POJ2309 -- BST

找找规律，实际上是二分查找的过程，只要找到了mid与输入的n相同的话，直接输出left和right就可以了。

代码如下：

#include <iostream>
using namespace std;

long long getroot(int n)
{
    long long root = 2;
    while(root*2 <= n)
        root *= 2;
    return root;
}

int main()
{
    int T;
    cin>>T;
    while(T --)
    {
        long long n,root;
        cin>>n;
        root = getroot(n);
        long long left = 1 , right = 2*root - 1;
        long long mid = (left + right)/2;
        while(mid != n)
        {
            if(mid > n) right = mid - 1;
            if(mid < n) left = mid + 1;
            mid = (right + left)/2;
        }
        
        cout<<left<<" "<<right<<endl;
        
    }
    return 0;
}