2015 Multi-University Training Contest 3 1008(DFS)

Solve this interesting problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1728    Accepted Submission(s): 517

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

 

 

题意：给定区间[L,R]，能否生成包含该区间的线段树，如果能，包含该段的线段树[0,n]的n的最小值是多少

分析：区间[L,R]可能是父区间的左孩子也可能是父区间的右孩子，所以有可能有四种情况，[L,R+(R-L+1)]，[L,R+(R-L+1)-1]，[L-(R-L+1),R],[L-(R-L+1)-1,R]

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<string>
#include<iostream>
#include<cstring>
#include<cmath>
#include<stack>
#include<queue>
#include<vector>
#include<map>
#include<stdlib.h>
#include<algorithm>
#define LL __int64
using namespace std;
const int INF=0x3f3f3f3f;
LL L,R,ans;

void DFS(LL l,LL r)
{
    LL len=r-l+1;  //区间长度
    if(r>=ans) return ;
    if(l==0)
    {
        ans=r;
        return ;
    }             //结束递归
    if(len>l) return ;  

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

    DFS(l,2*r-l);
    DFS(l,2*r-l+1);
}

int main()
{
    //freopen("in.txt","r",stdin);
    while(scanf("%I64d %I64d",&L,&R)!=EOF)
    {
        ans=INF;
        DFS(L,R);
        if(ans==INF) cout<<"-1"<<endl;
        else cout<<ans<<endl;
    }
    return 0;
}