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IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#108105#6411. Classical FFT Problemwhatever#WA 2ms3408kbC++178.4kb2023-05-23 16:26:162023-05-23 16:26:18

Judging History

你现在查看的是最新测评结果

  • [2023-08-10 23:21:45]
  • System Update: QOJ starts to keep a history of the judgings of all the submissions.
  • [2023-05-23 16:26:18]
  • 评测
  • 测评结果:WA
  • 用时:2ms
  • 内存:3408kb
  • [2023-05-23 16:26:16]
  • 提交

answer

#include <bits/stdc++.h>
using namespace std;
mt19937_64 rnd(chrono::steady_clock::now().time_since_epoch().count());
#define rep(i, a, b) for (int i = (a), I = (b); i <= I; ++i)
#define per(i, a, b) for (int i = (a), I = (b); i >= I; --i)
using i64 = long long;
using pii = pair<int, int>;
template<typename T> void up(T &x, T y) { if (x < y) x = y; }
template<typename T> void down(T &x, T y) { if (x > y) x = y; }

const int P = 998244353;
void add(int &x, int y) { (x += y) >= P && (x -= P); }
int Add(int x, int y) { return (x += y) >= P ? (x - P) : x; }
void sub(int &x, int y) { (x -= y) < 0 && (x += P); }
int Sub(int x, int y) { return (x -= y) < 0 ? (x + P) : x; }
void mul(int &x, int y) { x = 1ll * x * y % P; }
int Mul(int x, int y) { return 1ll * x * y % P; }

std::vector<int> rev, roots{0, 1};
int power(int a, int b) {
    int res = 1;
    for (; b; b >>= 1, a = 1ll * a * a % P)
        if (b & 1)
            res = 1ll * res * a % P;
    return res;
}
void dft(std::vector<int> &a) {
    int n = a.size();
    if (int(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 (int(roots.size()) < n) {
        int k = __builtin_ctz(roots.size());
        roots.resize(n);
        while ((1 << k) < n) {
            int e = power(3, (P - 1) >> (k + 1));
            for (int i = 1 << (k - 1); i < (1 << k); ++i) {
                roots[2 * i] = roots[i];
                roots[2 * i + 1] = 1ll * roots[i] * e % P;
            }
            ++k;
        }
    }
    for (int k = 1; k < n; k *= 2) {
        for (int i = 0; i < n; i += 2 * k) {
            for (int j = 0; j < k; ++j) {
                int u = a[i + j];
                int v = 1ll * a[i + j + k] * roots[k + j] % P;
                int x = u + v;
                if (x >= P)
                    x -= P;
                a[i + j] = x;
                x = u - v;
                if (x < 0)
                    x += P;
                a[i + j + k] = x;
            }
        }
    }
}
void idft(std::vector<int> &a) {
    int n = a.size();
    std::reverse(a.begin() + 1, a.end());
    dft(a);
    int inv = power(n, P - 2);
    for (int i = 0; i < n; ++i)
        a[i] = 1ll * a[i] * inv % P;
}
struct poly {
    std::vector<int> a;
    poly() {}
    poly(int a0) {
        if (a0)
            a = {a0};
    }
    poly(const std::vector<int> &a1) : a(a1) {
        while (!a.empty() && !a.back())
            a.pop_back();
    }
    int size() const {
        return a.size();
    }
    int operator[](int idx) const {
        if (idx < 0 || idx >= size())
            return 0;
        return a[idx];
    }
    poly mulxk(int k) const {
        auto b = a;
        b.insert(b.begin(), k, 0);
        return poly(b);
    }
    poly modxk(int k) const {
        k = std::min(k, size());
        return poly(std::vector<int>(a.begin(), a.begin() + k));
    }
    poly divxk(int k) const {
        if (size() <= k)
            return poly();
        return poly(std::vector<int>(a.begin() + k, a.end()));
    }
    friend poly operator+(const poly a, const poly &b) {
        std::vector<int> res(std::max(a.size(), b.size()));
        for (int i = 0; i < int(res.size()); ++i) {
            res[i] = a[i] + b[i];
            if (res[i] >= P)
                res[i] -= P;
        }
        return poly(res);
    }
    friend poly operator-(const poly a, const poly &b) {
        std::vector<int> res(std::max(a.size(), b.size()));
        for (int i = 0; i < int(res.size()); ++i) {
            res[i] = a[i] - b[i];
            if (res[i] < 0)
                res[i] += P;
        }
        return poly(res);
    }
    friend poly operator*(poly a, poly b) {
        int sz = 1, tot = a.size() + b.size() - 1;
        while (sz < tot)
            sz *= 2;
        a.a.resize(sz);
        b.a.resize(sz);
        dft(a.a);
        dft(b.a);
        for (int i = 0; i < sz; ++i)
            a.a[i] = 1ll * a[i] * b[i] % P;
        idft(a.a);
        return poly(a.a);
    }
    poly &operator+=(poly b) {
        return (*this) = (*this) + b;
    }
    poly &operator-=(poly b) {
        return (*this) = (*this) - b;
    }
    poly &operator*=(poly b) {
        return (*this) = (*this) * b;
    }
    poly deriv() const {
        if (a.empty())
            return poly();
        std::vector<int> res(size() - 1);
        for (int i = 0; i < size() - 1; ++i)
            res[i] = 1ll * (i + 1) * a[i + 1] % P;
        return poly(res);
    }
    poly integr() const {
        if (a.empty())
            return poly();
        std::vector<int> res(size() + 1);
        for (int i = 0; i < size(); ++i)
            res[i + 1] = 1ll * a[i] * power(i + 1, P - 2) % P;
        return poly(res);
    }
    poly inv(int m) const {
        poly x(power(a[0], P - 2));
        int k = 1;
        while (k < m) {
            k *= 2;
            x = (x * (2 - modxk(k) * x)).modxk(k);
        }
        return x.modxk(m);
    }
    poly log(int m) const {
        return (deriv() * inv(m)).integr().modxk(m);
    }
    poly exp(int m) const {
        poly x(1);
        int k = 1;
        while (k < m) {
            k *= 2;
            x = (x * (1 - x.log(k) + modxk(k))).modxk(k);
        }
        return x.modxk(m);
    }
    poly sqrt(int m) const {
        poly x(1);
        int k = 1;
        while (k < m) {
            k *= 2;
            x = (x + (modxk(k) * x.inv(k)).modxk(k)) * ((P + 1) / 2);
        }
        return x.modxk(m);
    }
    poly mulT(poly b) const {
        if (b.size() == 0)
            return poly();
        int n = b.size();
        std::reverse(b.a.begin(), b.a.end());
        return ((*this) * b).divxk(n - 1);
    }
    std::vector<int> eval(std::vector<int> x) const {
        if (size() == 0)
            return std::vector<int>(x.size(), 0);
        const int n = std::max(int(x.size()), size());
        std::vector<poly> q(4 * n);
        std::vector<int> ans(x.size());
        x.resize(n);
        std::function<void(int, int, int)> build = [&](int p, int l, int r) {
            if (r - l == 1) {
                q[p] = std::vector<int>{1, (P - x[l]) % P};
            } else {
                int m = (l + r) / 2;
                build(2 * p, l, m);
                build(2 * p + 1, m, r);
                q[p] = q[2 * p] * q[2 * p + 1];
            }
        };
        build(1, 0, n);
        std::function<void(int, int, int, const poly &)> work = [&](int p, int l, int r, const poly &num) {
            if (r - l == 1) {
                if (l < int(ans.size()))
                    ans[l] = num[0];
            } else {
                int m = (l + r) / 2;
                work(2 * p, l, m, num.mulT(q[2 * p + 1]).modxk(m - l));
                work(2 * p + 1, m, r, num.mulT(q[2 * p]).modxk(r - m));
            }
        };
        work(1, 0, n, mulT(q[1].inv(n)));
        return ans;
    }
};

