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Solve this interesting problem

Time 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

6 710 1310 11

 

Sample Output

7-112

 

Source

2015 Multi-University Training Contest 3

 

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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\n");        else        {            dfs(L, R);            if(flag)            printf("%I64d\n", flag);            else            printf("-1\n");        }    }    return 0;}