題目

對於 100% 的資料,$n, q ≤ 10^5$,$ai ≤ 10^9,$$ k ≤ 2^30$, 1 ≤ l ≤ r ≤ n.$

題解

每個數多被減30次。 線段樹維護，這一段全部是0？全部是1？既有0又有1？ 如果全是0，那麼直接return。 通過勢能分析，總複雜度為$O(n\ log^2(n))$

程式碼

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
 

#include<algorithm>
#define N 100010
#define LL long long
#define P(a) putchar(a)
#define mo 998244353
#define fo(i,a,b) for(i=a;i<=b;i++)
using namespace std;
LL tr[N],tr2[N];
LL Tr[30][N<<2];
int i,j,k,l,r,n,m,q,cs,op;
int O,opx,opy,opl,opr,ans;
LL Ans,a1,A,w;
int a[N],_2[31],_22[31];
int
 
 read(){
	int fh=0,rs=0;char ch=0;
	while((ch<'0'||ch>'9')&&(ch^'-'))ch=getchar();
	if(ch=='-')fh=1,ch=getchar();
	while(ch>='0'&&ch<='9')rs=(rs<<3)+(rs<<1)+(ch^'0'),ch=getchar();
	return fh?-rs:rs;
}
void write(int x){
	if(x>9)write(x/10);
	P(x%10+'0');
}
int lowbit
 
(int x){return x&(-x);}
void add(int x,LL z){
	for(;x<=n;x=x+lowbit(x))
	tr[x]=tr[x]+z;
}
LL qry(int x){
	LL rs=0;
	for(;x;x=x-lowbit(x))rs=rs+tr[x];
	return rs;
}
void add2(int x,LL z){
	for(;x<=n;x=x+lowbit(x))
	tr2[x]=(tr2[x]+z)%mo;
}
LL qry2(int x){
	LL rs=0;
	for(;x;x=x-lowbit(x))rs=(rs+tr2[x])%mo;
	return rs;
}
void update(int ps){
	if(Tr[opy][ps<<1]==1&&Tr[opy][(ps<<1)|1]==1){
		Tr[opy][ps]=1;
	}else
	if(Tr[opy][ps<<1]==0&&Tr[opy][(ps<<1)|1]==0){
		Tr[opy][ps]=0;
	}else Tr[opy][ps]=2;
}
void chx(int ps,int l,int r){
	if(Tr[opy][ps]==0)return;
	if(opl<=l&&r<=opr&&Tr[opy][ps]==1){
		int i;
		fo(i,l,r){
			add2(i,(mo-1ll*a[i]*a[i]%mo)%mo);
			add(i,-_2[opy]);
			a[i]=a[i]-_2[opy];
			add2(i,1ll*a[i]*a[i]%mo);
		}
		Tr[opy][ps]=0;
		return;
	}
	if(l==r)return;
	int wz=(l+r)>>1;
	if(opl<=wz)chx(ps<<1,l,wz);
	if(opr>wz)chx((ps<<1)|1,wz+1,r);
	update(ps);
}
void build(int ps,int l,int r){
	if(l==r){
		int i;
		fo(i,0,29)if(_2[i]&a[l]){
			Tr[i][ps]=1;
		}
		return;
	}
	int wz=(l+r)>>1;
	build(ps<<1,l,wz);
	build((ps<<1)|1,wz+1,r);
	fo(opy,0,29)update(ps);
}
int main(){
    O=1073741823;
    _2[0]=1;_22[0]=1;
	fo(i,1,30){
    	_2[i]=_2[i-1]<<1;
    	_22[i]=(1ll*_2[i]*_2[i])%mo;
	}
	n=read();
	fo(i,1,n){
		a[i]=read();
		add(i,a[i]);
		add2(i,(1ll*a[i]*a[i])%mo);
	}
	build(1,1,n);
	fo(j,0,29){
		opy=j;
		fo(i,1,n)if(a[i]&_2[j]){
			opx=i;
			chx(1,1,n);
		}
	}
	q=read();
	fo(cs,1,q){
		op=read();
		if(op==1){
			opl=read(),opr=read();k=read();
			fo(opy,0,29)if(!(k&_2[opy])){
				chx(1,1,n);
			}
		}else
		if(op==2){
			opl=read(),opr=read();
			Ans=qry(opr)-qry(opl-1);
			printf("%lld\n",Ans);
		}else{
			opl=read(),opr=read();
			Ans=(qry2(opr)-qry2(opl-1)+mo)%mo;
			Ans=(Ans*2*(opr-opl+1)%mo)%mo;
			A=(qry(opr)-qry(opl-1)+mo)%mo;
			A=(A*A%mo)*2%mo;
			ans=(Ans+A)%mo;
			write(ans),P('\n');
		}
	}
	return 0;
}