POJ - 2309 BST —— lowbit
2017-09-19 23:33
519 查看
BST
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
Sample Output
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <queue>
#include <stack>
#include <vector>
#define max_ 100010
#define inf 0x3f3f3f3f
#define ll long long
using namespace std;
int main(int argc, char const *argv[])
{
int n;
cin>>n;
while(n--)
{
int k;
cin>>k;
int s= (k&(-k)) -1;
cout<<k-s<<" "<<k+s<<endl;
}
return 0;
}
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 10348 | Accepted: 6334 |
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
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <queue>
#include <stack>
#include <vector>
#define max_ 100010
#define inf 0x3f3f3f3f
#define ll long long
using namespace std;
int main(int argc, char const *argv[])
{
int n;
cin>>n;
while(n--)
{
int k;
cin>>k;
int s= (k&(-k)) -1;
cout<<k-s<<" "<<k+s<<endl;
}
return 0;
}
相关文章推荐
- POJ 2309 BST 【lowbit()过】
- poj 2309 BST 使用树状数组的lowbit
- POJ 2309 BST
- POJ 2309 BST
- POJ 2309 BST
- poj 2309 BST (^ ^)
- POJ 2309 BST
- poj 2309 BST
- poj 2309 BST 使用树阵lowbit
- POJ 2309 BST
- POJ 2309 BST
- POJ 2309 BST 树状数组基本操作
- Poj 2309 BST
- poj2309——BST
- POJ 2309 BST
- POJ 2309 BST 二叉树性质 位运算
- poj 2309-BST解题报告
- POJ 2309 BST
- POJ2309 BST
- POJ 2309 BST