JOJ1040：Trees（卡特兰数+递归）

JOJ1040：Trees（卡特兰数+递归）

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) 

思路：
a数组表示节点数为j所能表示最大的数。
则第j个节点所能表示的数a[j]符合卡特兰数：
a[j] = a[0] * a[j - 1] + a[1] * a[j - 2] + ...... + a[j - 1] * a[0];
表示：有j个节点 = 左边0个节点的个数 * 右边j - 1个节点的个数 + ...... + 左边j - 1个节点的个数 * 右边0个节点的个数。

之后根据读入的n，判断出节点数，在再判断出左右的节点数和左右所代表的数。
然后调用递归。

#include  < cstdio >
#include  < cstring >
using   namespace  std;

int  a[ 25 ], b[ 25 ];

void  solve( int  n)
{
     int  t, i, j;
     if  (n  ==   0 )  return ;
     if  (n  ==   1 )
    {
        printf( " X " );
         return ;
    }
     for  (j  =   1 ;;  ++ j)
    {
         if  (b[j]  >=  n)
             break ;
    }
    n  =  n  -  b[j  -   1 ];
     for  (i  =   0 ; i  <  j;  ++ i)
    {
        t  =  a[i]  *  a[j  -   1   -  i];
         if  (n  >  t)
        {
            n  =  n  -  t;
        }
         else
             break ;
    }
     if  (i  !=   0 )
    {
        printf( " ( " );
        solve(b[i  -   1 ]  +   1   +  (n  -   1 ) /  a[j  -   1   -  i]);
        printf( " ) " );
    }
    printf( " X " );
     if  (i  !=  j  -   1 )
    {
        printf( " ( " );
        solve(b[j  -   2   -  i]  +   1   +  (n  -   1 )  %  a[j  -   1   -  i]);
        printf( " ) " );
    }
}

int  main()
{
     int  n;
     int  i, j;
    b[ 0 ]  =   0 ;
    a[ 0 ]  =  b[ 1 ]  =  a[ 1 ]  =   1 ;
     for  (i  =   2 ; i  <   20 ;  ++ i)
    {
        a[i]  =   0 ;
         for  (j  =   0 ; j  <  i;  ++ j)
        {
            a[i]  +=  a[j]  *  a[i  -  j  -   1 ];
        }
        b[i]  =  b[i  -   1 ]  +  a[i];
    }
     while  (scanf( " %d " ,  & n)  &&  n)
    {
        solve(n);
        printf( " \n " );
    }
     return   0 ;
}