poj 2309 BST(数学题)

本文介绍了一个利用Lowbit函数解决二叉搜索树中查询子树最大最小值的问题。通过输入节点数,程序可以快速计算出指定子树内的最小及最大数值。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

BST
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 8440 Accepted: 5093

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


==================================================================
这题就是Lowbit函数的使用,理解了Lowbit,这题就很容易了
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <math.h>

#define MAxx 1000

int Lowbit(int x)
{
    return x&(-x);
}
int main()
{
    int i,j,n,m;
    scanf("%d",&n);
    for(i=0;i<n;i++)
    {
        scanf("%d",&m);
        int cas=Lowbit(m);
        printf("%d %d\n",m-cas+1,m+cas-1);
    }
    return 0;
}

  

转载于:https://www.cnblogs.com/ccccnzb/p/3837742.html

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值