将$f(k)=sum_{i=0}^{m}a_{i}k^{i}$转换为$f(k)=sum_{i=0}^{m}b_{i}k^{underline{i}}$，其中$k^{underline{i}}=frac{k!}{(k-i)!}$
题目即求$sum_{k=0}^{n}c(n,k)x^{k}sum_{i=0}^{m}b_{i}cdot k^{underline{i}}$
调整枚举顺序，即$sum_{i=0}^{m}b_{i}sum_{k=0}^{n}c(n,k)x^{k}k^{underline{i}}$
观察发现$c(n,k)k^{underline{i}}=frac{n!}{k!(n-k)!}frac{k!}{(k-i)!}=frac{n!}{(k-i)!(n-k)!}=c(n-i,k-i)n^{underline{i}}$
代入原式，即$sum_{i=0}^{m}b_{i}n^{underline{i}}sum_{k=0}^{n}c(n-i,k-i)x^{k}$
令$k'=k-i$，即$sum_{i=0}^{m}b_{i}n^{underline{i}}x^{i}sum_{k'=0}^{n-i}c(n-i,k')x^{k'}$
观察发现右式即$(x+1)^{n-i}$二项式展开，那么即$sum_{i=0}^{m}b_{i}n^{underline{i}}x^{i}(x+1)^{n-i}$
那么问题变为如何求出$b_{i}$，即如何将多项式转换为下降幂多项式【luoguP5383】
根据第二类斯特林数的性质，有$x^{n}=sum_{i=0}^{n}c(x,i)i!S(n,i)=sum_{i=0}^{n}S(n,i)x^{underline{i}}$
那么就有$sum_{i=0}^{m}a_{i}x^{i}=sum_{i=0}^{m}a_{i}sum_{j=0}^{i}S(i,j)x^{underline{j}}=sum_{j=0}^{m}x^{underline{j}}sum_{i=j}^{m}a_{i}S(i,j)$，即$b_{j}=sum_{i=j}^{m}a_{i}S(i,j)$
本题由于$mle 1000$，仅需要根据$S$的递推式暴力求出$S$并$o(m^{2})$计算即可
View Code
1 #include<bits/stdc++.h> 2 using namespace std; 3 #define N 1005 4 int n,x,mod,m,ans,a[N],b[N],s[N][N]; 5 int ksm(int n,int m){ 6 if (!m)return 1; 7 int s=ksm(n,m>>1); 8 s=1LL*s*s%mod; 9 if (m&1)s=1LL*s*n%mod; 10 return s; 11 } 12 int main(){ 13 scanf("%d%d%d%d",&n,&x,&mod,&m); 14 for(int i=0;i<=m;i++)scanf("%d",&a[i]); 15 s[0][0]=1; 16 for(int i=1;i<=m;i++) 17 for(int j=1;j<=i;j++)s[i][j]=(s[i-1][j-1]+1LL*j*s[i-1][j])%mod; 18 for(int i=0;i<=m;i++) 19 for(int j=i;j<=m;j++)b[i]=(b[i]+1LL*a[j]*s[j][i])%mod; 20 int s=1; 21 for(int i=0;i<=m;i++){ 22 ans=(ans+1LL*b[i]*s%mod*ksm(x,i)%mod*ksm(x+1,n-i))%mod; 23 s=1LL*s*(n-i)%mod; 24 } 25 printf("%d",ans); 26 }