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：
{nm}=1m!∑i(mk)(m−k)n(−1)k\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 {nm}=m!1i∑(mk)(m−k)n(−1)k
因此：
Ans=∑i=0n∑j=0i∑k=0j(jk)(j−k)i(−1)k2jAns=\sum_{i=0}^n\sum_{j=0}^i\sum_{k=0}^j\tbinom{j}{k}(j-k)^i(-1)^k2^j Ans=i=0∑nj=0∑ik=0∑j(kj)(j−k)i(−1)k2j
进一步的：
Ans=∑j2jj!∑k∑i(−1)kk!(j−k)i(j−k)!Ans=\sum_{j}2^jj!\sum_{k}\sum_{i}\frac{(-1)^k}{k!}\frac{(j-k)^i}{(j-k)!} Ans=j∑2jj!k∑i∑k!(−1)k(j−k)!(j−k)i
发现后面就是一个卷积形式。
直接NumberTheoreticTransformNumber\;\;Theoretic\;\;TransformNumberTheoreticTransform计算即可。

时间复杂度O(nlgn)O(nlgn)O(nlgn)。

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 secondusing 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;
}