2013
11-11

# BST

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.

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).

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

2
8
10


1 15
9 11


//* @author  mekarlos@gmail.com
import java.util.Scanner;

public class Main {

public static void main(String[] args) {
Scanner scan=new Scanner(System.in);
long n=scan.nextInt();
long k=0,j,v,temp,l,r;
for(int i=0;i< n;i++){
k=scan.nextInt();
if(k%2==1)System.out.println(k+" "+k);
else{
j=(int)Math.floor(Math.log(k)/Math.log(2));
v=1<< j;
while(v!=k){
j--;
if(v>k)v-=(1<< j);
else v+=(1<< j);
}
l=k;
r=k;
l-=(1<< j)-1;
r+=(1<< j)-1;
System.out.println(l+" "+r);
}
}
}
}

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2. 我还有个问题想请教一下，就是感觉对于新手来说，递归理解起来有些困难，不知有没有什么好的方法或者什么好的建议？

3. 第2题，TCP不支持多播，多播和广播仅应用于UDP。所以B选项是不对的。第2题，TCP不支持多播，多播和广播仅应用于UDP。所以B选项是不对的。