BST Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 8565   Accepted: 5202 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 c…

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 c…

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 t…

题意:给出一棵二分搜索树,再给一个节点编号n,求以这个节点为根节点的子树叶子节点的最大值与最小值. 首先求n所在的层数,他的层数就是他的因子中2的个数(规律). n的左右各有num=2^i-1个数.最小值是n-num,最大值是n+num #include <iostream> #include <algorithm> #include <stdio.h> #include <string.h> #include <queue> using nam…

假设领悟了树阵lowbit,这个问题很简单,底部是奇数,使用lowbit(x)寻找x父亲,然后x父亲-1是的最大数量 至于lowbit问题是如何计算,寻找x父亲,事实上x+2^x二进制结束0的数量. #include<iostream> #include<stdio.h> using namespace std; typedef long long ll; ll lowbit(int x){ return x&(-x); } int main(){ ll n,a; cin&…

题意是给你一个满二叉树,给一个数字,求以这个数为根的树中最大值和最小值. 理解树状数组中的lowbit的用法. 说这个之前我先说个叫lowbit的东西,lowbit(k)就是把k的二进制的高位1全部清空,只留下最低位的1,比如10的二进制是1010,则lowbit(k)=lowbit(1010)=0010(2进制),介于这个lowbit在下面会经常用到,这里给一个非常方便的实现方式,比较普遍的方法lowbit(k)=k&-k,这是位运算,我们知道一个数加一个负号是把这个数的二进制取反+1,如-1…

思路:除以2^k,找到商为奇数的位置,k为层数,有2^(k+1)-1个节点 这里直接用位运算,x & -x 就求出 2^k 了. #include<iostream> using namespace std; long lowbit(long x) { return x & -x; } int main() { long n,x; cin>>n; while(n--) { cin>>x; cout<<x-lowbit(x)+1<<…

BST Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 9140   Accepted: 5580 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 c…

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 leve…

http://poj.org/problem?id=2309//找规律 可以看到每个根节点都可以将其在同一层的最左边的根节点整除,并且最大值为该节点加上最左边的节点值-1,最小值为////为该节点减去最左边的节点值-1 #include <iostream> #include<cmath> using namespace std; int main() { ]; s[] =; ;i<=;i++) s[i] = s[i-]<<; int t,n; cin>&g…