제출 #739183

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
739183		GrindMachine	Matching (CEOI11_mat)	C++17	63 / 100	308 ms	65536 KiB

mat

// Om Namah Shivaya

#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) 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

/*



*/

const int MOD = 1e9 + 7;
const int N = 1e6 + 5;
const int inf1 = int(1e9) + 5;
const ll inf2 = ll(1e18) + 5;

vector<int> ps = {5791, 8237, 3067, 6991, 3527};
vector<int> mods = {791139137, 433567597, 271149607, 561259969, 222708581};
ll pows[N][2];

void precalc() {
    mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

    shuffle(all(ps), rng);
    shuffle(all(mods), rng);

    rep(j, 2) {
        pows[0][j] = 1;
        rep1(i, N - 1) {
            pows[i][j] = pows[i - 1][j] * ps[j] % mods[j];
        }
    }
}

template<typename T>
struct segtree {
    // https://codeforces.com/blog/entry/18051

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

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

        data() {
            a.fill(0);
            cnt = 0;
        }
    };

    data neutral = data();

    data merge(data &left, data &right) {
        data curr;

        rep(j, 2) {
            curr.a[j] = (left.a[j] + right.a[j] * pows[left.cnt][j]) % mods[j];
        }

        curr.cnt = left.cnt + right.cnt;

        return curr;
    }

    void create(int i, T v) {
        rep(j, 2) {
            tr[i].a[j] = (v * ps[j]) % mods[j];
        }

        tr[i].cnt = (v > 0);
    }

    void modify(int i, T v) {
        rep(j, 2) {
            tr[i].a[j] = (v * ps[j]) % mods[j];
        }

        tr[i].cnt = (v > 0);
    }

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

    int n;
    vector<data> tr;

    segtree() {

    }

    segtree(int siz) {
        init(siz);
    }

    void init(int siz) {
        n = siz;
        tr.assign(2 * n, neutral);
    }

    void build(vector<T> &a, int siz) {
        rep(i, siz) create(i + n, a[i]);
        rev(i, n - 1, 1) tr[i] = merge(tr[i << 1], tr[i << 1 | 1]);
    }

    void pupd(int i, T v) {
        modify(i + n, v);
        for (i = (i + n) >> 1; i; i >>= 1) tr[i] = merge(tr[i << 1], tr[i << 1 | 1]);
    }

    data query(int l, int r) {
        data resl = neutral, resr = neutral;

        for (l += n, r += n; l <= r; l >>= 1, r >>= 1) {
            if (l & 1) resl = merge(resl, tr[l++]);
            if (!(r & 1)) resr = merge(tr[r--], resr);
        }

        return merge(resl, resr);
    }
};

void solve(int test_case)
{
    precalc();
    ll n, m; cin >> n >> m;
    vector<ll> p(n + 5), a(m + 5);
    rep1(i, n) cin >> p[i];
    rep1(i, m) cin >> a[i];

    vector<ll> b;
    rep1(i, m) b.pb(a[i]);

    sort(all(b));
    b.resize(unique(all(b)) - b.begin());

    rep1(i, m) {
        a[i] = lower_bound(all(b), a[i]) - b.begin();
    }

    array<ll, 2> want, toadd;
    want.fill(0), toadd.fill(0);

    rep1(i, n) {
        rep(j, 2) {
            want[j] = (want[j] + pows[i][j] * p[i]) % mods[j];
            toadd[j] = (toadd[j] + pows[i][j]) % mods[j];
        }
    }

    segtree<ll> st(m + 5);
    rep1(i, n) {
        st.pupd(a[i], i);
    }

    vector<ll> ans;

    if (st.query(0, m - 1).a == want) {
        ans.pb(1);
    }

    for (int i = 2; i <= m - n + 1; ++i) {
        rep(j, 2) {
            want[j] = (want[j] + toadd[j]) % mods[j];
        }

        st.pupd(a[i - 1], 0);
        st.pupd(a[i + n - 1], i + n - 1);

        if (st.query(0, m - 1).a == want) {
            ans.pb(i);
        }
    }

    cout << sz(ans) << endl;
    trav(x, ans) cout << x << " ";
    cout << endl;
}

int main()
{
    fastio;

    int t = 1;
    // cin >> t;

    rep1(i, t) {
        solve(i);
    }

    return 0;
}

• Subtask 1: 9 / 9
• Subtask 2: 9 / 9
• Subtask 3: 9 / 9
• Subtask 4: 9 / 9
• Subtask 5: 9 / 9
• Subtask 6: 9 / 9
• Subtask 7: 9 / 9
• Subtask 8: 0 / 9
• Subtask 9: 0 / 8
• Subtask 10: 0 / 10
• Subtask 11: 0 / 10