Christmas AtCoder

Christmas

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Time limit : 2sec / Memory limit : 1024MB

Score : 400 points

Problem Statement

In some other world, today is Christmas.

Mr. Takaha decides to make a multi-dimensional burger in his party. A level-L burger (L is an integer greater than or equal to 0) is the following thing:

• A level-0 burger is a patty.
• A level-L burger (L≥1) is a bun, a level-(L−1) burger, a patty, another level-(L−1) burger and another bun, stacked vertically in this order from the bottom.

For example, a level-1 burger and a level-2 burger look like BPPPB and BBPPPBPBPPPBB (rotated 90 degrees), where B and P stands for a bun and a patty.

The burger Mr. Takaha will make is a level-N burger. Lunlun the Dachshund will eat X layers from the bottom of this burger (a layer is a patty or a bun). How many patties will she eat?

Constraints

• 1≤N≤50
• 1≤X≤( the total number of layers in a level-N burger )
• N and X are integers.

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Input

Input is given from Standard Input in the following format:

N X

Output

Print the number of patties in the bottom-most X layers from the bottom of a level-N burger.

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Sample Input 1

Copy

2 7

Sample Output 1

Copy

4

There are 4 patties in the bottom-most 7 layers of a level-2 burger (BBPPPBPBPPPBB).

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Sample Input 2

Copy

1 1

Sample Output 2

Copy

0

The bottom-most layer of a level-1 burger is a bun.

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Sample Input 3

Copy

50 4321098765432109

Sample Output 3

Copy

2160549382716056

A level-50 burger is rather thick, to the extent that the number of its layers does not fit into a 32-bit integer.

第0级  P  第1级  B 第0级 P 第0级 B　　......

所以，暴力吧，别无选择。

切割降低级数，n级从中间切，判断在前面还是在后面，在前面就继续切，在后面就加上前面的B的个数，然后切后面

#include<iostream>
#include<algorithm>
using namespace std;
long long ttt(long long n,long long x,long long *arr,long long *brr)
{
	if(x<=n)		//提前结束并不会使下面循环x<=0 
	{
		return 0;	
	}
	if(n==0)
	{
		return 1;
	} 
	if(x==arr[n])	//提前跳出 
	{
	 	return brr[n];
	}
	while(1)
	{
		if(arr[n]/2==x)	//提前跳出
		{
			return brr[n-1];
		}
		else if(arr[n]/2>x)
		{
			n--;
			x--;
		}
		else
		{
			break;
		}
	}
	long long sum=brr[n-1]+ttt(n-1,x-arr[n-1]-2,arr,brr)+1;//加中间那个肉片 
	return sum;
	
}
int main()
{
	long long n,x;
	cin>>n>>x;
	long long arr[51];
	long long brr[51];
	brr[0]=1;
	arr[0]=1;
	for(int i=1;i<=n;i++)
	{
		arr[i]=arr[i-1]*2+3;
		brr[i]=brr[i-1]*2+1;
	}
	long long sum=ttt(n,x,arr,brr);
	cout<<sum;
}