BST POJ - 2309 思维题

Posted 2020-12-27 qldabiaoge

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("******
")
#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-k+1,n+k-1);
    }
    return 0;
}