（大数）Computer Transformation hdu1041

Computer Transformation

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

Total Submission(s): 8688????Accepted Submission(s): 3282

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Problem 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.

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

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Input

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

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Output

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

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Sample Input

2

3

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Sample Output

1

1

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用java，递推

递推：0->10 ?;

? ? ? ? ?1->01;

? ? ? ? ?00->1010;

? ? ? ? ?10->0110;

? ? ? ? ? 01->1001;

? ? ? ? ? 11->0101;

假设a[i]表示第i 步时候的00的个数，由上面的可以看到，00是由01 得到的，所以只要知道a[i-1]的01的个数就能够知道a[i]的00的个数了，那a[i-1]怎么求呢，同样看推导，01由1和00 得到，而第i步1的个数是2^(i-1),所以a[i]=2^(i-3)+a[i-2];（最后计算的是第i-2步情况）。

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import java.math.BigDecimal;
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[1001];
        while(in.hasNextInt()) {
            int n=in.nextInt();
            a[1]=BigInteger.valueOf(0);
            a[2]=BigInteger.valueOf(1);
            a[3]=BigInteger.valueOf(1);
            for(int i=4;i<=n;i++) {
                a[i]=BigInteger.valueOf(0);      //先进行初始化。
                int m=i-3;        //在大数的pow(m,n)中，n是int类型的，m是BigInteger类型的。             
                BigInteger q= new BigInteger("2");
                a[i]=a[i].add(q.pow(m));
                a[i]=a[i].add(a[i-2]);
            }
            System.out.println(a[n]);
        }
    }
}