「THUPC 2017」sum

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Problem

Description

给定一个长为 n 的数列 \{a_{n}\} ，定义数列 f ：

f_{k} = \sum_{i = 1}^{n} a_{i}^{k} \pmod{998244353}

现要求计算 f_{1}, f_{2}, \dots, f_{n} 的值。

Constraints

1 \le n \le 2 \times 10^{5}, 0 \le a_{i} \le 10^{9}

Solution

Analysis

设数列 \{f_{n}\} 的 OGF 为 F(x) ，那么有：

\begin{aligned} F(x) &= \sum_{i \ge 0} f_{i} x^{i} \\ &= \sum_{i \ge 0} x^{i} \sum_{j = 1}^{n} a_{j}^{i} \\ &= \sum_{i = 1}^{n} \sum_{j \ge 0} (a_{i} x)^{j} \\ &= \sum_{i = 1}^{n} \frac{1}{1 - a_{i} x} \end{aligned}

但是到这一步还是不能快速求出 F(x) 。考虑通过取对数来将求和变为求积，然后就可以分治 FFT 做了。

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

这样我们就得到了一个可以通过分治 FFT 快速求出的形式，直接做就行了。

时间复杂度 O\left(n \log^{2}{n}\right) 。

Code

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#include <cstdio>
#include <algorithm>

using i64 = long long;

namespace IOManager {
	constexpr int FILESZ = 131072;
	char buf[FILESZ];
	const char *ibuf = buf, *tbuf = buf;

	struct IOManager {
		inline char gc() {
			return (ibuf == tbuf) && (tbuf = (ibuf = buf) + fread(buf, 1, FILESZ, stdin), ibuf == tbuf) ? EOF : *ibuf++;
		}

		template <typename _Tp>
		inline operator _Tp() {
			_Tp s = 0u; char c = gc();
			for (; c < 48; c = gc());
			for (; c > 47; c = gc())
				s = (_Tp)(s * 10u + c - 48u);
			return s;
		}
	};
} IOManager::IOManager io;

namespace math {

using i64 = long long;
using u32 = unsigned int;
using u64 = unsigned long long;

constexpr int mod = 998244353;
constexpr int primitive_root = 3;

inline int sum(const int &x, const int &y) {
	return x + y < mod ? x + y : (x + y - mod);
}
inline int sub(const int &x, const int &y) {
	return x < y ? x - y + mod : (x - y);
}

inline void inc(int &x, const int &y) {
	x += y; (mod <= x) and (x -= mod);
}
inline void dec(int &x, const int &y) {
	x -= y; (x < 0) and (x += mod);
}

inline int fpow(int x, int n) {
	int pw = 1;
	for (; n; n >>= 1, x = (i64)x * x % mod)
		if (n & 1) pw = (i64)pw * x % mod;
	return pw;
}

template <class _Tp>
inline void swap(_Tp &x, _Tp &y) {
	_Tp z = x; x = y; y = z;
}

}

namespace poly {

using namespace math;

constexpr int maxn = 524288;

using poly_t = int[maxn];
using poly = int *const;

inline int calcpw2(const int &n) {
	for (int pw = 1; ; pw <<= 1)
		if (n <= pw) return pw;
}

poly_t wn, iwn, inv;

inline void init() {
	const int unit_root = fpow(primitive_root, (mod - 1) / maxn);
	wn[0] = iwn[0] = inv[1] = 1;
	for (int i = 1; i != maxn; ++i)
		wn[i] = (i64)wn[i - 1] * unit_root % mod;
	std::reverse_copy(wn + 1, wn + maxn, iwn + 1);
	for (int i = 2; i != maxn; ++i)
		inv[i] = (i64)inv[mod % i] * (mod - mod / i) % mod;
}

void DFT(poly &a, const int &n, const poly &omega = wn) {
	static poly_t curw; curw[0] = 1;
	for (int i = 0, j = 0; i != n; ++i) {
		if (i < j) swap(a[i], a[j]);
		for (int k = n >> 1; (j xor_eq k) < k; k >>= 1);
	}

