hdu_1041_Computer Transformation_201311051648

Computer Transformation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 5065    Accepted Submission(s): 1850

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

 

 

 #include <stdio.h>
 #include <string.h>

 int an1[],an2[];
 char s[][];

 int main()
 {
     int i,j,k,len,t,n,aa;
     memset(an1,,sizeof(an1));
     memset(an2,,sizeof(an2));
     memset(s,,sizeof(s));
     strcpy(s[],"");
     strcpy(s[],"");
     an1[]=;
     len=;t=;
     for(i=;i<=;i++)
     {
         t=;
         for(j=;j<=len;j++)
         {
             an2[j]=an1[j]*+t;
             if(an2[j]>)
             {
                 an2[j]-=;
                 t=;
             }
             else
             t=;
             if(j==len&&t==)
             len+=;
             an1[j]=an2[j];
         }
         if(i&)
         {an1[]=an2[]-=;}
         else
         {an1[]=an2[]+=;}
         for(k=,aa=j-;aa>=;aa--)
         {
             an1[aa]=an2[aa];
             s[i][k++]=an2[aa]+'';
         }
     }
     while(scanf("%d",&n)!=EOF)
     {
         printf("%s\n",s[n]);
     }
     return ;
 }
 //大数，找规律

找规律，大数问题