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ID题目提交者结果用时内存语言文件大小提交时间测评时间
#138918#1810. Generate the SequencesforeverlastingAC ✓5ms3772kbC++2010.8kb2023-08-12 14:17:202023-08-12 14:17:20

Judging History

你现在查看的是最新测评结果

  • [2023-08-12 14:17:20]
  • 评测
  • 测评结果:AC
  • 用时:5ms
  • 内存:3772kb
  • [2023-08-12 14:17:20]
  • 提交

answer

#include <bits/stdc++.h>

using LL = long long;
using ld = long double;
using ull = unsigned int;
using Pair = std::pair<int, int>;
//using unl = __int128;
#define inf 2'000'000'000'000'000'000ll
std::mt19937_64 rng(std::chrono::steady_clock::now().time_since_epoch().count());

template<int kcz>
struct ModInt {
#define T (*this)
    int x;

    ModInt() : x(0) {}

    ModInt(int y) : x(y >= 0 ? y : y + kcz) {}

    ModInt(LL y) : x(y >= 0 ? y % kcz : (kcz - (-y) % kcz) % kcz) {}

    inline int inc(const int &v) {
        return v >= kcz ? v - kcz : v;
    }

    inline int dec(const int &v) {
        return v < 0 ? v + kcz : v;
    }

    inline ModInt &operator+=(const ModInt &p) {
        x = inc(x + p.x);
        return T;
    }

    inline ModInt &operator-=(const ModInt &p) {
        x = dec(x - p.x);
        return T;
    }

    inline ModInt &operator*=(const ModInt &p) {
        x = (int) ((LL) x * p.x % kcz);
        return T;
    }

    inline ModInt inverse() const {
        int a = x, b = kcz, u = 1, v = 0, t;
        while (b > 0)t = a / b, std::swap(a -= t * b, b), std::swap(u -= t * v, v);
        return u;
    }

    inline ModInt &operator/=(const ModInt &p) {
        T *= p.inverse();
        return T;
    }

    inline ModInt operator-() const {
        return -x;
    }

    inline friend ModInt operator+(const ModInt &lhs, const ModInt &rhs) {
        return ModInt(lhs) += rhs;
    }

    inline friend ModInt operator-(const ModInt &lhs, const ModInt &rhs) {
        return ModInt(lhs) -= rhs;
    }

    inline friend ModInt operator*(const ModInt &lhs, const ModInt &rhs) {
        return ModInt(lhs) *= rhs;
    }

    inline friend ModInt operator/(const ModInt &lhs, const ModInt &rhs) {
        return ModInt(lhs) /= rhs;
    }

    inline bool operator==(const ModInt &p) const {
        return x == p.x;
    }

    inline bool operator!=(const ModInt &p) const {
        return x != p.x;
    }

    inline ModInt qpow(LL n) const {
        ModInt ret(1), mul(x);
        while (n > 0) {
            if (n & 1)ret *= mul;
            mul *= mul, n >>= 1;
        }
        return ret;
    }

    inline friend std::ostream &operator<<(std::ostream &os, const ModInt &p) {
        return os << p.x;
    }

    inline friend std::istream &operator>>(std::istream &is, ModInt &a) {
        LL t;
        is >> t, a = ModInt<kcz>(t);
        return is;
    }

    static int get_mod() {
        return kcz;
    }

    inline bool operator<(const ModInt &A) const {
        return x < A.x;
    }

    inline bool operator>(const ModInt &A) const {
        return x > A.x;
    }

#undef T
};

const int kcz = 998244353;
using Z = ModInt<kcz>;

namespace NTT {
    std::vector<int> rev;
    std::vector<Z> roots{0, 1};

    inline void dft(std::vector<Z> &a) {
        int n = (int) (a.size());
        if (rev.size() != n) {
            int k = __builtin_ctz(n) - 1;
            rev.resize(n);
            for (int i = 0; i < n; i++)rev[i] = rev[i >> 1] >> 1 | (i & 1) << k;
        }
        for (int i = 0; i < n; i++)if (rev[i] < i)std::swap(a[i], a[rev[i]]);
        if (roots.size() < n) {
            int k = __builtin_ctz(roots.size());
            roots.resize(n);
            while ((1 << k) < n) {
                Z e = Z(3).qpow((kcz - 1) >> (k + 1));
                for (int i = 1 << (k - 1); i < (1 << k); i++)
                    roots[i << 1] = roots[i], roots[i << 1 | 1] = roots[i] * e;
                k++;
            }
        }
        for (int k = 1; k < n; k <<= 1) {
            for (int i = 0; i < n; i += k << 1) {
                for (int j = 0; j < k; j++) {
                    Z u = a[i + j], v = a[i + j + k] * roots[k + j];
                    a[i + j] = u + v, a[i + j + k] = u - v;
                }
            }
        }
    }

