2020 CCPC网络赛 1013.Residual Polynomial

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$设t^{y}f(x)$表示$f(x)$求导$y$次。

由此构造得到$f_n(x)=f_1(x)(b_1t+c_1)….(b_it+c_i)$

可以得到$t^{y}f(x)$的系数就是后面这个，假设$t^yf(x)$的系数为$z_i$。

我们知道最后$x_i$的系数其实就是$ans_i=z_0a_i+z_1(i+1)a_{i+1}+z_2(i+2)(i+1)a_{2+1}…,ans_i\times(i)!=\sum_{j=0}^n z_jg_{j+i}$

$g_i=a_i\times(i)!$

上面就是卷积套路，翻转一下变成一般形式，就可以$NTT$了。

$HDU，G++(2900ms)$飘过，如果更快，特盘特殊点。

代码 
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#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define pii pair<int, int>
#define mk make_pair
const int N = 6e5 + 10;

const int mod = 998244353;
// const double pi = acos(-1.0);

const int P = 998244353, gi = 3;
int inc(int x, int y) { return (x + y >= mod) ? (x + y - mod) : (x + y); }
int del(int x, int y) { return (x - y < 0) ? (x - y + mod) : (x - y); }
int read()
{
    int x = 0, f = 1;
    char c = getchar();
    while (c < '0' || c > '9')
    {
        if (c == '-')
            f = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9')
        x = (x << 1) + (x << 3) + c - '0', c = getchar();
    return x * f;
}

int qpow(int a, int x, int mo)
{
    int res = 1;
    while (x)
    {
        if (x & 1)
            res = 1ll * res * a % mo;
        x >>= 1;
        a = 1ll * a * a % mo;
    }
    return res;
}

int rev[N];
void NTT(int *A, int n, int inv)
{
    for (int i = 0; i < n; i++)
        if (i < rev[i])
            swap(A[i], A[rev[i]]);
    for (int l = 1; l < n; l <<= 1)
    {
        int tt = qpow(gi, (P - 1) / (l << 1), P);
        int temp = (inv == 1 ? tt : qpow(tt, P - 2, P));
        for (int i = 0; i < n; i += (l << 1))
        {
            int omega = 1;
            for (int j = 0; j < l; j++, omega = 1ll * omega * temp % P)
            {
                int x = A[i + j], y = 1ll * omega * A[i + j + l] % P;
                A[i + j] = inc(x, y);
                A[i + j + l] = del(x, y);
            }
        }
    }
    int invv = qpow(n, P - 2, P);
    if (inv == -1)
        for (int i = 0; i < n; i++)
            A[i] = 1ll * A[i] * invv % P;
}

int Inv2;
int C[N], D[N];
int Ployinit(int n, int m)
{
    int ML = 1, bit = 0;
    while (ML < n + m - 1)
        ML <<= 1, bit++;
    return ML;
}
void NTTX(int *a, int n, int *b, int m)
{
    int ML = 1, bit = 0;
    while (ML < n + m - 1)
        ML <<= 1, bit++;
    for (int i = 0; i < ML; i++)
        rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (bit - 1));
    NTT(a, ML, 1);
    NTT(b, ML, 1);
    for (int i = 0; i < ML; i++)
        a[i] = 1ll * a[i] * b[i] % P;
    NTT(a, ML, -1);
}
int fac[N];
int ifac[N];

vector<int> g[N];

int p1[N], p2[N];
int solve(int l, int r)
{
    if (l == r)
    {
        return l;
    }
    int mid = (l + r) >> 1;
    int pa = solve(l, mid);
    int pb = solve(mid + 1, r);
    int len1 = g[pa].size();
    int len2 = g[pb].size();
    int ML = Ployinit(len1, len2);
    for (int i = 0; i < ML; i++)
        p1[i] = p2[i] = 0;
    for (int i = 0; i < len1; i++)
        p1[i] = g[pa][i];
    for (int i = 0; i < len2; i++)
        p2[i] = g[pb][i];

    NTTX(p1, len1, p2, len2);

    g[pa].clear();
    int li = len1 + len2 - 1;

    for (int i = 0; i < li; i++)
        g[pa].push_back(p1[i]);
    return pa;
}

int c[N], b[N], a[N];
int f[N];
int main()
{
    int T = read();
    while (T--)
    {
        int n = read();
        fac[0] = 1;
        for (int i = 1; i <= n; ++i)
            fac[i] = 1ll * fac[i - 1] * i % mod;
        ifac[n] = qpow(fac[n], mod - 2, mod);
        for (int i = n; i; --i)
            ifac[i - 1] = 1ll * ifac[i] * i % mod;
        memset(a, 0, sizeof(a));
        memset(f, 0, sizeof(f));
        memset(b, 0, sizeof(b));
        memset(c, 0, sizeof(c));
        for (int i = 0; i <= n; i++)
            a[i] = read();
        for (int i = 0; i <= n; i++)
            a[i] = 1ll * a[i] * fac[i] % mod;
        for (int i = 2; i <= n; i++)
            b[i] = read();
        for (int i = 2; i <= n; i++)
            c[i] = read();

        for (int i = 2; i <= n; i++)
        {
            g[i].clear();
            g[i].push_back(c[i]), g[i].push_back(b[i]);
        }
        int o = solve(2, n);
        reverse(g[o].begin(), g[o].end());
        int flen = g[o].size();

        for (int i = 0; i < g[o].size(); i++)
            f[i] = g[o][i];
        NTTX(f, flen, a, n + 1);
        for (int i = 0; i < n; ++i)
            printf("%lld ", (1ll * f[i + n - 1] * ifac[i]) % mod);
        printf("%lld\n", (1ll * f[n + n - 1] * ifac[n]) % mod);
    }
}