根節點 line ons ctu .cn span alt while algo

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.

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 Ith 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:

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

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

Sample Output

12
7
512
3
255
題目大意：給你一個D層深的樹，I個小球，每個小球從根節點開始走，一開始向左走，以後的球走到這個點都走與前一個到這個點的球的相反的方向，求最後每個球的位置編號.
分析：一開始看著是一道很水的模擬題，直接模擬一下每次怎麽走，判判方向就好了，可是由於數據量太大了，會超時。
      其實很容易就能發現規律了，每個球到當前點的方向取決於它是第幾個到這個點的，如果是奇數，就往左邊走，否則就往右邊走，那麽我們直接處理最後一個球就好了，那麽如果走左子樹，那麽它就是第(I+1)/2個到達的，否則就是I/2個到達的。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cmath>

using namespace std;

int T,D,I,res = 1;

int main()
{
    scanf("%d", &T);
    while (T--)
    {
        res = 1;
        scanf("%d%d", &D, &I);
        for (int i = 1; i <= D - 1; i++)
        {
            if (I % 2 == 1)
            {
                res *= 2;
                I = (I + 1) / 2;
            }
            else
            {
                res = res * 2 + 1;
                I /= 2;
            }
        }
        printf("%d\n", res);
    }

    return 0;
}

Uva679 Dropping Balls