仓鼠的数学题——生成函数

题面

　　洛谷P3711

解析

　　需要用到伯努利数

　　把伯努利数的式子代入问题：$$egin{align*}sum_{k=0}^nS_k(x)a_k&=sum_{k=0}^nfrac{a_k}{k+1}sum_{i=0}^kinom{k+1}{i}B_i x^{k+1-i}\&=sum_{k=0}^n a_k*k! sum_{i=0}^k frac{B_i}{i!} *frac{x^{k+1-i}}{(k+1-i)!} end{align*}$$

　　设$A_i=a_i * i!$，$C_i=frac{B_i}{i!}$，答案中$x^i$的系数为$frac{1}{i!}sum_{k=0}sum_{j=0}[k+1-j==i]A_kC_j$

　　设$D_i=C_{n+1-i}$，则$x^i$的系数为$frac{1}{i!}sum_{k=0}sum_{j=0}[k+j==n+i]A_kD_j$

　　卷积即可

　　$O(Nlog N)$

　代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
typedef long long ll;
const int maxn = 500005, mod = 998244353, g = 3;

inline int read()
{
    int ret, f=1;
    char c;
    while((c=getchar())&&(c<'0'||c>'9'))if(c=='-')f=-1;
    ret=c-'0';
    while((c=getchar())&&(c>='0'&&c<='9'))ret=(ret<<3)+(ret<<1)+c-'0';
    return ret*f;
}

int add(int x, int y)
{
    return x + y < mod? x + y: x + y - mod;
}

int rdc(int x, int y)
{
    return x - y < 0? x - y + mod: x - y;
}

ll qpow(ll x, int y)
{
    ll ret = 1;
    while(y)
    {
        if(y&1)
            ret = ret * x % mod;
        x = x * x % mod;
        y >>= 1;
    }
    return ret;
}

int n, lim, bit, rev[maxn<<1];
ll a[maxn<<1], fac[maxn], inv[maxn], b[maxn], B[maxn<<1], c[maxn<<1], ginv;

void init()
{
    ginv = qpow(g, mod - 2);
    fac[0] = 1;
    for(int i = 1; i <= n + 2; ++i)
        fac[i] = i * fac[i-1] % mod;
    inv[n+2] = qpow(fac[n+2], mod - 2);
    for(int i = n + 1; i >= 0; --i)
    {
        inv[i] = inv[i+1] * (i + 1) % mod;
        b[i] = inv[i+1];
    }

}

void NTT_init(int x)
{
    lim = 1;
    bit = 0;
    while(lim <= x)
    {
        lim <<= 1;
        ++ bit;
    }
    for(int i = 1; i < lim; ++i)
        rev[i] = (rev[i>>1] >> 1) | ((i & 1) << (bit - 1));
}

void NTT(ll *x, int y)
{
    for(int i = 1; i < lim; ++i)
        if(i < rev[i])
            swap(x[i], x[rev[i]]);
    ll wn, w, u, v;
    for(int i = 1; i < lim; i <<= 1)
    {
        wn = qpow((y == 1)? g: ginv, (mod - 1) / (i << 1));
        for(int j = 0; j < lim; j += (i << 1))
        {
            w = 1;
            for(int k = 0; k < i; ++k)
            {
                u = x[j+k];
                v = x[j+k+i] * w % mod;
                x[j+k] = add(u, v);
                x[j+k+i] = rdc(u, v);
                w = w * wn % mod;
            }
        }
    }
    if(y == -1)
    {
        ll iv = qpow(lim, mod - 2);
        for(int i = 0; i < lim; ++i)
            x[i] = x[i] * iv % mod;
    }
}

void get_inv(ll *x, ll *y, int len)
{
    if(len == 1)
    {
        x[0] = qpow(y[0], mod - 2);
        return ;
    }
    get_inv(x, y, (len + 1) >> 1);
    for(int i = 0; i < len; ++i)
        c[i] = y[i];
    NTT_init(len << 1);
    NTT(x, 1);
    NTT(c, 1);
    for(int i = 0; i < lim; ++i)
    {
        x[i] = rdc(add(x[i], x[i]), (c[i] * x[i] % mod) * x[i] % mod);
        c[i] = 0;
    }
    NTT(x, -1);
    for(int i = len; i < lim; ++i)
        x[i] = 0;
}

int main()
{
    n = read();
    init();
    for(int i = 0; i <= n; ++i)
        a[i] = read() * fac[i] % mod;
    printf("%lld ", a[0]);
    get_inv(B, b, n + 2);
    B[1] = qpow(2, mod - 2);
    reverse(B, B + n + 2);
    NTT_init((n << 1) + 5);
    NTT(a, 1);
    NTT(B, 1);
    for(int i = 0; i < lim; ++i)
        a[i] = a[i] * B[i] % mod;
    NTT(a, -1);
    for(int i = 1; i <= n + 1; ++i)
        printf("%lld ", a[i+n] * inv[i] % mod);
    return 0;
}