hdu 5524 Subtrees dfs

Subtrees

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)

Problem Description

There is a complete binary tree with N nodes.The subtree of the node i has Ai nodes.How many distinct numbers are there of Ai?

 

Input

There are multiple test cases, no more than 1000 cases.
For each case contains a single integer N on a line.(1≤N≤1018)

 

Output

The output of each case will be a single integer on a line:the number of subtrees that contain different nodes.

 

Sample Input

5
6
7
8

 

Sample Output

3
4
3
5

 

Source

BestCoder Round #61 (div.2)

 

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5524

 

题意：给你一颗n个节点的完全二叉树找出子树中个数不同的个数；

就是如果左子树和右子树相同只需要遍历一颗子树即可；

不同遍历两颗；

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
#define eps 1e-14
#define bug(x)  cout<<"bug"<<x<<endl;
const int N=2e3+,M=4e6+,inf=;
const ll INF=1e18+,mod=1e9+;

///   数组大小

set<ll>ans;
map<ll,ll>si;
ll n;
void dfs(ll x)
{
    if(x>n)return;
    ll l=x,r=x;
    while(l*<=n)l*=;
    while(r*+<=n)r=r*+;
    if(l==x&&r==x)
    {
        si[x]=;
        ans.insert();
        return;
    }
    if(l<=r)
    {
        dfs(x*);
        si[x]=*si[x*]+;
        ans.insert(si[x]);
        //cout<<x<<" "<<si[x]<<endl;
    }
    else
    {
        dfs(x*);
        dfs(x*+);
        si[x]=si[x*]+si[x*+]+;
        //cout<<x<<" "<<si[x]<<" "<<si[x*2]<<" "<<si[x*2+1]<<endl;
        ans.insert(si[x]);
    }
}
int main()
{
    while(~scanf("%lld",&n))
    {
        ans.clear();
        si.clear();
        dfs();
        printf("%d\n",ans.size());
    }
    return ;
}