HDUOJ-----Computer Transformation

 

 

 

Computer Transformation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 4842    Accepted Submission(s): 1769

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

做这道题，纯粹是一道大数的题，当然你需要推导出这个公式f[n]=f[n-1]+2*f[n-2];

#include<cstdio>
const int maxn=1000;
int arr[maxn+1][305]={0};
int len=1;
void LargeNum()
{
     arr[1][0]=1;
    for(int i=2;i<=maxn;i++)
    {
        int c=0;
        for(int j=0;j<len;j++)
        {
          arr[i][j]+=arr[i-1][j]+2*arr[i-2][j]+c;
          if(arr[i][len-1]>9)
              len++;
           c=arr[i][j]/10;
             arr[i][j]%=10;
        }
    }

}
int main( void )
{ 
    int n,i;
    LargeNum();
    while(scanf("%d",&n)==1)
    {
        if(n==1)puts("0");
        else
        {
        for(i=len;arr[n-1][i]==0;i--);
        for(int j=i;j>=0;j--)
              printf("%d",arr[n-1][j]);
        puts("");
        }
    }
    return 0;
}