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 https://blog.csdn.net/fengkuangdewoniudada/article/details/69660755
规律请看这篇博客
#include <cstdio>
#include <cstring>
#include <queue>
#include <cmath>
#include <algorithm>
#include <set>
#include <iostream>
#include <map>
#include <stack>
#include <string>
#include <vector>
#define pi acos(-1.0)
#define eps 1e-6
#define fi first
#define se second
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define bug printf("******\n")
#define mem(a,b) memset(a,b,sizeof(a))
#define fuck(x) cout<<"["<<x<<"]"<<endl
#define f(a) a*a
#define sf(n) scanf("%d", &n)
#define sff(a,b) scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf printf
#define FRE(i,a,b) for(i = a; i <= b; i++)
#define FREE(i,a,b) for(i = a; i >= b; i--)
#define FRL(i,a,b) for(i = a; i < b; i++)
#define FRLL(i,a,b) for(i = a; i > b; i--)
#define FIN freopen("DATA.txt","r",stdin)
#define lowbit(x) x&-x
#pragma comment (linker,"/STACK:102400000,102400000") using namespace std;
typedef long long LL; int main() {
int t, n;
scanf("%d", &t);
while(t--) {
scanf("%d", &n);
int k=lowbit(n);
printf("%d %d\n",n-k+,n+k-);
}
return ;
}

BST POJ - 2309 思维题的更多相关文章

  1. UVA 1394 And Then There Was One / Gym 101415A And Then There Was One / UVAlive 3882 And Then There Was One / POJ 3517 And Then There Was One / Aizu 1275 And Then There Was One (动态规划,思维题)

    UVA 1394 And Then There Was One / Gym 101415A And Then There Was One / UVAlive 3882 And Then There W ...

  2. Buy Tickets POJ - 2828 思维+线段树

    Buy Tickets POJ - 2828 思维+线段树 题意 是说有n个人买票,但是呢这n个人都会去插队,问最后的队列是什么情况.插队的输入是两个数,第一个是前面有多少人,第二个是这个人的编号,最 ...

  3. POJ推荐50题

    此文来自北京邮电大学ACM-ICPC集训队 此50题在本博客均有代码,可以在左侧的搜索框中搜索题号查看代码. 以下是原文: POJ推荐50题1.标记“难”和“稍难”的题目可以看看,思考一下,不做要求, ...

  4. zoj 3778 Talented Chef(思维题)

    题目 题意:一个人可以在一分钟同时进行m道菜的一个步骤,共有n道菜,每道菜各有xi个步骤,求做完的最短时间. 思路:一道很水的思维题, 根本不需要去 考虑模拟过程 以及先做那道菜(比赛的时候就是这么考 ...

  5. 最短路+线段交 POJ 1556 好题

    // 最短路+线段交 POJ 1556 好题 // 题意:从(0,5)到(10,5)的最短距离,中间有n堵墙,每堵上有两扇门可以通过 // 思路:先存图.直接n^2来暴力,不好写.分成三部分,起点 终 ...

  6. cf A. Inna and Pink Pony(思维题)

    题目:http://codeforces.com/contest/374/problem/A 题意:求到达边界的最小步数.. 刚开始以为是 bfs,不过数据10^6太大了,肯定不是... 一个思维题, ...

  7. ZOJ 3829 贪心 思维题

    http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3829 现场做这道题的时候,感觉是思维题.自己智商不够.不敢搞,想着队友智商 ...

  8. 洛谷P4643 [国家集训队]阿狸和桃子的游戏(思维题+贪心)

    思维题,好题 把每条边的边权平分到这条边的两个顶点上,之后就是个sb贪心了 正确性证明: 如果一条边的两个顶点被一个人选了,一整条边的贡献就凑齐了 如果分别被两个人选了,一作差就抵消了,相当于谁都没有 ...

  9. C. Nice Garland Codeforces Round #535 (Div. 3) 思维题

    C. Nice Garland time limit per test 1 second memory limit per test 256 megabytes input standard inpu ...

随机推荐

  1. MD5接口解密操作_接口签名校验

    很多HTTP接口在传参时,需要先对接口的参数进行数据签名加密如以下POST接口 http://localhost:8080/pinter/com/userInfo 参数为{"phoneNum ...

  2. @meida 媒体查询

    示例 @meida 媒体查询 在进行书写的时候需要考虑到加载顺序和样式权重使用meida响应式实现不同宽度布局示例 常用工具 https://mydevice.io 参考链接 https://deve ...

  3. Python3 集合

    1.集合的表示 集合是一个无序不重复的元素序列 创建空集合 set() 2.集合的运算 a={1,2,3} b={2,3,4} print(a-b) #a中包含b中不包含 print(a|b) #a中 ...

  4. Where to go from here

    Did you get through all of that content? Congratulations! You've learnt the fundamentals of algorith ...

  5. java实现几种简单的排序算法

    public class SimpleAri { public static void main(String[] args) { int[] t = {11, 21, 22, 1, 6, 10, 3 ...

  6. 3dContactPointAnnotationTool开发日志(十八)

      今天实现了tab效果,按tab键可以在status面板的各个输入框内来回切换,参考Unity3D - UGUI实现Tab键切换输入框.按钮(按Tab键切换高亮显示的UI)

  7. 第三章 持续集成jenkins工具使用之邮件配置

    1   Email Extension Plugin插件安装 持续集成很重要的一环就是及时将构建结果通知到对应的责任人,如:构建失败了,至少需要下发通知给造成本次构建失败的开发人员,如果包含自动化测试 ...

  8. java 基础 --匿名内部类-008

    不全代码 interface Inter(){void show();} class Outer{补全代码} class OuterDemo{ public static void main(Stri ...

  9. java基础--逻辑运算符-- 002

    1:int a = 10;int b = 20;boolean flag = (a == b) //falseboolean flag = (a = b) //报错,不兼容的类型 2: &, ...

  10. c++源文件到可执行文件的过程

    1.预处理(preprocessor):对#pragma.#include.#define.#ifdef/#endif.#ifndef/#endif,inline内联函数等进行处理 2.编译(comp ...