HDUOJ----1250 Hat's Fibonacci

Hat's Fibonacci

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5800    Accepted Submission(s): 1926

Problem Description

A Fibonacci sequence is calculated by adding the previous two members the sequence, with the first two members being both 1.
F(1) = 1, F(2) = 1, F(3) = 1,F(4) = 1, F(n>4) = F(n - 1) + F(n-2) + F(n-3) + F(n-4)
Your task is to take a number as input, and print that Fibonacci number.

 

Input

Each line will contain an integers. Process to end of file.

 

Output

For each case, output the result in a line.

 

Sample Input

100

 

Sample Output

4203968145672990846840663646 Note: No generated Fibonacci number in excess of 2005 digits will be in the test data, ie. F(20) = 66526 has 5 digits.

 

Author

戴帽子的

 

Recommend

Ignatius.L

 http://acm.hdu.edu.cn/showproblem.php?pid=1250

 

 大数题，我先是用打表的方法，发现数字到达7037的时候，数字规模才能达到2005 ...然后用7037*2005 ---->已经超过内存，所以这条路实不可取的。。。唯一的方法，只好暴力法，但又担心会超时，可是，喜感就是，竟然过了，还应458ms。。。。哎!！！。

贴下代码：

   #include<stdio.h>
   #include<string.h>
   #define maxn 2006
   int a[maxn],b[maxn],c[maxn],d[maxn],ans[maxn];
   int main()
   {
       int i,j,e,s,up,num;
       while(scanf("%d",&num)!=EOF)
       {
           if(num<=)
           printf("");
          else
          {
          memset(a,,sizeof a);
          memset(b,,sizeof b);
          memset(c,,sizeof c);
          memset(d,,sizeof d);
          memset(ans,,sizeof ans);
          a[]=b[]=c[]=d[]=;
       for(i=,up=;i<num;i++)
       {
          for(j=,e=;j<=up;j++)
          {
             s=a[j]+b[j]+c[j]+d[j]+e;
             ans[j]=s%;
             e=(s-ans[j])/;
             if(s>&&j==up) up++;
             a[j]=b[j];
             b[j]=c[j];
             c[j]=d[j];
             d[j]=ans[j];
          }
       }
          for(j=maxn;ans[j]==;j--);
          for(i=j;i>=;i--)
          {
           printf("%d",ans[i]);
          }
       }
          printf("\n");
      }
  return ;
  }