简单的数据结构题

首先考虑计算要求的式子，不妨设$l=1,r=n$。

${\sum }_{i=1}^n{a}_i^k{\prod }_{j\ne i}\frac{1-a_i{a}_j}{a_i-a_j}$

$={\sum }_{i=1}^n{a}_i^k{\prod }_{j\ne i}\frac{1}{a_i-a_j}{\prod }_{j\ne i}(1-a_i{a}_j)$

$={\sum }_{i=1}^n{a}_i^k{\prod }_{j\ne i}\frac{1}{a_i-a_j}{\sum }_{l=0}^{n-1}[x^{n-1-l}]{\prod }_{j\ne i}(x-a_i{a}_j)$

$={\sum }_{i=1}^n{a}_i^k{\prod }_{j\ne i}\frac{1}{a_i-a_j}{\sum }_{l=0}^{n-1}{a}_i^l[x^{n-1-l}]{\prod }_{j\ne i}(x-a_j)$

$={\sum }_{l=0}^{n-1}{\sum }_{i=1}^n{a}_i^{k+l}{\prod }_{j\ne i}\frac{1}{a_i-a_j}[x^{n-1-l}]{\prod }_{j\ne i}(x-a_j)$

$={\sum }_{l=0}^{n-1}{\sum }_{i=1}^n[x^{n-1-l}](a_i^{k+l}{\prod }_{j\ne i}\frac{1}{a_i-a_j})x^0{\prod }_{j\ne i}(x-a_j)$

$={\sum }_{l=0}^{n-1}{\sum }_{i=1}^n[x^{n-1-l}]a_i^{k+l}{\prod }_{j\ne i}\frac{x-a_j}{a_i-a_j}$

$={\sum }_{l=0}^{n-1}[x^{n-1-l}]{\sum }_{i=1}^n{a}_i^{k+l}{\prod }_{j\ne i}\frac{x-a_j}{a_i-a_j}$

---

根据拉格朗日插值：

$f(x)$是一个过点$(x_1,y_1),(x_2,y_2),...,(x_n,y_n)的函数$

记$M={\prod }_{i=1}^n(x-x_i)$，则

$f(x)\equiv {\sum }_{i=1}^n{y}_i{\prod }_{j\ne i}\frac{x-x_j}{x_i-x_j}(modM)$

---

函数$f(x)=x^{k+l}$过点$(a_1,a_1^{k+l}),(a_2,a_2^{k+l}),...,(a_n,a_n^{k+l})$

$\therefore x^{k+l}\equiv {\sum }_{i=1}^n{a}_i^{k+l}{\prod }_{j\ne i}\frac{x-a_j}{a_i-a_j}(mod(x-a_1)(x-a_2)...(x-a_n))$

$\therefore 原式={\sum }_{l=0}^{n-1}[x^{n-1-l}](x^{k+l}\mathrm{m}\mathrm{o}\mathrm{d}(x-a_1)(x-a_2)...(x-a_n))$

---

注意到$0\le k\le 2$，且$l\le n-1$，所以$k+l\ge n$的情况不多，分类讨论即可：

1.若$k+l\le n-1$（$x^{k+l}$的次数低于$mod$的多项式）：

$原式={\sum }_{l=1}^{n-1}[x^{n-1-l}]x^{k+l}={\sum }_{l=1}^{n-1}[n-1-l=k+l]$

解得$l=\frac{n-k-1}{2}$

则当$n-k-1\equiv 0(mod2)$且$0\le n-k-1$时原式为1，否则为0，
即当$n-k-1\equiv 1(mod2)$或$n=1,k=2$时原式为0，否则为1

2.若$k+l\ge n$：

$\because l\le n-1$ $\therefore k\ge 1$

2.1若$k=1$：

此时有$l=n-1$

$原式=[x^0](x^n\mathrm{m}\mathrm{o}\mathrm{d}(x-a_1)(x-a_2)...(x-a_n))$
$=[x^0](x^n-(x-a_1)(x-a_2)...(x-a_n))$
$=(-1{)}^{n+1}{a}_1{a}_2...a_n$

2.2若$k=2$：

此时有$l=n-2或l=n-1$

2.2.1若 $l=n-2$：

$\because l\ge 0$ $\therefore n\ne 1$

$原式=[x^1](x^n\mathrm{m}\mathrm{o}\mathrm{d}(x-a_1)(x-a_2)...(x-a_n))$
$=[x^1](x^n-(x-a_1)(x-a_2)...(x-a_n))$
$=(-1{)}^n{\sum }_{i=1}^n{\prod }_{j\ne i}{a}_j$

