【Codeforces gym 102220】I Temperature Survey（分治NTT）

传送门

考虑加一个$n+1$的
然后转成左上角走到右下角
直接没法分治$ntt$
考虑枚举将$a[mid]$作为矩阵划分出来
然后转移

那么就是左/上转移到右/下
设高为$n$，长为$m$
$上\to 下:$
$P(x)={\sum }_{i=0}^{m-1}{\sum }_{j=0}^i{a}_j\left(\genfrac{}{}{0ex}{}{i-j+n-1}{n-1}\right)x^i$
$左\to 右：$
$Q(x)={\sum }_{i=0}^{n-1}{\sum }_{j=0}^i{b}_j\left(\genfrac{}{}{0ex}{}{i-j+m-1}{m-1}\right)x^i$
$左\to 下：$
$P(x)={\sum }_{i=0}^{m-1}{\sum }_{j=0}^{n-1}{a}_j\left(\genfrac{}{}{0ex}{}{i-j+n-1}{i}\right)x^i$
$x^{n-1}P(x)={\sum }_{i=n-1}^{n+m-2}\frac{x^i}{(i-n+1)!}{\sum }_{j=0}^{n-1}\frac{a_j}{(n-j-1)!}(i-j)!$
$上\to 右$
$Q(x)={\sum }_{i=0}^{n=1}{\sum }_{j=0}^{m-1}{a}_j\left(\genfrac{}{}{0ex}{}{i+m-1-j}{i}\right)x^i$
$x^{m-1}Q(x)={\sum }_{i=m-1}^{n+m-2}\frac{x^i}{(i-m+1)!}{\sum }_{j=0}^{m-1}\frac{a_j}{(m-j-1)!}(i-j)!$

