2017/10/09

# UVA679题解

## question

``````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, either false 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
the false 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.

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 I 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:
2 ≤ D ≤ 20, and 1 ≤ I ≤ 524288.

Input
Contains l + 2 lines.
Line 1 l the number of test cases
Line 2 D1 I1 test case #1, two decimal numbers that are separated by one blank
...
Line k + 1 Dk Ik test case #k
Line l + 1 Dl Il 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

``````

这道题是典型的水题。如果通过暴力模拟来做的话可能会炸。毕竟最多要模拟524288个小球过2^20-1的节点，开数组可能会炸。

所以我们可以直接考虑只模拟最后一个小球下落的情况。 因为题意中说，每个小球经过节点时会改变节点的状态，所以我们可以认为，对于每个节点，第奇数次通过会向左走，第偶数次通过会向右走。同时，也不难看出，小球经过一个节点的序数等于其过父节点时的序数的1/2。

``````#include <iostream>
#include <cmath>
using namespace std;

int main()
{
int T;
while(cin>>T&&T!=-1)
{
for(int i=0;i<T;i++)
{
int d,I;
cin>>d>>I;
int po=1;
for(int i=0;i<d-1;i++)
{
//test
//	cout<<po<<" "<<I<<endl;
//
if(I%2==0)
{
po=po*2+1;
I=I>1?floor((float)I/2.0f+0.5):1;
}
else
{
po=po*2;
I=I>1?floor((float)I/2.0f+0.5):1;
}
}
cout<<po<<endl;
}
}
return 0;
}

``````

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