2.2.2若 $l=n-1$：

$原式=[x^0](x^{n+1}\mathrm{m}\mathrm{o}\mathrm{d}(x-a_1)(x-a_2)...(x-a_n))$
$=[x^0](x^{n+1}-\begin{array}{c}\underbrace{(x+a_1+a_2+...+a_n)}\\ 大除法得出\end{array}(x-a_1)(x-a_2)...(x-a_n))$
$=(-1{)}^{n+1}{\sum }_{i=1}^n{a}_i{\prod }_{j=1}^n{a}_j$

当$n=1,k=2$时，两处的特判可以相互抵消，于是可以不需要特判。

于是只要维护${\sum }_{i=1}^n{a}_i$ ，${\prod }_{i=1}^n{a}_i$ ，${\sum }_{i=1}^n{\prod }_{j\ne i}{a}_j$，在区间乘的操作下用线段树解决

#include<iostream>
#include<cstdio>
using namespace std;
const int mod=998244353;
const int N=300010;
int read(){int x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}return x*f;
}
struct Node{int sum,pro,ans;Node(){sum=0;pro=1;ans=0;}Node(int a,int b,int c){sum=a,pro=b,ans=c;}
}t[N<<2];
int n,m,a[N],laz[N<<2];
int add(int a,int b){return a+b>=mod?a+b-mod:a+b;}
int dec(int a,int b){return a-b<0?a-b+mod:a-b;}
int mul(int a,int b){return 1ll*a*b%mod;}
int power(int a,int b){int res=1;while(b){if(b&1) res=mul(res,a);a=mul(a,a);b>>=1;}return res;
}
Node merge(Node a,Node b){Node c;c.sum=add(a.sum,b.sum);c.pro=mul(a.pro,b.pro);c.ans=add(mul(a.ans,b.pro),mul(a.pro,b.ans));return c;
}
void pushup(int u){t[u]=merge(t[u<<1],t[u<<1|1]);
}
void build(int u,int l,int r){laz[u]=1;if(l==r){t[u]=Node(a[l],a[l],1);return;}int mid=(l+r)>>1;build(u<<1,l,mid);build(u<<1|1,mid+1,r);pushup(u);
}
void modify(int u,int l,int r,int x){t[u].sum=mul(t[u].sum,x);t[u].pro=mul(t[u].pro,power(x,r-l+1));t[u].ans=mul(t[u].ans,power(x,r-l));laz[u]=mul(laz[u],x);
}
void pushdown(int u,int l,int r){if(laz[u]!=1){int mid=(l+r)>>1;modify(u<<1,l,mid,laz[u]);modify(u<<1|1,mid+1,r,laz[u]);laz[u]=1;}
}
void update(int u,int l,int r,int a,int b,int x){if(a<=l&&r<=b){modify(u,l,r,x);return;}pushdown(u,l,r);int mid=(l+r)>>1;if(a<=mid) update(u<<1,l,mid,a,b,x);if(b>mid) update(u<<1|1,mid+1,r,a,b,x);pushup(u);
}
Node query(int u,int l,int r,int a,int b){if(a<=l&&r<=b) return t[u];pushdown(u,l,r);int mid=(l+r)>>1;Node res=Node(0,1,0);if(a<=mid) res=merge(res,query(u<<1,l,mid,a,b));if(b>mid) res=merge(res,query(u<<1|1,mid+1,r,a,b));return res; 
}
int main(){n=read();m=read();for(int i=1;i<=n;i++) a[i]=read();build(1,1,n);int opt,l,r,k;while(m--){opt=read();l=read();r=read();k=read();if(opt==1)update(1,1,n,l,r,k);else{int d=r-l+1,ans=0;if(!((d-k-1)&1)) ans++;Node now=query(1,1,n,l,r);if(k==1){if(d&1) ans=add(ans,now.pro);else ans=dec(ans,now.pro);}if(k==2){if(!(d&1)) ans=add(ans,now.ans);else ans=dec(ans,now.ans);if(d&1) ans=add(ans,mul(now.sum,now.pro));else ans=dec(ans,mul(now.sum,now.pro));}printf("%d\n",ans);}}return 0;
}