Binary tree (day 82)

1： subject

     1
    / \
   2   3
  / \ / \
 4  5 6  7
/\ /\ /\ /\
...     ... 
 As shown in the figure above , From positive integers  1,2,3,…  Formed an infinite （ full ） Binary tree .

 From any node to the root node （ The number is  1  The node of ） There is a unique path , For instance from  5  The path to the root node is  (5,2,1), from  4  The path to the root node is  (4,2,1), From the root  1  The path to the root node contains only one node  1, So the path is  (1).

 For two nodes  x  and  y, Suppose their paths to the root node are  (x1,x2,…,1)  and  (y1,y2,…,1), Then there must be two positive integers  i  and  j, Make from  xi  and  yj  Start , Yes  xi=yj,xi+1=yj+1,xi+2=yj+2,…
 The problem now is , Given  x  and  y, Ask their public parent , namely  xi（ That is to say  yj）.

 Input format 
1≤x,y≤231−1
 sample input ：
10 4
 sample output ：
2

2： Code implementation

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

int main()
{
    
    int x, y;
    cin >> x >> y;

    while (x != y)
    {
    
        if (x > y) x /= 2;
        else y /= 2;
    }

    cout << x << endl;

    return 0;
}