제출 #1079094

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1079094		Boas	디지털 회로 (IOI22_circuit)	C++17	18 / 100	3069 ms	4932 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;
vb a;
vi c_i;

/*struct val
{
  void swap()
  {
  }
  val operator+(const val &b)
  {
    val a = *this, res;
    return res;
  }
};

vector<val> tree;
vb lazy;

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

void push(int i)
{
}

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 m = (l + r) / 2;
  val res = operation(2 * i, l, m, ql, qr, op) +
            operation(2 * i + 1, m + 1, r, ql, qr, op);
  pull(i);
  return tree[i];
};*/

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;
}

vii calc(int i)
{
  if (i >= n)
  {
    return {{i, 1}};
  }
  int c = sz(children[i]);
  vi ones;
  vi prod, prefProd(c + 1, 1), sufProd(c + 1, 1);
  vvii vs;
  for (int j : children[i])
  {
    vii cur = calc(j);
    vs.pb(cur);
    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;
  }
  vii res;
  loop(c, j)
  {
    vii cur = vs[j];
    int mult = (prefProd[j] * sufProd[j + 1]) % MOD;
    for (auto [ix, v] : cur)
    {
      res.pb({ix, (v * mult) % MOD});
    }
  }
  return res;
}

void init(i32 N, i32 M, vi32 P, vi32 A)
{
  n = N;
  m = M;
  children = vvi(N);
  term01 = vi(N + M);
  a = vb(M);
  loop(sz(P), i)
  {
    if (i == 0)
      continue;
    children[P[i]].pb(i);
  }
  loop(sz(A), i)
  {
    a[i] = A[i];
  }
  calc01s(0);
  vii terms = calc(0);
  c_i = vi(m);
  for (auto [i, v] : terms)
    c_i[i - n] = v;
}

i32 count_ways(i32 L, i32 R)
{
  for (int i = L; i <= R; i++)
  {
    a[i - n].flip();
  }
  int res = 0;
  loop(m, i)
  {
    res += c_i[i] * a[i];
    res %= MOD;
  }
  return (i32)res;
}

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