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.
![](http://acm.hdu.edu.cn/data/images/C581-1008-1.jpg)
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.
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.
![](http://acm.hdu.edu.cn/data/images/C581-1008-1.jpg)
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
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; }