HOJ 2148&POJ 2680（DP递推，加大数运算）

Computer Transformation
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 4561 Accepted: 1738
Description

A sequence consisting of one digit,the number 1 is initially written into a computer. At each successive time step,the computer simultaneously tranforms each digit 0 into the sequence 1 0 and each digit 1 into the sequence 0 1. So,after the first time step,the sequence 0 1 is obtained; after the second,the sequence 1 0 0 1,after the third,the sequence 0 1 1 0 1 0 0 1 and so on.

How many pairs of consequitive zeroes will appear in the sequence after n steps?
Input

Every input line contains one natural number n (0 < n <= 1000).
Output

For each input n print the number of consequitive zeroes pairs that will appear in the sequence after n steps.
Sample Input

2
3
Sample Output

1
1
Source

这是我第一次用java进行大数运算

递推很简单，00只可能是上一个的01产生，上一个的01只可能是上上一个的00 1产生
hoj的测试机好像有问题，poj里面ac的代码提交不上去hoj

import java.math.BigInteger;
import java.util.Scanner;


/** * * @author chenyongkang */
public class Main {

    public static BigInteger a;
    public static BigInteger b[]=new BigInteger[1010];
    public static void init()
    {
        BigInteger x=BigInteger.valueOf(0);
        BigInteger y=BigInteger.valueOf(1);
        for(int i=0;i<=1000;i++)
            b[i]=BigInteger.valueOf(0);
        b[1]=x;b[2]=y;
        for(int i=3;i<=1000;i++)
        {
             a=BigInteger.valueOf(1);
            for(int j=1;j<=i-3;j++)
            {
                a=a.multiply(BigInteger.valueOf(2));
            }
           b[i]=b[i].add(b[i-2]);
           b[i]=b[i].add(a);
        }

    }
    /** * @param args the command line arguments */
    public static void main(String[] args) {
        // TODO code application logic here
           init();

        Scanner cin=new Scanner(System.in);
        while(cin.hasNext())
        {
            int n=cin.nextInt();
            System.out.println(b[n]);
        }
    }


}