P4091 [HEOI2016/TJOI2016]求和

首先,我们需要知道第二类斯特林数组的组合意义(即容斥)

(S^m_n = frac{1}{m!} sum_{k = 0}^{m}(-1)^k*C^k_m*(m - k)^n)

然后,题目中让我们求

(f(n) = sum_{i = 0}^nsum_{j = i}^nS^j_i*2^j*j!)

我们直接将(S^j_i)展开成上述的组合意义,同时(j)从(0)开始枚举（因为当(m > n)时,(S^m_n = 0)）

则有:

(f(n) = sum_{i = 0}^nsum_{j = 0}^nfrac{1}{j!}sum_{k = 0}^j(-1)^k*C^k_j *(j - k)^i * 2^j*j!)

我们发现(j!)可以约去,同时将(C^k_j)展开，(2^j)提到前边:

(=sum_{i = 0}^nsum_{j = 0}^n2^jsum_{k = 0}^j(-1)^k * frac{j!}{k! * (j - k)!}*(j - k) ^ i)

我们再讲(j!)提到前边,同时底数相同的化一下

(=sum_{i= 0}^nsum_{j = 0}^n2^j * j!sum_{k= 0}^j frac{(-1)^k}{k!}frac{(j - k)^i}{(j - k)!})

我们将第一个(sum)化到后面

(sum_{j = 0}^n2^j*j!sum_{k=0}^jfrac{(-1)^k}{k!}frac{sum_{i = 0}^n(j - k)^i}{(j- k)!})

根据等比数列求和公式

(sum_{i = 0}^n(j - k)^i = frac{1 - (j - k)^{n + 1}}{1 - (j - k)})

我们设

(f(i) = frac{(-1)^i}{i!},g(i) = frac{1 - i^{n + 1}}{(1 - i)*i!})

那么则有

上式(=sum_{j = 0}^n2^jj!sum_{k = 0}^j f(k)g(j - k))

发现,这是个卷积啊,直接上NTT就好了

#include<cstdio>
#include<iostream>
#include<cctype>
#include<cstring>
#include<algorithm>
#include<cmath>
#define LL long long
using namespace std;
const int N = 8e5 + 3;
const LL mod = 998244353;
LL n;
LL a[N],b[N];
LL fac[N];
LL p[N];
int r[N],l,limit = 1;
inline LL quick(LL a,LL b){
	LL res = 1;
	while(b){
		if(b & 1) res = res * a % mod;
		b >>= 1;
		a = a * a % mod;	
	}
	return res;
}
inline void nttle(LL *A,int type){
	for(int i = 0;i < limit;++i) 
		if(i < r[i]) swap(A[i],A[r[i]]);
	for(int mid = 1;mid < limit;mid <<= 1){
		LL Wn = (type == 1) ? quick(3,(mod - 1) / (mid << 1)) : 
		quick(3,mod - 1 - (mod - 1) / (mid << 1));
		for(int R = mid << 1,j = 0;j < limit;j += R){
			LL w = 1;
			for(int k = 0;k < mid;++k,w = w * Wn % mod){
				LL x = A[j + k],y = w * A[j + mid + k] % mod;
				A[j + k] = x + y;
				A[j + mid + k] = x - y;
				if(A[j + k] >= mod) A[j + k] -= mod;
				if(A[j + mid + k] < 0) A[j + mid + k] += mod;
			}
		}
	}
	if(type == -1){
		LL inv = quick(limit,mod - 2);
		for(int i = 0;i < limit;++i) A[i] = A[i] * inv % mod;
	}
}
int main(){
	scanf("%lld",&n);
	p[0] = fac[0] = 1;
	for(LL i = 1;i <= n;++i){
		p[i] = (p[i - 1] << 1) % mod;
		fac[i] = fac[i - 1] * i % mod; 
	}
	for(LL i = 0;i <= n;++i){
		if(i & 1) a[i] = (-1 + mod) * quick(fac[i],mod - 2) % mod;
		else a[i] = 1 * quick(fac[i],mod - 2) % mod;
		b[i] = (quick(i,n + 1) - 1 + mod) % mod * quick((i - 1 + mod) % mod * fac[i] % mod,mod - 2) % mod;
 	}
	b[0] = 1,b[1] = n + 1;
 //	for(int i = 0;i <= n;++i) printf("%lld ",a[i]);puts("");
 //	for(int i = 0;i <= n;++i) printf("%lld ",b[i]);puts("");
 	while(limit < 2 * (n + 1)) limit <<= 1,++l;
 	for(int i = 0;i < limit;++i) r[i] = (r[i >> 1] >> 1) | ((i & 1) << (l - 1));
	 //printf("%d ",r[i]);puts("");
 	nttle(a,1);nttle(b,1);
 	for(int i = 0;i < limit;++i) a[i] = a[i] * b[i] % mod;
 	nttle(a,-1);
 	LL ans = 0;
 //	for(int i = 0;i <= n;++i) printf("%lld ",a[i]);
 	for(int i = 0;i <= n;++i)
 		ans = (ans + p[i] * fac[i] % mod * a[i] % mod) % mod; 	
 	printf("%lld
",ans);
	return 0;	
}