題目鏈接

(BZOJ) https://www.lydsy.com/JudgeOnline/problem.php?id=4734
(UOJ) http://uoj.ac/problem/269

題解

似乎大家都是用神仙構造的做法構造了一個二項式反演，然而我只會拿Stirling數爆推QAQ……
首先考慮\(f(x)=x^m\)的情況，最后乘上一個系數求和即可。
\[\sum^n_{k=0}{n\choose k}k^mx^k(1-x)^{n-k}\\ =\sum^n_{k=0}\sum^m_{i=0}\begin{Bmatrix}m\\i\end{Bmatrix}k^{\underline i}x^k(1-x)^{n-k}\\ =\sum^m_{i=0}\begin{Bmatrix}m\\i\end{Bmatrix}\sum^n_{k=0}\frac{n!}{(n-k)!(k-i)!}x^k(1-x)^{n-k}\\ =\sum^m_{i=0}\begin{Bmatrix}m\\i\end{Bmatrix}n^{\underline i}\sum^n_{k=0}{n-i\choose k-i}x^k(1-x)^{n-k}\\ =\sum^m_{i=0}\begin{Bmatrix}m\\i\end{Bmatrix}n^{\underline i}x^i\]
設\(f(x)=\sum^m_{i=0}a_ix^i\), 則答案為\(\sum^m_{i=0}a_i\sum^i_{j=0}\begin{Bmatrix}i\\j\end{Bmatrix}n^{\underline j}x^j\)
嗯，直接求的話時間復雜度是\(O(m^2)\). 而這個做法看上去很難優化了，因為里面用到了斯特林數而且兩維都不固定，怎么辦？
斯特林數阻止了進一步的優化，因此剛才我們把冪轉成了斯特林數，現在再考慮把斯特林數轉回來！
首先要注意到一點就是斯特林數的基本公式\(\begin{Bmatrix}i\\j\end{Bmatrix}=\frac{1}{j!}\sum^j_{k=0}(-1)^{j-k}{j\choose k}k^i\)當\(i<j\)時也是適用的，此時等式兩邊都為\(0\).
于是有\[\sum^m_{i=0}a_i\sum^i_{j=0}\begin{Bmatrix}i\\j\end{Bmatrix}n^{\underline j}x^j\\ =\sum^m_{i=0}a_i\sum^i_{j=0}\sum^j_{k=0}(-1)^{j-k}{j\choose k}k^i{n\choose j}x^j\\ =\sum^m_{j=0}{n\choose j}x^j\sum^j_{k=0}(-1)^{j-k}{j\choose k}\sum^m_{k=0}a_ik^i\\=\sum^m_{j=0}{n\choose j}x^j\sum^j_{k=0}(-1)^{j-k}{j\choose k}f(k)\]
而題目里給定的恰好是\(f(x)\)在\(x=0,1,...,m\)處的點值！所以其實根本不需要插值！
推到這里做法就很顯然了: 用NTT對每個\(j\)求出\(g_j=\sum^m_{j=0}(-1)^{j-k}{j\choose k}f(k)\), 然后計算\(\sum^m_{j=0}{n\choose j}x^jg_j\)即可。
時間復雜度\(O(m\log m)\).

代碼

#include<bits/stdc++.h> #define llong long long using namespace std;inline int read() {int w=1,s=0;char ch=getchar();while(!isdigit(ch)) {if(ch=='-')w=-1;ch=getchar();}while(isdigit(ch)) {s=s*10+ch-'0';ch=getchar();}return w*s; }const int N = 1<<17; const int lgN = 17; const int P = 998244353; const int G = 3;llong quickpow(llong x,llong y) {llong cur = x,ret = 1ll;for(int i=0; y; i++){if(y&(1ll<<i)) {y-=(1ll<<i); ret = ret*cur%P;}cur = cur*cur%P;}return ret; } llong mulinv(llong x) {return quickpow(x,P-2);}namespace FFT {llong aux1[N+3],aux2[N+3],aux3[N+3],aux4[N+3],aux5[N+3];int fftid[N+3];llong sexp[N+3];void resize(int dgr1,int dgr2,llong poly[]) {if(dgr1>dgr2) swap(dgr1,dgr2); for(int i=dgr1; i<dgr2; i++) poly[i] = 0ll;}int getdgr(int n) {int dgr = 1; while(dgr<=n) dgr<<=1; return dgr;}void init_fftid(int dgr){int len = 0; for(int i=1; i<=lgN; i++) {if(dgr==(1<<i)) {len = i; break;}}fftid[0] = 0; for(int i=1; i<dgr; i++) fftid[i] = (fftid[i>>1]>>1)|((i&1)<<(len-1));}void ntt(int dgr,int coe,llong poly[],llong ret[]){init_fftid(dgr);if(poly==ret) {for(int i=0; i<dgr; i++) {if(i<fftid[i]) swap(ret[i],ret[fftid[i]]);}}else {for(int i=0; i<dgr; i++) ret[i] = poly[fftid[i]];}for(int i=1; i<dgr; i<<=1){llong tmp = quickpow(G,(P-1)/(i<<1)); if(coe==-1) tmp = mulinv(tmp);sexp[0] = 1ll; for(int j=1; j<i; j++) sexp[j] = sexp[j-1]*tmp%P;for(int j=0; j<dgr; j+=(i<<1)){for(llong *k=ret+j,*kk=sexp; k<ret+i+j; k++,kk++){llong x = (*k),y = k[i]*(*kk)%P;(*k) = x+y>=P?x+y-P:x+y; k[i] = x-y<0?x-y+P:x-y;}}}if(coe==-1) {llong tmp = mulinv(dgr); for(int i=0; i<dgr; i++) ret[i] = ret[i]*tmp%P;}}void polymul(int dgr,llong poly1[],llong poly2[],llong ret[]){ntt(dgr<<1,1,poly1,aux1); ntt(dgr<<1,1,poly2,aux2);for(int i=0; i<(dgr<<1); i++) ret[i] = aux1[i]*aux2[i]%P;ntt(dgr<<1,-1,ret,ret);} } using FFT::getdgr; using FFT::resize; using FFT::ntt; using FFT::polymul;llong fact[N+3],finv[N+3]; llong f[N+3],g[N+3],h[N+3]; int m; llong n,ax;int main() {fact[0] = 1ll; for(int i=1; i<=N; i++) fact[i] = fact[i-1]*i%P;finv[N] = quickpow(fact[N],P-2); for(int i=N-1; i>=0; i--) finv[i] = finv[i+1]*(i+1)%P;scanf("%lld%d%lld",&n,&m,&ax); int dgr = getdgr(m);for(int i=0; i<=m; i++) scanf("%lld",&f[i]),f[i] = f[i]*finv[i]%P;for(int i=0; i<=dgr; i++) g[i] = i&1?P-finv[i]:finv[i];polymul(dgr,f,g,h);for(int i=0; i<=m; i++) h[i] = h[i]*fact[i]%P;llong cur1 = 1ll,cur2 = 1ll,ans = 0ll;for(int i=0; i<=m&&i<=n; i++){llong tmp = cur1*cur2%P*h[i]%P;ans = (ans+tmp)%P;cur1 = cur1*mulinv(i+1ll)%P*(n-i)%P; cur2 = cur2*ax%P;}printf("%lld\n",ans);return 0; }

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