이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#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 n, m;
int add(int a, int b) {
return a + b < mod ? a + b : a + b - mod;
}
int mul (int a, int b) {
return 1LL * a * b % mod;
}
ve <int> G[N];
void calc(int u) {
if (!sz(G[u])) {
if(a[u]) {
dp[u][0] = 0;
dp[u][1] = 1;
} else {
dp[u][0] = 1;
dp[u][1] = 0;
}
return;
}
for (auto to : G[u]) calc(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));
}
}
void init(int N, int M, std::vector<int> P, std::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);
}
int count_ways(int L, int R) {
for (int i = L; i <= R; i++) a[i] = 1 - a[i];
calc(0);
return dp[0][1];
}
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