CF1278F Cards 加强版

题目
设 $p = \frac 1m$
$\begin{aligned} ans &= \sum_{i=0}^k \begin{Bmatrix}k\\i\end{Bmatrix}p^i\binom nii! \\ &=\sum_{i=0}^k \sum_{j=0}^i \binom ij\binom nip^i(-1)^j(i-j)^k\\ &=\sum_{j=0}^k(-1)^j\binom nj \sum_{i=j}^k \binom{n-j}{i-j}p^i(i-j)^k\\ &=\sum_{j=0}^k(-1)^jp^j\binom nj\sum_{i=0}^{k-j} \binom {n-j}ip^ii^k\\ &=\sum_{i=0}^ki^kp^i\sum_{j=0}^{k-i} (-p)^j\binom nj\binom{n-j}i\\ &=\sum_{i=0}^ki^kp^i\binom ni\sum_{j=0}^{k-i}(-p)^j\binom{n-i}{j}\end{aligned}$

考虑 $S_{i+1} = \sum_{j=0}^{k-i-1}(-p)^j\binom{n-i-1}j = \sum_{j=0}^{k-i-1}(-p)^j\left(\binom{n-i}j - \binom{n-i-1}{j-1}\right)\\=S_i-(-p)^k\binom {n-i}{k-i} + p\left(S_{i+1}-(-p)^{k-i-1}\binom {n-i-1}{k-i-1}\right)$

可以得到 $S_i = S_{i+1}(1-p) + (-p)^{k-i}\binom {n-i-1}{k-i}$
边界条件 $S_k = 1$。
所以 $ans = \sum_{i=0}^k i^kp^i\binom ni S_i$
$O(k)$线性筛预处理 $i^k$。
对于 $\binom ni$需要求 $n^{\underline i}$
对于 $\binom {n-i-1}{k-i}$需要求 $(n-k)^{\overline {k-i}}$
还有就是阶乘的逆元。
感觉可以搞搞模数搞拓展 $lucas$。

总复杂度 $O(k)$
$\mathcal AC \ Code$

#include<bits/stdc++.h>
#define maxn 10000007
#define rep(i,j,k) for(int i=(j),LIM=(k);i<=LIM;i++)
#define per(i,j,k) for(int i=(j),LIM=(k);i>=LIM;i--)
#define mod 998244353
using namespace std;

int Pow(int b,int k){ int r=1;for(;k;k>>=1,b=1ll*b*b%mod) if(k&1)r=1ll*r*b%mod;return r; }
int pr[maxn/8],cnt_pr,pwk[maxn],inv[maxn],s[maxn];
bool vis[maxn];
int n,m,K,p;

int main(){
	scanf("%d%d%d",&n,&m,&K);p = Pow(m , mod-2);
	pwk[1] = inv[0] = inv[1] = 1;
	rep(i,2,K){
		inv[i] = 1ll * (mod - mod / i) * inv[mod % i] % mod;
		if(!vis[i]) pr[cnt_pr++] = i , pwk[i] = Pow(i,K);
		for(int j=0;pr[j] * i <= K;j++){
			vis[i * pr[j]] = 1 , pwk[i * pr[j]] = 1ll * pwk[i] * pwk[pr[j]] % mod;
			if(i % pr[j] == 0) break;
		}
	}
	int C = 1 , pw = 1;
	rep(i,0,K){
		s[K-i] = (s[K-i+1] * (1ll - p) + 1ll * pw * C) % mod;
		pw = 1ll * pw * (-p) % mod;
		C = 1ll * C * (n-K+i) % mod * inv[i+1] % mod;
	}
	C = 1 , pw = 1;
	int ans = 0;
	rep(i,0,K){
		ans = (ans + 1ll * pwk[i] * pw % mod * C % mod * s[i]) % mod;
		pw = 1ll * pw * p % mod;
		C = 1ll * C * (n-i) % mod * inv[i+1] % mod;
	}
	printf("%d\n",(ans+mod)%mod);
}