結果

問題 No.206 数の積集合を求めるクエリ
コンテスト
ユーザー Mister
提出日時 2020-09-04 16:46:18
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 91 ms / 7,000 ms
コード長 5,277 bytes
コンパイル時間 1,067 ms
コンパイル使用メモリ 80,832 KB
最終ジャッジ日時 2025-01-14 04:33:11
ジャッジサーバーID
（参考情報） 		judge5 / judge2
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 3
other AC * 28
権限があれば一括ダウンロードができます

ソースコード

diff # 

#include <iostream>
#include <vector>

template <int MOD>
struct ModInt {
    using lint = long long;
    int val;

    // constructor
    ModInt(lint v = 0) : val(v % MOD) {
        if (val < 0) val += MOD;
    };

    // unary operator
    ModInt operator+() const { return ModInt(val); }
    ModInt operator-() const { return ModInt(MOD - val); }
    ModInt inv() const { return this->pow(MOD - 2); }

    // arithmetic
    ModInt operator+(const ModInt& x) const { return ModInt(*this) += x; }
    ModInt operator-(const ModInt& x) const { return ModInt(*this) -= x; }
    ModInt operator*(const ModInt& x) const { return ModInt(*this) *= x; }
    ModInt operator/(const ModInt& x) const { return ModInt(*this) /= x; }
    ModInt pow(lint n) const {
        auto x = ModInt(1);
        auto b = *this;
        while (n > 0) {
            if (n & 1) x *= b;
            n >>= 1;
            b *= b;
        }
        return x;
    }

    // compound assignment
    ModInt& operator+=(const ModInt& x) {
        if ((val += x.val) >= MOD) val -= MOD;
        return *this;
    }
    ModInt& operator-=(const ModInt& x) {
        if ((val -= x.val) < 0) val += MOD;
        return *this;
    }
    ModInt& operator*=(const ModInt& x) {
        val = lint(val) * x.val % MOD;
        return *this;
    }
    ModInt& operator/=(const ModInt& x) { return *this *= x.inv(); }

    // compare
    bool operator==(const ModInt& b) const { return val == b.val; }
    bool operator!=(const ModInt& b) const { return val != b.val; }
    bool operator<(const ModInt& b) const { return val < b.val; }
    bool operator<=(const ModInt& b) const { return val <= b.val; }
    bool operator>(const ModInt& b) const { return val > b.val; }
    bool operator>=(const ModInt& b) const { return val >= b.val; }

    // I/O
    friend std::istream& operator>>(std::istream& is, ModInt& x) noexcept {
        lint v;
        is >> v;
        x = v;
        return is;
    }
    friend std::ostream& operator<<(std::ostream& os, const ModInt& x) noexcept { return os << x.val; }
};

template <int MOD, int Root>
struct NumberTheoreticalTransform {
    using mint = ModInt<MOD>;
    using mints = std::vector<mint>;

    // the 2^k-th root of 1
    std::vector<mint> zetas;

    explicit NumberTheoreticalTransform() {
        int exp = MOD - 1;
        while (true) {
            mint zeta = mint(Root).pow(exp);
            zetas.push_back(zeta);
            if (exp & 1) break;
            exp /= 2;
        }
    }

    // ceil(log_2 n)
    static int clog2(int n) {
        int k = 0;
        while ((1 << k) < n) ++k;
        return k;
    }

    // 2-radix cooley-tukey algorithm without bit reverse
    // the size of f must be a power of 2
    void ntt(mints& f) const {
        int n = f.size();

        for (int m = n >> 1; m >= 1; m >>= 1) {
            auto zeta = zetas[clog2(m) + 1];
            mint zetapow(1);

            for (int p = 0; p < m; ++p) {
                for (int s = 0; s < n; s += (m << 1)) {
                    auto l = f[s + p],
                         r = f[s + p + m];

                    f[s + p] = l + r;
                    f[s + p + m] = (l - r) * zetapow;
                }
                zetapow = zetapow * zeta;
            }
        }
    }

    // the inverse of above function
    void intt(mints& f) const {
        int n = f.size();

        for (int m = 1; m <= (n >> 1); m <<= 1) {
            auto zeta = zetas[clog2(m) + 1].inv();
            mint zetapow(1);

            for (int p = 0; p < m; ++p) {
                for (int s = 0; s < n; s += (m << 1)) {
                    auto l = f[s + p],
                         r = f[s + p + m] * zetapow;

                    f[s + p] = l + r;
                    f[s + p + m] = l - r;
                }
                zetapow = zetapow * zeta;
            }
        }

        auto ninv = mint(n).inv();
        for (auto& x : f) x *= ninv;
    }

    mints convolute(mints f, mints g) const {
        int fsz = f.size(),
            gsz = g.size();

        // simple convolution in small cases
        if (std::min(fsz, gsz) < 8) {
            mints ret(fsz + gsz - 1, 0);
            for (int i = 0; i < fsz; ++i) {
                for (int j = 0; j < gsz; ++j) {
                    ret[i + j] += f[i] * g[j];
                }
            }
            return ret;
        }

        int n = 1 << clog2(fsz + gsz - 1);
        f.resize(n, mint(0));
        g.resize(n, mint(0));

        ntt(f);
        ntt(g);

        for (int i = 0; i < n; ++i) f[i] *= g[i];

        intt(f);

        f.resize(fsz + gsz - 1);
        return f;
    }
};

constexpr int MOD = 998244353;
using mint = ModInt<MOD>;
const NumberTheoreticalTransform<MOD, 3> NTT;

void solve() {
    int p, q, n;
    std::cin >> p >> q >> n;

    std::vector<mint> xs(n, 0), ys(n, 0);
    while (p--) {
        int x;
        std::cin >> x;
        xs[x - 1] = 1;
    }
    while (q--) {
        int x;
        std::cin >> x;
        ys[n - x] = 1;
    }

    auto zs = NTT.convolute(xs, ys);

    int m;
    std::cin >> m;
    for (int k = 0; k < m; ++k) {
        std::cout << zs[k + n - 1] << "\n";
    }
}

int main() {
    std::cin.tie(nullptr);
    std::ios::sync_with_stdio(false);

    solve();

    return 0;
}
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