Submission #1187569

#	Time	Username	Problem	Language	Result	Execution time	Memory
1187569		ByeWorld	Digital Circuit (IOI22_circuit)	C++20	4 / 100	618 ms	14892 KiB

circuit

#include "circuit.h"
#include <bits/stdc++.h>
#define ll long long
#define pb push_back
#define fi first
#define se second
using namespace std;
const int MAXN = 3e5+10;
const ll MOD = 1e9+2022;
ll sum(ll a, ll b){ return (a+b)%MOD;}
void chsum(ll &a, ll b){ a = (a+b)%MOD;}
ll mul(ll a, ll b){ return (a*b)%MOD; }
void chmul(ll &a, ll b){ a = (a*b)%MOD;}
void chmn(auto &a, auto b){ a = min(a, b); }
void chmx(auto &a, auto b){ a = max(a, b); }

int n, m, par[MAXN], a[MAXN];
ll dp[MAXN][3];
vector<int> adj[MAXN];

void dfs(int nw, int pa){
	for(auto nx : adj[nw]){
		dfs(nx, nw);
	}
	if(adj[nw].size() == 0){
		dp[nw][0] = dp[nw][1] = 0;
		if(a[nw] == 1) dp[nw][1] = 1;
		else dp[nw][0] = 1;
		return;
	}
	int le = adj[nw][0], ri = adj[nw][1];
	dp[nw][1] = sum(sum( mul(dp[le][1],dp[ri][0]), mul(dp[le][0], dp[ri][1])), 
				mul(2, mul(dp[le][1], dp[ri][1] ) ) ); // nyala

	dp[nw][0] = sum(sum( mul(dp[le][1],dp[ri][0]), mul(dp[le][0], dp[ri][1])), 
				mul(2, mul(dp[le][0], dp[ri][0] ) ) ); // mati
}
void bd(int nw){
	if(nw == 0) return;

	if(adj[nw].size() == 0){
		dp[nw][0] = dp[nw][1] = 0;
		if(a[nw] == 1) dp[nw][1] = 1;
		else dp[nw][0] = 1;
		bd(par[nw]);

		return;
	}
	int le = adj[nw][0], ri = adj[nw][1];
	dp[nw][1] = sum(sum( mul(dp[le][1],dp[ri][0]), mul(dp[le][0], dp[ri][1])), 
				mul(2, mul(dp[le][1], dp[ri][1] ) ) ); // nyala

	dp[nw][0] = sum(sum( mul(dp[le][1],dp[ri][0]), mul(dp[le][0], dp[ri][1])), 
				mul(2, mul(dp[le][0], dp[ri][0] ) ) ); // mati
	
	bd(par[nw]);
}
void init(int N, int M, std::vector<int> P, std::vector<int> A) {
	n = N; m = M;
	for(int i=1; i<=n+m; i++){
		par[i] = P[i-1]+1;
		adj[par[i]].pb(i);
	}
	for(int i=1; i<=m; i++){
		a[i+n] = A[i-1];
	}
	dfs(1, 0);
}

int count_ways(int L, int R) {
	int l = L+1, r = R+1;
	for(int i=l; i<=r; i++) a[i] = 1-a[i];
	bd(l);
	
	// for(int i=1; i<=3; i++){
	// 	cout << dp[i][0] << ' ' << dp[i][1] << " xx\n";
	// }
	return (int)dp[1][1];
}

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