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Solve this interesting problemTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1571 Accepted Submission(s): 454
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
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#include<cstdio>
#include<cmath>
#include<stdlib.h>
#include<map>
#include<set>
#include<time.h>
#include<vector>
#include<queue>
#include<string>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
#define eps 1e-8
#define INF 0x3f3f3f3f
#define LL long long
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
LL L, R;
LL flag;
void dfs(LL l, LL r)
{
if((flag && r >= flag)) return ;
if(l == 0)
{
flag == 0 ? flag = r : flag = min(flag, r);
return ;
}
long long t = r - l + 1;
if(t <= l)
{
dfs(l - t - 1, r);
dfs(l - t, r);
dfs(l, r + t - 1);
dfs(l, r + t);
}
}
int main()
{
while(~scanf("%I64d%I64d", &L, &R))
{
flag = 0;
if(R == 0)
printf("0
");
else
{
dfs(L, R);
if(flag)
printf("%I64d
", flag);
else
printf("-1
");
}
}
return 0;
}