UOJ269【清华集训2016】如何优雅地求和【数论，多项式】

题目描述：求

$$\sum_{k=0}^nf(k)\binom{n}{k}x^k(1-x)^{n-k}$$

输入$n$，$f(x)$的次数上界$m$，$x$，$f(0,1,\ldots,m)$，对$998244353$取模。

数据范围：$1\leq n\leq 10^9,1\leq m\leq 2*10^4,0\leq a_i,x<998244353$

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现在来讲正解：首先我们把$f(x)$用下降幂来表示，（看到这里的快点关掉，然后自己推一推）

$$\begin{align*}Ans&=\sum_{k=0}^n\sum_{i=0}^ma_i*\frac{k!}{(k-i)!}\frac{n!}{k!(n-k)!}x^k(1-x)^{n-k} \\&=\sum_{k=0}^n\sum_{i=0}^ma_i*\frac{n!}{(k-i)!(n-k)!}x^k(1-x)^{n-k} \\&=\sum_{i=0}^ma_in!\sum_{k=i}^n\frac{x^k}{(k-i)!}*\frac{(1-x)^{n-k}}{(n-k)!} \\&=\sum_{i=0}^ma_ix^in!(e^x*e^{1-x}[x^{n-i}]) \\&=\sum_{i=0}^ma_ix^i\frac{n!}{(n-i)!}\end{align*}$$

然后就是考虑如何点值转下降幂了

$$\begin{align*}f(i)&=\sum_{j=0}^ma_j\frac{i!}{(i-j)!} \\\frac{f(i)}{i!}&=\sum_{j=0}^ma_j*\frac{1}{(i-j)!}\end{align*}$$

所以$f(i)$的EGF等于$a$的OGF乘上$e^x$，所以$a$的OGF等于$f(i)$的EGF乘上$e^{-x}$，使用NTT优化计算，时间复杂度$O(m\log m)$

#include<bits/stdc++.h>
#define Rint register int
using namespace std;
typedef long long LL;
const int N = 1 << 16, mod = 998244353, G = 3, Gi = 332748118;
int n, m, x, A[N], B[N], fac[N], invfac[N], ans;
inline int kasumi(int a, int b){
    int res = 1;
    while(b){
        if(b & 1) res = (LL) res * a % mod;
        a = (LL) a * a % mod;
        b >>= 1;
    }
    return res;
}
inline void init(){
    fac[0] = 1;
    for(Rint i = 1;i <= m;i ++) fac[i] = (LL) i * fac[i - 1] % mod;
    invfac[m] = kasumi(fac[m], mod - 2);
    for(Rint i = m;i;i --) invfac[i - 1] = (LL) i * invfac[i] % mod;
}
int rev[N];
inline int calrev(int len){
    int limit = 1, L = -1;
    while(limit <= len){limit <<= 1; L ++;}
    for(Rint i = 0;i < limit;i ++)
        rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << L);
    return limit;
}
inline void NTT(int *A, int limit, int type){
    for(Rint i = 0;i < limit;i ++)
        if(i < rev[i]) swap(A[i], A[rev[i]]);
    for(Rint mid = 1;mid < limit;mid <<= 1){
        int Wn = kasumi(type == 1 ? G : Gi, (mod - 1) / (mid << 1));
        for(Rint j = 0;j < limit;j += (mid << 1)){
            int w = 1;
            for(Rint k = 0;k < mid;k ++, w = (LL) w * Wn % mod){
                int x = A[j + k], y = (LL) w * A[j + k + mid] % mod;
                A[j + k] = (x + y) % mod;
                A[j + k + mid] = (x - y + mod) % mod;
            }
        }
    }
    if(type == -1){
        int inv = kasumi(limit, mod - 2);
        for(Rint i = 0;i < limit;i ++)
            A[i] = (LL) A[i] * inv % mod;
    }
}
int main(){
    scanf("%d%d%d", &n, &m, &x);
    init();
    for(Rint i = 0;i <= m;i ++){
        scanf("%d", A + i);
        A[i] = (LL) A[i] * invfac[i] % mod;
        B[i] = invfac[i];
        if(i & 1) B[i] = mod - B[i];
    }
    int limit = calrev(m << 1);
    NTT(A, limit, 1); NTT(B, limit, 1);
    for(Rint i = 0;i < limit;i ++) A[i] = (LL) A[i] * B[i] % mod;
    NTT(A, limit, -1);
    int w = 1;
    for(Rint i = 0;i <= m;i ++){
        ans = (ans + (LL) w * A[i] % mod) % mod;
        w = (LL) w * (mod + n - i) % mod * x % mod;
    }
    printf("%d", ans);
}

 启发：推柿子遇到多项式时应考虑多项式的各种形式，如点值，普通/上升/下降幂等。