Uva 679 Dropping Balls

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 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.

#include<iostream>
#include<string.h>
#include<stdio.h>
#include<ctype.h>
#include<algorithm>
#include<stack>
#include<queue>
#include<set>
#include<math.h>
#include<vector>
#include<map>
#include<deque>
#include<list>
using namespace std;
const int maxd = 20;
int s[1<<maxd];//1^20,<<是二进制向左偏移的意思
int main()
{
int D,I;
while(scanf("%d%d",&D,&I)==2)
{
memset(s,0,sizeof(s));
int k,n=(1<<D)-1;
for(int i=0;i<I;i++)
{
k=1;
while(k<=n)
{
s[k]=!s[k];
k=s[k]?k*2:k*2+1;//二叉树
}
}
printf("%d\n",k/2);
}
return 0;
}

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