[LGP2791] 幼儿园篮球题

你猜猜题怎么出出来的？

显然第\(i\)场的答案为

\[\frac{1}{\binom{n_i}{m_i}\binom{n_i}{k_i}}\sum_{x=0}^{k_i}\binom{n_i}{m_i}\binom{m_i}{x}\binom{n_i-m_i}{k_i-x}x^L =\frac{1}{\binom{n_i}{k_i}}\sum_{x=0}^{k_i}\binom{m_i}{x}\binom{n_i-m_i}{k_i-x}x^L\\ \]

利用斯特林数进行变换

\[\sum_{x=0}^{k_i}\binom{m_i}{x}\binom{n_i-m_i}{k_i-x}x^L =\sum_{x=0}^{k_i}\binom{m_i}{x}\binom{n_i-m_i}{k_i-x}\sum_{y=0}^{x}y!\binom{x}{y}\left\{\begin{matrix}L\\y\end{matrix}\right\}\\ =\sum_{y=0}^{k_i}y!\left\{\begin{matrix}L\\y\end{matrix}\right\} \sum_{x=y}^{k_i}\binom{x}{y}\binom{m_i}{x}\binom{n_i-m_i}{k_i-x}\\ =\sum_{y=0}^{k_i}y!\left\{\begin{matrix}L\\y\end{matrix}\right\} \sum_{x=y}^{k_i}\binom{m_i}{y}\binom{m_i-y}{x-y}\binom{n_i-m_i}{k_i-x}\\ =\sum_{y=0}^{k_i}y!\binom{m_i}{y}\left\{\begin{matrix}L\\y\end{matrix}\right\} \sum_{x=0}^{k_i-y}\binom{m_i-y}{x}\binom{n_i-m_i}{k_i-y-x}\\ =\sum_{y=0}^{k_i}y!\binom{m_i}{y}\left\{\begin{matrix}L\\y\end{matrix}\right\} \binom{n_i-y}{k_i-y} \]

发现\(y\)的实际上界为\(\min(k_i,m_i,L)\)不过\(1e5\)，而询问仅\(2e2\)，预处理第\(L\)行斯特林数，询问复杂度\(O(L)\)可以接受。

斯特林数的预处理参见。

最终答案为

\[\sum_{i=1}^S\frac{k_i!(n_i-k_i)!}{n_i!}\sum_{y=0}^{k_i}y!\frac{m_i!(n_i-y)!}{y!(m_i-y)!(k_i-y)!(n_i-k_i)!}S(L,y)\\ =\sum_{i=1}^S\frac{k_i!m_i!}{n_i!}\sum_{y=0}^{k_i}\frac{(n_i-y)!}{(m_i-y)!(k_i-y)!}S(L,y) \]

显得十分的和谐。

此题卡常

#include <bits/stdc++.h>
#define IL inline 
#define ll long long 
using namespace std;
const int N=8e5+10;
const int M=2e7+10;
const int mod=998244353;

IL ll gi(){
    ll x=0,f=1;
    char ch=getchar();
    while(!isdigit(ch))f^=ch=='-',ch=getchar();
    while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
    return f?x:-x;
}

int lmt,w[N],rev[N];
IL int fpw(int x,int y) {
	int c=1;
	for(; y; y>>=1,x=(ll)x*x%mod) if(y&1) c=(ll)c*x%mod;
	return c;
}
IL void init(int n) {
	int l=0; lmt=1;
	while(lmt<=n) lmt<<=1, l++; 
	for(int i=0; i<lmt; ++i) rev[i]=(rev[i>>1]>>1)|((i&1)<<(l-1));
	int tmp=lmt>>1, wlmt=fpw(3,(mod-1)>>l); w[tmp]=1;
	for(int i=tmp+1; i<lmt; ++i) w[i]=(ll)w[i-1]*wlmt%mod;
	for(int i=tmp-1; i; --i) w[i]=w[i<<1];
	lmt=l;
}
IL void DFT(int*a,int len) {
	static unsigned ll tmp[N];
	int u=lmt-__builtin_ctz(len),T;
	for(int i=0; i<len; ++i) tmp[rev[i]>>u]=a[i];
	for(int m=1; m<len; m<<=1)
		for(int i=0,s=m<<1; i<len; i+=s)
			for(int j=0; j<m; ++j) 
				T=tmp[i+j+m]*w[j+m]%mod,tmp[i+j+m]=tmp[i+j]+mod-T,tmp[i+j]+=T;
	for(int i=0; i<len; ++i) a[i]=tmp[i]%mod;
}
IL void IDFT(int*a,int len) {
	reverse(a+1,a+len); DFT(a,len);
	ll T=mod-(mod-1)/len;
	for(int i=0; i<len; ++i) a[i]=T*a[i]%mod;
}
IL int getLen(int n) {return 1<<(32-__builtin_clz(n));}

int n,m,s,l,k,fiv[M],fac[M],f[N],g[N];
IL void getFac(int n) {
	fac[0]=1; 
	for(int i=1; i<=n; ++i) fac[i]=(ll)fac[i-1]*i%mod;
	fiv[n]=fpw(fac[n],mod-2); 
	for(int i=n-1; ~i; --i) fiv[i]=(ll)fiv[i+1]*(i+1)%mod;
}
int cnt,pri[N];
bool vis[N];
IL void getSti(int n) {
	ll fup=mod-1;
	for(int i=0; i<=n; ++i) {
		fup=mod-fup;
		f[i]=fup*fiv[i]%mod;
	}
	g[1]=1;
	for(int i=2; i<=n; ++i) {
		if(!vis[i]) pri[++cnt]=i,g[i]=fpw(i,n);
		for(int j=1; j<=cnt&&pri[j]*i<=n; ++j) {
			vis[pri[j]*i]=1;
			g[pri[j]*i]=(ll)g[pri[j]]*g[i]%mod;
			if(i%pri[j]==0) break; 
		}
	}
	for(int i=1; i<=n; ++i) g[i]=(ll)g[i]*fiv[i]%mod;
	int len=getLen(n<<1);
	DFT(f,len); DFT(g,len);
	for(int i=0; i<len; ++i) f[i]=(ll)f[i]*g[i]%mod;
	IDFT(f,len);
	for(int i=n+1; i<len; ++i) f[i]=0;
}

int main() {
	scanf("%d%d%d%d",&n,&m,&s,&l);
	init(l<<1); 
	getFac(max(n,l));
	getSti(l);
	while(s--) {
		scanf("%d%d%d",&n,&m,&k);
		int T=min(min(m,k),l),sum=0;
		for(int i=0; i<=T; ++i) sum=(sum+(ll)fac[n-i]*fiv[m-i]%mod*fiv[k-i]%mod*f[i]%mod)%mod;
		sum=(ll)sum*fac[k]%mod*fac[m]%mod*fiv[n]%mod;
		printf("%d\n",sum);
	}
	return 0;
}