Dropping Balls

Description

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 is false, 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 the true 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 be false, 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 is D, the maximum depth of FBT, and the second one is I, the I^th ball being dropped. You may assume the value of Iwill 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: 

Input

Contains l+2 lines.

Line 1  I the number of test cases Line 2  test case #1, two decimal numbers that are separatedby one blank ...   Line k+1 test case #k Line l+1 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

54 23 410 12 28 128-1

Sample Output

1275123255

很容易想到用递归的方法模拟小求掉落的路径，代码如下：

#include<cstdio>#include<cmath>#include<cstring>using namespace std;const int maxn=(1<<20);int num[maxn],D;int drop(int n,int deep){if(pow(2,deep-1)<=n&&n<=(pow(2,deep)-1))return n;num[n]=!num[n];if(num[n]) drop(2*n,D);else drop(2*n+1,D);}int main(){int Case;scanf("%d",&Case);while(Case--){int I,n;scanf("%d%d",&D,&I);memset(num,0,sizeof(num));for(int i=0;i<I;i++) n=drop(1,D);printf("%d\n",n);}return 0;}

但是此代码超时，因为如果测试数据的组数达到1000，而树的深度达到16，时间复杂度太大。很显然一个一个模拟的思路是行不通的。我们注意到，如果一个小球到达某一结点是第奇数个到达的，就会走向左子树，反之走向右子树。根据这一点我们可以直接模拟最后一个小球的掉落路径，代码如下：

#include<cstdio>#include<cmath>#include<cstring>using namespace std;const int maxn=(1<<20);int num[maxn],D;int drop(int n,int deep){if(pow(2,deep-1)<=n&&n<=(pow(2,deep)-1))return n;num[n]=!num[n];if(num[n]) drop(2*n,D);else drop(2*n+1,D);}int main(){int Case;scanf("%d",&Case);while(Case--){int I,n;scanf("%d%d",&D,&I);memset(num,0,sizeof(num));for(int i=0;i<I;i++) n=drop(1,D);printf("%d\n",n);}return 0;}