hdu 1023 卡特兰数《 大数》java

Train Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5372    Accepted Submission(s): 2911

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.

 

Author

Ignatius.L

 

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 java大数问题。

 //package hxltom;
 
 import java.io.*;
 import java.math.BigInteger;
 import java.util.*;
 
 public class Main {
 
     public static void main(String[] args) throws Exception{
         // TODO Auto-generated method stub
 
         Scanner cin  = new Scanner(System.in);
         BigInteger dp[] = new BigInteger[];
         dp[] = BigInteger.ONE;
         dp[] = BigInteger.ONE;
         for(int i=;i<=;i++)
         {
             dp[i] = dp[i-].multiply(BigInteger.valueOf(*i-)).divide(BigInteger.valueOf(i+));
         }
         int n;
         while(cin.hasNext())
         {
             n = cin.nextInt();
             System.out.println(dp[n]);
         }
     }
 
 }

另一种版本......

 //package hxltom;
 
 import java.io.*;
 import java.math.BigInteger;
 import java.util.*;
 
 public class Main {
 
     public static void main(String[] args) throws Exception{
         // TODO Auto-generated method stub
 
         Scanner cin  = new Scanner(System.in);
         BigInteger dp[] = new BigInteger[101];
         dp[0] = BigInteger.ONE;
         dp[1] = BigInteger.ONE;
         for(int i=2;i<=100;i++)
             dp[i] = BigInteger.ZERO;
         for(int i=2;i<=100;i++){
             for(int j=0;j<i;j++)
                 dp[i] = dp[i].add(dp[j].multiply(dp[i-j-1]));
         }
         int n;
         while(cin.hasNext())
         {
             n = cin.nextInt();
             System.out.println(dp[n]);
         }
     }
 
 }