♯p Subset Sum

2qbingxuan

2025-02-27

♯p Subset Sum : 5e5

Description

https://judge.yosupo.jp/problem/sharp_p_subset_sum

有 $N$ 個小於等於 $T$ 的正整數 $s_0,s_1,\dots,s_N$，對於所有 $t = 1,2,\dots,T$，計算有幾個 $I \subseteq \{0,1,\dots,N-1\}$ 使得 $\sum _ {i\in I} s_i = t$，輸出模 $998244353$ 的餘數

$N, T \leq 5 \times 10 ^ 5$

Solution

從 library-checker 學到的像是冷知識的東西。

簡單來說就是要算多項式
$$
f(x) = \prod _ {i=0} ^ {N-1} (1 + x^{s_i})
$$
的前 $T + 1$ 項係數。

看到連乘先想能不能多項式分治，但這邊總度數可能會很大。於是我們改對 $f$ 取 log。

$$
\begin{aligned}
\log f(x)
&= \sum _ {i=0} ^ {N-1} \log (1 + x^{s_i}) \\
&= \sum _ {i=0} ^ {N-1} \sum _ {j=1} ^ \infty \frac{(-1)^{j+1}}{j} (x^{s_i})^j \\
&= \sum _ {i=0} ^ {N-1} \sum _ {j=1} ^ \infty \frac{(-1)^{j+1}}{j} (x^{s_i})^j \\
&= \sum _ {k=1} ^ T a_k \sum _ {j=1} ^ \infty \frac{(-1)^{j+1}}{j} (x^k)^j \\
&= \sum _ {k=1} ^ T a_k \sum _ {j=1} ^ \infty \frac{(-1)^{j+1}}{j} x^{kj} \\
&\equiv \sum _ {k=1} ^ T a_k \sum _ {j=1} ^ {\lfloor T / k \rfloor} \frac{(-1)^{j+1}}{j} x^{kj} \pmod {x^{T+1}} \\
\end{aligned}
$$

其中 $a_k$ 代表有幾個 $s_i = k$。這邊用到 $\log (1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} \cdots$

照著這個式子我們就可以花 $\sum _ {1 \leq k \leq T} \lfloor T/k \rfloor$ 的時間（調和級數！）來得到 $\log f(x) \mod x^{T+1}$，再套個多項式 exp 的模板就可以在總時間 $\mathcal{O}(N + T\log T)$ 解決本題。

有關 exp、log 與乘法的關係，比較精準的說明請參考 maspy 的部落格

AC code

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// An AC a day keeps the doctor away.
#include <bits/stdc++.h>
using namespace std;

#define all(x) begin(x), end(x)
#ifdef CKISEKI
#include <experimental/iterator>
#define safe cerr<<__PRETTY_FUNCTION__<<" line "<<__LINE__<<" safe\n"
#define debug(a...) debug_(#a, a)
#define orange(a...) orange_(#a, a)
void debug_(auto s, auto ...a) {
  cerr << "\e[1;32m(" << s << ") = (";
  int f = 0;
  (..., (cerr << (f++ ? ", " : "") << a));
  cerr << ")\e[0m\n";
}
void orange_(auto s, auto L, auto R) {
  cerr << "\e[1;33m[ " << s << " ] = [ ";
  using namespace experimental;
  copy(L, R, make_ostream_joiner(cerr, ", "));
  cerr << " ]\e[0m\n";
}
#else
#define safe ((void)0)
#define debug(...) safe
#define orange(...) safe
#endif

const int mod = 998244353;
int add(int a, int b) {
  return a + b >= mod ? a + b - mod : a + b;
}
int sub(int a, int b) {
  return a - b < 0 ? a - b + mod : a - b;
}
int mul(int64_t a, int64_t b) {
  return static_cast<int>(a * b % mod);
}
int modpow(int e, int p) {
  int r = 1;
  while (p) {
    if (p & 1) r = mul(r, e);
    e = mul(e, e);
    p >>= 1;
  }
  return r;
}
int modinv(int x) {
  return modpow(x, mod - 2);
}

