Computer Transformation
Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 4561 Accepted: 1738 DescriptionA 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?
InputEvery input line contains one natural number n (0 < n <= 1000).
OutputFor each input n print the number of consequitive zeroes pairs that will appear in the sequence after n steps.
Sample Input2
3 Sample Output1
1 Source这是我第一次用java进行大数运算
递推很简单,00只可能是上一个的01产生,上一个的01只可能是上上一个的00 1产生
hoj的测试机好像有问题,poj里面ac的代码提交不上去hojimport java.math.BigInteger;import java.util.Scanner;/** * * @author chenyongkang */public class Main { public static BigInteger a; public static BigInteger b[]=new BigInteger[1010]; public static void init() { BigInteger x=BigInteger.valueOf(0); BigInteger y=BigInteger.valueOf(1); for(int i=0;i<=1000;i++) b[i]=BigInteger.valueOf(0); b[1]=x;b[2]=y; for(int i=3;i<=1000;i++) { a=BigInteger.valueOf(1); for(int j=1;j<=i-3;j++) { a=a.multiply(BigInteger.valueOf(2)); } b[i]=b[i].add(b[i-2]); b[i]=b[i].add(a); } } /** * @param args the command line arguments */ public static void main(String[] args) { // TODO code application logic here init(); Scanner cin=new Scanner(System.in); while(cin.hasNext()) { int n=cin.nextInt(); System.out.println(b[n]); } }}