Submission #1135016

#	Time	Username	Problem	Language	Result	Execution time	Memory
1135016		Lemser	Digital Circuit (IOI22_circuit)	C++20	18 / 100	3054 ms	5028 KiB

circuit

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

using ll = long long;
using ull = unsigned long long;
using lld = long double;
using ii = pair<int,int>;
using pll = pair<ll, ll>;

using vi = vector<int>;
using vll = vector<ll>;
using vii = vector<ii>;
using vpll = vector<pll>;
using vlld = vector<lld>;

// #define endl '\n'
#define all(x) x.begin(),x.end()
#define lsb(x) x&(-x)
#define gcd(a,b) __gcd(a,b)
#define sz(x) (int)x.size()
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define fls cout.flush()

#define fore(i,l,r) for(auto i=l;i<r;i++)
#define fo(i,n) fore(i,0,n)
#define forex(i,r,l) for(auto i=r; i>=l;i--)
#define ffo(i,n) forex(i,n-1,0)

bool cmin(ll &a, ll b) { if(b<a){a=b;return 1;}return 0; }
bool cmax(ll &a, ll b) { if(b>a){a=b;return 1;}return 0; }
void valid(ll in) { cout<<((in)?"Yes\n":"No\n"); }
ll lcm(ll a, ll b) { return (a/gcd(a,b))*b; }
ll gauss(ll n) { return (n*(n+1))/2; }

const int N = 1e5 + 7;
const int mod = 1e9 + 2022;
vector<vll> graph;
ll p[N], a[N], n, m, dp[N][2];

/*
dp[u][0] = #acomodos de parametros de los del subarbol de u para que u
sea 0
dp[u][1] = lo mismo para que u sea 1
*/

void init(int N, int M, std::vector<int> P, std::vector<int> A) {
  n = N;
  m = M; 
  graph = vector<vll>(N+M);
  fore (i, 1, N+M) graph[P[i]].pb(i);
  fo (i, N+M) p[i] = P[i];
  fore (i, N, N+M) a[i] = A[i-N];
}

void dfs (ll u) {
  if (u >= n) {
    dp[u][a[u]] = 1;
    dp[u][a[u]^1] = 0;
    return;
  }
  vll dpu(sz(graph[u])+1, 0);
  // dpu[i] = #formas de acomodar parametros para que u tenga i hijos
  // con 1's
  dpu[0] = 1;
  for (ll v: graph[u]) {
    dfs(v);
    // mezclar
    vll ndp(sz(graph[u])+1, 0);
    fo (i, sz(graph[u])+1) {
      (ndp[i] += dpu[i] * dp[v][0]) %= mod;
      if (i < sz(graph[u]))
        (ndp[i+1] += dpu[i] * dp[v][1]) %= mod;
    }
    dpu = ndp;
  }
  fore (i, 1, sz(graph[u])+1) {
    // poner de parametro i en u
    fore (j, i, sz(graph[u])+1) (dp[u][1] += dpu[j]) %= mod;
    fo (j, i) (dp[u][0] += dpu[j]) %= mod;
  }
}

int count_ways(int l, int r) {
  fore (i, l, r+1) a[i] ^= 1;
  fo (i, n+m) dp[i][0] = dp[i][1] = 0;
  dfs(0);
  return dp[0][1];
}

• 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