Implementing a binary tree with arrays

little monkey falling

Time Limit: 3000   ms | Memory Limit: 65535   KB

Difficulty: 3

describe

There is a binary tree with a maximum depth of D, and all leaves have the same depth. All nodes are numbered from left to right and from top to bottom as 1, 2, 3, ···, 2 to the D power minus 1. Put a little monkey at node 1 and it will run down. There is a switch on each inner node, which is initially closed. When a little monkey runs to a switch, its state will change. When reaching an inner node, if the switch is closed, the little monkey will turn left. Go, otherwise go right until you reach the leaf node.

Some little monkeys start running down from node 1, where will the last little monkey run?

enter

Enter the depth D of the leaves of the binary tree, and the number of little monkeys I, assuming that I does not exceed the number of leaves in the entire tree, D<=20. Finally, it ends with 0 0

output

Output the leaf number where the ith little monkey is located.

sample input

4 2 3 4 0 0

Sample output

12 7

#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;         //                 1
int main() // 2 3   
{                            //      4       5     6        7
	int d,s;                 //  8      9    10    11   12    13  14
	while(cin>>d>>s,s&&d)
	{	 
		 vector< int > vec(100000,0);
		int k=1;
		for(int i=0;i<s;i++)
		{
			k=1;
			for(int i=1;i<d;i++)
			{
					if (vec [k] == 0)
					{
						vec [k] = 1;
						k*=2;	                        
					}
					else
					{
						vec [k] = 0;
						k=k*2+1;
					}
			}	
		}
		cout<<k<<endl;
	}
}