제출 #1042478

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1042478		Alihan_8	디지털 회로 (IOI22_circuit)	C++17	0 / 100	433 ms	4964 KiB

circuit

#include "circuit.h"

#include <vector>

#include <bits/stdc++.h>

using namespace std;

#define ar array
#define pb push_back
#define ln '\n'

const int Mod = 1000002022;

vector <int> a, fa;

vector <vector<int>> adj;

vector <ar<int,2>> dp;

int n, m;

template <class F, class S>
void add(F &x, const S &y){
	x = (x + 0LL + y) % Mod;
	
	if ( x < 0 ) x += Mod; 
}

void init(int N, int M, std::vector<int> P, std::vector<int> A) {
	a = A, n = N, m = M, fa = P;
	
	adj.resize(n + m);
	
	for ( int i = 1; i < n + m; i++ ){
		adj[P[i]].pb(i);
	}
	
	vector <int> s(n + m);
	
	dp.resize(n + m);
	
	auto dfs = [&](auto dfs, int u) -> void{
		if ( u >= n ){
			dp[u][1] = a[u - n];
			dp[u][0] = a[u - n] ^ 1;
			
			return;
		}
		
		for ( auto &v: adj[u] ){
			dfs(dfs, v);
			
			s[u] += 1;
		}
		
		vector <int> f(s[u] + 1);
		
		f[0] = 1;
		
		for ( auto &v: adj[u] ){
			vector <int> nxt(s[u] + 1);
			
			for ( int j = 0; j <= s[u]; j++ ){
				add(nxt[j], f[j] * 1LL * dp[v][0] % Mod);
				
				if ( j > 0 ){
					add(nxt[j], f[j - 1] * 1LL * dp[v][1] % Mod);
				}
			}
			
			swap(f, nxt);
		}
		
		for ( int i = 0; i <= s[u]; i++ ){
			add(dp[u][1], i * 1LL * f[i] % Mod);
			add(dp[u][0], (s[u] - i) * 1LL * f[i] % Mod);
		}
	};
	
	dfs(dfs, 0);
}

int count_ways(int L, int R) {
	L -= n, R -= n;
	
	assert(L == R);
	
	for ( int i = L; i <= R; i++ ){
		a[i] ^= 1;
	}
	
	int u = L;
	
	while ( u != -1 ){
		if ( u >= n ){
			dp[u][1] = a[u - n];
			dp[u][0] = a[u - n] ^ 1;
		} else{		
			int s = (int)adj[u].size();
			
			vector <int> f(s + 1);
			
			f[0] = 1;
			
			for ( auto &v: adj[u] ){
				vector <int> nxt(s + 1);
				
				for ( int j = 0; j <= s; j++ ){
					add(nxt[j], f[j] * 1LL * dp[v][0] % Mod);
					
					if ( j > 0 ){
						add(nxt[j], f[j - 1] * 1LL * dp[v][1] % Mod);
					}
				}
				
				swap(f, nxt);
			}
			
			dp[u] = {0, 0};
			
			for ( int i = 0; i <= s; i++ ){
				add(dp[u][1], i * 1LL * f[i] % Mod);
				add(dp[u][0], (s - i) * 1LL * f[i] % Mod);
			}
		}
		
		u = fa[u];
	}
		
	return dp[0][1];
}

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