HDU1041 Computer Transformation（java）

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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.

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 consecutive zeroes pairs that will appear in the sequence after n steps.

Sample Input
2
3

Sample Output
1
1
题意是每步中将1变成01，0变成10，初始数字为1，求n步后结果串中有多少对00。
1
Step1:01。
Step2:1001。
Step3:01101001。
Step4:1001011001101001。
分析过程我们可以知道第n个串中的00来源有两个部分：
1.第n-2个串
2.第n-2个串中1的格式（1经过两步后变为1001）。
所以有有递推公式C(n)=C(n-2)+2^(n-3) 其中n>=4

import java.math.BigInteger;import java.util.Scanner;/** *  * @author  Jackie *  date    2015年10月22日 *  desc    ACM *  C(n)=C(n-2)+2^(n-3) 其中n>=4 */public class P1041 {    public static BigInteger[] midRes = new BigInteger[1001];    public static BigInteger[] result = new BigInteger[1001];    public static void main(String[] args) {        new P1041().run();    }    public void run(){        initial();        Scanner scanner = new Scanner(System.in);        int n;        while (scanner.hasNextInt()) {            n = scanner.nextInt();            System.out.println(result[n]);        }        scanner.close();    }    public void initial(){        result[1] = BigInteger.valueOf(0);        result[2] = BigInteger.valueOf(1);        result[3] = BigInteger.valueOf(1);        midRes[0] = BigInteger.valueOf(1);        midRes[1] = BigInteger.valueOf(2);        midRes[2] = BigInteger.valueOf(4);        midRes[3] = BigInteger.valueOf(8);        for (int i = 4; i <= 1000; i++) {            /**             *没有考虑到2^(i - 3)也是一个大数，使用long强制转换后会是远离值缩小             *所以可以使用一个大数数组存储中间值2^(i - 3);             *///          result[i] = result[i - 2].add(BigInteger.valueOf((long) Math.pow(2, i - 3)));            /**             *正确使用递推公式的方法             */            midRes[i] = midRes[i - 1].multiply(BigInteger.valueOf(2));            result[i] = result[i - 2].add(midRes[i - 3]);        }    }}

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