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]); } } } View Code
转载于:https://www.cnblogs.com/Weixu-Liu/p/9166488.html
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