Luogu 4721 【模板】分治 FFT

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 ? ‘\n‘ : ‘ ‘);

    return 0;
}

原文地址：https://www.cnblogs.com/CzxingcHen/p/10197696.html