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：

 1 #include<stdio.h>
 2 #include<string.h>
 3 #define maxn 2006    
 4 int a[maxn],b[maxn],c[maxn],d[maxn],ans[maxn];
 5 int main()
 6 {
 7     int i,j,e,s,up,num;
 8     while(scanf("%d",&num)!=EOF)
 9     {
10          if(num<=4)
11          printf("1");
12         else
13         {
14         memset(a,0,sizeof a);
15         memset(b,0,sizeof b);
16         memset(c,0,sizeof c);
17         memset(d,0,sizeof d);
18         memset(ans,0,sizeof ans);
19         a[0]=b[0]=c[0]=d[0]=1;
20      for(i=4,up=1;i<num;i++)
21      {
22         for(j=0,e=0;j<=up;j++)
23         {
24            s=a[j]+b[j]+c[j]+d[j]+e;
25            ans[j]=s%10;
26            e=(s-ans[j])/10;
27            if(s>9&&j==up) up++;
28            a[j]=b[j];
29            b[j]=c[j];
30            c[j]=d[j];
31            d[j]=ans[j];
32         }
33      }
34         for(j=maxn;ans[j]==0;j--);
35         for(i=j;i>=0;i--)
36         {
37          printf("%d",ans[i]);
38         }
39      }
40         printf("
");
41     }
42 return 0;
43 }View Code