分治 FFT（多项式求逆，分治fft）
题目
题解
方法一：多项式求逆
令\(g(0)=0\)，原式子可写成
\[f_i=\sum\limits_{j=0}^{i}{f_{i-1}g_j}
\]
把\(f\)，\(g\)看作多项式，等式右边即为\(f\times g\)，这说明有\(f=f\times g\)。除了\(i=0\)时，\((f\times g)_0 = 0\neq f_1\)。因此把它补上，就有
\[\begin{aligned}
f&\equiv f\times g + f_1 &\mod x^n \\
f(1-g)&\equiv f_1 &\mod x^n \\
f&\equiv f_1(1-g)^{-1} &\mod x^n
\end{aligned}
\]
求出\((1-g)^{-1}\)逆元即可。时间复杂度\(O(n\log n)\)。
#include <bits/stdc++.h>
#define endl '\n'
#define IOS std::ios::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define mp make_pair
#define seteps(N) fixed << setprecision(N)
typedef long long ll;
using namespace std;
/*-----------------------------------------------------------------*/
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
#define INF 0x3f3f3f3f
const int N = 1e6 + 10;
const int M = 998244353;
int rev[N];
inline ll qpow(ll a, ll b, ll m) {
ll res = 1;
while (b) {
if (b & 1)
res = (res * a) % m;
a = (a * a) % m;
b = b >> 1;
}
return res;
}
void change(ll y[], int len) { // 蝴蝶变换
for (int i = 0; i < len; ++i) {
rev[i] = rev[i >> 1] >> 1;
if (i & 1) {
rev[i] |= len >> 1;
}
}
for (int i = 0; i < len; ++i) {
if (i < rev[i]) {
swap(y[i], y[rev[i]]);
}
}
return;
}
void ntt(ll y[], int len, int on) { // -1逆变换
change(y, len);
for (int h = 2; h <= len; h <<= 1) {
ll gn = qpow(3, (M - 1) / h, M); // 原根为3
if (on == -1)
gn = qpow(gn, M - 2, M);
for (int j = 0; j < len; j += h) {
ll g = 1;
for (int k = j; k < j + h / 2; k++) {
ll u = y[k];
ll t = g * y[k + h / 2] % M;
y[k] = (u + t) % M;
y[k + h / 2] = (u - t + M) % M;
g = g * gn % M;
}
}
}
if (on == -1) {
ll inv = qpow(len, M - 2, M);
for (int i = 0; i < len; i++) {
y[i] = y[i] * inv % M;
}
}
}
int get(int x) {
int res = 1;
while(res < x) {
res <<= 1;
}
return res;
}
ll f[N], rf[N];
void solve(int len, ll x[], ll y[]) {
if(len == 1) {
y[0] = x[0];
return ;
}
solve(len >> 1, x, y);
for(int i = 0 ;i < (len << 1); i++) {
f[i] = rf[i] = 0;
}
for(int i = 0; i < len / 2; i++) {
rf[i] = y[i];
}
for(int i = 0; i < len; i++) {
f[i] = x[i];
}
ntt(f, len << 1, 1);
ntt(rf, len << 1, 1);
for(int i = 0; i < (len << 1); i++) {
rf[i] = rf[i] * (2 - rf[i] * f[i] % M + M) % M;
}
ntt(rf, len << 1, -1);
for(int i = 0; i < len; i++) y[i] = rf[i];
}
ll a[N], b[N];
int main() {
IOS;
int n;
cin >> n;
for(int i = 1; i < n; i++) {
cin >> a[i];
}
for(int i = 0; i < n; i++) {
a[i] = ((i == 0) - a[i] % M + M) % M;
}
int len = get(n);
solve(len, a, b);
for(int i = 0; i < n; i++) {
cout << b[i] << " \n"[i == n - 1];
}
}
方法二：分治fft
这个后一项的计算需要依赖前一项的贡献，所以用cdq分治计算，步骤大致上为：
- 计算左半部分；
- 转移左半部分的贡献到右半部分；
- 完成右半部分的计算。
注意转移贡献时，一个区间是左半部分区间[l,mid]，另外一个区间是[0,r-l+1]（转移起点是0），这样才能覆盖到右半部分区间[mid+1,r]
详见代码
#include <bits/stdc++.h>
#define endl '\n'
#define IOS std::ios::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define mp make_pair
#define seteps(N) fixed << setprecision(N)
typedef long long ll;
using namespace std;
/*-----------------------------------------------------------------*/
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
#define INF 0x3f3f3f3f
const int N = 3e5 + 10;
const int M = 998244353;
const double eps = 1e-5;
int rev[N];
inline ll qpow(ll a, ll b, ll m) {
ll res = 1;
while(b) {
if(b & 1) res = (res * a) % m;
a = (a * a) % m;
b = b >> 1;
}
return res;
}
void change(ll y[], int len) {
for(int i = 0; i < len; ++i) {
rev[i] = rev[i >> 1] >> 1;
if(i & 1) {
rev[i] |= len >> 1;
}
}
for(int i = 0; i < len; ++i) {
if(i < rev[i]) {
swap(y[i], y[rev[i]]);
}
}
return;
}
void fft(ll y[], int len, int on) {
change(y, len);
for(int h = 2; h <= len; h <<= 1) {
ll gn = qpow(3, (M - 1) / h, M);
if(on == -1) gn = qpow(gn, M - 2, M);
for(int j = 0; j < len; j += h) {
ll g = 1;
for(int k = j; k < j + h / 2; k++) {
ll u = y[k];
ll t = g * y[k + h / 2] % M;
y[k] = (u + t) % M;
y[k + h / 2] = (u - t + M) % M;
g = g * gn % M;
}
}
}
if(on == -1) {
ll inv = qpow(len, M - 2, M);
for(int i = 0; i < len; i++) {
y[i] = y[i] * inv % M;
}
}
}
int get(int x) {
int res = 1;
while(res < x) {
res <<= 1;
}
return res;
}
ll f[N], g[N];
ll tf[N], tg[N];
void solve(int l, int r) {
if(l == r) {
return ;
}
int len = (r - l + 1);
int mid = (l + r) / 2;
solve(l, mid);
for(int i = 0; i < len; i++) {
tf[i] = tg[i] = 0;
}
for(int i = l; i <= mid; i++) {
tf[i - l] = f[i];
}
for(int i = l; i <= r; i++) {
tg[i - l] = g[i - l];
}
fft(tf, len, 1);
fft(tg, len, 1);
for(int i = 0; i < len; i++) tf[i] = tf[i] * tg[i] % M;
fft(tf, len, -1);
for(int i = mid + 1; i <= r; i++) {
f[i] = (f[i] + tf[i - l]) % M;
}
solve(mid + 1, r);
}
int main() {
IOS;
int n;
cin >> n;
for(int i = 1; i < n; i++) {
cin >> g[i];
}
f[0] = 1;
solve(0, get(n) - 1);
for(int i = 0; i < n; i++) cout << f[i] << " \n"[i == n - 1];
}
