Solve this interesting problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1479    Accepted Submission(s): 423

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
 
Source
 
解题:搜索
 
 #include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const LL INF = 0x3f3f3f3f3f3f3f3f;
LL X,Y,ret;
void dfs(LL L,LL R) {
if(R >= ret || L < ) return;
if(L == ) {
ret = min(ret,R);
return;
}
if(R - L + > L) return;
dfs(*L - R - ,R);
dfs(*L - R - ,R);
dfs(L,*R - L);
dfs(L,*R - L + );
} int main() {
while(~scanf("%I64d%I64d",&X,&Y)) {
ret = INF;
dfs(X,Y);
printf("%I64d\n",ret == INF?-:ret);
}
return ;
}

2015 Multi-University Training Contest 3 hdu 5323 Solve this interesting problem的更多相关文章

  1. HDU 5323 SOLVE THIS INTERESTING PROBLEM 爆搜

    pid=5323" target="_blank" style="">链接 Solve this interesting problem Tim ...

  2. DFS+剪枝 HDOJ 5323 Solve this interesting problem

    题目传送门 /* 题意:告诉一个区间[L,R],问根节点的n是多少 DFS+剪枝:父亲节点有四种情况:[l, r + len],[l, r + len - 1],[l - len, r],[l - l ...

  3. 2015 Multi-University Training Contest 5 hdu 5349 MZL's simple problem

    MZL's simple problem Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Oth ...

  4. 2015 Multi-University Training Contest 8 hdu 5390 tree

    tree Time Limit: 8000ms Memory Limit: 262144KB This problem will be judged on HDU. Original ID: 5390 ...

  5. 2015 Multi-University Training Contest 8 hdu 5383 Yu-Gi-Oh!

    Yu-Gi-Oh! Time Limit: 2000ms Memory Limit: 65536KB This problem will be judged on HDU. Original ID:  ...

  6. 2015 Multi-University Training Contest 8 hdu 5385 The path

    The path Time Limit: 2000ms Memory Limit: 65536KB This problem will be judged on HDU. Original ID: 5 ...

  7. 2015 Multi-University Training Contest 3 hdu 5324 Boring Class

    Boring Class Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Tota ...

  8. 2015 Multi-University Training Contest 3 hdu 5317 RGCDQ

    RGCDQ Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submi ...

  9. 2015 Multi-University Training Contest 10 hdu 5406 CRB and Apple

    CRB and Apple Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)To ...

随机推荐

  1. C语言基本语法——函数

    1.什么是函数 2.函数语法 3.函数声明 4.函数调用 5.函数的形参与实参 6.return与exit关键字 7.递归函数 1.什么是函数 • 函数就是一连串语句被组合在一起,并指定了一个名字 • ...

  2. CefSharp获取页面Html代码的两种方式

    CefSharp在NuGet的简介是“The CefSharp Chromium-based browser component”,机翻的意思就是“基于Cefsharp Chromium的浏览器组件” ...

  3. STM32 关于头文件路径没添加错误问题(cannot open source input file "spi.h": No such file or directory)

    error:  #5: cannot open source input file "spi.h": No such file or directory 1.出现这种问题,首先要确 ...

  4. java字符文件的读写

    1.java文件读写,首先我们需要导入相应的包:java.io.*; 2.代码如下: package Demo1; import java.io.*; public class FileWirteTe ...

  5. 2015 Multi-University Training Contest 3 hdu 5318 The Goddess Of The Moon

    The Goddess Of The Moon Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/ ...

  6. 洛谷——P2483 [SDOI2010]魔法猪学院

    https://www.luogu.org/problem/show?pid=2483 题目描述 iPig在假期来到了传说中的魔法猪学院,开始为期两个月的魔法猪训练.经过了一周理论知识和一周基本魔法的 ...

  7. Android应用常规开发技巧——善用组件生命周期

    数据管理 对于仅仅读数据.一种经常使用的管理模式是在onCreate函数中进行数据的载入,直到组件的onDestory函数被调用时在进行释放. // 缓存仅仅读的数据 private Object r ...

  8. JVM的重排序

    重排序一般是编译器或执行时环境为了优化程序性能而採取的对指令进行又一次排序执行的一种手段.重排序分为两类:编译期重排序和执行期重排序,分别相应编译时和执行时环境. 在并发程序中,程序猿会特别关注不同进 ...

  9. caffe环境配置

    参考:http://blog.csdn.net/enjoyyl/article/details/47397505 http://blog.csdn.net/baobei0112/article/det ...

  10. 安卓获取百度地图的Api key

    1.进入开发中心 2.如何获取SHA1 3.如何获取包名