直接做即可
复杂度$O(nlo{g}^{2}n)$

#include<bits/stdc++.h>
using namespace std;
#define cs const
#define re register
#define pb push_back
#define pii pair<int,int>
#define ll long long
#define y1 shinkle
#define fi first
#define se second
#define bg begin
cs int RLEN=1<<20|1;
inline char gc(){
    static char ibuf[RLEN],*ib,*ob;
    (ib==ob)&&(ob=(ib=ibuf)+fread(ibuf,1,RLEN,stdin));
    return (ib==ob)?EOF:*ib++;
}
inline int read(){
    char ch=gc();
    int res=0;bool f=1;
    while(!isdigit(ch))f^=ch=='-',ch=gc();
    while(isdigit(ch))res=(res+(res<<2)<<1)+(ch^48),ch=gc();
    return f?res:-res;
}
inline ll readll(){
    char ch=gc();
    ll res=0;bool f=1;
    while(!isdigit(ch))f^=ch=='-',ch=gc();
    while(isdigit(ch))res=(res+(res<<2)<<1)+(ch^48),ch=gc();
    return f?res:-res;
}
inline int readstring(char *s){
	int top=0;char ch=gc();
	while(isspace(ch))ch=gc();
	while(!isspace(ch)&&ch!=EOF)s[++top]=ch,ch=gc();
	s[top+1]='\0';return top;
}
template<typename tp>inline void chemx(tp &a,tp b){a=max(a,b);}
template<typename tp>inline void chemn(tp &a,tp b){a=min(a,b);}
cs int mod=998244353;
inline int add(int a,int b){return (a+b)>=mod?(a+b-mod):(a+b);}
inline int dec(int a,int b){return (a<b)?(a-b+mod):(a-b);}
inline int mul(int a,int b){static ll r;r=(ll)a*b;return (r>=mod)?(r%mod):r;}
inline void Add(int &a,int b){a=(a+b)>=mod?(a+b-mod):(a+b);}
inline void Dec(int &a,int b){a=(a<b)?(a-b+mod):(a-b);}
inline void Mul(int &a,int b){static ll r;r=(ll)a*b;a=(r>=mod)?(r%mod):r;}
inline int ksm(int a,int b,int res=1){for(;b;b>>=1,Mul(a,a))(b&1)&&(Mul(res,a),1);return res;}
inline int Inv(int x){return ksm(x,mod-2);}
inline int fix(ll x){x%=mod;return (x<0)?x+mod:x;}
typedef vector<int> poly;
namespace Poly{
	cs int C=19,M=(1<<C)|5;
	int rev[M],*w[C+1],fac[M],ifac[M];
	inline void init_rev(int lim){
		for(int i=0;i<lim;i++)rev[i]=(rev[i>>1]>>1)|((i&1)*(lim>>1));
	}
	inline void init_w(){
		for(int i=1;i<=C;i++)w[i]=new int [(1<<(i-1))|1];
		int wn=ksm(3,(mod-1)/(1<<C));w[C][0]=1;
		for(int i=1,l=1<<(C-1);i<l;i++)w[C][i]=mul(w[C][i-1],wn);
		for(int i=C-1;i;i--)
		for(int j=0,l=1<<(i-1);j<l;j++)w[i][j]=w[i+1][j<<1];
		fac[0]=ifac[0]=1;
		for(int i=1;i<M;i++)fac[i]=mul(fac[i-1],i);
		ifac[M-1]=Inv(fac[M-1]);
		for(int i=M-2;i;i--)ifac[i]=mul(ifac[i+1],i+1);
	}
	inline void ntt(int *f,int lim,int kd){
		for(int i=0;i<lim;i++)if(i>rev[i])swap(f[i],f[rev[i]]);
		for(int mid=1,l=1,a0,a1;mid<lim;mid<<=1,l++)
		for(int i=0;i<lim;i+=mid<<1)
		for(int j=0;j<mid;j++)
		a0=f[i+j],a1=mul(f[i+j+mid],w[l][j]),f[i+j]=add(a0,a1),f[i+j+mid]=dec(a0,a1);
		if(kd==-1){
			reverse(f+1,f+lim);
			for(int i=0,iv=Inv(lim);i<lim;i++)Mul(f[i],iv);
		}
	}
	inline poly operator *(poly a,poly b){
		if(!a.size()||!b.size())return poly(0);
		int deg=a.size()+b.size()-1;
		if(a.size()<=20||b.size()<=20){
			poly c(deg,0);
			for(int i=0;i<a.size();i++)
			for(int j=0;j<b.size();j++)
			Add(c[i+j],mul(a[i],b[j]));
			return c;
		}int lim=1;while(lim<deg)lim<<=1;
		init_rev(lim);
		a.resize(lim),ntt(&a[0],lim,1);
		b.resize(lim),ntt(&b[0],lim,1);
		for(int i=0;i<lim;i++)Mul(a[i],b[i]);
		ntt(&a[0],lim,-1),a.resize(deg);return a;
	}
}
inline int Cb(int n,int m){
	using namespace Poly;
	return (n<m||n<0||m<0)?0:mul(fac[n],mul(ifac[m],ifac[n-m]));
}
cs int N=200005;
map<int,int> dp[N];
int a[N],n;
inline void trans(int l,int r,int dn,int up){
	using namespace Poly;
	if(l==1&&r==1&&up==n+1){
		for(int i=dn;i<=up;i++)dp[1][i]=1;
		return;
	}
	int n=up-dn+1,m=r-l+1;
	poly a(n),b(m);//a->left b->top
	poly p(n),q(m);//p->rigjt q->bottom
	for(int i=0;i<m;i++)b[i]=dp[l+i][up+1];
	for(int i=0;i<n;i++)a[i]=dp[l-1][up-i];
	poly f,g;
	f=a,g.resize(n);
	for(int i=0;i<n;i++)g[i]=Cb(i+m-1,m-1);
	f=f*g;
	for(int i=0;i<n;i++)Add(p[i],f[i]);
	f=b,g.resize(m);
	for(int i=0;i<m;i++)g[i]=Cb(i+n-1,n-1);
	f=f*g;
	for(int i=0;i<m;i++)Add(q[i],f[i]);
	f.resize(n),g.resize(n+m-1);
	for(int i=0;i<n;i++)f[i]=mul(a[i],ifac[n-i-1]);
	for(int i=0;i<n+m-1;i++)g[i]=fac[i];
	f=f*g;
	for(int i=0;i<m;i++)Add(q[i],mul(f[i+n-1],ifac[i]));
	f.resize(m),g.resize(n+m-1);
	for(int i=0;i<m;i++)f[i]=mul(b[i],ifac[m-i-1]);
	for(int i=0;i<n+m-1;i++)g[i]=fac[i];
	f=f*g;
	for(int i=0;i<n;i++)Add(p[i],mul(f[i+m-1],ifac[i]));
	for(int i=l;i<=r;i++)Add(dp[i][dn],q[i-l]);
	for(int i=up;i>dn;i--)Add(dp[r][i],p[up-i]);
}
void solve(int l,int r,int dn){
	if(l>r)return;
	int mid=(l+r)>>1,x=mid,y=mid;
	while(x>l&&a[x-1]==a[mid])x--;
	while(y<r&&a[y+1]==a[mid])y++;
	solve(l,x-1,a[mid]+1);
	trans(l,y,dn,a[mid]);
	solve(y+1,r,dn);
}
inline void solve(){
	n=read();
	for(int i=1;i<=n;i++)a[i]=read();
	a[n+1]=n+1;reverse(a+1,a+n+2);
	for(int i=0;i<=n+1;i++)dp[i].clear();
	solve(1,n+1,1);
	cout<<dp[n+1][1]<<'\n';
}
int main(){
	#ifdef Stargazer
	freopen("lx.in","r",stdin);
	#endif
	Poly::init_w();
	int T=read();
	while(T--)solve();
	return 0;
}