HDU 6061 RXD and functions（NTT）

最新推荐文章于 2021-07-09 15:27:21 发布

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最新推荐文章于 2021-07-09 15:27:21 发布

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分类专栏：多项式及生成函数

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本文介绍了一种通过快速数论变换(NTT)高效计算多项式变换的方法，并给出了具体的实现代码。文中详细阐述了变换过程及如何简化计算。

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RXD and functions

首先是有一个结论，对多项式做任意多次 transformation ，其结果跟做一次 transformation $\sum\limits_{i = 1} ^{m} a_i)$ 的结果是一样的，所以我们约定 $-\sum\limits_{i = 1} ^{m} a_i$ 。
$a)\\ \sum_{i = 0} ^{n} c_i(x + a) ^ i\\ \sum_{i = 0} ^{n} \sum_{j = 0} ^{i}c_i \frac{i!}{j!(i - j)!} x ^ j a ^ {i - j}\\ 我们考虑只求这个多项式的x ^ j系数\\ g_j = \sum_{i = j} ^{n} c_i \frac{i!}{j!(i - j)!}a ^{i - j}\\ g_j = \sum_{i = 0} ^{n - j} c_{i + j} \frac{(i + j)!}{j! i!} a ^ i\\ g_j = \frac{1}{j!} \sum_{i = 0} ^{n - j} (c_{i + j} (i + j)!) \times (\frac{a ^i}{i!})\\ 设f(n) = c_{n} n!, h(n) = \frac{a ^ n}{n!} \\ 不难发现后项像是多项式卷积的形式，考虑翻转\frac{a ^ i}{ i}，即i = n - i\\ g_j = \frac{1}{j!} \sum_{i = 0} ^{n - j} f(i + j) h(n - i)\\ 这就是一个完美的卷积形式了，于是可以NTT求解$

#include <bits/stdc++.h>

using namespace std;

const int N = 1e6 + 10, mod = 998244353;

int r[N];

int quick_pow(int a, int n) {
  int ans = 1;
  while (n) {
    if (n & 1) {
      ans = 1ll * a * ans % mod;
    }
    a = 1ll * a * a % mod;
    n >>= 1;
  }
  return ans;
}

void get_r(int lim) {
  for (int i = 0; i < lim; i++) {
    r[i] = (i & 1) * (lim >> 1) + (r[i >> 1] >> 1);
  }
}

void NTT(int *f, int lim, int rev) {
  for (int i = 0; i < lim; i++) {
    if (i < r[i]) {
      swap(f[i], f[r[i]]);
    }
  }
  for (int mid = 1; mid < lim; mid <<= 1) {
    int wn = quick_pow(3, (mod - 1) / (mid << 1));
    for (int len = mid << 1, cur = 0; cur < lim; cur += len) {
      int w = 1;
      for (int k = 0; k < mid; k++, w = 1ll * w * wn % mod) {
        int x = f[cur + k], y = 1ll * w * f[cur + mid + k] % mod;
        f[cur + k] = (x + y) % mod, f[cur + mid + k] = (x - y + mod) % mod;
      }
    }
  }
  if (rev == -1) {
    int inv = quick_pow(lim, mod - 2);
    reverse(f + 1, f + lim);
    for (int i = 0; i < lim; i++) {
      f[i] = 1ll * f[i] * inv % mod;
    }
  }
}

int a[N], c[N], A[N], B[N], fac[N], ifac[N], n, m, sum;

void init() {
  fac[0] = 1;
  for (int i = 1; i < N; i++) {
    fac[i] = 1ll * fac[i - 1] * i % mod;
  }
  ifac[N - 1] = quick_pow(fac[N - 1], mod - 2);
  for (int i = N - 2; i >= 0; i--) {
    ifac[i] = 1ll * ifac[i + 1] * (i + 1) % mod;
  }
}

int main() {
  // freopen("in.txt", "r", stdin);
  // freopen("out.txt", "w", stdout);
  // ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  init();
  while (scanf("%d", &n) != EOF) {
    for (int i = 0; i <= n; i++) {
      scanf("%d", &c[i]);
    }
    scanf("%d", &m);
    sum = 0;
    for (int i = 1; i <= m; i++) {
      scanf("%d", &a[i]);
      sum = (sum + a[i]) % mod;
    }
    sum = (mod - sum) % mod;
    int cur = 1;
    for (int i = 0; i <= n; i++) {
      A[i] = 1ll * c[i] * fac[i] % mod;
      B[n - i] = 1ll * ifac[i] * cur % mod;
      cur = 1ll * cur * sum % mod;
    }
    int lim = 1;
    while (lim <= 2 * n) {
      lim <<= 1;
    }
    get_r(lim);
    NTT(A, lim, 1);
    NTT(B, lim, 1);
    for (int i = 0; i < lim; i++) {
      A[i] = 1ll * A[i] * B[i] % mod;
    }
    NTT(A, lim, -1);
    for (int i = 0; i <= n; i++) {
      printf("%lld ", 1ll * ifac[i] * A[n + i] % mod);
    }
    puts("");
    for (int i = 0; i < lim; i++) {
      A[i] = B[i] = 0;
    }
  }
  return 0;
}