hdu 1041(递推,大数)

Computer Transformation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7210    Accepted Submission(s): 2628

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

 

Source

Southeastern Europe 2005

 

题意：给定初始值1，然后进行变换，变换的规则是 1->01 ,0->10,问进行n次变换后串中00的个数。

 

题解：

写出前几项：

              1

             01         第1次(项)

           1001       第2次（项）

       01101001    第3次 （项）

1001011001101001    第4次（项）

....

我们会发现只有当串中存在1001时会出现00

而要出现1001，则父串中要出现01

父串中出现01串有两种方法，一种是我所标注的红色部分，由祖父串中的1变过来

第二种方式就是当祖父串中有00时 父串变成 1010 时会出现 01

所以这样就可以得到递推公式：第n项00的个数 = 第 n-2项 1的个数+第n-2项 00 的个数。

f[1] = 0

f[2] = 1

f[n] = f[n-2]+2^(n-3)  ( n>2 )

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

 
public class Main {
    public static void main(String[] args) {
        BigInteger [] h = new BigInteger[1001];
        h[1] = new BigInteger("0");
        h[2] = new BigInteger("1");
        for(int i=3;i<=1000;i++){
            h[i] = h[i-2].add(pow(i-3));
        }
        Scanner sc =new Scanner (System.in);
        while(sc.hasNext()){
            int n =sc.nextInt();
            System.out.println(h[n]);
        }
    }

    private static BigInteger pow(int v) {
        BigInteger sum = BigInteger.valueOf(1);
        for(int i=0;i<v;i++){
            sum = sum.multiply(BigInteger.valueOf(2));
        }
        //System.out.println(sum);
        return sum;
    }
}