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, 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 isfalse, 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
題意 i個小球在一棵二叉樹上下落 第奇數個到達某個節點就往左 否則往右 求最後一個小球下落到最底層所在結點的序號
對於第i個小球 在根結點處 若i是奇數 則它是往左的第(i+1)/2個小球 若i是偶數 則它是往右的第n/2個小球
而對於一個結點k 它的左子結點序號爲2*k 右子節點序號爲2*k+1 每次更新i到到最底層就得出結果了
#include<cstdio>
int main()
{
int cas, ans, d, i;
while (scanf ("%d", &cas), cas + 1)
{
while (cas--)
{
ans = 1;
scanf ("%d%d", &d, &i);
while (--d)
{
if (i % 2)
ans = ans * 2, i = (i + 1) / 2;
else
ans = ans * 2 + 1, i = i / 2;
}
printf ("%d\n", ans);
}
}
return 0;
}