HDU 1023 Train Problem II （卡特兰数）

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Train Problem II

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7741 Accepted Submission(s): 4168

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

12310

Sample Output

12516796

Hint

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

Author

Ignatius.L

题意：求卡特兰数。

题解：卡特兰数公式：

　　　只要掌握卡特兰数的公式就行了，两个公式：

　　　1. 组合公式为 Cn = C(2n,n) / (n+1);
　　　2. 递推公式为 h(n ) = h(n-1)*(4*n-2) / (n+1)

1.Java代码：

<span style="font-size:18px;">import java.util.*;  import java.math.BigInteger;  public class Main {      public static void main(String[] args) {          BigInteger a[] = new BigInteger[105];          a[0] = BigInteger.ZERO;          a[1] = BigInteger.ONE;          for(int i=2;i<=102;i++) {              a[i] = a[i-1].multiply(BigInteger.valueOf(4*i-2)).divide(BigInteger.valueOf(i+1));          }                    Scanner in = new Scanner(System.in);                    while(in.hasNext()) {              int n=in.nextInt();              System.out.println(a[n]);          }      }  }</span>

2.C++代码：

<span style="font-size:18px;">#include <iostream>  #include <cstring>  #include <cstdio>  using namespace std;  const int MAXN = 105;  int a[MAXN][MAXN];  int b[MAXN];  ///可以作为模板  void Catalan()  {      int yu,len;      a[1][0] = 1;      a[1][1] = 1;      len = 1;      for(int i=2; i<MAXN; i++)      {          yu = 0;          for(int 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(int 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()  {      Catalan();      int n;      while(cin>>n)      {          for(int i=a[n][0]; i>0; i--)          cout<<a[n][i];          puts("");      }      return 0;  }  </span>

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