HDOJ 1130 How Many Trees?(卡特兰数+大数乘除法)

发布时间：2021-05-27 08:51 所属栏目：125 来源：网络整理

导读：How Many Trees? Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 3380????Accepted Submission(s): 1958 Problem Description A binary search tree is a binary tree with root k such that a

How Many Trees?

Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 3380????Accepted Submission(s): 1958

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?
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Input The input will contain a number 1 <= i <= 100 per line representing the number of elements of the set.
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Output You have to print a line in the output for each entry with the answer to the previous question.
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Sample Input

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Sample Output

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题意：给出一个n，每个节点的编号为1~n，问n个节点的二叉搜索树有多少种？
题解：和hdoj 1023一模一样啊，卡特兰数+大数。不了解的点击：?点击打开链接?
代码如下：

//h(n)=h(n-1)*(4*n-2)/(n+1);
#include<cstdio>
#include<cstring>
using namespace std;
int a[110][110];
//a[n][0]表示n的卡特兰数的长度，存储是反向的，a[n][1]表示个位数 

void ktl()//打表 
{
	int i,j,len;
	int c,s;
	a[1][0]=1;
	a[1][1]=1;
	a[2][0]=1;
	a[2][1]=2;
	len=1;
	for(i=3;i<101;++i)
	{
		c=0;
		for(j=1;j<=len;++j)
		{
			s=a[i-1][j]*(4*i-2)+c;
			c=s/10;
			a[i][j]=s%10;
		}
		while(c)
		{
			a[i][++len]=c%10;
			c/=10;
		}
		for(j=len;j>0;--j)
		{
			s=a[i][j]+c*10;
			a[i][j]=s/(i+1);
			c=s%(i+1);
		}
		while(!a[i][len])
			len--;
		a[i][0]=len;
	}
}

int main()
{
	ktl();
	int n;
	while(scanf("%d",&n)!=EOF)
	{
		for(int i=a[n][0];i>0;--i)
			printf("%d",a[n][i]);
		printf("\n");
	}
	return 0;
}

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