丢个好看一点的(poly)板子

多项式求逆

ln

exp

#include<bits/stdc++.h>
#define N 800050
#define sz(x) ((int)x.size())
#define poly vector<int>
#define mod 998244353
using namespace std;
int add(int x, int y) { x += y;
    if(x >= mod) x -= mod;
    return x;
}
int sub(int x, int y) { x -= y;
    if(x < 0) x += mod;
    return x;
}
int mul(int x, int y) {
    return 1ll * x * y % mod;
}
int qpow(int x, int y) {
    int ret = 1;
    for(; y; y >>= 1, x = mul(x, x)) if(y & 1) ret = mul(ret, x);
    return ret;
}
const int G = 3;
const int Ginv = qpow(G, mod - 2);
int rev[N << 1];
void ntt(int *a, int n, int o) {
    for(int i = 0; i < n; i ++) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) * (n >> 1));
    for(int i = 0; i < n; i ++) if(rev[i] > i) swap(a[i], a[rev[i]]);
    for(int len = 2; len <= n; len <<= 1) {
        int w0 = qpow((o == 1)? G : Ginv, (mod - 1) / len);
        for(int j = 0; j < n; j += len) {
            int wn = 1;
            for(int k = j; k < j + (len >> 1); k ++, wn = mul(wn, w0)) {
                int X = a[k], Y = mul(a[k + (len >> 1)], wn);
                a[k] = add(X, Y), a[k + (len >> 1)] = sub(X, Y);
            }   
        }
    }
    int ninv = qpow(n, mod - 2);
    if(o == -1)
        for(int i = 0; i < n; i ++) a[i] = mul(a[i], ninv);
}
poly operator + (const poly &A, const poly & B) {
    poly C = A; C.resize(max(sz(A), sz(B)));
    for(int i = 0; i < sz(B); i ++) C[i] = add(C[i], B[i]);
    return C;
}
poly operator - (const poly &A, const poly & B) {
    poly C = A; C.resize(max(sz(A), sz(B)));
    for(int i = 0; i < sz(B); i ++) C[i] = sub(C[i], B[i]);
    return C;
}
#define clr(a, n) (memset(a, 0, sizeof(int) * n))
int a[N << 1], b[N << 1], lim;
poly operator * (const poly & A, const poly & B) {
    for(int i = 0; i < sz(A); i ++) a[i] = A[i];
    for(int i = 0; i < sz(B); i ++) b[i] = B[i];
    poly C; C.resize(min(lim, sz(A) + sz(B) - 1));
    int len = 1;
    for(; len <= sz(A) + sz(B) - 1; len <<= 1);
    ntt(a, len, 1), ntt(b, len, 1);
    for(int i = 0; i < len; i ++) a[i] = mul(a[i], b[i]);
    ntt(a, len, -1);
    for(int i = 0; i < sz(C); i ++) C[i] = a[i];
    clr(a, len), clr(b, len);
    return C;
}
poly operator * (const int & a, const poly & A) {
    poly C; C.resize(sz(A));
    for(int i = 0; i < sz(A); i ++) C[i] = mul(A[i], a);
    return C;
}
void pINV(poly &A, poly &B, int n) {
    if(n == 1) B.push_back(qpow(A[0], mod - 2));
    else {
        pINV(A, B, (n + 1) / 2);
        poly C = A; C.resize(n);
        B = 2 * B - B * B * C;
        B.resize(n);
    }
    
}
poly INV(poly A) {
    poly B; pINV(A, B, sz(A));
    return B;
}
int inv[N];
void init(int n) {
    inv[1] = 1;
    for(int i = 2; i <= n; i ++)
        inv[i] = sub(0, mul(mod / i, inv[mod % i]));
}
poly qiudao(const poly A) {
    poly B;
    for(int i = 1; i < sz(A); i ++) B.push_back(mul(i, A[i]));
    return B;
}
poly jifen(const poly A) {
    poly B; B.resize(sz(A));
    for(int i = 1; i < sz(A); i ++) B[i] = mul(A[i - 1], inv[i]);
    B[0] = 0;
    return B;
}
poly ln(const poly A) {
    return jifen(qiudao(A) * INV(A));
}
void pexp(poly &A, poly & B, int n) {
    if(n == 1) B.push_back(1);
    else {
        pexp(A, B, (n + 1) / 2);
        poly lnB; lnB = B; lnB.resize(n);
        lnB = ln(lnB);
        for(int i = 0; i < sz(lnB); i ++) lnB[i] = sub(A[i], lnB[i]);
        lnB[0] = add(lnB[0], 1);
        B = B * lnB;
        B.resize(n);
    }
}
poly exp(poly A) {
    poly C; pexp(A, C, sz(A));
    return C;
}
poly f;
int n;
int main() {
    scanf("%d", &n); lim = n; init(n);
    f.resize(n);
    for(int i = 0; i < n; i ++) scanf("%d", &f[i]);
    f = exp(f);
    for(int i = 0; i < n; i ++) printf("%d ", f[i]);
    return 0;
}