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


[align=left]Problem Description[/align]
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?

[align=left]Input[/align]
Every input line contains one natural number n (0 < n ≤1000).

[align=left]Output[/align]
For each input n print the number of consecutive zeroes pairs that will appear in the sequence after n steps.

[align=left]Sample Input[/align]

2
3

[align=left]Sample Output[/align]

1
1

[align=left]Source[/align]
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
= 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
);
}
}

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;
}
}