Solve this interesting problem

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1020 Accepted Submission(s): 279

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 7 10 13 10 11

Sample Output

7 -1 12

Source

2015 Multi-University Training Contest 3

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 1 #include <stdio.h>
 2 #include <string.h>
 3 #include <assert.h>
 4 #include <algorithm>
 5 const long long inf=(long long) 1e30;
 6
 7 long long n;
 8
 9 /*int look(int x)
10 {
11     if(x%2==0)
12         return 2;
13     else
14         return 1;
15 }*/
16
17 void dfs(long long l,long long r)
18 {
19     if(l<0 || r>inf)
20         return;
21     if(l==0)
22     {
23         if(r<n)
24             n=r;
25         return ;
26     }
27     if(l<r-l+1)
28         return;
29     if(r-l+1!=1)
30         dfs(l,r+(r-l+1)-1);
31     dfs(l-(r-l+1),r);
32     dfs(l-(r-l+1)-1,r);
33     dfs(l,r+(r-l+1));
34     return;
35 }
36
37 int main()
38 {
39     int L,R;
40     while(scanf("%d %d",&L,&R)!=EOF)
41     {
42         assert(0 <= L && L <= (int)1e9);
43         assert(0 <= R && R <= (int)1e9);
44         assert(L <= R);
45         assert(L / (R - L + 1) <= 2015);
46         n=inf;
47         dfs((long long) L,(long long) R);
48         if(n==inf)
49             printf("-1\n");
50         else
51             printf("%I64d\n",n);
52     }
53     return 0;
54 }

View Code

转载于:https://www.cnblogs.com/cyd308/p/4685222.html

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