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Solve this interesting problem Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1571????Accepted Submission(s): 454 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 6 7 10 13 10 11 ? Sample Output 7 -1 12 ? Source 2015 Multi-University Training Contest 3 ? Recommend wange2014???|???We have carefully selected several similar problems for you:??5326?5325?5324?5322?5321? ?

#include<cstdio>
#include<cmath>
#include<stdlib.h>
#include<map>
#include<set>
#include<time.h>
#include<vector>
#include<queue>
#include<string>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
#define eps 1e-8
#define INF 0x3f3f3f3f
#define LL long long
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))


LL L, R;
LL flag;

void dfs(LL l, LL r)
{
    if((flag && r >= flag))  return ;
    if(l == 0)
        {
            flag == 0 ? flag = r : flag = min(flag, r);
            return ;
        }
    long long  t = r - l + 1;
    if(t <= l)
    {
        dfs(l - t - 1, r);
        dfs(l - t, r);
        dfs(l, r + t - 1);
        dfs(l, r + t);
    }
}

int main()
{
    while(~scanf("%I64d%I64d", &L, &R))
    {
        flag = 0;
        if(R == 0)
        printf("0\n");
        else
        {
            dfs(L, R);
            if(flag)
            printf("%I64d\n", flag);
            else
            printf("-1\n");
        }
    }
    return 0;
}