template <int mod, int G, int maxn> struct NTT {
  static_assert (maxn == (maxn & -maxn));
  int roots[maxn];
  NTT () {
    int r = modpow(G, (mod - 1) / maxn);
    for (int i = maxn >> 1; i; i >>= 1) {
      roots[i] = 1;
      for (int j = 1; j < i; j++)
        roots[i + j] = mul(roots[i + j - 1], r);
      r = mul(r, r);
      // for (int j = 0; j < i; j++) // FFT (tested)
      //   roots[i+j] = polar<llf>(1, PI * j / i);
    }
  }
  // n must be 2^k, and 0 <= F[i] < mod
  template <typename T>
  void operator()(int F[], T n, bool inv = false) {
    for (T i = 0, j = 0; i < n; i++) {
      if (i < j) swap(F[i], F[j]);
      for (T k = n>>1; (j^=k) < k; k>>=1);
    }
    for (T s = 1; s < n; s *= 2) {
      for (T i = 0; i < n; i += s * 2) {
        for (T j = 0; j < s; j++) {
          int a = F[i+j], b = mul(F[i+j+s], roots[s+j]);
          F[i+j] = add(a, b);   // a + b
          F[i+j+s] = sub(a, b); // a - b
        }
      }
    }
    if (inv) {
      int iv = modinv(int(n));
      for (T i = 0; i < n; i++) F[i] = mul(F[i], iv);
      reverse(F + 1, F + n);
    }
  }
};

using lld = int64_t;

int modpow(lld e, int p, int m) {
  lld r = 1;
  while (p) {
    if (p & 1) r = r * e % m;
    e = e * e % m;
    p >>= 1;
  }
  return (int)r;
}

int get_root(int n, int P = mod) { // ensure 0 <= n < p
  if (P == 2 or n == 0) return n;
  auto check = [&](lld x) {
    return modpow(int(x), (P - 1) / 2, P); };
  if (check(n) != 1) return -1;
  mt19937 rnd(7122); lld z = 1, w;
  while (check(w = (z * z - n + P) % P) != P - 1)
    z = rnd() % P;
  const auto M = [P, w](auto &u, auto &v) {
    auto [a, b] = u; auto [c, d] = v;
    return make_pair((a * c + b * d % P * w) % P,
        (a * d + b * c) % P);
  };
  pair<lld, lld> r(1, 0), e(z, 1);
  for (int q = (P + 1) / 2; q; q >>= 1, e = M(e, e))
    if (q & 1) r = M(r, e);
  return int(r.first); // sqrt(n) mod P where P is prime
}
NTT<mod, 3, (1<<20)> ntt;

