Hdu 5323 2015多校对抗赛三

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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): 2029 Accepted Submission(s): 617

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

Author

ZSTU

给定一个区间 [l,r] ，求一个最小的线段树存在此区间节点，输出最小线段树大小即可。

暴力搜索，对于一个区间节点，他的父亲节点可能存在四种状态：


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

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

剪枝有：答案剪枝，区间可行性剪枝（线段树右节点长度不能大于左节点长度）

#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
#define LL long long
#define inf 1000000000000ll
LL ans;
void dfs(LL l,LL r)
{if(ans<=r)return;if(l==0) {if(ans>r)ans=r;return;}if(r-l+1>l) return;dfs(2*l-1-r,r);dfs(2*l-2-r,r);dfs(l,2*r-l);dfs(l,2*r-l+1);
}
int main()
{LL l,r;while(~scanf("%lld%lld",&l,&r)){ans = inf;dfs(l,r);if(ans == inf) puts("-1");else printf("%lld\n",ans);}
}

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