Solve this interesting problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 422    Accepted Submission(s): 98

Problem DescriptionHave 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. InputThe 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 OutputFor each test, output one line contains one integer. If there is no such n, just output -1. Sample Input6 7
10 13
10 11 Sample Output7
-1
12 Source2015 Multi-University Training Contest 3

它的上一个节点有(2*l-2-r,r),(2*l-1-r,r),(l,2*r-l+1),(l,2*r – l)四种情况，直接搜索，加上几个特殊判断即可。

#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;
}