Submission #1074863

#	Time	Username	Problem	Language	Result	Execution time	Memory
1074863		c2zi6	디지털 회로 (IOI22_circuit)	C++17	18 / 100	3060 ms	8636 KiB

circuit

#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
#define ff first
#define ss second
#define pb push_back
#define all(a) (a).begin(), (a).end()
#define replr(i, a, b) for (int i = int(a); i <= int(b); ++i)
#define reprl(i, a, b) for (int i = int(a); i >= int(b); --i)
#define rep(i, n) for (int i = 0; i < int(n); ++i)
#define mkp(a, b) make_pair(a, b)
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef vector<PII> VPI;
typedef vector<VI> VVI;
typedef vector<VVI> VVVI;
typedef vector<VPI> VVPI;
typedef pair<ll, ll> PLL;
typedef vector<ll> VL;
typedef vector<PLL> VPL;
typedef vector<VL> VVL;
typedef vector<VVL> VVVL;
typedef vector<VPL> VVPL;
template<class T> T setmax(T& a, T b) {if (a < b) return a = b; return a;}
template<class T> T setmin(T& a, T b) {if (a < b) return a; return a = b;}
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
template<class T>
using indset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
#include "circuit.h"

const ll MOD = 1000002022;

bool test6;

int n, sz;
VVI gp;
VI val;

VL dpa; // sev
VL dpb; // spitak
VI depth;

VL contr;

void dfs2(int u = 0) {
    for (int v : gp[u]) {
        depth[v] = depth[u]+1;
        dfs2(v);
    }
}

void init(int N, int M, VI P, VI A) {
    n = N + M;
    sz = M;
    gp = VVI(n);
    rep(u, n) if (u) gp[P[u]].pb(u);
    val = VI(n);
    rep(u, M) val[u+N] = A[u];
    depth = VI(n);
    test6 = true;
    rep(u, n) if (gp[u].size() != 2) test6 = false;
    if (test6) {
        dfs2();
        contr = VL(n);
        VL twopow(n+1);
        twopow[0] = 1;
        replr(i, 1, n) twopow[i] = twopow[i-1]*2 % MOD;
        rep(u, n) contr[u] = twopow[N - depth[u]];
    }
}

VL subsets(VL& a, VL& b) {
    int n = a.size();
    VVL dp(n+1, VL(n+1));
    dp[0][0] = 1;
    replr(j, 1, n) dp[j][0] = dp[j-1][0]*b[j-1]%MOD;
    replr(i, 1, n) replr(j, 1, n) {
        dp[i][j] = dp[i-1][j-1]*a[i-1]%MOD + dp[i-1][j]*b[i-1]%MOD;
        dp[i][j] %= MOD;
    }
    VL ret(n+1);
    replr(i, 0, n) ret[i] = dp[n][i];
    return ret;
}

void dfs1(int u = 0) {
    for (int v : gp[u]) dfs1(v);
    dpa[u] = dpb[u] = 0;
    if (gp[u].size() == 0) {
        if (val[u]) dpa[u] = 1;
        else dpb[u] = 1;
        return;
    }
    VL a, b;
    for (int v : gp[u]) {
        a.pb(dpa[v]);
        b.pb(dpb[v]);
    }
    int c = gp[u].size();
    VL s = subsets(a, b);
    replr(i, 0, c) {
        dpa[u] += s[i]*i % MOD;
        dpa[u] %= MOD;
        dpb[u] += s[i]*(c-i) % MOD;
        dpb[u] %= MOD;
    }
}

int count_ways(int L, int R) {
    if (test6) {
        replr(i, L, R) val[i] = !val[i];
        ll ans = 0;
        rep(i, n) if (val[i]) ans += contr[i];
        return ans;
    } else {
        replr(i, L, R) val[i] = !val[i];
        dpa = dpb = VL(n);
        dfs1();
        return dpa[0];
    }
    return -1488;
}

• 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