How Many Trees?
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3382 Accepted Submission(s): 1960
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).
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
2
3
Sample Output
1 2 5
题意：由 n个结点组成二叉树的种数
卡特兰数+BigInteger
import java.math.BigInteger;
import java.util.Scanner; public class Main {
public static void main(String[] args) {
BigInteger [] h = new BigInteger[101];
h[1] = new BigInteger("1");
for(int i=2;i<=100;i++){
h[i] = h[i-1].multiply(BigInteger.valueOf(4*i-2)).divide(BigInteger.valueOf(i+1));
}
Scanner sc =new Scanner (System.in);
while(sc.hasNext()){
int n =sc.nextInt();
System.out.println(h[n]);
}
}
}
hdu:1131 由n个带编号的结点组成二叉树的个数
思路：卡特兰数乘上编号的全排列
import java.math.BigInteger;
import java.util.Scanner; public class Main {
public static void main(String[] args) {
BigInteger [] h = new BigInteger[101];
h[1] = new BigInteger("1");
for(int i=2;i<=100;i++){
h[i] = h[i-1].multiply(BigInteger.valueOf(4*i-2)).divide(BigInteger.valueOf(i+1));
}
Scanner sc =new Scanner (System.in);
while(sc.hasNext()){
int n =sc.nextInt();
if(n==0)break;
System.out.println(h[n].multiply(fac(n)));
}
} private static BigInteger fac(int n) {
BigInteger sum = BigInteger.valueOf(1);
for(int i=1;i<=n;i++) sum=sum.multiply(BigInteger.valueOf(i));
return sum;
} }