AcCoder Contest-115 D

D - 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

2 7

Sample Output 1

4

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

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Sample Input 2

1 1

Sample Output 2

0

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

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Sample Input 3

50 4321098765432109

Sample Output 3

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.

　　

　　题意： 一个L级别的汉堡由 一个b ，一个L-1级别的汉堡， 一个p， 一个L-1级别的汉堡， 一个b 共五层组成。

　　0 级别的汉堡单独由 p 组成。

　　例如 1级别的汉堡由 b, 1-1, p, 1-1, b 也就是bpppb

　　2级别的汉堡由 b, 2-1, p, 2-1, b 组成 也就是b(bpppb)p(bpppb)b

　　问： 现在有一个N级别的汉堡，从最底层（最顶层）开始吃 吃掉x层后，一共吃掉了多少层p

　　

#include<cstdio>
#include<algorithm>
using namespace std;

long long arr[55];
long long stu[55];
long long ans = 0;
int n;
long long x;
void f(int d){
    if(x==arr[d]){//如果吃x层会整好吃完级别为 d 的汉堡
        ans += stu[d];
        return ;
    }else if(x){
        x--;// 吃掉最底层的 b
        if(x>arr[d-1]){// x 落在了上面的 L-1 级的汉堡
            x-=arr[d-1];//吃掉下面的 L-1 级的汉堡
            ans += stu[d-1];// L-1级别的汉堡有stu[l-1]层 p
            x--;// 吃掉中间的 p
            ans++;
            if(x)f(d-1);// 进入上面的 L-1 级的汉堡
            //这里可以不进行if判断直接进入 因为进入后还会判断x非零
        } else f(d-1);// 落点在下面的 L-1 级汉堡
    }
}
int main(){
    arr[0] = 1;
    arr[1] = 5;
    arr[2] = 13;
    stu[0] = 1;
    stu[1] = 3;
    stu[2] = 7;
    for(int i=3;i<51;i++){
        //arr[i] = 1+arr[i-1]+1+arr[i-1]+1;
        arr[i] = arr[i-1]*2+3;// 级别为 i 的汉堡的层数 b+p
        stu[i] = stu[i-1]*2+1;// 级别为 i 的汉堡的 p 的数量
    }
    scanf("%d%lld",&n,&x);
    f(n);
    printf("%lld
",ans);
    return 0;
}