luogu2791 幼儿园篮球题 第二类斯大林数（特）卡常NTT

我真的是***的
原本应该是乘以$i^L$
但是因为
$$i^L=\sum _{j=0}^L\left\{\genfrac{}{}{0ex}{}{L}{j}\right\}\left(\genfrac{}{}{0ex}{}{i}{j}\right)j!$$
于是这题显然就是
$$\sum _{i=0}^{k_i}\left(\genfrac{}{}{0ex}{}{n-m}{k_i-i}\right)\left(\genfrac{}{}{0ex}{}{m}{i}\right)\sum _{j=0}^L\left\{\genfrac{}{}{0ex}{}{L}{j}\right\}\left(\genfrac{}{}{0ex}{}{i}{j}\right)j!$$
然后期望就是整体除以$\left(\genfrac{}{}{0ex}{}{n}{k_i}\right)$
首先交换一下求和顺序
就会变成
$$\sum _{j=0}^L\left\{\genfrac{}{}{0ex}{}{L}{j}\right\}j!\sum _{i=0}^{k_i}\left(\genfrac{}{}{0ex}{}{n-m}{k_i-i}\right)\left(\genfrac{}{}{0ex}{}{m}{i}\right)\left(\genfrac{}{}{0ex}{}{i}{j}\right)$$
而
$$\left(\genfrac{}{}{0ex}{}{m}{i}\right)\left(\genfrac{}{}{0ex}{}{i}{j}\right)=\left(\genfrac{}{}{0ex}{}{m}{j}\right)\left(\genfrac{}{}{0ex}{}{m-j}{i-j}\right)$$
于是就变成了
$$\sum _{j=0}^L\left\{\genfrac{}{}{0ex}{}{L}{j}\right\}j!\left(\genfrac{}{}{0ex}{}{m}{j}\right)\sum _{i=0}^{k_i}\left(\genfrac{}{}{0ex}{}{n-m}{k_i-i}\right)\left(\genfrac{}{}{0ex}{}{m-j}{i-j}\right)$$
后面发现上面的和为$n-j$，下面的和为$k_i-j$，都与$i$无瓜
于是就是一个范德蒙德求和的形式
就可以变成
$$\sum _{j=0}^L\left\{\genfrac{}{}{0ex}{}{L}{j}\right\}j!\left(\genfrac{}{}{0ex}{}{m}{j}\right)\left(\genfrac{}{}{0ex}{}{n-j}{k_i-j}\right)$$
前面的第二类斯大林数可以预处理，可以利用
$$\left\{\genfrac{}{}{0ex}{}{L}{j}\right\}=\sum _{k=0}^m\frac{(-1{)}^k}{k!}\frac{(j-k{)}^L}{(j-k)!}$$
直接NTT预处理
这题比较卡常
如果直接线性预处理逆元然后做前缀积会T飞。
只有反过来用$n!$的逆元乘回来才可以（*的我卡常卡了一早上没想到这里慢破天际）

/* ***********************************************
Author        :BPM136
Created Time  :2019/8/14 9:17:59
File Name     :2791.cpp
************************************************ */

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <ctime>

#define USE_CIN_COUT ios::sync_with_stdio(0)

using namespace std;

typedef long long ll;

namespace fastIO{  
    #define BUF_SIZE 100000  
    #define OUT_SIZE 100000  
    //fread->read  
    bool IOerror=0;  
    inline char nc(){  
		static char buf[BUF_SIZE],*p1=buf+BUF_SIZE,*pend=buf+BUF_SIZE;  
		if (p1==pend){  
			p1=buf; pend=buf+fread(buf,1,BUF_SIZE,stdin);  
			if (pend==p1){IOerror=1;return -1;}  
			//{printf("IO error!\n");system("pause");for (;;);exit(0);}  
		}  
		return *p1++;  
	}  
	inline bool blank(char ch){return ch==32||ch==10||ch==13||ch==9;}  
	inline bool enter(char ch){return ch==10||ch==13;}
	inline void read(int &x){  
		bool sign=0; char ch=nc(); x=0;  
		for (;blank(ch);ch=nc());  
		if (IOerror)return;  
		if (ch==45)sign=1,ch=nc();  
		for (;ch>=48&&ch<=57;ch=nc())x=x*10+ch-48;  
		if (sign)x=-x;  
	}  
	inline void read(ll &x){  
		bool sign=0; char ch=nc(); x=0;  
		for (;blank(ch);ch=nc());  
		if (IOerror)return;  
		if (ch==45)sign=1,ch=nc();  
		for (;ch>=48&&ch<=57;ch=nc())x=x*10+ch-48;  
		if (sign)x=-x;  
	}  
#undef OUT_SIZE  
#undef BUF_SIZE  
}
using fastIO::read;

template<typename T>
void print(T x) {
	static char s[100], *s1; s1 = s;
	if (!x) *s1++ = '0';
	if (x < 0) putchar('-'), x = -x;
	while(x) *s1++ = (x % 10 + '0'), x /= 10;
	while(s1-- != s) putchar(*s1);
}