	poly endpos = a + n, wendpos = omega + maxn;
	for (int l = 1, tp = maxn >> 1; l != n; l <<= 1, tp >>= 1) {
		for (int *i = curw, *w = omega + tp; w != wendpos; w += tp)
			*++i = *w;
		for (int *i = a, z; i != endpos; i += l + l)
			for (int j = 0; j != l; ++j)
				z = (i64)i[j + l] * curw[j] % mod,
				i[j + l] = sub(i[j], z),
				inc(i[j], z);
	}
}

void IDFT(poly &a, const int &n) {
	DFT(a, n, iwn);
	const int invn = fpow(n, mod - 2);
	for (int i = 0; i != n; ++i)
		a[i] = (i64)a[i] * invn % mod;
}

inline void derivative(const poly &h, const int &n, poly &f) {
	for (int i = 1; i != n; ++i)
		f[i - 1] = (i64)h[i] * i % mod;
	f[n - 1] = 0;
}

inline void integral(const poly &h, const int &n, poly &f) {
	for (int i = n - 1; i; --i)
		f[i] = (i64)h[i - 1] * inv[i] % mod;
	f[0] = 0;
}

void polyinv(const poly &h, const int &n, poly &f) {
	static poly_t inv_t;
	std::fill(f, f + n + n, 0);
	std::fill(inv_t, inv_t + n + n, 0);
	f[0] = fpow(h[0], mod - 2);
	for (int t = 2; t <= n; t <<= 1) {
		const int t2 = t << 1;
		std::copy(h, h + t, inv_t);

		DFT(f, t2); DFT(inv_t, t2);
		for (int i = 0; i != t2; ++i)
			f[i] = (i64)f[i] * sub(2, (i64)f[i] * inv_t[i] % mod) % mod;
		IDFT(f, t2);

		std::fill(f + t, f + t2, 0);
	}
}

void polyln(const poly &h, const int &n, poly &f) {
	if (h[0] != 1) throw "in function polyln(): the constant term not equal to 1";
	static poly_t ln_t;
	const int t = n << 1;

	derivative(h, n, ln_t);
	std::fill(ln_t + n, ln_t + t, 0);
	polyinv(h, n, f);

	DFT(ln_t, t); DFT(f, t);
	for (int i = 0; i != t; ++i)
		ln_t[i] = (i64)ln_t[i] * f[i] % mod;
	IDFT(ln_t, t);

	integral(ln_t, n, f);
}

inline void mul(poly &a, const int &an, poly &b, const int &bn, poly &c) {
	static poly_t f, g;
	const int n = calcpw2(an + bn + 1);
	f[0] = 1; std::copy(a, a + an, f + 1); std::fill(f + an + 1, f + n, 0);
	g[0] = 1; std::copy(b, b + bn, g + 1); std::fill(g + bn + 1, g + n, 0);

	DFT(f, n); DFT(g, n);
	for (int i = 0; i != n; ++i)
		f[i] = (i64)f[i] * g[i] % mod;
	IDFT(f, n);

	std::copy(f + 1, f + an + bn + 1, c);
}

}

using math::mod;

poly::poly_t a, f;

void calc(const int &l, const int &r) {
	if (l == r) return void(f[l] = mod - a[l]);
	const int m = (l + r) >> 1;
	calc(l, m); calc(m + 1, r);
	poly::mul(f + l, m - l + 1, f + m + 1, r - m, f + l);
}

int main() {
	poly::init();
	for (int T = io; T; --T) {
		const int n = io;
		for (int i = 1; i <= n; ++i)
			a[i] = io, (mod <= a[i]) and (a[i] -= mod);

		f[0] = 1; calc(1, n);
		poly::polyln(f, poly::calcpw2(n + 1), a);

		int ans = 0;
		for (int i = 1; i <= n; ++i)
			ans xor_eq (mod - (i64)a[i] * i % mod);
		printf("%d\n", ans);
	}

	return 0;
}