#define fi(l, r) for (size_t i = (l); i < (r); i++)
using S = vector<int>;
auto Mul(auto a, auto b, size_t sz) {
  a.resize(sz), b.resize(sz);
  ntt(a.data(), sz); ntt(b.data(), sz);
  fi(0, sz) a[i] = mul(a[i], b[i]);
  return ntt(a.data(), sz, true), a;
}
S Newton(const S &v, int init, auto &&iter) {
  S Q = { init };
  for (int sz = 2; Q.size() < v.size(); sz *= 2) {
    S A{begin(v), begin(v) + min(sz, int(v.size()))};
    A.resize(sz * 2), Q.resize(sz * 2);
    iter(Q, A, sz * 2); Q.resize(sz);
  }
  return Q.resize(v.size()), Q;
}
S Inv(const S &v) { // v[0] != 0
  return Newton(v, modinv(v[0]),
    [](S &X, S &A, int sz) {
      ntt(X.data(), sz), ntt(A.data(), sz);
      for (int i = 0; i < sz; i++)
        X[i] = mul(X[i], sub(2, mul(X[i], A[i])));
      ntt(X.data(), sz, true); });
}
S Dx(S A) {
  fi(1, A.size()) A[i - 1] = mul(i, A[i]);
  return A.empty() ? A : (A.pop_back(), A);
}
S Sx(S A) {
  A.insert(A.begin(), 0);
  fi(1, A.size()) A[i] = mul(modinv(int(i)), A[i]);
  return A;
}
S Ln(const S &A) { // coef[0] == 1; res[0] == 0
  auto B = Sx(Mul(Dx(A), Inv(A), bit_ceil(A.size()*2)));
  return B.resize(A.size()), B;
}
S Exp(const S &v) { // coef[0] == 0; res[0] == 1
  return Newton(v, 1,
    [](S &X, S &A, int sz) {
      auto Y = X; Y.resize(sz / 2); Y = Ln(Y);
      fi(0, Y.size()) Y[i] = sub(A[i], Y[i]);
      Y[0] = add(Y[0], 1); X = Mul(X, Y, sz); });
}
S Pow(S a, lld M) { // period mod*(mod-1)
  assert(!a.empty() && a[0] != 0);
  const auto imul = [&a](int s) {
    for (int &x: a) x = mul(x, s); }; int c = a[0];
  imul(modinv(c)); a = Ln(a); imul(int(M % mod));
  a = Exp(a); imul(modpow(c, int(M % (mod - 1))));
  return a; // mod x^N where N=a.size()
}
S Sqrt(const S &v) { // need: QuadraticResidue
  assert(!v.empty() && v[0] != 0);
  const int r = get_root(v[0]); assert(r != -1);
  return Newton(v, r,
    [](S &X, S &A, int sz) {
      auto Y = X; Y.resize(sz / 2);
      auto B = Mul(A, Inv(Y), sz);
      for (int i = 0, inv2 = mod / 2 + 1; i < sz; i++)
        X[i] = mul(inv2, add(X[i], B[i])); });
}
S Mul(auto &&a, auto &&b) {
  const auto n = a.size() + b.size() - 1;
  auto R = Mul(a, b, bit_ceil(n));
  return R.resize(n), R;
}
S MulT(S a, S b, size_t k) {
  assert(b.size()); reverse(all(b)); auto R = Mul(a, b);
  R = vector(R.begin() + b.size() - 1, R.end());
  return R.resize(k), R;
}
S Eval(const S &f, const S &x) {
  if (f.empty()) return vector(x.size(), 0);
  const int n = int(max(x.size(), f.size()));
  auto q = vector(n * 2, S(2, 1)); S ans(n);
  fi(0, x.size()) q[i + n][1] = sub(0, x[i]);
  for (int i = n - 1; i > 0; i--)
    q[i] = Mul(q[i << 1], q[i << 1 | 1]);
  q[1] = MulT(f, Inv(q[1]), n);
  for (int i = 1; i < n; i++) {
    auto L = q[i << 1], R = q[i << 1 | 1];
    q[i << 1 | 0] = MulT(q[i], R, L.size());
    q[i << 1 | 1] = MulT(q[i], L, R.size());
  }
  for (int i = 0; i < n; i++) ans[i] = q[i + n][0];
  return ans.resize(x.size()), ans;
}
pair<S, S> DivMod(const S &A, const S &B) {
  assert(!B.empty() && B.back() != 0);
  if (A.size() < B.size()) return {{}, A};
  const auto sz = A.size() - B.size() + 1;
  S X = B; reverse(all(X)); X.resize(sz);
  S Y = A; reverse(all(Y)); Y.resize(sz);
  S Q = Mul(Inv(X), Y);
  Q.resize(sz); reverse(all(Q)); X = Mul(Q, B); Y = A;
  fi(0, Y.size()) Y[i] = sub(Y[i], X[i]);
  while (Y.size() && Y.back() == 0) Y.pop_back();
  while (Q.size() && Q.back() == 0) Q.pop_back();
  return {Q, Y};
} // empty means zero polynomial
int LinearRecursionKth(S a, S c, int64_t k) {
  const auto d = a.size(); assert(c.size() == d + 1);
  const auto sz = bit_ceil(2 * d + 1), o = sz / 2;
  S q = c; for (int &x: q) x = sub(0, x); q[0]=1;
  S p = Mul(a, q); p.resize(sz); q.resize(sz);
  for (int r; r = (k & 1), k; k >>= 1) {
    fill(d + all(p), 0); fill(d + 1 + all(q), 0);
    ntt(p.data(), sz); ntt(q.data(), sz);
    for (size_t i = 0; i < sz; i++)
      p[i] = mul(p[i], q[(i + o) & (sz - 1)]);
    for (size_t i = 0, j = o; j < sz; i++, j++)
      q[i] = q[j] = mul(q[i], q[j]);
    ntt(p.data(), sz, true); ntt(q.data(), sz, true);
    for (size_t i = 0; i < d; i++) p[i] = p[i << 1 | r];
    for (size_t i = 0; i <= d; i++) q[i] = q[i << 1];
  } // Bostan-Mori
  return mul(p[0], modinv(q[0]));
} // a_n = \sum c_j a_(n-j), c_0 is not used

signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);

  int N, T;
  cin >> N >> T;
  vector<int> cnt(T + 1);

  for (int i = 0; i < N; i++) {
    int x;
    cin >> x;
    cnt[x] += 1;
  }

  vector<int> inv(T + 1);
  inv[1] = 1;
  for (int i = 2; i <= T; i++)
    inv[i] = mul(inv[mod % i], mod - mod / i);

  vector<int> a(T + 1);
  for (int i = 1; i <= T; i++) {
    for (int j = 1; i * j <= T; j++) {
      if (j & 1)
        a[i * j] = add(a[i * j], mul(cnt[i], inv[j]));
      else
        a[i * j] = sub(a[i * j], mul(cnt[i], inv[j]));
    }
  }

  a = Exp(a);
  for (int i = 1; i <= T; i++)
    cout << a[i] << (i+1==N ? '\n' : ' ');
}