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): 362    Accepted Submission(s): 208

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

 

先計算 f [i]，保存有 i 個節點的樹可有多少個，在每個節點試探左兒子有多少個節點，右兒子有多少節點，然後進行一些細節的計算。

比如得到左兒子節點數後，就知道左兒子是那種樹（節點數相同是一種樹），這時再計算左兒子應該是這種樹中的第幾棵。

雖然已經儘量追求代碼整潔了，但是代碼還是比較醜。。。

#include <cstdio>
#include <cstdlib>
#include <cassert>
#include <cstring>
#include <algorithm>
#include <numeric>

#define MAXN 100

long long f [MAXN];

void print_tree (long long n) {
#ifdef _DEBUG
    printf ("n = %lld\n", n);
#endif
    if (n == 0) {
        return;
    } else if (n == 1) {
        printf ("X");
        return;
    }

    long long nodes = 0;
    long long count = 0;
    while (count <= n) {
        count += f [nodes++];
    }
    count -= f [--nodes];

    long long l_nodes = 0;
    long long r_nodes = nodes - 1 - l_nodes;
    long long d = 0;
    while (count + d <= n) {
        d += f [l_nodes++] * f [r_nodes--];
    }
    d -= f [--l_nodes] * f [++r_nodes];

    long long t = n - count - d;
    long long l_n = std::accumulate (f, f + l_nodes, 0) + t / f [r_nodes];
    long long r_n = std::accumulate (f, f + r_nodes, 0) + t % f [r_nodes];

    if (l_n) {
        printf ("(");
        print_tree (l_n);
        printf (")");
    }

    printf ("X");

    if (r_n) {
        printf ("(");
        print_tree (r_n);
        printf (")");
    }
}

int main () {
    memset (f, 0, sizeof (f));
    long long sum = 0;
    f [0] = f [1] = 1;
    for (int i=2; i<MAXN; ++i) {
        for (int k=0; k<i; ++k) {
            f [i] += f [k] * f [i - k - 1];
        }
        if ((sum += f [i]) > 5000000000LL) {
            break;
        }
    }
#ifdef _DEBUG
    for (int i=0; i<20; ++i) {
        printf ("%lld%s", f [i], i==19 ? "\n" : " ");
    }
#endif

    long long n;
    while (scanf ("%lld", &n) == 1 && n) {
        print_tree (n);
        printf ("\n");
    }
    return 0;
}