제출 #837172

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
837172		tengiz05	디지털 회로 (IOI22_circuit)	C++17	18 / 100	3092 ms	3096 KiB

circuit

#include "circuit.h"

#include <vector>
#include <array>
#include <iostream>
using namespace std;
constexpr int P = 1E9 + 2022;
int norm(int x) {
    if (x < 0) x += P;
    if (x >= P) x -= P;
    return  x;
}
struct Z {
    int x;
    Z(int x = 0) : x(norm(x)) {}
    int val() { return x; }
    friend Z operator+(const Z &a, const Z &b) {
        return Z(a.x + b.x);
    }
    friend Z operator-(const Z &a, const Z &b) {
        return Z(a.x - b.x);
    }
    friend Z operator*(const Z &a, const Z &b) {
        return Z(int(a.x * 1LL * b.x % P));
    }
    void operator += (const Z &o) {
        *this = *this + o;
    }
};

int n, m;
vector<int> a;
vector<vector<int>> e;

void init(int N, int M, std::vector<int> P, std::vector<int> A) {
    n = N;
    m = M;
    e.resize(n);
    a = A;
    for (int i = 1; i < n + m; i++) {
        e[P[i]].push_back(i);
    }
}

int count_ways(int L, int R) {
    R++;
    L -= n;
    R -= n;
    for (int i = L; i < R; i++) {
        a[i] ^= 1;
    }
    // for (int i = 0; i < m; i++) cout << a[i];
    // cout << "\n";
    vector<array<Z, 2>> res(n + m);
    for (int i = 0; i < m; i++) {
        res[n + i][a[i]] = 1;
    }
    for (int i = n - 1; i >= 0; i--) {
        vector<Z> dp(e[i].size() + 1);
        dp[0] = 1;
        for (int j : e[i]) {
            vector<Z> ndp(e[i].size() + 1);
            for (int k = e[i].size(); k >= 0; k--) {
                if (k < int(e[i].size())) {
                    ndp[k + 1] += dp[k] * res[j][1];
                }
                ndp[k] += dp[k] * res[j][0];
            }
            dp = ndp;
        }
        // cout << "node " << i << ": ";
        // for (int i = 0; i < int(dp.size()); i++) {
            // cout << dp[i].val() << " ";
        // }
        // cout << "\n";
        int c = e[i].size();
        res[i][0] += dp[0] * c;
        for (int j = 1; j < int(dp.size()); j++) {
            res[i][1] += dp[j] * j;
            res[i][0] += dp[j] * (c - j);
        }
    }
    return res[0][1].val();
}

• 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