[洛谷P4721]【模板】分治 FFT

题目大意：给定长度为$n-1$的数组$g_{[1,n)}$，求$f_{[0,n)}$，要求：

$$
f_i=\sum_{j=1}^if_{i-j}g_j\\
f_0=1
$$

题解：直接求复杂度是$O(n^2)$，明显不可以通过此题

分治$FFT$，可以用$CDQ$分治，先求出$f_{[l,mid)}$，可以发现这部分对区间的$f_{[mid,r)}$的贡献是$f_{[l,mid)}*g_{[0,r-l)}$，卷出来加到对应位置就行了，复杂度$O(n\log_2^2n)​$

卡点：无

 

C++ Code：

#include <algorithm>
#include <cstdio>
#include <cctype>
namespace std {
	struct istream {
#define M (1 << 21 | 3)
		char buf[M], *ch = buf - 1;
		inline istream() {
#ifndef ONLINE_JUDGE
			freopen("input.txt", "r", stdin);
#endif
			fread(buf, 1, M, stdin);
		}
		inline istream& operator >> (int &x) {
			while (isspace(*++ch));
			for (x = *ch & 15; isdigit(*++ch); ) x = x * 10 + (*ch & 15);
			return *this;
		}
#undef M
	} cin;
	struct ostream {
#define M (1 << 21 | 3)
		char buf[M], *ch = buf - 1;
		int w;
		inline ostream& operator << (int x) {
			if (!x) {
				*++ch = '0';
				return *this;
			}
			for (w = 1; w <= x; w *= 10);
			for (w /= 10; w; w /= 10) *++ch = (x / w) ^ 48, x %= w;
			return *this;
		}
		inline ostream& operator << (const char x) {*++ch = x; return *this;}
		inline ~ostream() {
#ifndef ONLINE_JUDGE
			freopen("output.txt", "w", stdout);
#endif
			fwrite(buf, 1, ch - buf + 1, stdout);
		}
#undef M
	} cout;
}

#define maxn 131072 | 3
const int mod = 998244353, G = 3;

namespace Math {
	inline int pw(int base, int p) {
		static int res;
		for (res = 1; p; p >>= 1, base = static_cast<long long> (base) * base % mod) if (p & 1) res = static_cast<long long> (res) * base % mod;
		return res;
	}
	inline int inv(int x) {return pw(x, mod - 2);}
}

int n;
int f[maxn], g[maxn];
namespace Poly {
#define N 131072 | 3
	int s, lim, ilim, rev[N];
	int Wn[N + 1];
	inline void reduce(int &x) {x += x >> 31 & mod;}
	inline void clear(register int *l, const int *r) {
		if (l >= r) return ;
		while (l != r) *l++ = 0;
	}
	inline void init(const int n) {
		s = -1, lim = 1; while (lim <= n) lim <<= 1, s++; ilim = Math::inv(lim);
		for (int i = 1; i < lim; i++) rev[i] = rev[i >> 1] >> 1 | (i & 1) << s;
		const int t = Math::pw(G, (mod - 1) / lim);
		*Wn = 1; for (register int *i = Wn; i != Wn + lim; ++i) *(i + 1) = static_cast<long long> (*i) * t % mod;
	}

	inline void NTT(int *A, const int op = 1) {
		for (register int i = 1; i < lim; i++) if (i < rev[i]) std::swap(A[i], A[rev[i]]);
		for (register int mid = 1; mid < lim; mid <<= 1) {
			const int t = lim / mid >> 1;
			for (register int i = 0; i < lim; i += mid << 1) {
				for (register int j = 0; j < mid; j++) {
					const int W = op ? Wn[t * j] : Wn[lim - t * j];
					const int X = A[i + j], Y = static_cast<long long> (A[i + j + mid]) * W % mod;
					reduce(A[i + j] += Y - mod), reduce(A[i + j + mid] = X - Y);
				}
			}
		}
		if (!op) for (int i = 0; i < lim; i++) A[i] = static_cast<long long> (A[i]) * ilim % mod;
	}

	int A[N], B[N];
	void CDQ_NTT(const int l, const int r) {
		if (r - l < 2) return ;
		const int mid = l + r >> 1;
		CDQ_NTT(l, mid); init(r - l);
		std::copy(f + l, f + mid, A); clear(A + mid - l, A + lim);
		std::copy(g, g + r - l, B); clear(B + r - l, B + lim);
		NTT(A), NTT(B);
		for (int i = 0; i < lim; i++) A[i] = static_cast<long long> (A[i]) * B[i] % mod;
		NTT(A, 0);
		for (int i = mid; i < r; i++) reduce(f[i] += A[i - l] - mod);
		CDQ_NTT(mid, r);
	}
#undef N
}

int main() {
	std::cin >> n;
	for (int i = 1; i < n; i++) std::cin >> g[i];
	*f = 1;
	Poly::CDQ_NTT(0, n);
	for (int i = 0; i < n; i++) std::cout << f[i] << ' ';
	std::cout << '\n';
	return 0;
}

　　

来源：https://www.cnblogs.com/Memory-of-winter/p/10127918.html