（大数）Computer Transformation hdu1041

Posted 2020-11-17 weixu-liu

Computer Transformation

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

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

 

 

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

 

 

 

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）。

 

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]);
        }
    }
}

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