loj#6261. 一个人的高三楼（NTT+组合数学）

题面

传送门

题解

统计\(k\)阶前缀和，方法和这题一样

然后这里\(n\)比较大，那么把之前的柿子改写成

\[s_{j,k}=\sum_{i=1}^ja_i{j-i+k-1\choose j-i}=\sum_{i=1}^na_i{(j-i+k-1)^{\underline{j-i}}\over (j-i)!}
\]

就可以化成卷积形式了

//minamoto
#include<bits/stdc++.h>
#define R register
#define ll long long
#define fp(i,a,b) for(R int i=(a),I=(b)+1;i<I;++i)
#define fd(i,a,b) for(R int i=(a),I=(b)-1;i>I;--i)
#define go(u) for(int i=head[u],v=e[i].v;i;i=e[i].nx,v=e[i].v)
using namespace std;
char buf[1<<21],*p1=buf,*p2=buf;
inline char getc(){return p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++;}
int read(){
    R int res,f=1;R char ch;
    while((ch=getc())>'9'||ch<'0')(ch=='-')&&(f=-1);
    for(res=ch-'0';(ch=getc())>='0'&&ch<='9';res=res*10+ch-'0');
    return res*f;
}
ll readll(){
    R ll res,f=1;R char ch;
    while((ch=getc())>'9'||ch<'0')(ch=='-')&&(f=-1);
    for(res=ch-'0';(ch=getc())>='0'&&ch<='9';res=res*10+ch-'0');
    return res*f;
}
char sr[1<<21],z[20];int C=-1,Z=0;
inline void Ot(){fwrite(sr,1,C+1,stdout),C=-1;}
void print(R int x){
    if(C>1<<20)Ot();if(x<0)sr[++C]='-',x=-x;
    while(z[++Z]=x%10+48,x/=10);
    while(sr[++C]=z[Z],--Z);sr[++C]='\n';
}
const int N=(1<<18)+5,P=998244353;
inline int add(R int x,R int y){return x+y>=P?x+y-P:x+y;}
inline int dec(R int x,R int y){return x-y<0?x-y+P:x-y;}
inline int mul(R int x,R int y){return 1ll*x*y-1ll*x*y/P*P;}
int ksm(R int x,R int y){
	R int res=1;
	for(;y;y>>=1,x=mul(x,x))if(y&1)res=mul(res,x);
	return res;
}
int r[N],rt[2][N],inv[N],a[N],c[N];
int lim,d,ilim,n,k;
void Pre(){
	lim=1,d=0;while(lim<(n<<1))lim<<=1,++d;ilim=ksm(lim,P-2);
	inv[0]=inv[1]=1;fp(i,2,n)inv[i]=mul(P-P/i,inv[P%i]);
	c[0]=1;fp(i,1,n-1)c[i]=1ll*c[i-1]*(i+k-1)%P*inv[i]%P;
	fp(i,1,lim-1)r[i]=(r[i>>1]>>1)|((i&1)<<(d-1));
	for(R int t=(P-1)>>1,i=1,x,y;i<lim;i<<=1,t>>=1){
		x=ksm(3,t),y=ksm(332748118,t),rt[0][i]=rt[1][i]=1;
		fp(k,1,i-1)
			rt[1][i+k]=mul(rt[1][i+k-1],x),
			rt[0][i+k]=mul(rt[0][i+k-1],y);
	}
}
void NTT(int *A,int ty){
	fp(i,0,lim-1)if(i<r[i])swap(A[i],A[r[i]]);
	for(R int mid=1;mid<lim;mid<<=1)
		for(R int j=0,t;j<lim;j+=(mid<<1))
			fp(k,0,mid-1)
				A[j+k+mid]=dec(A[j+k],t=mul(rt[ty][mid+k],A[j+k+mid])),
				A[j+k]=add(A[j+k],t);
	if(!ty)fp(i,0,lim-1)A[i]=mul(A[i],ilim);
}
int main(){
//	freopen("testdata.in","r",stdin);
	n=read(),k=readll()%P;
	fp(i,0,n-1)a[i]=read();
	Pre();
	NTT(a,1),NTT(c,1);
	fp(i,0,lim-1)a[i]=mul(a[i],c[i]);
	NTT(a,0);
	fp(i,0,n-1)print(a[i]);
	return Ot(),0;
}