1008 Solve this interesting problem
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)/2+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
题解
比赛的时候没看到，无非就是四种状态的DFS啊……大失误。每个区间的父节点有四种，只要排出来就好；
吾日三省吾身，
- l<0否？
- l=0否？
- l>0的话r超了没有？
- 左边是不是爆了？
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31#include <iostream>
using namespace std;
const long long maxn = 1e18 + 5;
long long mins;
void (long long l,long long r)
{
if(l < 0) return;
if(l == 0) {mins=min(mins,r);return;}
if(r - l + 1 > l||r >= mins) return;
dfs(2 * l - r - 2, r);
dfs(2 * l - r - 1, r);
dfs(l, 2 * r - l);
dfs(l, 2 * r - l + 1);
}
int main()
{
long long l, r;
while(cin>>l>>r)
{
mins = maxn;
dfs(l, r);
if(mins == maxn) cout<<"-1"<<endl;
else cout<<mins<<endl;
}
return 0;
}
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