Submission #364289

#	Time	Username	Problem	Language	Result	Execution time	Memory
364289		dolphingarlic	Fish (IOI08_fish)	C++14	17 / 100	578 ms	17628 KiB

fish

#include <bits/stdc++.h>
typedef long long ll;
using namespace std;

int n, k, m, segtree[2000000];
tuple<int, int, int> f[500000];
pair<int, int> mx[500000];
int nxt[500000], ord[500000], idx[500000];

void increment(int pos, int node = 1, int l = 0, int r = k - 1) {
    if (l == r) segtree[node] = (segtree[node] + 1) % m;
    else {
        int mid = (l + r) / 2;
        if (pos > mid) increment(pos, node * 2 + 1, mid + 1, r);
        else increment(pos, node * 2, l, mid);
        segtree[node] = (segtree[node * 2] * segtree[node * 2 + 1]) % m;
    }
}

int query(int a, int b, int node = 1, int l = 0, int r = k - 1) {
    if (l > b || r < a) return 1;
    if (l >= a && r <= b) return segtree[node];
    int mid = (l + r) / 2;
    return (query(a, b, node * 2, l, mid) * query(a, b, node * 2 + 1, mid + 1, r)) % m;
}

int main() {
    cin.tie(0)->sync_with_stdio(0);
    cin >> n >> k >> m;
    fill(segtree, segtree + 4 * k, 1);
    for (int i = 0; i < n; i++) {
        int l, t;
        cin >> l >> t;
        t--;
        mx[t] = max(mx[t], {l, i});
        nxt[t] = max(nxt[t], l);
        f[i] = {l, t, i};
    }
    for (int i = 0; i < n; i++)
        if (mx[get<1>(f[i])].first < get<0>(f[i]) * 2)
            nxt[get<1>(f[i])] = min(nxt[get<1>(f[i])], get<0>(f[i]));

    sort(f, f + n);
    iota(ord, ord + k, 0);
    sort(ord, ord + k, [](int A, int B) { return mx[A] < mx[B]; });
    for (int i = 0; i < k; i++) idx[ord[i]] = i;

    int ans = 0;
    for (int i = 0, j = 0, l = 0; i < n; i++) {
        while (get<0>(f[i]) >= get<0>(f[j]) * 2) increment(idx[get<1>(f[j++])]);
        if (get<2>(f[i]) == mx[get<1>(f[i])].second) {
            ans = (ans + query(0, idx[get<1>(f[i])])) % m;
            while (l < k && mx[ord[l]].first < nxt[get<1>(f[i])] * 2) l++;
            if (l - idx[get<1>(f[i])] > 1)
                ans = (ans + query(0, idx[get<1>(f[i])] - 1) * (query(idx[get<1>(f[i])] + 1, l - 1) + m - 1)) % m;
        }
    }
    cout << ans;
    return 0;
}

• Subtask 1: 4 / 4
• Subtask 2: 4 / 4
• Subtask 3: 0 / 4
• Subtask 4: 4 / 4
• Subtask 5: 0 / 4
• Subtask 6: 0 / 5
• Subtask 7: 0 / 6
• Subtask 8: 0 / 6
• Subtask 9: 0 / 6
• Subtask 10: 0 / 6
• Subtask 11: 0 / 7
• Subtask 12: 0 / 7
• Subtask 13: 0 / 7
• Subtask 14: 0 / 5
• Subtask 15: 0 / 5
• Subtask 16: 5 / 5
• Subtask 17: 0 / 5
• Subtask 18: 0 / 5
• Subtask 19: 0 / 5