Computer Transformation
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6543 Accepted Submission(s): 2378
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?
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
23
Sample Output
11
Source
Southeastern Europe 2005
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首先解释一下题目：
题目的大概意思很简单，就是说电脑里面存了一个初始的数字1，后面的所有规律和操作都从1开始，然后每一次产生一个变换，变换的法则如下
即 在这个数列中所有的1变换成01，所有的0变换成10，这样说可能还不明确，从1 开始，下一次则得到01（1->01），再下次则得到1001（上一次的0->10,1->01）然后得到01101001，然后依次类推.......问你经过n次变换有多少组00！（consequitive意为连续，相同的意思！）
参考网友：星夜&永恒的推理：
00只能由01推到,即01->1001->00
01只能由1,00推到,即1->01,00->1010->01.
现设a[n]表示n秒时1的个数，
b[n]表示n秒时00的个数，
c[n]表示n秒时01的个数，
由题知0,1过一秒都会产生一个1，
所以a[n+1]=2^n.....0秒1个数，1秒2*1个数，2秒2*1*2个数，...n秒2*1*2*2*2..*2个数=2^n.
b[n+1]=c[n];
c[n+1]=a[n]+b[n],
所以b[n]=c[n-1]=a[n-2]+b[n-2]=2^(n-3)+b[n-2].
其实本题多写几个样例就能发现另一个规律b[n]=2*b[n-2]+b[n-1].
注意本题大数相加.
01只能由1,00推到,即1->01,00->1010->01.
现设a[n]表示n秒时1的个数，
b[n]表示n秒时00的个数，
c[n]表示n秒时01的个数，
由题知0,1过一秒都会产生一个1，
所以a[n+1]=2^n.....0秒1个数，1秒2*1个数，2秒2*1*2个数，...n秒2*1*2*2*2..*2个数=2^n.
b[n+1]=c[n];
c[n+1]=a[n]+b[n],
所以b[n]=c[n-1]=a[n-2]+b[n-2]=2^(n-3)+b[n-2].
其实本题多写几个样例就能发现另一个规律b[n]=2*b[n-2]+b[n-1].
注意本题大数相加.
/* * 依据递推式：b[n]=2*b[n-2]+b[n-1] */import java.io.*;import java.math.BigInteger;import java.util.*;public class Main{ public static void main(String[] args) { // TODO Auto-generated method stub Scanner input = new Scanner(System.in); BigInteger a[] = new BigInteger[1001]; BigInteger b[] = new BigInteger[1001]; a[0] = BigInteger.ONE; b[2] = BigInteger.ONE; b[0] = BigInteger.ZERO; b[1] = BigInteger.ZERO; for (int i = 1; i < 1001; i++) { a[i] = a[i - 1].multiply(BigInteger.valueOf(2)); //这里算a[i]的但是这里的a[i]=2^(i-2) if (i >= 3) //所以下面算b[i]时候要i要多减去一个1 { b[i] = a[i - 3].add(b[i - 2]); } } while (input.hasNext()) { int n = input.nextInt(); System.out.println(b[n]); } }}