Submission #688139

#	Time	Username	Problem	Language	Result	Execution time	Memory
688139		finn__	Fish (IOI08_fish)	C++17	21 / 100	307 ms	65536 KiB

fish

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

class SegmentTree
{
    vector<unsigned> t;
    size_t l, m;

public:
    SegmentTree(size_t _n, size_t _m)
    {
        l = 1 << (32 - __builtin_clz(_n));
        m = _m;
        t = vector<unsigned>(2 * l, 1);
    }

    void increment(size_t i)
    {
        i += l;
        t[i] = (t[i] + 1) % m;

        while (i >>= 1)
            t[i] = (t[2 * i] * t[2 * i + 1]) % m;
    }

    unsigned range_prod(size_t i, size_t j)
    {
        if (j > l)
            return 1;
        unsigned x = 1;
        i += l, j += l;

        while (i <= j)
        {
            if (i & 1)
                x = (x * t[i++]) % m;
            if (!(j & 1))
                x = (x * t[j--]) % m;
            i >>= 1;
            j >>= 1;
        }

        return x;
    }
};

int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);

    size_t n, k, m;
    cin >> n >> k >> m;

    vector<pair<unsigned, unsigned>> fish(n);
    vector<pair<unsigned, unsigned>> largest(k, {0, 0});
    vector<unsigned> F(k);

    for (auto &[l, g] : fish)
    {
        cin >> l >> g;
        g--;
        largest[g] = max(largest[g], {l, g});
        F[g] = max(F[g], l);
    }

    sort(fish.begin(), fish.end());
    sort(largest.begin(), largest.end());
    vector<unsigned> c(k);
    vector<vector<unsigned>> e(k, vector<unsigned>(k));
    unsigned ans = 0;
    SegmentTree tree(k, m);
    size_t p = 0;

    for (size_t i = 0; i < k; i++)
    {
        while (p < n && 2 * fish[p].first <= largest[i].first)
        {
            c[fish[p].second]++;
            p++;
        }

        for (size_t j = 0; j < k; j++)
        {
            e[largest[i].second][j] = c[j] % m;
        }
    }

    for (size_t i = 0; i < k; i++)
    {
        unsigned x = 1, y = 1;
        for (size_t j = 0; j < k; j++)
        {
            if (!(e[j][i] >= e[i][i] + 1) && F[j] > F[i])
                y = (y * (e[i][j] + 1)) % m;
            else if (F[i] > F[j])
                x = (x * (e[i][j] + 1)) % m, y = (y * (e[i][j] + 1)) % m;
        }
        ans = (ans + x * e[i][i]) % m;
        ans = (ans + y) % m;
    }

    cout << ans << '\n';
}

• Subtask 1: 4 / 4
• Subtask 2: 4 / 4
• Subtask 3: 4 / 4
• Subtask 4: 4 / 4
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• Subtask 19: 0 / 5