제출 #1079172

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1079172		Boas	디지털 회로 (IOI22_circuit)	C++17	62 / 100	3010 ms	47712 KiB

circuit

#include "circuit.h"

// #pragma GCC optimize("trapv")

#include <bits/stdc++.h>
using namespace std;

#define int long long

#define pb push_back
#define loop(x, i) for (int i = 0; i < x; i++)
#define rev(x, i) for (int i = (int)x - 1; i >= 0; i--)
#define ALL(x) begin(x), end(x)
#define sz(x) (int)x.size()

typedef signed i32;
typedef array<int, 2> ii;
typedef vector<ii> vii;
typedef vector<vii> vvii;
typedef vector<bool> vb;
typedef vector<int> vi;
typedef vector<i32> vi32;
typedef vector<vi> vvi;
typedef vector<vvi> vvvi;

constexpr int MOD = 1'000'002'022;

vvi children;
vi term01;
int n, m, treeN = 1;
vi vs;

struct val
{
  int v = 0, totalV = 0;
  void swap()
  {
    v = (totalV - v + MOD) % MOD;
  }
  val operator+(const val &b)
  {
    val a = *this, res;
    res.v = a.v + b.v;
    res.totalV = a.totalV + b.totalV;
    res.totalV %= MOD;
    res.v %= MOD;
    return res;
  }
};

vector<val> tree;
vb lazy;

void pull(int i)
{
  tree[i] = tree[2 * i] + tree[2 * i + 1];
}

void push(int i)
{
  if (!lazy[i])
    return;
  tree[i].swap();
  if (i < treeN)
  {
    lazy[2 * i].flip();
    lazy[2 * i + 1].flip();
  }
  lazy[i] = 0;
}

val operation(int i, int l, int r, int ql, int qr, int op)
{
  push(i);
  if (r < ql || qr < l)
    return {};
  if (ql <= l && r <= qr)
  {
    if (op == 1)
    {
      lazy[i] = 1;
      push(i);
    }
    return tree[i];
  }
  int mid = (l + r) / 2;
  val res = operation(2 * i, l, mid, ql, qr, op) +
            operation(2 * i + 1, mid + 1, r, ql, qr, op);
  pull(i);
  return res;
};

int calc01s(int i)
{
  if (i >= n)
  {
    return term01[i] = 1;
  }
  int all = sz(children[i]);
  for (int j : children[i])
  {
    int v = calc01s(j);
    all *= v;
    all %= MOD;
  }
  return term01[i] = all;
}

vi calc(int i)
{
  if (i >= n)
  {
    vs[i - n] = 1;
    return {i - n};
  }
  int c = sz(children[i]);
  vi prod, prefProd(c + 1, 1), sufProd(c + 1, 1);
  vvi leafs;
  for (int j : children[i])
  {
    leafs.pb(move(calc(j)));
    prod.pb(term01[j]);
  }
  loop(c, j)
  {
    prefProd[j + 1] = prefProd[j] * prod[j];
    prefProd[j + 1] %= MOD;
  }
  rev(c, j)
  {
    sufProd[j] = sufProd[j + 1] * prod[j];
    sufProd[j] %= MOD;
  }
  vi res;
  int largest = 0;
  loop(c, j)
  {
    if (sz(leafs[j]) > sz(leafs[largest]))
    {
      largest = j;
    }
  }
  res = leafs[largest];
  loop(c, j)
  {
    int mult = (prefProd[j] * sufProd[j + 1]) % MOD;
    for (int ix : leafs[j])
    {
      vs[ix] = (vs[ix] * mult) % MOD;
      if (j != largest)
        res.pb(ix);
    }
  }
  return res;
  // return move(res);
}

void init(i32 N, i32 M, vi32 P, vi32 A)
{
  n = N;
  m = M;
  while (treeN < m)
    treeN *= 2;
  tree.resize(2 * treeN);
  lazy.resize(2 * treeN);
  children = vvi(N);
  term01 = vi(N + M);
  vs = vi(M);
  loop(sz(P), i)
  {
    if (i == 0)
      continue;
    children[P[i]].pb(i);
  }
  calc01s(0);
  calc(0);
  loop(m, i)
  {
    tree[treeN + i].totalV = vs[i];
    if (A[i])
      tree[treeN + i].v = vs[i];
  }
  rev(treeN, i) pull(i);
}

i32 count_ways(i32 L, i32 R)
{
  operation(1, 0, treeN - 1, L - n, R - n, 1);
  return (i32)operation(1, 0, treeN - 1, 0, m - 1, 0).v;
}

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