HOJ 2148&POJ 2680（DP递推，加大数运算）
HOJ 2148&POJ 2680（DP递推，加大数运算）
Computer Transformation Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 4561 Accepted: 1738 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 consequitive zeroes pairs that will appear in the sequence after n steps. Sample Input
2 3 Sample Output
1 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hoj
import 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]);
}
}
}