1, tm+1, tr, i, x);
	}
	t[v].sum = (t[v*2].sum * t[v*2+1].sum) % MOD;
	t[v].ans = max(t[v*2].ans, (t[v*2].sum * t[v*2+1].ans)) % MOD;
}

void updatey(ll v, ll tl, ll tr, ll i, ll y) {
	if (tl == tr) {
		t[v].ans = (a[tl] * y) % MOD;
		b[tl] = y;
		return;
	}
	ll tm = (tl + tr) / 2;
	if (i <= tm) {
		updatey(v*2, tl, tm, i, y);
	}
	else {
		updatey(v*2+1, tm+1, tr, i, y);
	}
	t[v].sum = (t[v*2].sum * t[v*2+1].sum) % MOD;
	t[v].ans = max(t[v*2].ans, (t[v*2].sum * t[v*2+1].ans)) % MOD;
}

int init(int N, int X[], int Y[]) {
	cin >> N;
	for (int i=0; i < N; i++) {
		cin >> X[i];
	}
	for (int i=0; i < N; i++) {
		cin >> Y[i];
	}
	n = (ll)N;
	for (int i=0; i < N; i++) {
		a.pb(X[i]);
		b.pb(Y[i]);
	}
	build(1, 0, n-1);
	return t[1].ans;
}

int updateX(int pos, int val) {
	ll i = (ll)pos, x = (ll)val;
	updatex(1, 0, n-1, i, x);
	return t[1].ans;
}

int updateY(int pos, int val) {
	ll i = (ll)pos, y = (ll)val;
	updatey(1, 0, n-1, i, y);
	return t[1].ans;
}

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