제출 #807523

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
807523		alex_2008	디지털 회로 (IOI22_circuit)	C++17	4 / 100	814 ms	4644 KiB

circuit

#include "circuit.h"
#include <cmath>
#include <algorithm>
#include <vector>
#include <iostream>
using namespace std;
typedef long long ll;
int NN, MM;
const int N = 2e5 + 10, P = 1000002022;
int p[N];
int lazy[4 * N];
int a[N];
pair <ll, ll> tree[4 * N];
bool ch = true;
void pushh(int v, int tl, int tr) {
	if (lazy[v]) {
		lazy[v] = 0;
		if (tl != tr) {
			lazy[2 * v] = (lazy[2 * v] + 1) % 2;
			lazy[2 * v + 1] = (lazy[2 * v + 1] + 1) % 2;
			swap(tree[2 * v].first, tree[2 * v].second);
			swap(tree[2 * v + 1].first, tree[2 * v + 1].second);
		}
	}
}
void build_tree(int v, int tl, int tr) {
	if (tl == tr) {
		if (a[tl] == 1) tree[v] = { 1, 0 };
		else tree[v] = { 0, 1 };
	}
	else {
		int tm = (tl + tr) / 2;
		build_tree(2 * v, tl, tm);
		build_tree(2 * v + 1, tm + 1, tr);
		ll a = tree[2 * v].first, b = tree[2 * v].second, c = tree[2 * v + 1].first, d = tree[2 * v + 1].second;
		tree[v].first = 2 * a * c + a * d + b * c;
		tree[v].second = 2 * b * d + a * d + b * c;
		tree[v].first %= P;
		tree[v].second %= P;
	}
}
void update(int v, int tl, int tr, int l, int r) {
	if (tl > r || tr < l) return;
	if (tl >= l && tr <= r) {
		lazy[v] = (lazy[v] + 1) % 2;
		swap(tree[v].first, tree[v].second);
		return;
	}
	int tm = (tl + tr) / 2;
	update(2 * v, tl, tm, l, r);
	update(2 * v + 1, tm + 1, tr, l, r);
	ll a = tree[2 * v].first, b = tree[2 * v].second, c = tree[2 * v + 1].first, d = tree[2 * v + 1].second;
	tree[v].first = 2 * a * c + a * d + b * c;
	tree[v].second = 2 * b * d + a * d + b * c;
	tree[v].first %= P;
	tree[v].second %= P;
}
void init(int N, int M, std::vector<int> P, std::vector<int> A) {
	NN = N, MM = M;
	for (int i = 0; i < N + M - 1; i++) {
		p[i] = P[i];
		if (i > 0 && p[i] != (i - 1) / 2) ch = false;
	}
	for (int i = 0; i < M; i++) {
		a[i] = A[i];
	}
	if (M != N + 1) ch = false;
	int u = log2(M);
	if ((1 << u) != M) ch = false;
	build_tree(1, 0, M - 1);
}
int count_ways(int L, int R) {
	update(1, 0, MM - 1, L - NN, R - NN);
	return tree[1].first;
}

• Subtask 1: 0 / 2
• Subtask 2: 0 / 7
• Subtask 3: 0 / 9
• Subtask 4: 4 / 4
• Subtask 5: 0 / 12
• Subtask 6: 0 / 27
• Subtask 7: 0 / 28
• Subtask 8: 0 / 11