hdu1100

#include <iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>

using namespace std;

int n,op[40],sum[40];
void init()
{
    memset(op,0,sizeof(op));
    op[1]=sum[1]=1;op[0]=1;sum[0]=0;
    for(int i=2;i<20;i++)
    {
        for(int j=0;j<i;j++)
            op[i]+=op[i-1-j]*op[j];
        sum[i]=sum[i-1]+op[i];
    }
}
void dfs(int n)
{
    if(n==0)
        return ;
    int num=lower_bound(sum,sum+19,n)-sum;
    num--;
    n-=sum[num];
    int nl=0;
    while(n>op[nl]*op[num-nl])
    {
         n-=op[nl]*op[num-nl];
         nl++;
    }
    int dt=0,lp=1;
    if(nl)
    {
        dt=1+(n-1)/op[num-nl];
        n-=(dt-1)*(op[num-nl]);
        printf("(");
        dfs(dt+sum[nl-1]);
        printf(")");
    }
    printf("X");
    if(num-nl)
    {
        printf("(");
        dfs(n+sum[num-nl-1]);
        printf(")");
    }
}
int main()
{
    init();
    while(cin>>n&&n)
    {
    dfs(n);printf("\n");
    }
    return 0;
}

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.

此題剛開始看的時候以爲是個樹，其實這是一個數學問題，此題的遞推規則其實和卡特蘭數十分相似。

每次先對n計算有多少nodes，然後減掉多餘的求出在這麼多nodes的條件下的排列序號，此時再根據卡特蘭數的遞推公式可以輕鬆的求出左右子數的個數是多少，然後便是求解左樹子排列序號以及右樹子的排列序號，從而可以使用dfs完美求解