链接：http://acm.hdu.edu.cn/showproblem.php?pid=5323

Solve this interesting problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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=ncontains
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

题意：告诉你一个线段树有 一个区间 l到r，如果有这个种线段树  问根节点  0-n，  n最小是多少。如果没有输出-1

做法：搜索剪枝，主要那个剪枝右边界比较难想 比较重要。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
#define inf 100000000000
typedef long long ll;
ll ans;

void dfs(ll l,ll r)
{
	if(l<0) return ;//越界
	if(l==0) {ans=min(ans,r); return;}//符合要求 取最小值
	if(l<(r-l+1)) return ;
	//l 这个区间 右边空余的长度，   r-l+1  是这个区间的长度
	//要对称，所以左边空白区间  必须是 这个区间的 某一倍数，可以相等，但是必须要大于等于 当前这个区间。
	//用于限制r的增长
	ll len=r-l+1;
	dfs(l-len,r);
	dfs(l-len-1,r);

	dfs(l,r+len);
	if(l!=r)
	dfs(l,r+len-1);
}
int main()
{
    ll l,r;
    while(scanf("%I64d%I64d",&l,&r)!=EOF){
		ans=inf;
		dfs(l,r);
		if(ans==inf)
			puts("-1");
		else
		printf("%I64d\n",ans);
	}
	return 0;
}

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