hdu 5323

Solve this interesting problem

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
  
  

 

题目大意：给出区间[l,r] 要求找出线段树最上面的区间的r

思        路：dfs+剪枝

                      

                    刚看到题目的时候想到可能会爆栈，但是查了下资料，系统栈的内存是一定的，但是根据储存量不同栈的深度也不相同。

代码如下：

/*踏实！！努力！！*/
#include<iostream>
#include<stdio.h>
#include<cmath>
#include<cstring>
#include<map>
#include<queue>
#include<stack>
using namespace std;
int flag;
void dfs(int l,int r)
{
    if(flag&&r>=flag) return ;
    if(l==0){
        if(flag) flag=min(r,flag);
        else flag=r;
        return ;
    }
    if(l<r-l+1) return ;
    int mid=r-l+1;
    dfs(l-mid-1,r);
    dfs(l-mid,r);
    dfs(l,r+mid);
    dfs(l,r+mid-1);
    return ;
}
int main()
{
    int l,r;
    while(scanf("%d%d",&l,&r)!=EOF){
        flag=0;
        if(r==0) {printf("0\n"); continue;}
        dfs(l,r);
        if(flag) printf("%d\n",flag);
        else printf("-1\n");
    }
    return 0;
}