hdu_1100_Trees_Made_to_Order

Trees Made to Order

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 465    Accepted Submission(s): 271

Problem Description

We can number binary trees using the following scheme: 
The empty tree is numbered 0.
The single-node tree is numbered 1.
All binary trees having m nodes have numbers less than all those having m+1 nodes.
Any binary tree having m nodes with left and right subtrees L and R is numbered n such that all trees having m nodes numbered > n have either
Left subtrees numbered higher than L, or
A left subtree = L and a right subtree numbered higher than R.

The first 10 binary trees and tree number 20 in this sequence are shown below:



Your job for this problem is to output a binary tree when given its order number.

 

Input

Input consists of multiple problem instances. Each instance consists of a single integer n, where 1 <= n <= 500,000,000. A value of n = 0 terminates input. (Note that this means you will never have to output the empty tree.)

 

Output

For each problem instance, you should output one line containing the tree corresponding to the order number for that instance. To print out the tree, use the following scheme:

A tree with no children should be output as X.
A tree with left and right subtrees L and R should be output as (L')X(R'), where L' and R' are the representations of L and R.
If L is empty, just output X(R').
If R is empty, just output (L')X.

 

Sample Input

1 20 31117532 0

 

Sample Output

X ((X)X(X))X (X(X(((X(X))X(X))X(X))))X(((X((X)X((X)X)))X)X)

 

Source

East Central North America 2001

 

Recommend

JGShining

個人理解：

要解決問題，主要是要得出左子樹和右子樹的序數。所以要首先得出整棵樹在有相同節點數的樹中排第幾。另外就是要注意序數增長時，左右兩個子樹的變化規律。

#include <cstdio>
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
#define PB push_back

typedef long long ll;

const int maxn=500000000;
vector<ll> sum,tree;

void init() {
    sum.PB(0);
    tree.PB(1);
    while(*sum.rbegin()<maxn) {
        tree.PB(0);
        for(int i=0;i<tree.size()-1;i++)
            *tree.rbegin()+=tree[i]*tree[tree.size()-i-2];
        sum.PB(*sum.rbegin()+*tree.rbegin());
    }
}

void dfs(int p) {
    if(!p)
        return ;
    int m=lower_bound(sum.begin(),sum.end(),p)-sum.begin();
    p-=sum[m-1];//該棵樹在相同節點數的樹中的序數
    int k=0,lp=1,rp;
    while(p>tree[m-1-k]*tree[k]&&m-1-k-1>=0) {
        p-=tree[m-1-k]*tree[k];
        k++;
    }//求左子樹節點數
    if(k) {
        while(p>tree[m-1-k]) {
            p-=tree[m-1-k];
            lp++;
        }
        lp+=sum[k-1];//左子樹序數
        printf("(");
        dfs(lp);
        printf(")");
    }
    printf("X");
    if(m-1-k) {
        rp=p+sum[m-1-k-1];//右子樹序數
        printf("(");
        dfs(rp);
        printf(")");
    }
}

int main() {
    init();
    int n;
    while(cin >> n,n) {
        dfs(n);
        cout << endl;
    }
    return 0;
}