题目：

分析：

线段树

代码：

 

// Problem: F - Two Sequence Queries
// Contest: AtCoder - SuntoryProgrammingContest2024（AtCoder Beginner Contest 357）
// URL: https://atcoder.jp/contests/abc357/tasks/abc357_f
// Memory Limit: 1024 MB
// Time Limit: 5000 ms
// 
// Powered by CP Editor (https://cpeditor.org)#include<bits/stdc++.h>using namespace std;
#define int long longconst int N = 2e6+5;
const int mod=998244353;int n,q;
int a[N],b[N];
int taga[N],tagb[N];
int sum[N],suma[N],sumb[N];void addtag(int p,int pl,int pr,int d,int k){if(k==1){(suma[p]+=1ll*d*(pr-pl+1)%mod)%=mod;(sum[p]+=1ll*sumb[p]*d%mod)%=mod;(taga[p]+=d)%=mod;}else{(sumb[p]+=1ll*d*(pr-pl+1)%mod)%=mod;(sum[p]+=1ll*suma[p]*d%mod)%=mod;(tagb[p]+=d)%=mod;}
}void pushup(int p){(suma[p]=(suma[p*2]+suma[p*2+1])%mod)%=mod;(sumb[p]=(sumb[p*2]+sumb[p*2+1])%mod)%=mod;(sum[p]=(sum[p*2]+sum[p*2+1])%mod)%=mod;
}void pushdown(int p,int pl,int pr){int mid=(pl+pr)/2;if(taga[p]){addtag(p*2,pl,mid,taga[p],1);addtag(p*2+1,mid+1,pr,taga[p],1);taga[p]=0;}if(tagb[p]){addtag(p*2,pl,mid,tagb[p],2);addtag(p*2+1,mid+1,pr,tagb[p],2);tagb[p]=0;}
}void build(int p,int pl,int pr){if(pl==pr){suma[p]=a[pl]%mod;sumb[p]=b[pl]%mod;sum[p]=1ll*a[pl]*b[pl]%mod;taga[p]=tagb[p]=0;return;}int mid=(pl+pr)/2;build(p*2,pl,mid);build(p*2+1,mid+1,pr);pushup(p);return;
}void update(int l,int r,int p,int pl,int pr,int d,int k){if(l<=pl&&pr<=r){addtag(p,pl,pr,d,k);return;}int mid=(pl+pr)/2;pushdown(p,pl,pr);if(l<=mid){update(l,r,p*2,pl,mid,d,k);}if(r>mid){update(l,r,p*2+1,mid+1,pr,d,k);}pushup(p);return;
}int query(int l,int r,int p,int pl,int pr){if(l<=pl&&pr<=r){return sum[p];}int mid=(pl+pr)/2;pushdown(p,pl,pr);int ans=0;if(l<=mid){(ans+=query(l,r,p*2,pl,mid))%=mod;}if(r>mid){(ans+=query(l,r,p*2+1,mid+1,pr))%=mod;}return ans;
}signed main(){ios::sync_with_stdio(false);cin.tie(nullptr);cin>>n>>q;for(int i=1;i<=n;i++){cin>>a[i];}for(int i=1;i<=n;i++){cin>>b[i];}build(1,1,n);while(q--){int op,l,r,x;cin>>op;if(op==1){cin>>l>>r>>x;update(l,r,1,1,n,x,1);}else if(op==2){cin>>l>>r>>x;update(l,r,1,1,n,x,2);}else if(op==3){cin>>l>>r;cout<<query(l,r,1,1,n)<<"\n";}}return 0;	}