hdu 1041 Computer Transformation

Computer Transformation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8419    Accepted Submission(s): 3165

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

 

Source

Southeastern Europe 2005 

 

大整数的题目   已经不想说什么了   

#include <stdio.h>
#include <string.h>
#include <iostream>
using namespace std;
int ans[1000][100];
int main()
{
    int n;
    memset(ans,0,sizeof(ans));
    for(int i=2; i<=1000; i++)
    {
        for(int j=0; j<100; j++)
            ans[i][j]=ans[i-1][j];
        int t=0;
        if(i&1)
        {
            for(int j=0; j<100; j++)
            {
                t=t+ans[i][j]*2;
                ans[i][j]=t%1000000;
                t/=1000000;
            }
            ans[i][0]-=1;
        }
        else
        {
            for(int j=0; j<100; j++)
            {
                t=t+ans[i][j]*2;
                ans[i][j]=t%1000000;
                t/=1000000;
            }
            ans[i][0]+=1;
        }
    }
    while(scanf("%d",&n)!=EOF)
    {
        int k=99;
        if(n==1)
        {
            puts("0");
            continue;
        }
        while(ans[n][k]==0) k--;
        printf("%d",ans[n][k]);
        k--;
        for(int j=k; j>=0; j--)
            printf("%06d",ans[n][j]);
        putchar(10);
    }
    return 0;
}