POJ2084 catalan數

Game of Connections

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 6323		Accepted: 3258

This is a small but ancient game. You are supposed to write down the numbers 1, 2, 3, . . . , 2n - 1, 2n consecutively in clockwise order on the ground to form a circle, and then, to draw some straight line segments to connect them into number pairs. Every number must be connected to exactly one another. 
And, no two segments are allowed to intersect. 
It's still a simple game, isn't it? But after you've written down the 2n numbers, can you tell me in how many different ways can you connect the numbers into pairs? Life is harder, right?

Each line of the input file will be a single positive number n, except the last line, which is a number -1. 
You may assume that 1 <= n <= 100.

For each n, print in a single line the number of ways to connect the 2n numbers into pairs.

2
3
-1

2
5

Shanghai 2004 Preliminary

#include <iostream>
#include<cstdio>
#include<cstring>
using namespace std;

#define MAX 100
#define BASE 10000

void multiply(int a[],int Max,int b)
{
    int array=0;
    for(int i=Max-1; i>=0; i--)
    {
        array+=b*a[i];
        a[i]=array%BASE;
        array/=BASE;
    }
}
void divide(int a[],int Max,int b)
{
    int div=0;
    for(int i=0; i<Max; i++)
    {
        div=div*BASE+a[i];
        a[i]=div/b;
        div%=b;
    }
}
int main()
{
    int n,i,a[101][MAX];
    memset(a[1],0,MAX*sizeof(int));
    for(i=2,a[1][MAX-1]=1; i<101; i++)
    {
        memcpy(a[i],a[i-1],MAX*sizeof(int));
        multiply(a[i],MAX,4*i-2);
        divide(a[i],MAX,i+1);
    }
    while(cin>>n)
    {
        if(n==-1)
            break;
        for(i=0; i<MAX&&a[n][i]==0; i++);
        cout<<a[n][i++];
        for(; i<MAX; i++)
            printf("%04d",a[n][i]);
        cout<<endl;
    }
    return 0;
}

再來看看catalan數的應用 我在維基 百度的百科和博客上看到的總結一下

 

C_n表示通過連結頂點而將n + 2邊的凸多邊形分成三角形的方法個數。下圖中為n = 4的情況：

C_n表示對{1, ..., n}依序進出棧的置換個數。一個置換w是依序進出棧的當S(w) = (1, ..., n), 其中S(w)遞歸定義如下：令w = unv，其中n為w的最大元素，u和v為更短的數列；再令S(w) = S(u)S(v)n，其中S為所有含一個元素的數列的單位元。