POJ2309 BST

 

Time Limit: 1000MS

Memory Limit: 65536KB

64bit IO Format: %lld & %llu

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 <= 2 ³¹ - 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

POJ Monthly,Minkerui

 

用树状数组的lowbit处理即可。(x&-x)可以取出x二进制表示下的最后一个1，即可知x的管辖半径。

/*by SilverN*/
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;
int read(){
    int x=0,f=1;char ch=getchar();
    while(ch<'0' || ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0' && ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int lowbit(int x){return x&-x;}
int T;
int n;
int main(){
    T=read();
    while(T--){
        n=read();
        printf("%d %d
",n-lowbit(n)+1,n+lowbit(n)-1);
    }
    return 0;
}