/////eval多点求值,interpolation插值
//不可传入一个系数为负数的多项式，可能在减法的时候会爆mod
//常数稍微有点大
//G是TT的原根
//NTT前记得getRevRoot

int const TT = 998244353, G = 3;
int const MAXN = 2e7 + 5;
int const MAXM = 3e5 + 5;
int const mod = TT;

int fac[MAXN], inv[MAXN];
int N, M, S, L;

inline int KSM (int a, int k) {
	int ret = 1 % TT;
	for (; k; k >>= 1, a = 1ll * a * a % TT)
		if (k & 1)
			ret = 1ll * ret * a % TT;
	return ret;
}

inline int add (int x, int y) {
    if (x + y >= TT)
        return x + y - TT;
    else
        return x + y;
}

inline int sub (int x, int y) {
    if (x - y >= 0)
        return x - y;
    else
        return x - y + TT;
}

int rev[MAXM << 2], rt[MAXM << 2];

inline void getRevRoot (int n) {
	int m = log(n) / log(2) + 1e-7;
	for (int i = 1; i < n; ++i)
		rev[i] = rev[i >> 1] >> 1 | (i & 1) << (m - 1);
	for (int len = 1, uni; len < n; len <<= 1) {
		uni = KSM(G, (TT ^ 1) / (len << 1));
		rt[len] = 1;
		for (int i = 1; i < len; ++i)
			rt[i + len] = 1ll * rt[i + len - 1] * uni % TT;
	}
}

inline void NTT (int* f, int n) {
	for (int i = 0; i < n; ++i) 
		if (i < rev[i])
			swap(f[i], f[rev[i]]);
	for (int len = 1; len < n; len <<= 1)
		for (int i = 0; i < n; i += len << 1)
			for (int j = 0, x, y; j < len; ++j) {
				x = f[i + j];
				y = 1ll * f[i + j + len] * rt[j + len] % TT;
				f[i + j] = add(x, y);
				f[i + j + len] = sub(x, y);
			}
}

int f[MAXM << 2], g[MAXM << 2];

inline void NTT() {
	int n = 1, m = (L + 1) * 2 - 1, ivn;
	while (n < m)
		n <<= 1;
	ivn = KSM(n, TT - 2);
	getRevRoot(n);
	NTT(f, n);
	NTT(g, n);
	for (int i = 0; i < n; ++i)
		f[i] = 1ll * f[i] * g[i] % TT;
	reverse(f + 1, f + n);
	NTT(f, n);
	for (int i = 0; i < m; ++i)
		f[i] = 1ll * f[i] * ivn % TT;
}

inline void init() {
	inv[0] = inv[1] = fac[0] = fac[1] = 1;
	for (int i = 2; i <= N; ++i) 
		fac[i] = 1ll * fac[i - 1] * i % mod;
	inv[N] = KSM(fac[N], mod - 2);
	for (int i = N - 1; i >= 1; --i)
		inv[i] = 1ll * inv[i + 1] * (i + 1) % mod;
	for (int i = 0; i <= L; ++i) {
		f[i] = ((i & 1) ? TT - inv[i] : inv[i]);
		g[i] = 1ll * KSM(i, L) * inv[i] % mod;
	}
	NTT();
}

int main() {
	read(N);
	read(M);
	read(S);
	read(L);
	init();
	int ans;
	int n, m, K;
	while (S--) {
		read(n);
		read(m);
		read(K);
		ans = 0;
		int lim = min(m, min(K, L));
		for (int j = 0; j <= lim; ++j) {
			ans += 1ll * f[j] * inv[m - j] % mod * fac[n - j] % mod * inv[K - j] % mod;
			if (ans >= mod)
				ans -= mod;
		}
		ans = 1ll * ans * fac[K] % mod * inv[n] % mod * fac[m] % mod;
		print(ans);
		putchar('\n');
	}
    return 0;
}