HDOJ 1130 How Many Trees?(卡特兰数+大数乘除法)

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?

 

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

123

 

Sample Output

125

 

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;}