【二叉树、二分】小猴子下落

//直观想法，用二叉树、深搜方法模拟每一个小猴子跑一遍
#include<iostream>
#include<vector>
using namespace std;
struct node{
    int num;
    int light;
    node* left;
    node* right;
};
int dfs(node* root,node* now){
    now->light= - now->light;
    if(now->left==NULL){
        return now->num;
    }
    if(now->light==1) dfs(root,now->left);
    else dfs(root,now->right);
}        
int main(){
    int d,i;
    while(cin>>d>>i){
        int size=(1<<d)-1;
        node tree[size];
        for(int i=0;i<size;i++){
            tree[i].num=i+1;
            tree[i].light=-1;
            if(i<((1<<d-1)-1)){
                tree[i].left=&tree[2*i+1];
                tree[i].right=&tree[2*i+2];
            }
            else tree[i].left=tree[i].right=NULL;
        }
        int tem;
        for(int j=0;j<i;j++){
           tem = dfs(tree,tree);
        }
        if(!d | !i) {
            cout<<endl;
            continue;
        }
        cout<<tem<<endl;
    }
    return 0;
}

// 观察到每一个节点是按到达该层的猴子奇偶来判断向左或右的，类似二分法。
int main(){
    int d,i;
    while(cin>>d>>i){
        if(d==0&&i==0) break;
        int answer=1;
        for(int j=1;j<d;j++){
            if(i%2){
                answer*=2;
                i=i/2+1;
            }
            else{
                answer=answer*2+1;
                i=i/2;
            }
        }
        cout<<answer<<endl;
    }
    return 0;
}