HDOJ 5524 Subtrees

Subtrees

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 105    Accepted Submission(s): 59

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

5678

 

Sample Output

3435

 

Source

BestCoder Round #61 (div.2)

题解:一颗完全二叉树，左右子树都会为完全二叉树，其中必然有一个最后一层是满的。对于最后一层是满的完全二叉树，每一层的节点的子树形态都是相同的，只统计logN种，然后递归处理另一颗子树。最后对记录下的所有子树根据节点数判重.

判重的时候我们可以用一个map就轻松实现了. 整个算法的复杂度log(N).

或者明确一点的说, N个节点构成的完全二叉树, 除去根节点, 我们该如何将剩下的N - 1个节点分成两个完全二叉数, 这就是解这道题的关键.

#include <cstdio>#include <cstring>#include <map>using namespace std;typedef long long LL;map<LL, bool> mp;LL fun(LL x) {    if (x <= 0) return 0;    LL t = 1;    for (LL i = 1; (t - 1) * 2 < x; ++i)         t *= 2;if ((t - 1) * 2 == x || (t - 1) + (t - 1) / 2 <= x)return t - 1;elsereturn t / 2 - 1;}void solve(LL n) {    mp[n] = true;    LL x = fun(n - 1);    if (x > 0 && mp[x] == false)solve(x);    x = n - x - 1;    if (x > 0 && mp[x] == false)solve(x);}int main() {    LL n;    while (~scanf("%lld", &n)) {        mp.clear();solve(n);        printf("%ld\n", mp.size());    }    return 0;}