    inline void idft(std::vector<Z> &a) {
        int n = (int) (a.size());
        reverse(a.begin() + 1, a.end()), dft(a);
        Z inv = Z(n).inverse();
        for (int i = 0; i < n; i++)a[i] = a[i] * inv;
    }
}
struct Poly : public std::vector<Z> {
#define T (*this)
    using vector<Z>::vector;

    inline int deg() const {
        return (int) (size());
    }

    inline Z operator[](const int &idx) const {
        if (idx < 0 || idx >= deg())return Z(0);
        return at(idx);
    }

    inline Z &operator[](const int &idx) {
        return at(idx);
    }

    inline Poly &operator^=(const Poly &b) {
        if (b.deg() < deg())resize(b.deg());
        for (int i = 0, sz = deg(); i < sz; i++)T[i] *= b[i];
        return T;
    }

    inline Poly &operator<<=(const int &k) {
        return insert(begin(), k, Z(0)), T;
    }

    inline Poly operator<<(const int &r) const {
        return Poly(T) <<= r;
    }

    inline Poly operator>>(const int &r) const {
        return r >= deg() ? Poly() : Poly(begin() + r, end());
    }

    inline Poly &operator>>=(const int &r) {
        return T = T >> r;
    }

    inline Poly mod(const int &k) const {
        return k < deg() ? Poly(begin(), begin() + k) : T;
    }

    inline friend Poly operator*(const Z &a, Poly b) {
        for (auto &x: b)x *= a;
        return b;
    }

    inline friend Poly operator*(Poly b, const Z &a) {
        for (auto &x: b)x *= a;
        return b;
    }

    inline friend Poly operator*(Poly a, Poly b) {
        if (a.empty() || b.empty())return {};
        int sz = 1, tot = a.deg() + b.deg() - 1;
        while (sz < tot)sz <<= 1;
        a.resize(sz), b.resize(sz);
        NTT::dft(a), NTT::dft(b);
        for (int i = 0; i < sz; i++)a[i] *= b[i];
        NTT::idft(a), a.resize(tot);
        return a;
    }

    inline Poly &operator*=(const Poly &b) {
        return T = T * b;
    }

    inline friend Poly operator+(const Poly &a, const Poly &b) {
        int n = (int) std::max(a.size(), b.size());
        Poly c;
        c.resize(n);
        for (int i = 0, sz = (int) a.size(); i < sz; i++)c[i] = a[i];
        for (int i = 0, sz = (int) b.size(); i < sz; i++)c[i] += b[i];
        return c;
    }

    inline Poly &operator+=(const Poly &b) {
        return T = T + b;
    }

    inline friend Poly operator-(const Poly &a, const Poly &b) {
        int n = (int) std::max(a.size(), b.size());
        Poly c;
        c.resize(n);
        for (int i = 0, sz = (int) a.size(); i < sz; i++)c[i] = a[i];
        for (int i = 0, sz = (int) b.size(); i < sz; i++)c[i] -= b[i];
        return c;
    }

    inline Poly &operator-=(const Poly &b) {
        return T = T - b;
    }

    inline Poly derivation() const {
        if (T.empty())return {};
        int n = (int) (T.size());
        Poly c;
        c.resize(n - 1);
        for (int i = 0; i < n - 1; i++)c[i] = T[i + 1] * (i + 1);
        return c;
    }

    inline Poly integration() const {
        int n = (int) (T.size());
        Poly c;
        c.resize(n + 1);
        for (int i = 0; i < n; i++)c[i + 1] = T[i] * Z(i + 1).inverse();
        return c;
    }

    inline Poly inv(const int &m) const {
        Poly c{T[0].inverse()};
        int k = 1;
        while (k < m)k <<= 1, c = (c * (Poly{2} - T.mod(k) * c)).mod(k);
        return c.mod(m);
    }

    inline Poly log(const int &m) const {
        return (derivation() * inv(m)).integration().mod(m);
    }

    inline Poly exp(const int &m) const {
        Poly x{1};
        int k = 1;
        while (k < m)k <<= 1, x = (x * (Poly{1} - x.log(k) + mod(k))).mod(k);
        return x.mod(m);
    }

