【模板】多项式求逆

代码如下

#include <bits/stdc++.h>

using namespace std;

typedef long long LL;

const LL mod = 998244353, g = 3, ig = 332748118;

inline LL fpow(LL a, LL b) {
    LL ret = 1 % mod;
    for (; b; b >>= 1, a = a * a % mod) {
        if (b & 1) {
            ret = ret * a % mod;
        }
    }
    return ret;
}

void ntt(vector<LL> &v, vector<int> &rev, int opt) {
    int tot = v.size();
    for (int i = 0; i < tot; i++) {
        if (i < rev[i]) {
            swap(v[i], v[rev[i]]);
        }
    }
    for (int mid = 1; mid < tot; mid <<= 1) {
        LL wn = fpow(opt == 1 ? g : ig, (mod - 1) / (mid << 1));
        for (int j = 0; j < tot; j += mid << 1) {
            LL w = 1;
            for (int k = 0; k < mid; k++) {
                LL x = v[j + k], y = v[j + mid + k] * w % mod;
                v[j + k] = (x + y) % mod, v[j + mid + k] = (x - y + mod) % mod;
                w = w * wn % mod;
            }
        }
    }
    if (opt == -1) {
        LL itot = fpow(tot, mod - 2);
        for (int i = 0; i < tot; i++) {
            v[i] = v[i] * itot % mod;
        }
    }
}

void solve(int n, vector<LL> &a, vector<LL> &b) {
    if (n == 1) {
        b[0] = fpow(a[0], mod - 2);
        return;
    }
    int mid = (n + 1) >> 1;
    solve(mid, a, b);
    int bit = 0, tot = 1;
    while (tot <= 2 * n) {
        tot <<= 1;
        bit++;
    }
    vector<int> rev(tot);
    for (int i = 0; i < tot; i++) {
        rev[i] = (rev[i >> 1] >> 1) | (i & 1) << (bit - 1);
    }
    vector<LL> foo(tot), bar(tot);
    for (int i = 0; i < n; i++) {
        foo[i] = a[i];
    }
    for (int i = 0; i < mid; i++) {
        bar[i] = b[i];
    }
    ntt(foo, rev, 1), ntt(bar, rev, 1);
    for (int i = 0; i < tot; i++) {
        bar[i] = bar[i] * (2 - foo[i] * bar[i] % mod + mod) % mod;
    }
    ntt(bar, rev, -1);
    for (int i = 0; i < n; i++) {
        b[i] = bar[i];
    }
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    int n;
    cin >> n;
    vector<LL> a(n), b(n);
    for (int i = 0; i < n; i++) {
        cin >> a[i];
    }
    solve(n, a, b);
    for (int i = 0; i < n; i++) {
        cout << b[i] << " ";
    }
    return 0;
} 

【模板】多项式求逆

原文：https://www.cnblogs.com/wzj-xhjbk/p/11601151.html