LOJ#2409. 「THUPC 2017」小 L 的计算题 / Sum（生成函数）

题意

给定一个长为 $n$ 的序列 $\{a_i\}$ 对于 $k \in [1, n]$ 求

f_{k} = \sum_{i = 1}^{n} a_{i}^{k} (\mod 998244353)

$f_k = \sum_{i = 1}^{n} a_i^k \pmod {998244353}$

$n \le 2 \times 10^5$

题解

不会牛顿恒等式TAT，参考了这位大佬的博客。

我们令 $F(x)$ 为 $f_k$ 的生成函数，我们有

\begin{aligned} F (x) & = \sum_{k} (\sum_{i = 1}^{n} a_{i}^{k}) x^{k} \\ = \sum_{i = 1}^{n} \sum_{k} a_{i}^{k} x^{k} \\ = \sum_{i = 1}^{n} \frac{1}{1 - a_{i} x} \end{aligned}

$\begin{aligned} F(x) &= \sum_{k} (\sum_{i = 1}^{n} a_i^k) x^k\\ &=\sum_{i = 1}^n \sum_k a_i^kx^k\\ &=\sum_{i = 1}^{n} \frac{1}{1 - a_ix}\\ \end{aligned}$

这几步都比较基础，但看起来还是挺不可做的，我们用一些神奇的操作。

\begin{aligned} F (x) & = \sum_{i = 1}^{n} (1 + \frac{a_{i} x}{1 - a_{i} x}) \\ = n - x \sum_{i = 1}^{n} \frac{- a_{i}}{1 - a_{i} x} \\ = n - x \sum_{i = 1}^{n} \ln^{'} (1 - a_{i} x) \\ = n - x \ln^{'} (\prod_{i = 1}^{n} (1 - a_{i} x)) \end{aligned}

$\begin{aligned} F(x) &= \sum_{i = 1}^{n} (1 + \frac{a_ix}{1 - a_ix})\\ &= n - x \sum_{i= 1}^{n} \frac{-a_i}{1 - a_ix}\\ &= n - x \sum_{i = 1}^{n} \ln'(1 - a_ix)\\ &= n - x \ln'(\prod_{i = 1}^n(1 - a_ix)) \end{aligned}$

我们先分治求出 $\prod_{i = 1}^{n}(1 - a_ix)$ 然后套 $\ln$ 的板子即可。 $\mathcal O(n \log^2 n)$

代码

#include <bits/stdc++.h>

#define For(i, l, r) for (register int i = (l), i##end = int(r); i <= i##end; ++i)
#define Fordown(i, r, l) for (register int i = (r), i##end = (int)(l); i >= i##end; --i)
#define Rep(i, r) for (register int i = (0), i##end = int(r); i < i##end; ++i)
#define Set(a, v) memset(a, v, sizeof(a))
#define Cpy(a, b) memcpy(a, b, sizeof(a))
#define debug(x) cout << #x << ": " << (x) << endl

using namespace std;

typedef vector<int> VI;

template<typename T> inline bool chkmin(T &a, T b) { return b < a ? a = b, 1 : 0; }
template<typename T> inline bool chkmax(T &a, T b) { return b > a ? a = b, 1 : 0; }

inline int read() {
	int x(0), sgn(1); char ch(getchar());
	for (; !isdigit(ch); ch = getchar()) if (ch == '-') sgn = -1;
	for (; isdigit(ch); ch = getchar()) x = (x * 10) + (ch ^ 48);
	return x * sgn;
}

void File() {
#ifdef zjp_shadow
	freopen ("2409.in", "r", stdin);
	freopen ("2409.out", "w", stdout);
#endif
}

const int Mod = 998244353;

inline int fpm(int x, int power) {
	int res(1);
	for (; power; power >>= 1, x = 1ll * x * x % Mod)
		if (power & 1) res = 1ll * res * x % Mod;
	return res;
}

int findlen(int l) {
	int res = 1; while (res <= l) res <<= 1; return res;
}

template<int Maxn>
struct Poly {

	const int g = 3, invg = fpm(g, Mod - 2);

	int rev[Maxn], W[Maxn], len;

