Computer Transformation(hdoj 1041)

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

step 1：01
step 2：1001
step 3：0110 1001
step 4：1001 0110 0110 1001
step 5：0110 1001 1001 0110 1001 0110 0110 1001

/*找规律，0 1 1 3 5 11 21 43 85，找出递推关系f(n)=2*f(n-2)+f(n-1),*/
#include<stdio.h>
int a[1001][305]={{0},{0},{1},{1}};
int b[1001]={1};
int calc()/*由于n可以取到1000，2^1000超出long long，必须要用数组按位存储，所以考大数*/
{
    int i=4;
    for(i=4;i<1001;i++)
    {
        int c,j;
        for(c=0,j=0;j<305;j++)/*利用b数组存储2*f(n-2)*/
        {
            b[j]=b[j]*2+c;/*c为余数*/
            c=b[j]/10;
            b[j]=b[j]%10;
        }
        for(c=0,j=0;j<305;j++)/* 2*f(n-2)+f(n-1) */
        {
            a[i][j]=b[j]+a[i-2][j]+c;
            c=a[i][j]/10;
            a[i][j]=a[i][j]%10;
        }
    }
}
int main()
{
    int n,i,k=0;
    calc();
    while(~scanf("%d",&n))
    {
        if(n==1)
            printf("0");
        else
        {
            k=0;
            for(i=300;i>=0;i--)
            {
                if(a[n][i]||k)
                {    
                    printf("%d",a[n][i]);
                    k=1;
                }
            }
        }
        printf("
");
    }
}