    inline Poly pow(const int &k, const int &m) const {
        int i = 0;
        while (i < T.size() && T[i] == Z(0))i++;
        if (i == T.size() || (LL) i * k >= m)return Poly(m);
        Z v = T[i];
        auto g = (T >> i) * (v.inverse());
        return ((g.log(m - i * k) * Z(k)).exp(m - i * k) << (i * k)) * v.qpow(k);
    }

    inline Poly sqrt(const int &m) const {
        Poly x{1};
        int k = 1;
        while (k < m)k <<= 1, x = (x + (mod(k) * x.inv(k)).mod(k)) * Z(2).inverse();
        return x.mod(m);
    }

    inline Poly rev() const {
        return Poly(rbegin(), rend());
    }

    inline Poly mulT(const Poly &b) const {
        return T * b.rev() >> (b.deg() - 1);
    }

    inline vector <Z> eval(vector <Z> x) const {
        if (T.empty())return vector<Z>(x.size(), Z(0));
        int n = std::max((int) (x.size()), (int) (T.size()));
        vector <Poly> q(4 * n);
        vector <Z> ans(x.size());
        x.resize(n);
        std::function<void(int, int, int)> build = [&](int rt, int l, int r) {
            if (l == r) {
                q[rt] = {Z(1), -x[l]};
                return;
            }
            int mid = (l + r) >> 1;
            build(rt << 1, l, mid), build(rt << 1 | 1, mid + 1, r);
            q[rt] = q[rt << 1] * q[rt << 1 | 1];
        };
        build(1, 0, n - 1);
        std::function<void(int, int, int, const Poly &)> work = [&](int rt, int l, int r, const Poly &num) {
            if (l == r) {
                if (l < (int) (ans.size()))ans[l] = num[0];
                return;
            }
            int mid = (l + r) >> 1;
            work(rt << 1, l, mid, num.mulT(q[rt << 1 | 1]).mod(mid - l + 1));
            work(rt << 1 | 1, mid + 1, r, num.mulT(q[rt << 1]).mod(r - mid));
        };
        work(1, 0, n - 1, mulT(q[1].inv(n)));
        return ans;
    }

#undef T
};

inline Z divAt(Poly F, Poly G, LL k) {
    int i;
    for (; k; k >>= 1) {
        Poly R = G;
        // R=G(-x)
        int sz = (int) (R.size());
        for (i = 1; i < sz; i += 2)R[i] = -R[i];
        F *= R, G *= R, sz = (int) (F.size());
        for (i = (int) (k & 1); i < sz; i += 2)F[i >> 1] = F[i];
        F.resize(i / 2), sz = (int) (G.size());
        for (i = 0; i < sz; i += 2)G[i >> 1] = G[i];
        G.resize(i / 2);
    }
    return F.empty() ? Z(0) : (F[0] * (G[0].inverse()));
}

inline Z getAn(Poly F, const Poly &A, const LL &n, const int &k) {
    F = Poly{1} - F;
    Poly f = A * F;;
    return divAt(f.mod(k), F, n);
}

inline void solve(const int &Case) {
    int n, m;
    std::cin >> n >> m;
    std::vector<Z> inv(n + 1);
    inv[0] = inv[1] = 1;
    for (int i = 2; i <= n; i++)inv[i] = inv[kcz % i] * (kcz - kcz / i);
    Poly g(n + 1);
    g[0] = 1;
    for (int i = 1; i <= n; i++)g[i] = g[i - 1] * (m - 1 - i) * inv[i];
    g = g.integration();
    g[1] += 1;
    g = g.exp(n + 1);
    Z ans = g[n];
    for (int i = 1; i <= n; i++)ans *= i;
    std::cout << ans << '\n';
}

int main() {
//    freopen("1.in", "r", stdin);
//    freopen("1.out", "w", stdout);
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    std::cout.tie(nullptr);
    int T = 1;
//    std::cin >> T;
    for (int i = 1; i <= T; i++)solve(i);
    return 0;
}

详细

Test #1:

score: 100
Accepted
time: 0ms
memory: 3552kb

input:

2 3

output:

5

result:

ok answer is '5'

Test #2:

score: 0
Accepted
time: 3ms
memory: 3700kb

input:

1024 52689658

output:

654836147

result:

ok answer is '654836147'

Test #3:

score: 0
Accepted
time: 1ms
memory: 3496kb

input:

