제출 #936583

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
936583		GrindMachine	Digital Circuit (IOI22_circuit)	C++17	100 / 100	824 ms	49720 KiB

circuit

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace std;
using namespace __gnu_pbds;

template<typename T> using Tree = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
typedef long long int ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;

#define fastio ios_base::sync_with_stdio(false); cin.tie(NULL)
#define pb push_back
#define endl '\n'
#define sz(a) (int)a.size()
#define setbits(x) __builtin_popcountll(x)
#define ff first
#define ss second
#define conts continue
#define ceil2(x,y) ((x+y-1)/(y))
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define yes cout << "Yes" << endl
#define no cout << "No" << endl

#define rep(i,n) for(int i = 0; i < n; ++i)
#define rep1(i,n) for(int i = 1; i <= n; ++i)
#define rev(i,s,e) for(int i = s; i >= e; --i)
#define trav(i,a) for(auto &i : a)

template<typename T>
void amin(T &a, T b) {
    a = min(a,b);
}

template<typename T>
void amax(T &a, T b) {
    a = max(a,b);
}

#ifdef LOCAL
#include "debug.h"
#else
#define debug(x) 42
#endif

/*

refs:
edi (took hints from the solution to the two inputs subtask)

*/

const int MOD = 1000002022;
const int N = 2e5 + 5;
const int inf1 = int(1e9) + 5;
const ll inf2 = ll(1e18) + 5;

#include "circuit.h"

template<typename T>
struct lazysegtree {
    /*=======================================================*/

    struct data {
        array<ll,2> a;
    };

    struct lazy {
        ll a;
    };

    data d_neutral = {{0,0}};
    lazy l_neutral = {0};

    void merge(data &curr, data &left, data &right) {
        rep(j,2){
            curr.a[j] = (left.a[j]+right.a[j])%MOD;
        }
    }

    void create(int x, int lx, int rx, T v) {
        tr[x].a[v.ff] = v.ss;
    }

    void modify(int x, int lx, int rx, T v) {
        lz[x].a = 1;
    }

    void propagate(int x, int lx, int rx) {
        ll v = lz[x].a;
        if(!v) return;

        swap(tr[x].a[0],tr[x].a[1]);

        if(rx-lx > 1){
            lz[2*x+1].a ^= 1;
            lz[2*x+2].a ^= 1;
        }

        lz[x] = l_neutral;
    }

    /*=======================================================*/

    int siz = 1;
    vector<data> tr;
    vector<lazy> lz;

    lazysegtree() {

    }

    lazysegtree(int n) {
        while (siz < n) siz *= 2;
        tr.assign(2 * siz, d_neutral);
        lz.assign(2 * siz, l_neutral);
    }

    void build(vector<T> &a, int n, int x, int lx, int rx) {
        if (rx - lx == 1) {
            if (lx < n) {
                create(x, lx, rx, a[lx]);
            }

            return;
        }

        int mid = (lx + rx) / 2;

        build(a, n, 2 * x + 1, lx, mid);
        build(a, n, 2 * x + 2, mid, rx);

        merge(tr[x], tr[2 * x + 1], tr[2 * x + 2]);
    }

    void build(vector<T> &a, int n) {
        build(a, n, 0, 0, siz);
    }

    void rupd(int l, int r, T v, int x, int lx, int rx) {
        propagate(x, lx, rx);

        if (lx >= r or rx <= l) return;
        if (lx >= l and rx <= r) {
            modify(x, lx, rx, v);
            propagate(x, lx, rx);
            return;
        }

        int mid = (lx + rx) / 2;

        rupd(l, r, v, 2 * x + 1, lx, mid);
        rupd(l, r, v, 2 * x + 2, mid, rx);

        merge(tr[x], tr[2 * x + 1], tr[2 * x + 2]);
    }

    void rupd(int l, int r, T v) {
        rupd(l, r + 1, v, 0, 0, siz);
    }

    data query(int l, int r, int x, int lx, int rx) {
        propagate(x, lx, rx);

        if (lx >= r or rx <= l) return d_neutral;
        if (lx >= l and rx <= r) return tr[x];

        int mid = (lx + rx) / 2;

        data curr;
        data left = query(l, r, 2 * x + 1, lx, mid);
        data right = query(l, r, 2 * x + 2, mid, rx);

        merge(curr, left, right);
        return curr;
    }

    data query(int l, int r) {
        return query(l, r + 1, 0, 0, siz);
    }
};

vector<ll> adj[N];
vector<ll> sub_ways(N);

void dfs1(ll u){
    sub_ways[u] = max((ll)sz(adj[u]),1ll);
    trav(v,adj[u]){
        dfs1(v);
        sub_ways[u] = sub_ways[u]*sub_ways[v]%MOD;
    }
}

ll n,m;
vector<int> a;
vector<ll> contrib;

void dfs2(ll u, ll val){
    auto children = adj[u];
    if(children.empty()){
        contrib[u-n] = val;
        return;
    }

    ll siz = sz(children);
    children.insert(children.begin(),0);
    vector<ll> pref(siz+5), suff(siz+5);
    pref[0] = 1, suff[siz+1] = 1;
    rep1(i,siz){
        ll v = children[i];
        pref[i] = pref[i-1]*sub_ways[v]%MOD;
    }
    rev(i,siz,1){
        ll v = children[i];
        suff[i] = suff[i+1]*sub_ways[v]%MOD;
    }

    rep1(i,siz){
        ll v = children[i];
        ll new_val = val*pref[i-1]%MOD*suff[i+1]%MOD;
        dfs2(v,new_val);
    }
}

lazysegtree<pll> st;

void init(int n_, int m_, std::vector<int> P, std::vector<int> A) {
    n = n_, m = m_;
    a = A;
    contrib = vector<ll>(m);

    rep(i,n+m){
        if(P[i] != -1){
            adj[P[i]].pb(i);
        }
    }

    dfs1(0);
    dfs2(0,1);

    vector<pll> b;
    rep(i,m) b.pb({a[i],contrib[i]});

    st = lazysegtree<pll>(m);
    st.build(b,m);
}

int count_ways(int l, int r) {
    l -= n, r -= n;
    st.rupd(l,r,{1,1});
    ll ans = st.query(0,m-1).a[1];
    return ans;
}

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