POJ 2309 BST(二叉搜索树)

BST

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8657		Accepted: 5277

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

算法分析：
         此题目的意思是，按照原则建成的二叉树，每次我输入一个根节点的值，从当前根节点一直往左跑，跑到左边最底层，此时的该点，一定是以输入值为根节点的二叉树的 
    最小值，同理往右边跑，一直跑到底，则是最大值。此题目符合树状数组的结构特点。

#include <iostream>
#include <stdio.h>
#include <string>
#include <string.h>
#include <iomanip>
#include <algorithm>

using namespace std;

int main()
{
    int t;
    int n;int m, dd;

    scanf("%d", &t);
    while(t--)
    {
        scanf("%d", &m);
        dd= m&(-m) ;
        dd--;
        //printf("%d
", dd);
        printf("%d %d
", m-dd, m+dd );
    }
    return 0;
}