点击打开题目
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 3718 Accepted Submission(s): 2088
Problem Description
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
Source
UVA
题意:给你一个数字 n ,问从1~n 可以构成多少种二叉树。
这种组合数学的思想还不是很清晰,但是知道是卡特兰数后就好办了,用 java 的 BigDecimal 打表就行了。
卡特兰数的递推公式:
F ( i ) = F ( i - 1 ) * ( 4 * i - 2 ) / ( n + 1 )
代码如下:
import java.math.BigDecimal;
import java.util.Scanner;
public class Main
{
public static void main(String[] args)
{
Scanner sc = new Scanner(System.in);
BigDecimal one = new BigDecimal(1);
BigDecimal two = new BigDecimal(2);
BigDecimal four = new BigDecimal(4); //三个常数
BigDecimal f[] = new BigDecimal [111];
f[0] = f[1] = new BigDecimal(1);
f[2] = new BigDecimal(2);
for (int i = 3 ; i <= 100 ; i++)
{
BigDecimal t = new BigDecimal(i);
f[i] = f[i-1].multiply(four.multiply(t).subtract(two)).divide(t.add(one));
}
while (sc.hasNext())
{
int n = sc.nextInt();
System.out.println(f[n]);
}
}
}