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

---

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 =  << , mod = , G = , Gi = ;
 int n, m, x, A[N], B[N], fac[N], invfac[N], ans;
 inline int kasumi(int a, int b){
     int res = ;
     while(b){
         if(b & ) res = (LL) res * a % mod;
         a = (LL) a * a % mod;
         b >>= ;
     }
     return res;
 }
 inline void init(){
     fac[] = ;
     for(Rint i = ;i <= m;i ++) fac[i] = (LL) i * fac[i - ] % mod;
     invfac[m] = kasumi(fac[m], mod - );
     for(Rint i = m;i;i --) invfac[i - ] = (LL) i * invfac[i] % mod;
 }
 int rev[N];
 inline int calrev(int len){
     int limit = , L = -;
     while(limit <= len){limit <<= ; L ++;}
     for(Rint i = ;i < limit;i ++)
         rev[i] = (rev[i >> ] >> ) | ((i & ) << L);
     return limit;
 }
 inline void NTT(int *A, int limit, int type){
     for(Rint i = ;i < limit;i ++)
         if(i < rev[i]) swap(A[i], A[rev[i]]);
     for(Rint mid = ;mid < limit;mid <<= ){
         int Wn = kasumi(type ==  ? G : Gi, (mod - ) / (mid << ));
         for(Rint j = ;j < limit;j += (mid << )){
             int w = ;
             for(Rint k = ;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 == -){
         int inv = kasumi(limit, mod - );
         for(Rint i = ;i < limit;i ++)
             A[i] = (LL) A[i] * inv % mod;
     }
 }
 int main(){
     scanf("%d%d%d", &n, &m, &x);
     init();
     for(Rint i = ;i <= m;i ++){
         scanf("%d", A + i);
         A[i] = (LL) A[i] * invfac[i] % mod;
         B[i] = invfac[i];
         if(i & ) B[i] = mod - B[i];
     }
     int limit = calrev(m << );
     NTT(A, limit, ); NTT(B, limit, );
     for(Rint i = ;i < limit;i ++) A[i] = (LL) A[i] * B[i] % mod;
     NTT(A, limit, -);
     int w = ;
     for(Rint i = ;i <= m;i ++){
         ans = (ans + (LL) w * A[i] % mod) % mod;
         w = (LL) w * (mod + n - i) % mod * x % mod;
     }
     printf("%d", ans);
 }

UOJ269

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。