	void NTT(int *P, int opt) {
		Rep (i, len) if (i < rev[i]) swap(P[i], P[rev[i]]);
		for (int i = 2, p; p = i >> 1, i <= len; i <<= 1) {
			W[0] = 1; W[1] = fpm(~opt ? g : invg, (Mod - 1) / i);
			For (k, 2, p - 1) W[k] = 1ll * W[k - 1] * W[1] % Mod;
			for (int j = 0; j < len; j += i) Rep (k, p) {
				int u = P[j + k], v = 1ll * P[j + k + p] * W[k] % Mod;
				P[j + k] = (u + v) % Mod; P[j + k + p] = (u - v + Mod) % Mod;
			}
		}
		if (!~opt) {
			int invn = fpm(len, Mod - 2);
			Rep (i, len) P[i] = 1ll * P[i] * invn % Mod;
		}
	}

	void Prepare(int lc) {
		int cnt = -1;
		for (len = 1; len <= lc; len <<= 1) ++ cnt;
		Rep (i, len) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << cnt);
	}

	int A[Maxn], B[Maxn], C[Maxn];
	void Mult(int *a, int *b, int *c, int la, int lb) {
		int lc = la + lb; Prepare(lc);
		if (lc <= 400) {
			static int tmp[410] = {0};
			For (i, 0, la) if (a[i]) For (j, 0, lb)
				tmp[i + j] = (tmp[i + j] + 1ll * a[i] * b[j]) % Mod;
			For (i, 0, lc) c[i] = tmp[i], tmp[i] = 0; return;
		}
		Rep (i, len) A[i] = i <= la ? a[i] : 0; NTT(A, 1);
		Rep (i, len) B[i] = i <= lb ? b[i] : 0; NTT(B, 1);
		Rep (i, len) C[i] = 1ll * A[i] * B[i] % Mod; NTT(C, -1);
		For (i, 0, lc) c[i] = C[i];
	}

	VI Mult(VI a, VI b) {
		static int ta[Maxn], tb[Maxn], tc[Maxn]; VI c;
		Rep (i, a.size()) ta[i] = a[i];
		Rep (i, b.size()) tb[i] = b[i];
		Mult(ta, tb, tc, a.size() - 1, b.size() - 1);
		Rep (i, a.size() + b.size() - 1) c.push_back(tc[i]); return c;
	}

	void Inv(int *f, int *g, int lf) {
		if (lf == 1) return void(g[0] = fpm(f[0], Mod - 2));
		Inv(f, g, lf >> 1); Prepare(lf << 1);
		Rep (i, len) A[i] = i < lf ? f[i] : 0; NTT(A, 1); 
		Rep (i, len) B[i] = i < lf ? g[i] : 0; NTT(B, 1);
		Rep (i, len) C[i] = 1ll * A[i] * B[i] % Mod * B[i] % Mod; NTT(C, -1);
		Rep (i, lf) g[i] = (g[i] * 2ll + Mod - C[i]) % Mod;
	}

	int inv[Maxn];
	Poly() {
		inv[1] = 1;
		For (i, 2, Maxn - 1)
			inv[i] = 1ll * inv[Mod % i] * (Mod - Mod / i) % Mod;
	}

	void Der(int *f, int *g, int lf) {
		For (i, 1, lf) g[i - 1] = 1ll * i * f[i] % Mod; g[lf] = 0;
	}

	void Int(int *f, int *g, int lf) {
		g[0] = 0;
		For (i, 1, lf + 1)
			g[i] = 1ll * f[i - 1] * inv[i] % Mod;
	}

	void Ln(int *f, int *g, int lf) {
		static int der[Maxn], tmp[Maxn];
		Der(f, der, lf); Inv(f, tmp, lf);
		Mult(der, tmp, tmp, lf, lf); Int(tmp, g, lf);
	}

};

const int N = 1 << 20;

Poly<N> T;

int n, a[N], f[N], g[N], ans[N];

VI Solve(int l, int r) {
	if (l == r) return {1, Mod - a[l]};
	int mid = (l + r) >> 1;
	return T.Mult(Solve(l, mid), Solve(mid + 1, r));
}

int main () {

	File();

	int cases = read();

	while (cases --) {
		For (i, 1, n = read()) a[i] = read() % Mod;

		VI res = Solve(1, n);
		For (i, 0, n) f[i] = res[i];

		T.Ln(f, g, findlen(n)); T.Der(g, f, n);
		ans[0] = n; For (i, 1, n) ans[i] = Mod - f[i - 1];

		int Ans = 0;
		For (i, 1, n) Ans ^= ans[i];
		printf ("%d\n", Ans);
	}

	return 0;

}

__EOF__

本文作者：zjp_shadow
本文链接：https://www.cnblogs.com/zjp-shadow/p/11025360.html
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