A binary search tree is a binary tree with root k such that any node v reachable from its left has label (v) <label (k) and any node w reachable from its right has label (w) > label (k). It is a search structure which can find a node with label x in O(n log n) average time, where n is the size of the tree (number of vertices).

Given a number n, can you tell how many different binary search trees may be constructed with a set of numbers of size n such that each element of the set will be associated to the label of exactly one node in a binary search tree?
Input
The input will contain a number 1 <= i <= 100 per line representing the number of elements of the set.
Output
You have to print a line in the output for each entry with the answer to the previous question.
Sample Input
1
2
3
Sample Output
1
2
5

分析：

题意：
给你从1~n这n个整数，问分别用这n个数做根节点构建二叉搜索树，能有多少种构建二叉搜索树的方法？

解析：
看到给出的样例输入和输出，有点像卡塔兰数，又写了n=4的情况，还是与卡塔兰数符合，卡塔兰数的前几项分别是：

越看越像卡塔兰数，我就把昨晚的HDU1023的代码提交了上去试试，果然过了！

实话说，我也不知道这个题为啥卡塔兰数，看着像就写完了！！！

代码：

#include<iostream>
#include<cstdio>
#define N 105

using namespace std;

int dp[N][N];

int main()
{
	int n;
	dp[1][0]=dp[1][1]=1;
	for(int i=2;i<=100;i++)
	{
		int len=dp[i-1][0],v=0;
		int k=(4*i-2);
		for(int j=1;j<=len;j++)
		{
			dp[i][j]=k*dp[i-1][j]+v;
			v=dp[i][j]/10;
			dp[i][j]%=10;
		}
		while(v)
		{
			dp[i][++len]=v%10;
			v/=10;
		}
		for(int j=len;j>0;j--)
		{
			v=v*10+dp[i][j];
			dp[i][j]=v/(i+1);
			v%=(i+1);
		}
		while(!dp[i][len])
		{
			len--;
		}
		dp[i][0]=len;
	}
	while(~scanf("%d",&n))
	{
		for(int i=dp[n][0];i>0;i--)
		{
			printf("%d",dp[n][i]);
		}
		printf("\n");
	}
	return 0;
}