HDU——1130 How Many Trees?

How Many Trees?

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3952    Accepted Submission(s): 2238

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

 

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题目大意：

给定一个数n，你能说出有多少个不同的二进制搜索树可以用一组n个大小的数字来构造，这样集合的每个元素都将与二进制搜索树中的一个节点的标签相关联？

思路：

看题意，准是有很多大佬已经想到这个题就是让求一个二叉树有几种形态，那不就是一个裸的卡特兰数吗？！

然后我就信心满满的写了个卡特兰数，结果交上去接着就wa了。。。。

靠，卡特兰数到52就暴long long了！！！

唉，要用高精啊。。。。

代码：

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define N 110
using namespace std;
int n,len,ans,sum,tmp,a[N][N],b[N];
int catelan()
{
    len=; a[][]=b[]=;
    ;i<;i++)
    {
        ;j<len;j++)
         a[i][j]=a[i-][j]*(*i-);
        sum=;
        ;j<len;j++)
        {
            tmp=sum+a[i][j];
            a[i][j]=tmp%;
            sum=tmp/;
        }
        while(sum)
        {
            a[i][len++]=sum%;
            sum/=;
        }
        ;j>=;j--)
        {
            tmp=sum*+a[i][j];
            a[i][j]=tmp/(i+);
            sum=tmp%(i+);
        }
        ])
         len--;
        b[i]=len;
    }
}
int main()
{
    catelan();
    while(~scanf("%d",&n))
    {
        ;i>=;i--)
         printf("%d",a[n][i]);
        printf("\n");
    }
    ;
}