Problem Description

Have you learned something about segment tree? If not, don’t worry, I will explain it for you.
Segment Tree is a kind of binary tree, it can be defined as this:
- For each node u in Segment Tree, u has two values: Lu and Ru.
- If Lu=Ru,
u is a leaf node. 
- If Lu≠Ru,
u has two children x and y,with Lx=Lu,Rx=⌊Lu+Ru2⌋,Ly=⌊Lu+Ru2⌋+1,Ry=Ru.
Here is an example of segment tree to do range query of sum.

Given two integers L and R, Your task is to find the minimum non-negative n satisfy that: A Segment Tree with root node's value Lroot=0 and Rroot=n contains
a node u with Lu=L and Ru=R.

 

Input

The input consists of several test cases. 
Each test case contains two integers L and R, as described above.
0≤L≤R≤109
LR−L+1≤2015

 

Output

For each test, output one line contains one integer. If there is no such n, just output -1.

 

Sample Input

 

Sample Output

 

Source

2015 Multi-University Training Contest 3

#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

typedef long long ll;

using namespace std;

ll maxn = 0x3f3f3f3f;

ll ma;

void dfs(ll l,ll r)

{

    if(l == 0)

    {

        ma = min(r,ma);

        return;

    }

    if(l > r)

        return ;

    if(r >= 1e18)

        return ;

    if(r - l + 1 > l)

        return ;

    if(l == r)

    {

        ma = r;

        return ;

    }

    dfs(2*l-2-r,r);

    dfs(2*l-1-r,r);

    dfs(l,2*r-l+1);

    dfs(l,2*r - l);

    return ;

}

int main()

{

    ll a,b;

    //freopen("8.txt","r",stdin);

    while(~scanf("%I64d%I64d",&a,&b))

    {

        ma = maxn;

        dfs(a,b);

        if(ma == maxn)

            printf("-1\n");

        else

            printf("%I64d\n",ma);

    }

    return 0;

}