题解 CF1401F 【Reverse and Swap】

Solution- $CF1401F$

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$\mathrm{S}\mathrm{o}\mathrm{l}$

A not difficult ds question

We consider putting the sequence on the line segment tree, $2^{The}$ number of${}^n$ is equivalent to the number of leaf nodes of the line segment tree.

For $1\sim$The operation of$4$ is also$2,3\; is$ more difficult to maintain, we consider$2$ operation, assuming the$i$ node isdd on theline segment tree$d$ layer ($1$ node is$n$ layer). Then the inversion operation is equivalent to inverting the left and right sons of all the layers below respectively, so we set$0\sim$Just mark layer$d$ . For$3$ operation is equivalent to$d+The\; 1-$ layer branding machine can realize the exchange of adjacent blocks.

For the marked layer, we can do the reverse recursion, and the rest is the sum of the interval of the line segment tree. You can see the code implementation for details.

Time complexity $O(2^{18}\times 18)$

$\mathrm{C}\mathrm{o}\mathrm{d}\mathrm{e}$

const int mo=998244353;
const int N=1<<18;

int n,m,S,a[N+5],t[N*4+5],flg[21];

inline void Fix(int x,int l,int r,int p,int val,int d)
{
    
    
	if(p>r||p<l) return;
	if(l==r)
	{
    
    
		t[x]=val;
		return;
	}
	int mid=l+r>>1;
	if(flg[d]) Fix(x<<1|1,l,mid,p,val,d-1),Fix(x<<1,mid+1,r,p,val,d-1);
	else Fix(x<<1,l,mid,p,val,d-1),Fix(x<<1|1,mid+1,r,p,val,d-1);
	t[x]=t[x<<1]+t[x<<1|1];
}

inline int Ask(int x,int l,int r,int ll,int rr,int d)
{
    
    
	if(ll>r||rr<l) return 0;
	if(ll<=l&&r<=rr) return t[x];
	int mid=l+r>>1;
	int ans=0;
	if(flg[d]) ans=Ask(x<<1,mid+1,r,ll,rr,d-1)+Ask(x<<1|1,l,mid,ll,rr,d-1);//如果打上标记，那么对于左儿子要在[mid+1,r]里递归下去
	else ans=Ask(x<<1,l,mid,ll,rr,d-1)+Ask(x<<1|1,mid+1,r,ll,rr,d-1); 
	return ans;
}

signed main()
{
    
    
	io.read(n),io.read(m);
	S=(1ll<<n);
	For(i,1,S) 
	{
    
    
		io.read(a[i]);
		Fix(1,1,S,i,a[i],n);
	}
	for (;m--;)
	{
    
    
		int type,x,y;
		io.read(type);
		io.read(x);
		if((type^2)&&(type^3)) io.read(y);
		if(type==1) Fix(1,1,S,x,y,n);
		if(type==3) flg[x+1]^=1;
		if(type==2) For(i,0,x) flg[i]^=1;
		if(type==4) io.write(Ask(1,1,S,x,y,n)),puts("");
	}
	return 0;
}