标签：pw int rev Transform Chirp 1LL choose ipw

推导

\[f(c^i)=\sum_{j=0}^{n-1}a_jc^{ij} \]

\[ij={i+j\choose 2}-{i\choose 2}-{j\choose 2} \]

\[\begin{aligned} f(c^i)&=\sum_{j=0}^{n-1}a_jc^{{i+j\choose 2}-{i\choose 2}-{j\choose 2}}\\ &=c^{-{i\choose 2}}\sum_{j=0}^{n-1}a_jc^{-{j\choose 2}}c^{i+j\choose 2} \end{aligned} \]

Code

#include <bits/stdc++.h>
using std::swap; using std::reverse;
const int N = 1000005, M = 40005, P = 998244353, inv2 = P+1>>1;
int n, c, m, L, a[N*4], b[N*4];
int inc(int a, int b) { return (a += b) >= P ? a - P : a; }
int qpow(int a, int b) {
	int t = 1;
	for (; b; b >>= 1, a = 1LL * a * a % P)
		if (b & 1) t = 1LL * t * a % P;
	return t;
}
int W[N*4], pw[N*4], ipw[N*4];
void prework(int n) {
	for (int i = 1; i < n; i <<= 1)
		for (int j = W[i] = 1, Wn = qpow(3, (P-1)/i>>1); j < i; j++)
			W[i+j] = 1LL * W[i+j-1] * Wn % P;
	pw[0] = ipw[0] = pw[1] = ipw[1] = 1; pw[2] = c, ipw[2] = qpow(c, P-2);
	for (int i = 3; i < n; i++)
		pw[i] = 1LL * pw[i-1] * pw[2] % P, ipw[i] = 1LL * ipw[i-1] * ipw[2] % P;
	for (int i = 3; i < n; i++)
		pw[i] = 1LL * pw[i-1] * pw[i] % P, ipw[i] = 1LL * ipw[i-1] * ipw[i] % P;
}
void fft(int *a, int n, int op) {
	static int rev[N*4] = {0};
	for (int i = 1; i < n; i++)
		if ((rev[i] = rev[i>>1]>>1|(i&1?n>>1:0)) > i) swap(a[i], a[rev[i]]);
	for (int q = 1; q < n; q <<= 1)
		for (int p = 0; p < n; p += q << 1)
			for (int i = 0, t; i < q; i++)
				t = 1LL * W[q+i] * a[p+q+i] % P, a[p+q+i] = inc(a[p+i], P-t), a[p+i] = inc(a[p+i], t);
	if (op) return;
	for (int i = 0, inv = qpow(n, P-2); i < n; i++) a[i] = 1LL * a[i] * inv % P;
	reverse(a + 1, a + n);
}
int getsz(int n) { int x = 1; while (x < n) x <<= 1; return x; }
int main() {
	scanf("%d%d%d", &n, &c, &m);
	for (int i = 0; i < n; i++) scanf("%d", &a[i]);
	int L = getsz(n + m); prework(L);
	for (int i = 0; i < n; i++) a[i] = 1LL * a[i] * ipw[i] % P;
	reverse(a, a + n + 1);
	for (int i = 0; i < n + m; i++) b[i] = pw[i];
	fft(a, L, 1), fft(b, L, 1);
	for (int i = 0; i < L; i++) a[i] = 1LL * a[i] * b[i] % P;
	fft(a, L, 0);
	for (int i = 0; i < m; i++) printf("%d ", 1LL * a[n + i] * ipw[i] % P);
	return 0;
}

标签：pw,int,rev,Transform,Chirp,1LL,choose,ipw
来源： https://www.cnblogs.com/ac-evil/p/14418335.html

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