Hdu 5323 2015多校对抗赛三

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=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

 

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);
}
}