POJ2084 catalan数

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;
}

再来看看catalan数的应用 我在维基 百度的百科和博客上看到的总结一下

 

C_n表示通过连结顶点而将n + 2边的凸多边形分成三角形的方法个数。下图中为n = 4的情况：

C_n表示对{1, ..., n}依序进出栈的置换个数。一个置换w是依序进出栈的当S(w) = (1, ..., n), 其中S(w)递归定义如下：令w = unv，其中n为w的最大元素，u和v为更短的数列；再令S(w) = S(u)S(v)n，其中S为所有含一个元素的数列的单位元。