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

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