前置知识

https://www.luogu.com/article/mtmz3rvx

本题解法

• 操作一，最多输 20 个数，可以直接遍历 rope。
• 操作二，可以用 substr 进行拼接。
• 操作三，参考「前置知识」提到的文章，用两个 rope 维护正反即可。移动到序列末尾直接截取就行了。但是要注意的是，做任何带有修改的操作时一定要同步更新正反 rope。
• 操作四，还是可以用 substr 拼接。

std：

#include<bits/stdc++.h>
#include<ext/rope>
#define rep(i,a,b) for(int i(a);i<=(b);++i)
#define req(i,a,b) for(int i(a);i>=(b);--i)
using namespace std;
char buf[1<<23],*p1=buf,*p2=buf,ubuf[1<<23],*u=ubuf;
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)
template<typename TP> inline TP read(TP &num)
{
	TP x=0;
	int f=0;
	char ch=getchar();
	while(ch<48||ch>57) f|=ch=='-',ch=getchar();
	while(48<=ch&&ch<=57) x=(x<<1)+(x<<3)+(ch^48),ch=getchar();
	return num=f?-x:x;
}
template<typename ...Args> inline void read(Args &...args)
{
	(read(args),...);
}
template<typename TP> inline void write(TP x)
{
	(x<0)?(putchar('-'),x=-x):0;
	(x>9)?(write(x/10),0):0;
	putchar((x%10)^48);
}
template<typename TP> inline void writeln(TP x)
{
	write<TP>(x);
	puts("");
}
using namespace __gnu_cxx;
namespace Solve
{
	int n,q,opt,x,y,a[200001];
	rope<int> tree,rtree;
	inline void tree_reverse(int l,int r)
	{
		rope<int> tmp=tree.substr(tree.begin()+l,tree.begin()+r);
		tree=tree.substr(tree.begin(),tree.begin()+l)+tree.substr(tree.begin()+r,tree.end())+rtree.substr(rtree.begin()+n-r,rtree.begin()+n-l);
		rtree=tmp+rtree.substr(rtree.begin(),rtree.begin()+n-r)+rtree.substr(rtree.begin()+n-l,rtree.end());
	}
	inline void main()
	{
		read(n,q);
		rep(i,1,n) read(a[i]),tree.append(a[i]);
		req(i,n,1) rtree.append(a[i]);
		rep(i,1,q)
		{
			read(opt,x,y);
			if(opt==1)
			{
				rep(i,x-1,y-1) write(tree[i]),putchar(" \n"[i==y-1]);
			}
			else if(opt==4)
			{
				rope<int> treel=tree.substr(tree.begin(),tree.begin()+x-1),
						  treer=tree.substr(tree.begin()+x,tree.end()),
						  rtreel=rtree.substr(rtree.begin(),rtree.begin()+n-x),
						  rtreer=rtree.substr(rtree.begin()+n-x+1,rtree.end());
				tree=treel+y+treer,rtree=rtreel+y+rtreer;
			}
			else if(opt==3) tree_reverse(x-1,y);
			else if(opt==2)
			{
				int ps=tree[x-1],pt=tree[y-1];
				rope<int> treel=tree.substr(tree.begin(),tree.begin()+x-1),
						  treem=tree.substr(tree.begin()+x,tree.begin()+y-1),
						  treer=tree.substr(tree.begin()+y,tree.end()),
						  rtreel=rtree.substr(rtree.begin(),rtree.begin()+n-y),
						  rtreem=rtree.substr(rtree.begin()+n-y+1,rtree.begin()+n-x),
						  rtreer=rtree.substr(rtree.begin()+n-x+1,rtree.end());
				tree=treel+treem+pt+ps+treer,rtree=rtreel+ps+pt+rtreem+rtreer;
			}
		}
		printf("Final sequence: ");
		for(auto i:tree) write(i),putchar(32);
	}
}
signed main()
{
	Solve::main();
	return 0;
}

若想测试较强的完整数据，请在洛谷查看本题：https://www.luogu.com.cn/problem/T442649。