bzoj#4555. [Tjoi2016&Heoi2016]求和-优快云博客

本文链接：https://blog.youkuaiyun.com/xmr_pursue_dreams/article/details/103500049

bzoj#4555. [Tjoi2016&Heoi2016]求和

题目描述

Solution

有一个关于第二类斯特林数的公式：
$\left\{ \begin{aligned} n \\ m \end{aligned} \right\}= \frac{1}{m!}\sum_i \left ( \begin{aligned} m \\ k \end{aligned} \right )(m-k)^n(-1)^k$
因此：
$Ans=\sum_{i=0}^n\sum_{j=0}^i\sum_{k=0}^j\tbinom{j}{k}(j-k)^i(-1)^k2^j$
进一步的：
$Ans=\sum_{j}2^jj!\sum_{k}\sum_{i}\frac{(-1)^k}{k!}\frac{(j-k)^i}{(j-k)!}$
发现后面就是一个卷积形式。
直接 $Number\;\;Theoretic\;\;Transform$ 计算即可。

时间复杂度 $O (n l g n)$ 。

Code

#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <ctime>
#include <cassert>
#include <string.h>
//#include <unordered_set>
//#include <unordered_map>
//#include <bits/stdc++.h>

#define MP(A,B) make_pair(A,B)
#define PB(A) push_back(A)
#define SIZE(A) ((int)A.size())
#define LEN(A) ((int)A.length())
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define fi first
#define se second

using namespace std;

template<typename T>inline bool upmin(T &x,T y) { return y<x?x=y,1:0; }
template<typename T>inline bool upmax(T &x,T y) { return x<y?x=y,1:0; }

typedef long long ll;
typedef unsigned long long ull;
typedef long double lod;
typedef pair<int,int> PR;
typedef vector<int> VI;

const lod eps=1e-11;
const lod pi=acos(-1);
const int oo=1<<30;
const ll loo=1ll<<62;
const int mods=998244353;
const int G=3;
const int Gi=(mods+1)/G;
const int MAXN=600005;
const int INF=0x3f3f3f3f;//1061109567
/*--------------------------------------------------------------------*/
inline int read()
{
	int f=1,x=0; char c=getchar();
	while (c<'0'||c>'9') { if (c=='-') f=-1; c=getchar(); }
	while (c>='0'&&c<='9') { x=(x<<3)+(x<<1)+(c^48); c=getchar(); }
	return x*f;
}
int f[MAXN],g[MAXN],rev[MAXN],fac[MAXN],Limit,L;
int upd(int x,int y) { return x+y>=mods?x+y-mods:x+y; }
int quick_pow(int x,int y)
{
	int ret=1;
	for (;y;y>>=1)
	{
		if (y&1) ret=1ll*ret*x%mods;
		x=1ll*x*x%mods;
	}
	return ret;
}
void Number_Theoretic_Transform(int *A,int type)
{
	for (int i=0;i<Limit;i++) 
		if (i<rev[i]) swap(A[i],A[rev[i]]);
	for (int mid=1;mid<Limit;mid<<=1)
	{
		int Wn=quick_pow(type==1?G:Gi,(mods-1)/(mid<<1));
		for (int j=0;j<Limit;j+=(mid<<1))
			for (int k=j,w=1;k<j+mid;w=1ll*w*Wn%mods,k++)
			{
				int x=A[k],y=1ll*w*A[k+mid]%mods;
				A[k]=upd(x,y),A[k+mid]=upd(x,mods-y);
			}
	}
	if (type==-1)
		for (int inv=quick_pow(Limit,mods-2),i=0;i<Limit;i++) A[i]=1ll*A[i]*inv%mods;			
}
int main()
{
	int n=read();
	fac[0]=1;
	for (int i=1;i<=n;i++) fac[i]=1ll*fac[i-1]*i%mods;
	f[n]=quick_pow(fac[n],mods-2);
	for (int i=n-1;i>=0;i--) f[i]=1ll*f[i+1]*(i+1)%mods;
	
	f[0]=g[0]=1,f[1]=n+1,g[1]=mods-1;
	for (int i=2;i<=n;i++) 
	{
		g[i]=(i&1)?mods-f[i]:f[i]; 
		f[i]=1ll*upd(quick_pow(i,n+1),mods-1)*quick_pow(i-1,mods-2)%mods*f[i]%mods;
	}
	
	Limit=1,L=0;
	while (Limit<=n<<1) Limit<<=1,L++;
	for (int i=0;i<Limit;i++) rev[i]=(rev[i>>1]>>1)|((i&1)<<(L-1));
	Number_Theoretic_Transform(f,1);
	Number_Theoretic_Transform(g,1);
	for (int i=0;i<Limit;i++) f[i]=1ll*f[i]*g[i]%mods;
	Number_Theoretic_Transform(f,-1);
	
//	for (int i=0;i<=n;i++) cout<<i<<" "<<f[i]<<endl;
	
	int ans=0;
	for (int i=0,t=1;i<=n;i++,t=upd(t,t)) ans=upd(ans,1ll*fac[i]*t%mods*f[i]%mods);
	printf("%d\n",ans);
	return 0;
}