卡特兰数（Train Problem II ）

                                             Train Problem II

As we all know the Train Problem I, the boss of the Ignatius Train Station want to know if all the trains come in strict-increasing order, how many orders that all the trains can get out of the railway.

Input

The input contains several test cases. Each test cases consists of a number N(1<=N<=100). The input is terminated by the end of file.

Output

For each test case, you should output how many ways that all the trains can get out of the railway.

Sample Input

1
2
3
10

Sample Output

1
2
5
16796

它的一般项公式为:

Hn=1n+1Cn2n=(2n)!(n+1)!n!

递推公式为:

Hn+1=∑i=0nHiHn−i=2(2n+1)n+2Hn(n≥0)

//ｈ( n ) = ( ( 4*n-2 )/( n+1 )*h( n-1 ) );

#include<stdio.h>

//*******************************
//打表卡特兰数
//第 n个 卡特兰数存在a[n]中，a[n][0]表示长度；
//注意数是倒着存的，个位是 a[n][1] 输出时注意倒过来。
//*********************************
int a[105][100];
void ktl()
{
    int i,j,yu,len;
    a[2][0]=1;
    a[2][1]=2;
    a[1][0]=1;
    a[1][1]=1;
    len=1;
    for(i=3;i<101;i++)
    {
        yu=0;
        for(j=1;j<=len;j++)
        {
            int t=(a[i-1][j])*(4*i-2)+yu;
            yu=t/10;
            a[i][j]=t%10;
        }
        while(yu)
        {
            a[i][++len]=yu%10;
            yu/=10;
        }
        for(j=len;j>=1;j--)
        {
            int t=a[i][j]+yu*10;
            a[i][j]=t/(i+1);
            yu = t%(i+1);
        }
        while(!a[i][len])
        {
            len--;
        }
        a[i][0]=len;
    }

}
int main()
{
    ktl();
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        for(int i=a[n][0];i>0;i--)
        {
            printf("%d",a[n][i]);
        }
        puts("");
    }
    return 0;
}