CodeForces960DFull Binary Tree Queries(二叉树的操作)

Full Binary Tree Queries

You have a full binary tree having infinite levels.

Each node has an initial value. If a node has value x, then its left child has value 2·x and its right child has value 2·x + 1.

The value of the root is 1.

You need to answer Q queries.

There are 3 types of queries:

1. Cyclically shift the values of all nodes on the same level as node with value X by K units. (The values/nodes of any other level are not affected).
2. Cyclically shift the nodes on the same level as node with value X by K units. (The subtrees of these nodes will move along with them).
3. Print the value of every node encountered on the simple path from the node with value X to the root.

Positive K implies right cyclic shift and negative K implies left cyclic shift.

It is guaranteed that atleast one type 3 query is present.

Input

The first line contains a single integer Q (1 ≤ Q ≤ 105).

Then Q queries follow, one per line:

• Queries of type 1 and 2 have the following format: T X K (1 ≤ T ≤ 2; 1 ≤ X ≤ 1018; 0 ≤ |K| ≤ 1018), where T is type of the query.
• Queries of type 3 have the following format: 3 X (1 ≤ X ≤ 1018).

Output

For each query of type 3, print the values of all nodes encountered in descending order.

Examples

input

Copy

5
3 12
1 2 1
3 12
2 4 -1
3 8

output

Copy

12 6 3 1 
12 6 2 1 
8 4 2 1 

input

Copy

5
3 14
1 5 -3
3 14
1 3 1
3 14

output

Copy

14 7 3 1 
14 6 3 1 
14 6 2 1 

Note

Following are the images of the first 4 levels of the tree in the first test case:

题意：无限长度的二叉树,每次操作1把包含数x的那一层整体移动k个位置，正往右，负往左，操作2把包含数x的那一层带着子树移动k个位置.操作3打印从x到根沿线的所有数.

思路：对于每一层我们可以记录旋转了多少次,因为最多有60+层.对于操作2它的子树就分别移动k^2次,k^4次...，对于查询操作,我们可以先找到x的位置，然后依次除以2就是上层的对应位置,根据旋转次数打印相应数字即可.

代码：

#include<bits/stdc++.h>
#define mem(a,b) memset(a,b,sizeof(a))
#define mod 1000000007
using namespace std;
typedef long long ll;
const int maxn = 1e6+5;
const double esp = 1e-7;
const int ff = 0x3f3f3f3f;
map<int,int>::iterator it;

ll d[110];
ll c[110];

void init()
{
	c[1] = 1;
	for(int i = 2;i< 63;i++)
		c[i] = c[i-1]*2;
}

ll get_deep(ll x)
{
	ll tmp = 1,cnt = 0;
	while(tmp<= x)
		tmp*= 2,cnt++;
	return cnt;
}

ll get_pos(ll x)//找x的当前位置 
{
	ll tmp = 1,cnt = 0;
	while(tmp<= x)
		tmp*= 2,cnt++;
	ll pos = x-tmp/2;
	
	return (pos+d[cnt])%c[cnt];
}

int main()
{
	init();
	int t;
	cin>>t;
	
	while(t--)
	{
		ll o,x,k,deep,pos;
		scanf("%lld",&o);	
		
		if(o == 1)
		{
			scanf("%lld %lld",&x,&k);
			deep = get_deep(x);
			k%= c[deep];
			d[deep] = (d[deep]+k+c[deep])%c[deep];
		}
		else if(o == 2)
		{
			scanf("%lld %lld",&x,&k);
			deep = get_deep(x);
			while(deep<= 62)//子树旋转次数 
			{
				k%= c[deep];
				d[deep] = (d[deep]+k+c[deep])%c[deep];
				k*= 2;
				deep++;
			}
		}
		else
		{
			scanf("%lld",&x);
			deep = get_deep(x);
			pos = get_pos(x);
			while(deep>= 1)
			{
				ll tmp = (pos+c[deep]-d[deep])%c[deep];//逆向求哪个数到了这个位置 
				printf("%lld ",c[deep]+tmp);
				deep--;
				pos = pos/2;
			}
			printf("\n");
		}
	}
	
	return 0;
}