POJ 2309 BST

BST

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8565		Accepted: 5202

Description

Consider an infinite full binary search tree (see the figure below), the numbers in the nodes are 1, 2, 3, .... In a subtree whose root node is X, we can get the minimum number in this subtree by repeating going down the left node until the last level, and
we can also find the maximum number by going down the right node. Now you are given some queries as "What are the minimum and maximum numbers in the subtree whose root node is X?" Please try to find answers for there queries. 

Input

In the input, the first line contains an integer N, which represents the number of queries. In the next N lines, each contains a number representing a subtree with root number X (1 <= X <= 2³¹ - 1).

Output

There are N lines in total, the i-th of which contains the answer for the i-th query.

Sample Input

2
8
10

Sample Output

1 15
9 11

Source

field=source&key=POJ+Monthly" style="text-decoration:none">POJ Monthly,Minkerui

lowbit的作用。计算x相应的二进制数中第一个1的位置k，返回权值2k。

这个函数的作用就是求出t这个数的二进制存储下，最高的非0bit所表示的大小。
即满足2^k<=t的最大的2^k，当中k为非负整数。

1、min和max为奇数，否则min和max非叶子，还能够向下拓展

2、依据满二叉树的性质，x的左右子树的个数都为2的k次方减1个节点

3、依据二叉树搜索的性质，左子树编号的区间为[min,x-1],右子树的编号区间为[x+1,max]

由此得出min=x-(2^k-1),max=x+(2^k-1)

#include <iostream>
using namespace std;
int lowbit(int x)
{
    return x&(-x);
}
int main()
{
    int N;
    cin>>N;

    int c;
    for(c=1;c<=N;c++)
    {
        int n;
        cin>>n;

        int min,max;
        min=n-lowbit(n)+1;
        max=n+lowbit(n)-1;

        cout<<min<<" "<<max<<" "<<endl;
    }
    return 0;
}