const int N = 140000;
int n, a[N], b[N], k;
int fac[N], ifac[N];
void init(int n) {
    fac[0] = 1;
    rep(i, 1, n) fac[i] = Mul(fac[i - 1], i);
    ifac[n] = power(fac[n], P - 2);
    per(i, n, 1) ifac[i - 1] = Mul(ifac[i], i);
}
int binom(int n, int m) {
    if (n < 0 || m < 0 || n < m) return 0;
    return Mul(fac[n], Mul(ifac[m], ifac[n - m]));
}

poly solve(int l, int r, int *a) {
    if (l == r) return poly({P - a[l], 1});
    int mid = (l + r) >> 1;
    return solve(l, mid, a) * solve(mid + 1, r, a); 
}

int solve(int *a) {
    int ans = 0;
    poly f = solve(1, k, a);
    vector<int> x(a[k + 1] + 1);
    iota(x.begin(), x.end(), 0);
    poly g = f.eval(x);
    rep(j, 0, a[k + 1]) {
        int coef = binom(a[k + 1], j);
        if (j & 1) coef = Sub(0, coef);
        add(ans, Mul(coef, g[j]));
    }
    return ans;
}

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

    cin >> n;
    init(n);
    rep(i, 1, n) cin >> a[i];
    reverse(a + 1, a + n + 1);
    for (int i = 1, j = n; i <= n; ++i) {
        while (j && a[j] < i) --j;
        b[i] = j;
    }

    k = 0;
    while (k < n && a[k + 1] >= k + 1) ++k;
    cout << k << " ";

    int ans = Add(solve(a), solve(b));
    sub(ans, fac[k]);
    cout << ans << "\n";

    return 0;
}

Details

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Test #1:

score: 100
Accepted
time: 2ms
memory: 3408kb

input:

3
1 2 3

output:

2 6

result:

ok 2 number(s): "2 6"

Test #2:

score: -100
Wrong Answer
time: 2ms
memory: 3384kb

input:

1
1

output:

1 998244350

result:

wrong answer 2nd numbers differ - expected: '1', found: '998244350'