2015 多校联赛 ——HDU5323(搜索)

Solve this interesting problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 422    Accepted Submission(s): 98

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.

2015 多校联赛 ——HDU5323(搜索)

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

它的上一个节点有(2*l-2-r,r),(2*l-1-r,r),(l,2*r-l+1),(l,2*r - l)四种情况,直接搜索,加上几个特殊判断即可。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
typedef long long ll;
using namespace std;
ll maxn = 0x3f3f3f3f;
ll ma; void dfs(ll l,ll r)
{
if(l == 0)
{
ma = min(r,ma);
return;
}
if(l > r)
return ;
if(r >= 1e18)
return ;
if(r - l + 1 > l)
return ;
if(l == r)
{
ma = r;
return ;
}
dfs(2*l-2-r,r);
dfs(2*l-1-r,r);
dfs(l,2*r-l+1);
dfs(l,2*r - l);
return ;
} int main()
{
ll a,b;
//freopen("8.txt","r",stdin);
while(~scanf("%I64d%I64d",&a,&b))
{
ma = maxn;
dfs(a,b); if(ma == maxn)
printf("-1\n");
else
printf("%I64d\n",ma);
}
return 0;
}

  

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