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

 

先计算 f [i]，保存有 i 个节点的树可有多少个，在每个节点试探左儿子有多少个节点，右儿子有多少节点，然后进行一些细节的计算。

比如得到左儿子节点数后，就知道左儿子是那种树（节点数相同是一种树），这时再计算左儿子应该是这种树中的第几棵。

虽然已经尽量追求代码整洁了，但是代码还是比较丑。。。

#include 
#include 
#include 
#include 
#include 
#include 

#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 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;
}