Codeforces 461C. Appleman and a Sheet of Paper

每次只把短的部分往长的部分折叠，直接用树状数组爆搞就可以了。

每次长度都缩小一些暴力的复杂度不是太高，启发式暴力？？？？

C. Appleman and a Sheet of Paper

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Appleman has a very big sheet of paper. This sheet has a form of rectangle with dimensions 1 × n. Your task is help Appleman with folding of such a sheet. Actually, you need to perform q queries. Each query will have one of the following types:

1. Fold the sheet of paper at position pi. After this query the leftmost part of the paper with dimensions 1 × pi must be above the rightmost part of the paper with dimensions 1 × ([current width of sheet] - pi).
2. Count what is the total width of the paper pieces, if we will make two described later cuts and consider only the pieces between the cuts. We will make one cut at distance li from the left border of the current sheet of paper and the other at distance ri from the left border of the current sheet of paper.

Please look at the explanation of the first test example for better understanding of the problem.

Input

The first line contains two integers: n and q (1  ≤ n ≤ 105; 1 ≤ q ≤ 105) — the width of the paper and the number of queries.

Each of the following q lines contains one of the described queries in the following format:

• "1 pi" (1 ≤ pi < [current width of sheet]) — the first type query.
• "2 li ri" (0 ≤ li < ri ≤ [current width of sheet]) — the second type query.

Output

For each query of the second type, output the answer.

Sample test(s)

input

7 4
1 3
1 2
2 0 1
2 1 2

output

4
3

input

10 9
2 2 9
1 1
2 0 1
1 8
2 0 8
1 2
2 1 3
1 4
2 2 4

output

7
2
10
4
5

Note

The pictures below show the shapes of the paper during the queries of the first example:

After the first fold operation the sheet has width equal to 4, after the second one the width of the sheet equals to 2.

#include 
#include 
#include 
#include 

using namespace std;

const int maxn=101000;

int a[maxn],n,q;

inline int lowbit(int x)
{
	return x&(-x);
}

void add(int p,int v)
{
	for(int i=p;i<=n;i+=lowbit(i))
		a[i]+=v;
}

int sum(int p)
{
	int sum=0;
	for(int i=p;i;i-=lowbit(i))
		sum+=a[i];
	return sum;
}

int main()
{
	scanf("%d%d",&n,&q);
	int flag=0;
	for(int i=1;i<=n;i++) add(i,1);
	int beg=1,end=n;
	while(q--)	
	{
		int k,p,l,r;
		scanf("%d",&k);
		if(k==1)
		{
			scanf("%d",&p);
			if(p<=(end-beg+1)/2)
			{
				if(flag==0)
				{
					for(int i=p+beg-1,j=p+beg;i>=beg;i--,j++)
					{
						add(j,sum(i)-sum(i-1));
					}
					beg=beg+p;
				}
				else
				{
					for(int i=end-p,j=end-p+1;j<=end;i--,j++)
					{
						add(i,sum(j)-sum(j-1));
					}
					end=end-p;
				}
			}
			else
			{
				if(flag==0)
				{
					for(int i=beg+p-1,j=p+beg;j<=end;j++,i--)
					{
						add(i,sum(j)-sum(j-1));
					}
					end=beg+p-1;
				}
				else
				{
					for(int i=end-p,j=end-p+1;i>=beg;i--,j++)
					{
						add(j,sum(i)-sum(i-1));
					}
					beg=end-p+1;
				}
				flag=flag^1;
			}
		}
		else if(k==2)
		{
			scanf("%d%d",&l,&r);
			l++;
			if(flag==0)
			{
				l=beg+l-1; r=beg+r-1;
				printf("%d\n",sum(r)-sum(l-1));
			}
			else
			{
				l=end-l+1; r=end-r+1;
				printf("%d\n",sum(l)-sum(r-1));
			}
		}
	}
	return 0;
}