洛谷 P6031

洛谷题面传送门

u1s1 这个推式子其实挺套路的吧，可惜有一步没推出来看了题解

[egin{aligned} res&=sumlimits_{i=0}^ni^kdbinom{n}{i}(dfrac{1}{m})^i(dfrac{m-1}{m})^{n-i}\ &=sumlimits_{i=0}^nsumlimits_{j=1}^kegin{Bmatrix}k\jend{Bmatrix}i^{underline{j}}dbinom{n}{i}(dfrac{1}{m})^i(dfrac{m-1}{m})^{n-i}\ &=(dfrac{m-1}{m})^{n}sumlimits_{i=0}^nsumlimits_{j=1}^kegin{Bmatrix}k\jend{Bmatrix}i^{underline{j}}dbinom{n}{i}(dfrac{1}{m-1})^i\ &=(dfrac{m-1}{m})^{n}sumlimits_{i=0}^nsumlimits_{j=1}^kegin{Bmatrix}k\jend{Bmatrix}i^{underline{j}}dbinom{n}{i}(dfrac{1}{m-1})^i\ &=(dfrac{m-1}{m})^{n}sumlimits_{j=1}^kegin{Bmatrix}k\jend{Bmatrix}j!sumlimits_{i=0}^ndbinom{i}{j}dbinom{n}{i}(dfrac{1}{m-1})^i\ &=(dfrac{m-1}{m})^{n}sumlimits_{j=1}^kegin{Bmatrix}k\jend{Bmatrix}j!dbinom{n}{j}(dfrac{1}{m-1})^jsumlimits_{i=0}^{n-j}dbinom{n-j}{i-j}(dfrac{1}{m-1})^i\ &=(dfrac{m-1}{m})^{n}sumlimits_{j=1}^kegin{Bmatrix}k\jend{Bmatrix}j!dbinom{n}{j}(dfrac{1}{m-1})^j(dfrac{m}{m-1})^{n-j}\ &=sumlimits_{j=1}^kegin{Bmatrix}k\jend{Bmatrix}j!dbinom{n}{j}(dfrac{1}{m})^j\ &=sumlimits_{j=1}^ksumlimits_{l=0}^jdbinom{j}{l}l^k(-1)^{j-l}dbinom{n}{j}(dfrac{1}{m})^j\ &=sumlimits_{l=0}^jl^ksumlimits_{j=1}^kdbinom{j}{l}(-1)^{j-l}dbinom{n}{j}(dfrac{1}{m})^j\ &=sumlimits_{l=0}^kl^kdbinom{n}{l}sumlimits_{j=l}^k(-1)^{j-l}dbinom{n-l}{j-l}(dfrac{1}{m})^j\ end{aligned} ]

前面那东西我们要枚举 (l)，复杂度自然有一个 (k)，后面那东西看似可以二项式定理，但是由于上界为 (k) 而不是 (n)，实则不能，因此考虑按照套路进行递推，设 (f_l)​ 表示

[f_l=sumlimits_{j=l}^k(-1)^{j-l}dbinom{n-l}{j-l}(dfrac{1}{m})^j ]

那么

[egin{aligned} f_{l-1}&=sumlimits_{j=l-1}^k(-1)^{j-l+1}dbinom{n-l+1}{j-l+1}(dfrac{1}{m})^j\ &=-sumlimits_{j=l-1}^k(-1)^{j-l}dbinom{n-l+1}{j-l+1}(dfrac{1}{m})^j\ &=-sumlimits_{j=l-1}^k(-1)^{j-l}dbinom{n-l}{j-l+1}(dfrac{1}{m})^j-sumlimits_{j=l-1}^k(-1)^{j-l}dbinom{n-l}{j-l}(dfrac{1}{m})^j\ &=-sumlimits_{j=l}^{k+1}(-1)^{j-1-l}dbinom{n-l}{j-l}(dfrac{1}{m})^{j-1}-sumlimits_{j=l-1}^k(-1)^{j-l}dbinom{n-l}{j-l}(dfrac{1}{m})^j\ &=msumlimits_{j=l}^{k+1}(-1)^{j-l}dbinom{n-l}{j-l}(dfrac{1}{m})^{j}-sumlimits_{j=l-1}^k(-1)^{j-l}dbinom{n-l}{j-l}(dfrac{1}{m})^j\ &=m(sumlimits_{j=l}^{k}(-1)^{j-l}dbinom{n-l}{j-l}(dfrac{1}{m})^{j}+(-1)^{k+1-l}dbinom{n-l}{k+1-l}(dfrac{1}{m})^{k+1})-sumlimits_{j=l}^k(-1)^{j-l}dbinom{n-l}{j-l}(dfrac{1}{m})^j\ &=m(f_l+(-1)^{k+1-l}dbinom{n-l}{n-k-1}(dfrac{1}{m})^{k+1})-f_l end{aligned} ]

从后往前一遍递推即可。( ext{id}_k) 可以一遍线筛筛出，复杂度 (mathcal O(k))

至于我是哪里被卡住了呢……在第一个公式的第七个等于号那边，我上来就想拆开来递推，然后发现递推完了要套好几层，然后就 quit 了。事实证明一般只有只带一个组合数的式子才能递推求解。

那么问题就来了，为什么我的 (mathcal O(k)) 跑不过 (mathcal O(klog k))/yun/yun

const int MAXK=1e7+1;
const int MOD=998244353;
int n,m,k;
int qpow(int x,int e){
	int ret=1;
	for(;e;e>>=1,x=1ll*x*x%MOD) if(e&1) ret=1ll*ret*x%MOD;
	return ret;
}
int fac[MAXK+5],ifac[MAXK+5],f[MAXK+5],inv[MAXK+5];
void init_fac(int n){
	for(int i=(fac[0]=ifac[0]=inv[0]=inv[1]=1)+1;i<=n;i++) inv[i]=1ll*inv[MOD%i]*(MOD-MOD/i)%MOD;
	for(int i=1;i<=n;i++) fac[i]=1ll*fac[i-1]*i%MOD,ifac[i]=1ll*ifac[i-1]*inv[i]%MOD;
}
int pwk[MAXK+5],pr[MAXK/10+5],prcnt=0;
bitset<MAXK+5> vis;
void sieve(int n){
	pwk[1]=1;
	for(int i=2;i<=n;i++){
		if(!vis[i]) pwk[i]=qpow(i,k);
		for(int j=1;j<=prcnt&&pr[j]*i<=n;j++){
			vis[pr[j]*i]=1;pwk[pr[j]*i]=1ll*pwk[pr[j]]*pwk[i]%MOD;
			if(i%pr[j]==0) break;
		}
	}
}
int main(){
	init_fac(MAXK);int res=0;
	scanf("%d%d%d",&n,&m,&k);sieve(k+1);
	int ivm=qpow(m,MOD-2),pw=qpow(ivm,k+1);
	for(int l=k+1,cur=1;l>=2;l--){
		int t=1ll*cur*pw%MOD;
//		printf("%d
",cur);
		if((k+1-l)&1) f[l-1]=(1ll*m*(f[l]-t+MOD)%MOD-f[l]+MOD)%MOD;
		else f[l-1]=(1ll*m*(f[l]+t)%MOD-f[l]+MOD)%MOD;
		cur=1ll*cur*(n-l+1+MOD)%MOD*inv[k+2-l]%MOD;
	}
	for(int l=1,cur=n;l<=k;l++){
		res=(res+1ll*f[l]*pwk[l]%MOD*cur)%MOD;
		cur=1ll*cur*(n-l)%MOD*inv[l+1]%MOD;
	} printf("%d
",res);
	return 0;
}