C

You have n boxes in a line on the table numbered 1 . . . n from left to right. Your task is to simulate 4
kinds of commands:
• 1 X Y : move box X to the left to Y (ignore this if X is already the left of Y )
• 2 X Y : move box X to the right to Y (ignore this if X is already the right of Y )
• 3 X Y : swap box X and Y
• 4: reverse the whole line.
Commands are guaranteed to be valid, i.e. X will be not equal to Y .
For example, if n = 6, after executing 1 1 4, the line becomes 2 3 1 4 5 6. Then after executing
2 3 5, the line becomes 2 1 4 5 3 6. Then after executing 3 1 6, the line becomes 2 6 4 5 3 1.
Then after executing 4, then line becomes 1 3 5 4 6 2

Input

There will be at most 10 test cases. Each test case begins with a line containing 2 integers n, m
(1 ≤ n, m ≤ 100, 000). Each of the following m lines contain a command.

Output

For each test case, print the sum of numbers at odd-indexed positions. Positions are numbered 1 to n
from left to right.

Sample Input

6 4
1 1 4
2 3 5
3 1 6
4
6 3
1 1 4
2 3 5
3 1 6
100000 1
4

Sample Output

Case 1: 12
Case 2: 9
Case 3: 2500050000

四种操作，1把x移到y左边，2把x移到y右边，3交换xy，4逆转链表；

用数组模拟链表，用LEFT和RIGHT两个数组储存每一个数字左右两边的数字是什么

如果有经过一次4，那个做个标记，后面有操作1，2就要交换，1要变成2，2要变成1，3没影响，对他们的左右连接改动就好了，和链表的操作一样

吐槽一句，刘汝佳有毒啊，他书上的代码会wa，后面还是自己写出来过的

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include <iomanip>
#include<cmath>
#include<float.h> 
#include<string.h>
#include<algorithm>
#define sf scanf
#define pf printf
#define mm(x,b) memset((x),(b),sizeof(x))
#include<vector>
#include<queue>
//#include<map>
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=a;i>=n;i--)
typedef long long ll;
typedef long double ld;
typedef double db;
const ll mod=1e9+100;
const db e=exp(1);
using namespace std;
const double pi=acos(-1.0);
int n,LEFT[100010],RIGHT[100010];
void link(int x,int y)
{
	LEFT[y]=x;
	RIGHT[x]=y;
}
int main()
{
	int m,kase=0;
	while(sf("%d%d",&n,&m)!=EOF)
	{
		rep(i,1,n+1)
		{
			LEFT[i]=i-1;
			RIGHT[i]=(i+1)%(n+1);
		}
		RIGHT[0]=1;LEFT[0]=n;
		int op,x,y,temp=0;
		while(m--)
		{
			sf("%d",&op);
			if(op==4) temp=!temp;
			else
			{
				sf("%d%d",&x,&y);
				if(op==3&&RIGHT[y]==x)	swap(x,y);
				if(op!=3&&temp) op=3-op;
				int lx=LEFT[x],rx=RIGHT[x],ly=LEFT[y],ry=RIGHT[y];
				if(op==1)
				{
					if(LEFT[y]!=x)
					{
						if(RIGHT[y]==x)
						{
							link(y,rx);link(ly,x);link(x,y);
						}else
						{
							link(lx,rx);link(x,y);link(ly,x);
						}
					}
				}else if(op==2)
				{
					if(RIGHT[y]!=x)
					{
						if(LEFT[y]==x)
						{
							link(lx,y);link(y,x);link(x,ry);
						}else
						{
							link(lx,rx);link(y,x);link(x,ry);
						}
					}
				}else if(op==3)
				{
					if(RIGHT[x]==y)
					{
						link(lx,y);link(y,x);link(x,ry);
					}else if(RIGHT[y]==x)
					{
						link(ly,x);link(x,y);link(y,rx);
					}else
					{
						link(lx,y);link(y,rx);link(ly,x);link(x,ry);
					}
				}
			}
		}
		int b=0;
		ll ans=0;
		rep(i,1,n+1)
		{
			b=RIGHT[b];
			if(i%2==1) ans+=b;
		}
		if(temp&&n%2==0) ans=(ll)n*(n+1)/2-ans;
		pf("Case %d: %lld
",++kase,ans);
	}
	return 0;
}