poj——2084 Game of Connections

　　　　　　　　　　　　　　　　　　　　　　Game of Connections

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 8664		Accepted: 4279

Description

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?

Input

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.

Output

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

Sample Input

2
3
-1

Sample Output

2
5

Source

Shanghai 2004 Preliminary

题目大意：一圆环上有2n个点，求两两连线且不交叉的方法数。

思路：

卡特兰数

代码：

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#define N 110
using namespace std;
int n,len,tmp,sum,b[N],a[N][N];
int read()
{
    ,f=; char ch=getchar();
    ; ch=getchar();}
    +ch-'; ch=getchar();}
    return x*f;
}
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();
    )
    {
        n=read();
        ) break;
        ;i>=;i--)
         printf("%d",a[n][i]);
        printf("\n");
    }
    ;
}