（母函数 Catalan数 大数乘法 大数除法） Train Problem II hdu1023

Train Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 10372    Accepted Submission(s): 5543

Problem Description

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

Hint

The result will be very large, so you may not process it by 32-bit integers.

注意：

这是卡特兰数，用大数。使用的公式为：

h(n)=h(n-1)*(4*n-2)/(n+1)。

用JAVA更简单：

import java.math.BigInteger;
import java.util.Scanner;
public class Main {
    public static void main(String[] args) {
        Scanner in = new Scanner (System.in);
        BigInteger a[]=new BigInteger[101];
        a[0]=BigInteger.ZERO;
        a[1]=BigInteger.ONE;
        for(int i=2;i<=100;i++)
            a[i]=a[i-1].multiply(BigInteger.valueOf(4*i-2)).divide(BigInteger.valueOf(i+1));
        while(in.hasNextInt()) {
            int n=in.nextInt();
            System.out.println(a[n]);
        }
    }
}

接下来是C++：

#include <iostream>
#include <cstdio>
using namespace std;
int a[105][105];
int i,j,n;
void ctl()         //打表。
{
    a[2][0]=1;
    a[2][1]=2;
    a[1][0]=1;
    a[1][1]=1;
    int len=1,t,yu=0;       //注意初始化。
    for(i=3;i<101;i++)         //先打表求出1到100的所有Catalan数。
    {
        for(j=1;j<=len;j++)              //大数乘法。
        {
            t=a[i-1][j]*(4*i-2)+yu;
            yu=t/10;
            a[i][j]=t%10;       //求每位数的确定的数。
        }
        while(yu)         //进位。一直到yu为0为止。
        {
            a[i][++len]=yu%10;
            yu/=10;
        }
        for(j=len;j>=1;j--)              //大数除法。联系手工除法步骤。
        {
            t=a[i][j]+yu*10;
            a[i][j]=t/(i+1);           //可以看做手工除法的商。
            yu=t%(i+1);              //可以看做手工除法的余数。
        }
        while(!a[i][len])        //去掉前边的0。
        {
            len--;
        }
        a[i][0]=len;
    }
}
int main()
{
    ctl();
    int n;
    while(~scanf("%d",&n))
    {
        for(int i=a[n][0];i>0;i--)
        {
            printf("%d",a[n][i]);
        }   
        puts("");
    }   
    return 0;
}