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UVa 679 - Dropping Balls

A number of K balls are dropped one by one from the root of a fully binary tree structure FBT. Each time the ball being dropped first visits a non-terminal node. It then keeps moving down, either follows the path of the left subtree, or follows the path of the right subtree, until it stops at one of the leaf nodes of FBT. To determine a ball‘s moving direction a flag is set up in every non-terminal node with two values, eitherfalse or true. Initially, all of the flags are false. When visiting a non-terminal node if the flag‘s current value at this node isfalse, then the ball will first switch this flag‘s value, i.e., from thefalse to the true, and then follow the left subtree of this node to keep moving down. Otherwise, it will also switch this flag‘s value, i.e., from thetrue to the false, but will follow the right subtree of this node to keep moving down. Furthermore, all nodes of FBT are sequentially numbered, starting at 1 with nodes on depth 1, and then those on depth 2, and so on. Nodes on any depth are numbered from left to right.


For example, Fig. 1 represents a fully binary tree of maximum depth 4 with the node numbers 1, 2, 3, ..., 15. Since all of the flags are initially set to befalse, the first ball being dropped will switch flag‘s values at node 1, node 2, and node 4 before it finally stops at position 8. The second ball being dropped will switch flag‘s values at node 1, node 3, and node 6, and stop at position 12. Obviously, the third ball being dropped will switch flag‘s values at node 1, node 2, and node 5 before it stops at position 10.


Fig. 1: An example of FBT with the maximum depth 4 and sequential node numbers.


Now consider a number of test cases where two values will be given for each test. The first value isD, the maximum depth of FBT, and the second one is I, the Ith ball being dropped. You may assume the value ofI will not exceed the total number of leaf nodes for the given FBT.

Please write a program to determine the stop position P for each test case.


For each test cases the range of two parameters D and I is as below:

\begin{displaymath}2 \le D \le 20, \mbox{ and } 1 \le I \le 524288.
\end{displaymath}


Input 

Contains l+2 lines.


Line 1 		 I the number of test cases 
Line 2 		 $D_1 \ I_1$
test case #1, two decimal numbers that are separatedby one blank 
... 		 		 
Line k+1 $D_k \ I_k$
test case #k 
Line l+1 $D_l \ I_l$
test case #l 
Line l+2 -1 		 a constant -1 representing the end of the input file

Output 

Contains l lines.


Line 1 		 the stop position P for the test case #1 
... 		 
Line k the stop position P for the test case #k 
... 		 
Line l the stop position P for the test case #l

Sample Input 

5
4 2
3 4
10 1
2 2
8 128
-1

Sample Output 

12
7
512
3
255


先考虑小球落到根节点,第一个会落到左子树,第二个落到右子树,第三个左子树。。I是偶数就落到左边,奇数就落到右边,且落到左边的个数与右边的个数各为一半。


#include <stdio.h>
#include <string.h>
#include <math.h>
#include <iostream>
#include <queue>
#include <algorithm>
#define mem(f) memset(f,0,sizeof(f))
#define M 100005
#define mod 1000000007
#define lson o<<1, l, m
#define rson o<<1|1, m+1, r
using namespace std;
typedef long long LL;
const int MAX = 0x3f3f3f3f;
const int maxn = 2111111;

int n, m, d;

int main()
{
    while(~scanf("%d", &n) && n != -1) {
        while(n--) {
            scanf("%d%d", &d, &m);
            int k = 1;
            for(int i = 0; i < d-1; i++) {
                if(m%2) {
                    k = 2*k;
                    m = (m+1)/2;
                } else {
                    k = 2*k+1;
                    m = m/2;
                }
            }
            printf("%d\n", k);
        }
    }
    return 0;
}



UVa 679 - Dropping Balls