hdu 1130,hdu 1131(卡特兰数,大数)
2016-04-08 11:21
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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
[align=left]Problem Description[/align]
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?
[align=left]Input[/align]
The input will contain a number 1 <= i <= 100 per line representing the number of elements of the set.
[align=left]Output[/align]
You have to print a line in the output for each entry with the answer to the previous question.
[align=left]Sample Input[/align]
1
2
3
[align=left]Sample Output[/align]
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 ); } } }
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 .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; } }
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