Submission #798024

#	Time	Username	Problem	Language	Result	Execution time	Memory
798024		gesgha	Digital Circuit (IOI22_circuit)	C++17	61 / 100	3050 ms	28968 KiB

circuit

#include "circuit.h"
#include <bits/stdc++.h>
 
#define fr(i, a, b) for(int i = a; i <= b; i++)
#define rf(i, a, b) for(int i = a; i >= b; i--)
#define fe(x, y) for (auto& x : y)
 
#define fi first
#define se second
#define pb push_back
 
#define all(x) x.begin(), x.end()
#define pw(x) (1LL << (x))
#define sz(x) (int)x.size()
 
using namespace std;
 
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
#define fbo find_by_order
#define ook order_of_key
template <typename T>
using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
 
template <typename T>
using ve = vector <T>;
 
template <typename T>
bool umx (T& a, T b) { return a < b ? a = b, 1 : 0; }
 
template <typename T>
bool umn (T& a, T b) { return a > b ? a = b, 1 : 0; }
 
using ll = long long;
using ld = long double;
using pll = pair <ll, ll>;
using pii = pair <int, int>;
using ull = unsigned long long;
 
const int oo = 1e9;
const ll OO = 1e18;
const int N = 2e5 + 10;
const int M = 5e3 + 100;
const int mod = 1e9 + 2022;

int a[N];
int dp[N][2];
int dp1[N][2];
ve <int> G[N];
bool ok = 1;


int n, m;

int add(int a, int b) {
  return a + b < mod ? a + b : a + b - mod;
}

int sub(int a, int b) {
  return a - b >= 0 ? a - b : a - b + mod;
}

int mul (int a, int b) {
  return 1LL * a * b % mod;
}
int bpow(int a, int n) {
  int res = 1;
  for (; n; n >>= 1, a = mul(a, a)) if(n&1) res = mul(res, a);
  return res;
}
int tin[N];
int tout[N];
int tim;
int mn[N];
int mx[N];
bool isupper(int u, int v) {
  return tin[u] <= tin[v] && tout[v] <= tout[u];
}

int dead[N];

int S[4 * N];
int res[4 * N];
void build(int v, int tl, int tr) {
  if (tl == tr) {
    S[v] = bpow(2, n - dead[tl + n]);
    if (a[tl + n]) res[v] = bpow(2, n - dead[tl + n]);
    // cout << tl << " - " << res[v] << "\n";
    return;
  }
  int tm = (tl + tr) >> 1;
  build(v << 1, tl, tm);
  build(v << 1 | 1, tm + 1, tr);
  S[v] = add(S[v << 1], S[v << 1 | 1]);
  res[v] = add(res[v << 1], res[v << 1 | 1]);
}

int rev[N * 4];

void push(int v) {
  if (!rev[v]) return;
  for (auto to : {(v << 1), (v << 1 | 1)}) {
    rev[to] ^= 1;
    res[to] = sub(S[to], res[to]);
  }
  rev[v] = 0;
}

void upd(int v, int tl, int tr, int L, int R) {
  // cout << tl << " " << tr << " " << L << " " << R << "\n";
  if (L > R) return;
  if (tl == L && tr == R) {
    res[v] = sub(S[v], res[v]);
    rev[v] ^= 1;
    return;
  }
  int tm = (tl + tr) >> 1;
  push(v);
  upd(v << 1, tl, tm, L, min(R, tm));
  upd(v << 1 | 1, tm + 1, tr, max(L, tm + 1), R);
  res[v] = add(res[v << 1], res[v << 1 | 1]);
}

void dfs(int u) {
  tin[u] = tim++;
  if (!sz(G[u])) {
    mn[u] = u;
    mx[u] = u;
    if(a[u]) {
      dp[u][0] = 0;
      dp[u][1] = 1;
    } else {
      dp[u][0] = 1;
      dp[u][1] = 0;
    }
    return;
  }
  mx[u] = -oo;
  mn[u] = oo;
  for (auto to : G[u]) {
    dead[to] = dead[u] + 1;
    dfs(to);
    mx[u] = max(mx[u], mx[to]);
    mn[u] = min(mn[u], mn[to]);
  }
  ve <ve <int>> d(sz(G[u]) + 1);

  fe (x, d) x.resize(sz(G[u]) + 1);
  d[0][0] = 1;
  for (int i = 0; i < sz(G[u]); i++) {
    int to = G[u][i];
    for (int cnt_al = 0; cnt_al <= i; cnt_al++) {
      d[i + 1][cnt_al + 1] = add(d[i + 1][cnt_al + 1], mul(d[i][cnt_al], dp[to][1]));
      d[i + 1][cnt_al] = add(d[i + 1][cnt_al], mul(d[i][cnt_al], dp[to][0]));
    }
  }
  dp[u][0] = dp[u][1] = 0;
  for (int cnt_al = 0; cnt_al <= sz(G[u]); cnt_al++) {
    dp[u][1] = add(dp[u][1], mul(d[sz(G[u])][cnt_al], max(0, cnt_al)));
    dp[u][0] = add(dp[u][0], mul(d[sz(G[u])][cnt_al], sz(G[u]) - cnt_al));
  }
  tout[u] = tim;
}




void init(int N, int M, vector<int> P, vector<int> A) {
  n = N;
  m = M;
  for (int i = 0; i < M; i++) a[i + N] = A[i];
  for (int i = 1; i < N + M; i++) G[P[i]].pb(i);
  for (int i = 0; i < N; i++) ok &= bool(sz(G[i]) == 2);
  dfs(0);

  if (ok) {
      build(1, 0, m - 1);
  }
}

int count_ways(int L, int R) {

  if (!ok) {
    for (int i = L; i <= R; i++) {
      a[i] = 1 - a[i];
    }
    dfs(0);
  } else {
    L -= n;
    R -= n;
    upd(1, 0, m - 1, L, R);
    return res[1];
  }
  
  return dp[0][1];
}

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