Submission #475414

#	Time	Username	Problem	Language	Result	Execution time	Memory
475414		SamAnd	Energetic turtle (IZhO11_turtle)	C++17	100 / 100	885 ms	66200 KiB

turtle

#include <bits/stdc++.h>
using namespace std;
#define m_p make_pair
#define all(x) (x).begin(),(x).end()
#define sz(x) ((int)(x).size())
#define fi first
#define se second
typedef long long ll;
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
mt19937 rnf(2106);
const int N = 600005, K = 22;

int ast(int x, int n, int M)
{
    int ans = 1;
    while (n)
    {
        if (n % 2 == 1)
            ans = (ans * 1LL * x) % M;
        x = (x * 1LL * x) % M;
        n /= 2;
    }
    return ans;
}

int M;

vector<pair<int, int> > p;
vector<int> pp;
vector<vector<int> > r;

vector<int> g[N];
vector<int> u[N];
void pre()
{
    int x = M;
    for (int i = 2; i * i <= x; ++i)
    {
        if (x % i == 0)
        {
            p.push_back(m_p(i, 0));
            pp.push_back(1);
            while (x % i == 0)
            {
                x /= i;
                p.back().se++;
                pp.back() *= i;
            }
        }
    }
    if (x > 1)
    {
        p.push_back(m_p(x, 1));
        pp.push_back(x);
    }

    for (int i = 0; i < sz(p); ++i)
    {
        r.push_back({});
        for (int j = 0; j < sz(p); ++j)
        {
            if (i == j)
            {
                r.back().push_back(0);
            }
            else
            {
                r.back().push_back(ast(pp[i], pp[j] - pp[j] / p[j].fi - 1, pp[j]));
            }
        }
    }

    g[0].assign(sz(p), 1);
    u[0].assign(sz(p), 0);
    for (int i = 1; i < N; ++i)
    {
        u[i].resize(sz(p));
        for (int j = 0; j < sz(p); ++j)
            u[i][j] = u[i - 1][j];

        g[i].resize(sz(p));
        for (int j = 0; j < sz(p); ++j)
        {
            int x = i;
            while (x % p[j].fi == 0)
            {
                u[i][j]++;
                x /= p[j].fi;
            }
            g[i][j] = (g[i - 1][j] * 1LL * x) % pp[j];
        }
    }
}

map<pair<int, int>, int> memo;
int C(int n, int k)
{
    if (memo.find(m_p(n, k)) != memo.end())
        return memo[m_p(n, k)];

    static vector<int> a;
    a.resize(sz(p));
    for (int j = 0; j < sz(p); ++j)
    {
        int yans = (g[n][j] * 1LL * ast((g[k][j] * 1LL * g[n - k][j]) % pp[j], pp[j] - pp[j] / p[j].fi - 1, pp[j])) % pp[j];
        int uu = u[n][j] - u[n - k][j] - u[k][j];
        if (uu >= p[j].se)
            yans = 0;
        else
            yans = (yans * 1LL * ast(p[j].fi, uu, pp[j])) % pp[j];
        a[j] = yans;
    }

    static vector<int> x;
    x.resize(sz(p));
    for (int j = 0; j < sz(p); ++j)
    {
        x[j] = a[j];
        for (int k = 0; k < j; ++k)
            x[j] = ((x[j] - x[k]) * 1LL * r[k][j]) % pp[j];
    }

    int ans = 0;
    for (int j = sz(p) - 1; j >= 0; --j)
    {
        ans = (ans + x[j]) % M;
        if (j - 1 >= 0)
            ans = (ans * 1LL * pp[j - 1]) % M;
    }

    return memo[m_p(n, k)] = ans;
}

int n, m, k, t;
pair<int, int> a[K];

int q[K];
int yans[K];

void solv()
{
    cin >> n >> m >> k >> t >> M;
    for (int i = 0; i < k; ++i)
        cin >> a[i].fi >> a[i].se;

    pre();

    sort(a, a + k);
    for (int x = 0; x < (1 << k); ++x)
    {
        vector<pair<int, int> > v;
        v.push_back(m_p(0, 0));
        for (int i = 0; i < k; ++i)
        {
            if ((x & (1 << i)))
                v.push_back(a[i]);
        }
        v.push_back(m_p(n, m));

        bool z = true;
        for (int i = 0; i < sz(v) - 1; ++i)
        {
            if (!(v[i].fi <= v[i + 1].fi && v[i].se <= v[i + 1].se))
            {
                z = false;
                break;
            }
        }
        if (!z)
            continue;

        int ans = 1;
        for (int i = 0; i < sz(v) - 1; ++i)
            ans = (ans * 1LL * C(v[i + 1].fi - v[i].fi + v[i + 1].se - v[i].se, v[i + 1].fi - v[i].fi)) % M;
        q[sz(v) - 2] = (q[sz(v) - 2] + ans) % M;
    }

    for (int i = k; i >= 0; --i)
    {
        yans[i] = q[i];
        for (int j = i + 1; j <= k; ++j)
        {
            yans[i] = (yans[i] - yans[j] * 1LL * C(j, i)) % M;
        }
    }

    int ans = 0;
    for (int i = 0; i <= t; ++i)
        ans = (ans + yans[i]) % M;

    ans = (ans + M) % M;

    cout << ans << "\n";
}

int main()
{
    #ifdef SOMETHING
    freopen("input.txt", "r", stdin);
    //freopen("output.txt", "w", stdout);
    #endif // SOMETHING
    ios_base::sync_with_stdio(false), cin.tie(0);

    int tt = 1;
    //cin >> tt;
    while (tt--)
        solv();
    return 0;
}

• Subtask 1: 100 / 100