1 2

output:

2

result:

ok answer is '2'

Test #4:

score: 0
Accepted
time: 1ms
memory: 3616kb

input:

1 3

output:

2

result:

ok answer is '2'

Test #5:

score: 0
Accepted
time: 1ms
memory: 3584kb

input:

1 100000000

output:

2

result:

ok answer is '2'

Test #6:

score: 0
Accepted
time: 1ms
memory: 3544kb

input:

2 2

output:

4

result:

ok answer is '4'

Test #7:

score: 0
Accepted
time: 0ms
memory: 3496kb

input:

2 4

output:

6

result:

ok answer is '6'

Test #8:

score: 0
Accepted
time: 1ms
memory: 3564kb

input:

2 5

output:

7

result:

ok answer is '7'

Test #9:

score: 0
Accepted
time: 1ms
memory: 3604kb

input:

2 100000000

output:

100000002

result:

ok answer is '100000002'

Test #10:

score: 0
Accepted
time: 1ms
memory: 3496kb

input:

3 2

output:

8

result:

ok answer is '8'

Test #11:

score: 0
Accepted
time: 1ms
memory: 3548kb

input:

3 3

output:

14

result:

ok answer is '14'

Test #12:

score: 0
Accepted
time: 1ms
memory: 3548kb

input:

3 4

output:

22

result:

ok answer is '22'

Test #13:

score: 0
Accepted
time: 1ms
memory: 3580kb

input:

3 5

output:

32

result:

ok answer is '32'

Test #14:

score: 0
Accepted
time: 1ms
memory: 3540kb

input:

3 100000000

output:

446563791

result:

ok answer is '446563791'

Test #15:

score: 0
Accepted
time: 2ms
memory: 3756kb

input:

3000 2

output:

21292722

result:

ok answer is '21292722'

Test #16:

score: 0
Accepted
time: 5ms
memory: 3712kb

input:

3000 3

output:

172222927

result:

ok answer is '172222927'

Test #17:

score: 0
Accepted
time: 5ms
memory: 3752kb

input:

3000 100000000

output:

736503947

result:

ok answer is '736503947'

Test #18:

score: 0
Accepted
time: 5ms
memory: 3660kb

input:

2522 61077387

output:

857454425

result:

ok answer is '857454425'

Test #19:

score: 0
Accepted
time: 1ms
memory: 3628kb

input:

426 7215704

output:

799491736

result:

ok answer is '799491736'

Test #20:

score: 0
Accepted
time: 2ms
memory: 3676kb

input:

772 72289915

output:

848141383

result:

ok answer is '848141383'

Test #21:

score: 0
Accepted
time: 3ms
memory: 3652kb

input:

1447 83321470

output:

160422285

result:

ok answer is '160422285'

Test #22:

score: 0
Accepted
time: 5ms
memory: 3756kb

input:

2497 64405193

output:

355300540

result:

ok answer is '355300540'

Test #23:

score: 0
Accepted
time: 2ms
memory: 3696kb

input:

775 9385367

output:

470172346

result:

ok answer is '470172346'

Test #24:

score: 0
Accepted
time: 2ms
memory: 3576kb

input:

982 72596758

output:

7144187

result:

ok answer is '7144187'

Test #25:

score: 0
Accepted
time: 2ms
memory: 3640kb

input:

417 26177178

output:

776374896

result:

ok answer is '776374896'

Test #26:

score: 0
Accepted
time: 3ms
memory: 3772kb

input:

1932 19858856

output:

285834553

result:

ok answer is '285834553'

Test #27:

score: 0
Accepted
time: 5ms
memory: 3772kb

input:

2728 23009122

output:

433516287

result:

ok answer is '433516287'

Test #28:

score: 0
Accepted
time: 3ms
memory: 3768kb

input:

1857 22578508

output:

243488639

result:

ok answer is '243488639'

Test #29:

score: 0
Accepted
time: 2ms
memory: 3664kb

input:

2918 69623276

output:

546299707

result:

ok answer is '546299707'

Test #30:

score: 0
Accepted
time: 3ms
memory: 3704kb

input:

1679 21332149

output:

217000656

result:

ok answer is '217000656'

Test #31:

score: 0
Accepted
time: 1ms
memory: 3768kb

input:

1340 6251797

output:

267221018

result:

ok answer is '267221018'

Test #32:

score: 0
Accepted
time: 2ms
memory: 3624kb

input:

868 64770398

output:

652067665

result:

ok answer is '652067665'