【CF923E】Perpetual Subtraction

by AmanoKumiko

Description

有一个整数\(x\)，范围在\([0,n]\)之间

初值为\(i\)的概率为\(p_i\)

然后会有\(m\)次操作

每次\(x\)会等概率变成\([0,x]\)中的整数

对于每个\(i\in [0,n]\)，求出\(m\)次操作后\(x\)为\(i\)的概率

对\(998244353\)取模

Input

第一行两个数\(n,m\)

第二行\(n+1\)个数读入\(p\)

Output

一行一个数表示答案

Sample Input

9 350
3 31 314 3141 31415 314159 3141592 31415926 314159265 649178508

Sample Output

822986014 12998613 84959018 728107923 939229297 935516344 27254497 413831286 583600448 442738326

Data Constraint

\(1\le n\le 10^5,1\le m\le 10^{18}\)

Solution

需要对积分比较敏感呢

令\(f_{i,j}\)表示\(i\)次操作后\(x\)为\(j\)的概率

那么有

\[f_{i,j}=\sum_{k=j}^n\frac{f_{i-1,k}}{k+1}\\ \]

用生成函数观察性质

\[\begin{align} F_{i}(x)&=\sum_{j=0}^nf_{i,j}x^j\\ &=\sum_{j=0}^nx^j\sum_{k=j}^n\frac{f_{i-1,k}}{k+1}\\ &=\sum_{k=0}^n\frac{f_{i-1,k}}{k+1}\sum_{j=0}^kx^j\\ &=\sum_{k=0}^n\frac{f_{i-1,k}}{k+1}\frac{x^{k+1}-1}{x-1}\\ &=\frac{1}{x-1}\sum_{k=0}^nf_{i-1,k}\int_{1}^{x}t^{k}dt\\ &=\frac{1}{x-1}\int_{1}^{x}F_{i-1}(t)dt\\ \end{align} \]

这里从\(1\)开始积分不好处理

我们希望的是\(\int_0^xf(t)dt=\int f(x)dx\)

那么设\(G_{i}(x)=F_{i}(x+1)\)同时\(G_{i}(x)=\sum_{j=0}^ng_{i,j}x^j\)

\[\begin{align} G_{i}(x)&=\frac{1}{x}\int_{0}^{x}F_{i-1}(t+1)d(t+1)\\ &=\sum_{k=0}^{n}\frac{g_{i-1,k}x^{k}}{k+1}\\ \end{align} \]

所以\(g_{i,j}=\frac{g_{i-1,j}}{j+1}\)
即\(g_{m,j}=\frac{g_{0,j}}{(j+1)^m}\)

众所周知，多项式平移可以用二项式定理展开变为差卷积

\[\begin{align} G_{0}(x)&=F_{0}(x+1)\\ &=\sum_{i=0}^np_i(x+1)^i\\ &=\sum_{i=0}^np_i\sum_{j=0}^i\binom{i}{j}x^j\\ &=\sum_{j=0}^n\frac{x^j}{j!}\sum_{i=j}^n\frac{i!p_i}{(i-j)!} \end{align} \]

\[\begin{align} F_{m}(x)&=G_{m}(x-1)\\ &=\sum_{i=0}^ng_{m,i}(x-1)^i\\ &=\sum_{i=0}^ng_{m,i}\sum_{j=0}^i\binom{i}{j}x^j(-1)^{i-j}\\ &=\sum_{j=0}^n\frac{x^j}{j!}\sum_{i=j}^n\frac{g_{m,i}i!(-1)^{i-j}}{(i-j)!} \end{align} \]

Code

#include
using namespace std;
#define F(i,a,b) for(int i=a;i<=b;i++)
#define Fd(i,a,b) for(int i=a;i>=b;i--)
#define N 500010
#define mo 998244353
#define LL long long

vectorIs;

int rev[N],G1[N],G2[N],fac[N],ifac[N],inv[N];

int mod(int x){return x>=mo?x-mo:x;}

int mi(int x,int y){
	if(!y)return 1;
	if(y==1)return x;
	return y%2?1ll*x*mi(1ll*x*x%mo,y/2)%mo:mi(1ll*x*x%mo,y/2);
}

void init(){
	Is.push_back(1);
	fac[0]=ifac[0]=1;
	F(i,1,N-10)fac[i]=1ll*fac[i-1]*i%mo,inv[i]=(i==1?1:1ll*mo/i*mod(mo-1ll*inv[mo%i]%mo)%mo);
	ifac[N-10]=mi(fac[N-10],mo-2);
	Fd(i,N-11,1)ifac[i]=1ll*ifac[i+1]*(i+1)%mo;
	for(int l=1;l<=N-10;l<<=1)G1[l]=mi(3,(mo-1)/(l*2)),G2[l]=mi(G1[l],mo-2);
}

void BRT(int x){F(i,0,x-1)rev[i]=(rev[i>>1]>>1)|((i&1)?(x>>1):0);}

struct poly{
	vectorval;
	void ins(int x){val.push_back(x);}
	void clear(){vector().swap(val);}
	int sz(){return val.size();}
	void rsz(int x){val.resize(x);}
	void shrink(){for(;sz()&&!val.back();val.pop_back());}
	poly modxn(int x){
		if(val.size()<=x)return (poly){val};
		else return (poly){vector(val.begin(),val.begin()+x)};
	}
	poly I(){return (poly){Is};}
	int operator[](int x)const{
		if(x<0||x>=val.size())return 0;
		return val[x];
	}
	void NTT(int x){
		F(i,0,sz()-1)if(rev[i]y.sz())swap(x,y);
			poly ret;
			ret.rsz(x.sz()+y.sz());
			F(i,0,ret.sz()-1){
				for(int j=0;j<=i&&j=sz())return f;
		if(tail*k>=Len)return f;
		f.rsz(Len);
		int Mul=mi(val[tail],mo-2);
		F(i,0,min(Len-1,sz()-tail-1))f.val[i]=1ll*val[i+tail]*Mul%mo;
		Mul=mi(val[tail],k);
		f=f.Ln(Len);
		F(i,0,Len-1)f.val[i]=1ll*f[i]*k%mo;
		f=f.Exp(Len);
		Fd(i,Len-1,tail*k)f.val[i]=1ll*f[i-tail*k]*Mul%mo;
		F(i,0,tail*k-1)f.val[i]=0;
		return f;
	}
};

int n,p[N];
LL m;

int main(){
	init();
	scanf("%d%lld",&n,&m);
	F(i,0,n)scanf("%d",&p[i]);
	poly f,g,G0,Fm;
	F(i,0,n)f.ins(ifac[i]);
	F(i,0,n)g.ins(1ll*p[n-i]*fac[n-i]%mo);
	f*=g;G0.rsz(n+1);
	F(i,0,n)G0.val[i]=1ll*f[n-i]*ifac[i]%mo*mi(mi(i+1,m%(mo-1)),mo-2)%mo;
	f.clear();g.clear();
	F(i,0,n)f.ins((i&1)?mo-ifac[i]:ifac[i]);
	F(i,0,n)g.ins(1ll*G0[n-i]*fac[n-i]%mo);
	f*=g;Fm.rsz(n+1);
	F(i,0,n)Fm.val[i]=1ll*f[n-i]*ifac[i]%mo;
	F(i,0,n)printf("%d ",Fm[i]);
	return 0;
}

生成函数多项式 CodeForces

【CF923E】Perpetual Subtraction