HDU1130 卡特兰数

How Many Trees?

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3397    Accepted Submission(s): 1964

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). 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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http://blog.csdn.net/sunshine_yg/article/details/47685737
卡特兰数 




#include  
#include  
#include<string>  
#include  
#include  
using namespace std;  
const int base = 10000;  
const int N = 100 + 2;  
int katelan[N][N];  
int main()  
{  
    katelan[1][1] = 1;  
    katelan[2][1] = 2;  
    katelan[3][1] = 5;  
    for (int i = 4; i <= 100; i++)  
    {  
        for (int j =1; j<100; j++)//大数乘法  
        {  
            katelan[i][j] += katelan[i - 1][j] * (4 * i - 2);  
            katelan[i][j + 1] += katelan[i][j] / base;  
            katelan[i][j] %= base;  
        }  
        int temp;  
        for (int j = 100; j > 0; j--)//大数除法  
        {  
            temp=katelan[i][j] % (i + 1);  
            katelan[i][j-1]+=temp* base;  
            katelan[i][j] /= (i + 1);  
        }  
    }  
    int n;  
    while (cin >> n)  
    {  
        int i = 100;  
        while (katelan[n][i] == 0)i--;  
        cout << katelan[n][i--];  
        while (i > 0)  
            printf("%04d", katelan[n][i--]);  
        cout << endl;  
    }  
    return 0;  
}