HDU——1134 Game of Connections

　　　　　　　　Game of Connections

　　　　　　Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
　　　　　　　　Total Submission(s): 4844    Accepted Submission(s): 2811

Problem 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

Asia 2004, Shanghai (Mainland China), Preliminary

 

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题意：

这是一个小而古老的游戏。你应该把数字1，2，3，.，2n-1，2n连续地按顺时针顺序在地上形成一个圆圈，然后画出一些直线线段，将它们连接成数对。每一个数字都必须相互联系。没有两个区段被允许相交。 这仍然是个简单的游戏，不是吗？但是，在你写下了2n的数字之后，你能告诉我，你可以用多少种不同的方式将数字连接成对？生活更艰难，对吧？

思路：

看题意，这道题是不是圆的划分？！

对，就是圆的划分，这样我们又可以用卡特兰数来做了。。

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");
    }
    ;
}