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

6 710 1310 11

 

Sample Output

7-112

 

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 1000000000000llLL 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);    }}