Luogu 4721 模板分治 FFT
Posted czxingchen
tags:
篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了Luogu 4721 模板分治 FFT相关的知识,希望对你有一定的参考价值。
还不会这题的多项式求逆的算法。
发现每一项都是一个卷积的形式,那么我们可以使用$NTT$来加速,直接做是$O(n^2logn)$的,我们考虑如何加速转移。
可以采用$cdq$分治的思想,对于区间$[l, r]$中的数,先计算出$[l, mid]$中的数对$[mid + 1, r]$中的数的贡献,然后直接累加到右边去。
容易发现,这样子每一次需要用向量$[l,l + 1, l + 2, dots, mid]$卷上$g$中$[1, 2, dots, r - l]$。
时间复杂度$O(nlog^2n)$,感觉这东西跑得并不慢鸭。
Code:
#include <cstdio> #include <cstring> using namespace std; typedef long long ll; const int N = 3e5 + 5; const ll P = 998244353LL; int n, lim, pos[N]; ll f[N], g[N], a[N], b[N]; template <typename T> inline void read(T &X) { X = 0; char ch = 0; T op = 1; for (; ch > ‘9‘|| ch < ‘0‘; ch = getchar()) if (ch == ‘-‘) op = -1; for (; ch >= ‘0‘ && ch <= ‘9‘; ch = getchar()) X = (X << 3) + (X << 1) + ch - 48; X *= op; } template <typename T> inline void swap(T &x, T &y) { T t = x; x = y; y = t; } inline ll fpow(ll x, ll y) { ll res = 1LL; for (; y > 0; y >>= 1) { if (y & 1) res = res * x % P; x = x * x % P; } return res; } inline void prework(int len) { int l = 0; for (lim = 1; lim <= len; lim <<= 1, ++l); for (int i = 0; i < lim; i++) pos[i] = (pos[i >> 1] >> 1) | ((i & 1) << (l - 1)); } inline void ntt(ll *c, int opt) { for (int i = 0; i < lim; i++) if (i < pos[i]) swap(c[i], c[pos[i]]); for (int i = 1; i < lim; i <<= 1) { ll wn = fpow(3, (P - 1) / (i << 1)); if (opt == -1) wn = fpow(wn, P - 2); for (int len = i << 1, j = 0; j < lim; j += len) { ll w = 1; for (int k = 0; k < i; k++, w = w * wn % P) { ll x = c[j + k], y = c[j + k + i] * w % P; c[j + k] = (x + y) % P, c[j + k + i] =(x - y + P) % P; } } } if (opt == -1) { ll inv = fpow(lim, P - 2); for (int i = 0; i < lim; i++) c[i] = c[i] * inv % P; } } void solve(int l, int r) { if (l == r) { a[l] = (a[l] + b[l]) % P; return; } int mid = ((l + r) >> 1); solve(l, mid); prework(r - l + 1); for (int i = 0; i < lim; i++) g[i] = f[i] = 0; for (int i = l; i <= mid; i++) f[i - l] = a[i]; for (int i = 1; i <= r - l; i++) g[i - 1] = b[i]; ntt(f, 1), ntt(g, 1); for (int i = 0; i < lim; i++) f[i] = f[i] * g[i] % P; ntt(f, -1); for (int i = mid + 1; i <= r; i++) a[i] = (a[i] + f[i - l - 1]) % P; solve(mid + 1, r); } int main() { read(n); n--; for (int i = 1; i <= n; i++) read(b[i]); a[0] = 1; solve(1, n); for (int i = 0; i <= n; i++) printf("%lld%c", a[i], i == n ? ‘ ‘ : ‘ ‘); return 0; }
以上是关于Luogu 4721 模板分治 FFT的主要内容,如果未能解决你的问题,请参考以下文章